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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 136965, 2986]*) (*NotebookOutlinePosition[ 283009, 8159]*) (* CellTagsIndexPosition[ 282931, 8153]*) (*WindowFrame->Normal*) Notebook[{ Cell["42nd International Mathematical Olympiad", "Title"], Cell["\<\ Washington, DC, United States of America July 8\[Dash]9, 2001\ \>", "Subtitle"], Cell[CellGroupData[{ Cell["Problems", "Section"], Cell[TextData[{ StyleBox["Each problem is worth seven points.", "TI"], " " }], "Text"], Cell[CellGroupData[{ Cell["Problem 1", "SubsectionIcon"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], " be an acute\[Hyphen]angled triangle with circumcentre ", Cell[BoxData[ \(TraditionalForm\`O\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`P\)]], " on ", Cell[BoxData[ \(TraditionalForm\`B C\)]], " be the foot of the altitude from ", Cell[BoxData[ \(TraditionalForm\`A\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "Suppose that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B C A \[GreaterEqual] \[Angle] A B C + 30\^\[EmptySmallCircle]\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "Prove that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] C A B + \[Angle] C O P < \(\(90\^\[EmptySmallCircle]\)\(.\)\)\)]], " " }], "Text", ZeroWidthTimes->True] }, Closed]], Cell[CellGroupData[{ Cell["Problem 2", "SubsectionIcon"], Cell["Prove that ", "Text"], Cell[BoxData[ \(TraditionalForm\`a\/\@\(a\^2 + 8 b c\) + b\/\@\(b\^2 + 8 c a\) + c\/\@\(c\^2 + 8 a b\) \[GreaterEqual] 1\)], "DisplayFormula"], Cell[TextData[{ "for all positive real numbers ", Cell[BoxData[ \(TraditionalForm\`a, b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], ". " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Problem 3", "SubsectionIcon"], Cell["\<\ Twenty\[Hyphen]one girls and twenty\[Hyphen]one boys took part in a \ mathematical contest. \ \>", "Text"], Cell["\<\ \[Bullet] Each contestant solved at most six problems. \[Bullet] For each girl and each boy, at least one problem was solved by \ both of them. \ \>", "IndentedText"], Cell["\<\ Prove that there was a problem that was solved by at least three \ girls and at least three boys. \ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Problem 4", "SubsectionIcon"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`n\)]], " be an odd integer greater than 1, and let ", Cell[BoxData[ \(TraditionalForm\`k\_1, k\_2, \[Ellipsis], k\_n\)]], " be given integers. For each of the ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " permutations ", Cell[BoxData[ \(TraditionalForm\`a = \((a\_1, a\_2, \[Ellipsis], a\_n)\)\)]], " of ", Cell[BoxData[ \(TraditionalForm\`1, 2, \[Ellipsis], n\)]], ", let " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{"S", \((a)\), "=", UnderoverscriptBox["\[Sum]", \(i = 1\), "n", LimitsPositioning->False], \(k\_i\), \(a\_i\), "."}], TraditionalForm]], "DisplayFormula"], Cell[TextData[{ "Prove that there are two permutations ", Cell[BoxData[ \(TraditionalForm\`b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b \[NotEqual] c\)]], ", such that ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " is a divisor of ", Cell[BoxData[ \(TraditionalForm\`S(b) - S(c)\)]], ". " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Problem 5", "SubsectionIcon"], Cell[TextData[{ "In a triangle ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], ", let ", Cell[BoxData[ \(TraditionalForm\`A P\)]], " bisect ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B A C\)]], ", with ", Cell[BoxData[ \(TraditionalForm\`P\)]], " on ", Cell[BoxData[ \(TraditionalForm\`B C\)]], ", and let ", Cell[BoxData[ \(TraditionalForm\`B Q\)]], " bisect ", Cell[BoxData[ \(TraditionalForm\`\[Angle] A B C\)]], ", with ", Cell[BoxData[ \(TraditionalForm\`Q\)]], " on ", Cell[BoxData[ \(TraditionalForm\`C A\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "It is known that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B A C = 60\^\[EmptySmallCircle]\)]], " and that ", Cell[BoxData[ \(TraditionalForm\`A B + B P = A Q + Q B\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "What are the possible angles of triangle ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], "? " }], "Text", ZeroWidthTimes->True] }, Closed]], Cell[CellGroupData[{ Cell["Problem 6", "SubsectionIcon"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`a, b, c, d\)]], " be integers with ", Cell[BoxData[ \(TraditionalForm\`a > b > c > d > 0\)]], ". Suppose that " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a c + b d = \((b + d + a - c)\) \(\((b + d - a + c)\)\(.\)\)\)], "DisplayFormula"], Cell[TextData[{ "Prove that ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " is not prime. " }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problems with Solutions", "Section", PageBreakAbove->True], Cell[CellGroupData[{ Cell["Problem 1", "SubsectionIcon", ZeroWidthTimes->True], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], " be an acute\[Hyphen]angled triangle with circumcentre ", Cell[BoxData[ \(TraditionalForm\`O\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`P\)]], " on ", Cell[BoxData[ \(TraditionalForm\`B C\)]], " be the foot of the altitude from ", Cell[BoxData[ \(TraditionalForm\`A\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "Suppose that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B C A \[GreaterEqual] \[Angle] A B C + 30\^\[EmptySmallCircle]\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "Prove that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] C A B + \[Angle] C O P < \(\(90\^\[EmptySmallCircle]\)\(.\)\)\)]], " " }], "Text", ZeroWidthTimes->True], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon", ZeroWidthTimes->True], Cell["Solution 1", "MathCaption", ZeroWidthTimes->True], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = \[Angle] C A B\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Beta] = \[Angle] A B C\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Gamma] = \[Angle] B C A\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\[Delta] = \[Angle] C O P\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`K\)]], " and ", Cell[BoxData[ \(TraditionalForm\`Q\)]], " be the reflections of ", Cell[BoxData[ \(TraditionalForm\`A\)]], " and ", Cell[BoxData[ \(TraditionalForm\`P\)]], ", respectively, across the perpendicular bisector of ", Cell[BoxData[ \(TraditionalForm\`B C\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`R\)]], " denote the circumradius of ", Cell[BoxData[ \(TraditionalForm\`\[EmptyUpTriangle] A B C\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`O A = \(O B = \(O C = \(O K = R\)\)\)\)]], ". Furthermore, we have ", Cell[BoxData[ \(TraditionalForm\`Q P = K A\)]], " because ", Cell[BoxData[ \(TraditionalForm\`K Q P A\)]], " is a rectangle. Now note that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] A O K = \(\[Angle] A O B - \[Angle] K O B = \(\[Angle] A O B - \[Angle] A O C = 2 \[Gamma] - 2 \[Beta] \[GreaterEqual] 60\^\[EmptySmallCircle]\)\)\)]], ". 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\(TraditionalForm\`K A \[GreaterEqual] R\)]], " and ", Cell[BoxData[ \(TraditionalForm\`Q P \[GreaterEqual] R\)]], ". Therefore, using the Triangle Inequality, we have ", Cell[BoxData[ \(TraditionalForm\`O P + R = \(O Q + O C > Q C = Q P + P C \[GreaterEqual] R + P C\)\)]], ". It follows that ", Cell[BoxData[ \(TraditionalForm\`O P > P C\)]], ", and hence in ", Cell[BoxData[ \(TraditionalForm\`\[EmptyUpTriangle] C O P\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Angle] P C O > \[Delta]\)]], ". Now since ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = \(\(1\/2\) \[Angle] B O C = \(\(1\/2\) \((180\^\[EmptySmallCircle] - 2 \[Angle] P C O)\) = 90\^\[EmptySmallCircle] - \[Angle] P C O\)\)\)]], ", it indeed follows that ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] + \[Delta] < 90\^\[EmptySmallCircle]\)]], ". " }], "Text", ZeroWidthTimes->True], Cell["Solution 2", "MathCaption", ZeroWidthTimes->True], Cell[TextData[{ "As in the previous solution, it is enough to show that ", Cell[BoxData[ \(TraditionalForm\`O P > P C\)]], ". To this end, recall that by the (Extended) Law of Sines, ", Cell[BoxData[ \(TraditionalForm\`A B = 2 R sin \[Gamma]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`A C = 2 R sin \[Beta]\)]], ". Therefore, we have " }], "Text", ZeroWidthTimes->True], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["BP", FontSlant->"Italic"], "-", StyleBox["PC", FontSlant->"Italic"]}], "=", \(ABcos\[Beta] - A C cos \[Gamma] = \(2 \( R( sin \[Gamma] cos \[Beta] - sin \[Beta] cos \[Gamma])\) = 2 R \(\( sin(\[Gamma] - \[Beta])\)\(.\)\)\)\)}], TraditionalForm]], "DisplayFormula"], Cell["It follows from this and from ", "Text", ZeroWidthTimes->True], Cell[BoxData[ \(TraditionalForm\`30\^\[EmptySmallCircle] \[LessEqual] \[Gamma] - \ \[Beta] < \[Gamma] < 90\^\[EmptySmallCircle]\)], "DisplayFormula", ZeroWidthTimes->True], Cell[TextData[{ "that ", Cell[BoxData[ \(TraditionalForm\`B P - P C \[GreaterEqual] R\)]], ". Therefore, we obtain that ", Cell[BoxData[ \(TraditionalForm\`R + O P = B O + O P > B P \[GreaterEqual] R + P C\)]], ", from which ", Cell[BoxData[ \(TraditionalForm\`O P > O C\)]], ", as desired. " }], "Text", ZeroWidthTimes->True], Cell["Solution 3", "MathCaption", ZeroWidthTimes->True], Cell[TextData[{ "We first show that ", Cell[BoxData[ \(TraditionalForm\`R\^2 > C P\[CenterDot]C B\)]], ". To this end, since ", Cell[BoxData[ \(TraditionalForm\`C B = 2 R sin \[Alpha]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`C P = \(A C cos \[Gamma] = 2 R sin \[Beta] cos \[Gamma]\)\)]], ", it suffices to show that ", Cell[BoxData[ \(TraditionalForm\`1\/4 > sin \[Alpha] sin \[Beta] cos \[Gamma]\)]], ". We note that ", Cell[BoxData[ \(TraditionalForm\`1 > sin \[Alpha] = \(sin(\[Gamma] + \[Beta]) = sin \[Gamma] cos \[Beta] + sin \[Beta] cos \[Gamma]\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`1\/2 \[LessEqual] sin(\[Gamma] - \[Beta]) = sin \[Gamma] cos \[Beta] - sin \[Beta] cos \[Gamma]\)]], " since ", Cell[BoxData[ \(TraditionalForm\`30\^\[EmptySmallCircle] \[LessEqual] \[Gamma] - \ \[Beta] < 90\^\[EmptySmallCircle]\)]], ". It follows that ", Cell[BoxData[ \(TraditionalForm\`1\/4 > sin \[Beta] cos \[Gamma]\)]], " and that ", Cell[BoxData[ \(TraditionalForm\`1\/4 > sin \[Alpha] sin \[Beta] cos \[Gamma]\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "Now we choose a point ", Cell[BoxData[ \(TraditionalForm\`J\)]], " on ", Cell[BoxData[ \(TraditionalForm\`B C\)]], " so that ", Cell[BoxData[ \(TraditionalForm\`C J\[CenterDot]C P = R\^2\)]], ". It follows from this and from ", Cell[BoxData[ \(TraditionalForm\`R\^2 > C P\[CenterDot]C B\)]], " that ", Cell[BoxData[ \(TraditionalForm\`C J > C B\)]], ", so that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] O B C > \[Angle] O J C\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`O C/C J = P C/C O\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Angle] J C O = \[Angle] O C P\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`\[EmptyUpTriangle] J C O \[TildeFullEqual] \[EmptyUpTriangle] O C P\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Angle] O J C = \(\[Angle] P O C = \[Delta]\)\)]], ". It follows that ", Cell[BoxData[ \(TraditionalForm\`\[Delta] < \[Angle] O B C = 90\^\[EmptySmallCircle] - \[Alpha]\)]], " or ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] + \[Delta] < 90\^\[EmptySmallCircle]\)]], ". " }], "Text", ZeroWidthTimes->True], Cell["Solution 4", "MathCaption", ZeroWidthTimes->True], Cell[TextData[{ "On the one hand, as in the third solution, we have ", Cell[BoxData[ \(TraditionalForm\`R\^2 > C P\[CenterDot]C B\)]], ". On the other hand, the power of ", Cell[BoxData[ \(TraditionalForm\`P\)]], " with respect to the circumcircle of ", Cell[BoxData[ \(TraditionalForm\`\[EmptyUpTriangle] A B C\)]], " is ", Cell[BoxData[ \(TraditionalForm\`B P\[CenterDot]P C = R\^2 - O P\^2\)]], ". From these two equations we find that " }], "Text", ZeroWidthTimes->True], Cell[BoxData[ \(TraditionalForm\`\(\(O P\^2 = \(R\^2 - B P\[CenterDot]P C > P C\[CenterDot]C B - B P\[CenterDot]P C = P C\^2\)\)\(,\)\)\)], "DisplayFormula", ZeroWidthTimes->True], Cell[TextData[{ "from which ", Cell[BoxData[ \(TraditionalForm\`O P > P C\)]], ". Therefore, as in the first solution, we conclude that ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] + \[Delta] < 90\^\[EmptySmallCircle]\)]], ". " }], "Text", ZeroWidthTimes->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problem 2", "SubsectionIcon"], Cell["Prove that ", "Text"], Cell[BoxData[ \(TraditionalForm\`a\/\@\(a\^2 + 8 b c\) + b\/\@\(b\^2 + 8 c a\) + c\/\@\(c\^2 + 8 a b\) \[GreaterEqual] 1\)], "DisplayFormula"], Cell[TextData[{ "for all positive real numbers ", Cell[BoxData[ \(TraditionalForm\`a, b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon"], Cell["First we shall prove that ", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(a\/\@\(a\^2 + 8 b c\) \[GreaterEqual] a\^\(4\/3\)\/\(a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\)\)\)\(,\)\)\)], \ "DisplayFormula"], Cell["or equivalently, that ", "Text"], Cell[BoxData[ \(TraditionalForm\`\((a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\))\)\^2 \ \[GreaterEqual] \(\(\(a\^\(2\/3\)\)( a\^2 + 8 b c)\)\(.\)\)\)], "DisplayFormula"], Cell["The AM\[Hyphen]GM inequality yields ", "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\ \))\)\^2 - \((a\^\(4\/3\))\)\^2\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((b\^\(4\/3\) + c\^\(4\/3\))\) \((a\^\(4\/3\) + a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\))\)\)]], " " }]]}, {Cell[""], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` \[GreaterEqual] \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`2 \( b\^\(2\/3\)\) c\^\(2\/3\)\[CenterDot]4 \( a\^\(2\/3\)\) \(b\^\(1\/3\)\) c\^\(1\/3\)\)]], " " }]]}, {Cell[""], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`8 \( a\^\(2\/3\)\) b \( \(c\)\(.\)\)\)]]]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", CellTags->"\@@FnameCK"], Cell["Thus ", "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\ \))\)\^2\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` \[GreaterEqual] \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((a\^\(4\/3\))\)\^2 + 8 \( a\^\(2\/3\)\) b c\)]], " " }]]}, {Cell[""], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(\(\(a\^\(2\/3\)\)( a\^2 + 8 b c)\)\(,\)\)\)]], " " }]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", CellTags->"\@@FnameCK"], Cell["so ", "Text"], Cell[BoxData[ \(TraditionalForm\`a\/\@\(a\^2 + 8 b c\) \[GreaterEqual] a\^\(4\/3\)\/\(\(a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\)\)\(.\)\)\)], \ "DisplayFormula"], Cell["Similarly, we have ", "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`b\/\@\(b\^2 + 8 c a\)\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` \[GreaterEqual] \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`b\^\(4\/3\)\/\(a\^\(4\/3\) + b\^\(4\/3\) \ + c\^\(4\/3\)\)\)]], " and" }]]}, {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`c\/\@\(c\^2 + 8 a b\)\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` \[GreaterEqual] \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(\(c\^\(4\/3\)\/\(a\^\(4\/3\) + b\^\(4\/3\) + c\^\(4\/3\)\)\)\(.\)\)\)]], " " }]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", CellTags->"\@@FnameCK"], Cell["Adding these three inequalities yields ", "Text"], Cell[BoxData[ \(TraditionalForm\`a\/\@\(a\^2 + 8 b c\) + b\/\@\(b\^2 + 8 c a\) + c\/\@\(c\^2 + 8 a b\) \[GreaterEqual] 1. \)], "DisplayFormula"], Cell[TextData[{ StyleBox["Comment.", "TI"], " It can be shown that for any ", Cell[BoxData[ \(TraditionalForm\`a, b, c > 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Lambda] \[GreaterEqual] 8\)]], ", the following inequality holds: " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a\/\@\(a\^2 + \[Lambda] b c\) + b\/\@\(b\^2 + \[Lambda] c a\) + c\/\@\(c\^2 + \[Lambda] a b\) \[GreaterEqual] 3\/\(\(\@\(1 + \[Lambda]\)\)\(.\)\)\)], "DisplayFormula"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problem 3", "SubsectionIcon"], Cell["\<\ Twenty\[Hyphen]one girls and twenty\[Hyphen]one boys took part in a \ mathematical contest. \ \>", "Text"], Cell["\<\ \[Bullet] Each contestant solved at most six problems. \[Bullet] For each girl and each boy, at least one problem was solved by \ both of them. \ \>", "IndentedText"], Cell["\<\ Prove that there was a problem that was solved by at least three \ girls and at least three boys. \ \>", "Text"], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon"], Cell["Solution 1", "MathCaption", CounterAssignments->{{"NumberedEquation", 0}}], Cell[TextData[{ "We introduce the following symbols: ", Cell[BoxData[ \(TraditionalForm\`G\)]], " is the set of girls at the competition and ", Cell[BoxData[ \(TraditionalForm\`B\)]], " is the set of boys, ", Cell[BoxData[ \(TraditionalForm\`P\)]], " is the set of problems, ", Cell[BoxData[ \(TraditionalForm\`P(g)\)]], " is the set of problems solved by ", Cell[BoxData[ \(TraditionalForm\`g \[Element] G\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`P(b)\)]], " is the set of problems solved by ", Cell[BoxData[ \(TraditionalForm\`b \[Element] B\)]], ". Finally, ", Cell[BoxData[ \(TraditionalForm\`G(p)\)]], " is the set of girls that solve ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], " and ", Cell[BoxData[ \(TraditionalForm\`B(p)\)]], " is the set of boys that solve ", Cell[BoxData[ \(TraditionalForm\`p\)]], ". In terms of this notation, we have that for all ", Cell[BoxData[ \(TraditionalForm\`g \[Element] G\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b \[Element] B\)]], ", " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\((a)\)\ \ \[VerticalSeparator] P(g) \[VerticalSeparator] \(\(\[LessEqual]\)\(6 \(\(,\)\(\ \)\)\)\) \ \[VerticalSeparator] P(b) \[VerticalSeparator] \(\(\[LessEqual]\)\(6 \(\(,\)\(\ \ \ \ \)\) \ \((b)\)\ \ \(P(g)\) \[Intersection] P(b)\)\(\[NotEqual]\)\(\(\[Diameter]\)\(.\)\)\)\)], "DisplayFormula"], Cell[TextData[{ "We wish to prove that some ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], " satisfies ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], ". To do this, we shall assume the contrary and reach a contradiction by \ counting (two ways) all ordered triples ", Cell[BoxData[ \(TraditionalForm\`\((p, q, r)\)\)]], " such that ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P(g) \[Intersection] P(b)\)]], ". With ", Cell[BoxData[ \(TraditionalForm\`T = {\((p, g, b)\) \(\(:\)\(\ \)\) p \[Element] P(g) \[Intersection] P(b)}\)]], ", condition (b) yields " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{"\[VerticalSeparator]", "T", "\[VerticalSeparator]", "=", UnderscriptBox["\[Sum]", \(g \[Element] G\), LimitsPositioning->False], UnderscriptBox["\[Sum]", \(b \[Element] B\), LimitsPositioning->False], "\[VerticalSeparator]", "P", \((g)\), "\[Intersection]", "P", \((b)\), "\[VerticalSeparator]", "\[GreaterEqual]", "\[VerticalSeparator]", "G", "\[VerticalSeparator]", "\[CenterDot]", "\[VerticalSeparator]", "B", "\[VerticalSeparator]", "=", \(21\^2\), "."}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ "Assume that no ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], " satisfies ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], ". We begin by noting that " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] P\), LimitsPositioning->False], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "=", UnderscriptBox["\[Sum]", \(g \[Element] G\), LimitsPositioning->False], "\[VerticalSeparator]", "P", \((g)\), "\[VerticalSeparator]", "\[LessEqual]", "6", "\[VerticalSeparator]", "G", \(\(\[VerticalSeparator]\)\(\ \ \ \ \)\), "and", " ", UnderscriptBox["\[Sum]", \(p \[Element] P\), LimitsPositioning->False], "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]", "\[LessEqual]", "6", "\[VerticalSeparator]", "B", "\[VerticalSeparator]", "."}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ "(", StyleBox["Note.", "TI"], " The equality in (2) is obtained by a standard double\[Hyphen]counting \ technique: Let ", Cell[BoxData[ \(TraditionalForm\`\[Chi](g, p) = 1\)]], " if ", Cell[BoxData[ \(TraditionalForm\`g\)]], " solves ", Cell[BoxData[ \(TraditionalForm\`p\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Chi](g, p) = 0\)]], " otherwise, and interchange the orders of summation in ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] P\), LimitsPositioning->True], UnderscriptBox["\[Sum]", \(g \[Element] G\), LimitsPositioning->True], "\[Chi]", \((g, p)\)}], TraditionalForm]]], ".) Let " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(P\_+\) = {p \[Element] P \(\(:\)\(\ \)\) \[VerticalSeparator] G(p) \[VerticalSeparator] \(\(\[GreaterEqual]\)\(3\)\)}, \n\(P\_-\) \ = \(\({p \[Element] P \(\(:\)\(\ \)\) \[VerticalSeparator] G(p) \[VerticalSeparator] \(\(\[LessEqual]\)\(2\)\)}\)\(.\)\)\)], \ "DisplayFormula"], Cell[TextData[{ StyleBox["Claim.", "TB"], " ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[GreaterEqual]", "\[VerticalSeparator]", "G", "\[VerticalSeparator]"}], TraditionalForm]]], "; thus ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[LessEqual]", "5", "\[VerticalSeparator]", "G", "\[VerticalSeparator]"}], TraditionalForm]]], ". Also ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->True], "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]", "\[GreaterEqual]", "\[VerticalSeparator]", "B", "\[VerticalSeparator]"}], TraditionalForm]]], "; thus ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->True], \(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(5\)\)\(\ \[VerticalSeparator]\)\(B\)\(\[VerticalSeparator]\)\)}], TraditionalForm]]], ". " }], "Text"], Cell[TextData[{ StyleBox["Proof.", "TI"], " Let ", Cell[BoxData[ \(TraditionalForm\`g \[Element] G\)]], " be arbitrary. By the Pigeonhole Principle, conditions (a) and (b) imply \ that ", Cell[BoxData[ \(TraditionalForm\`g\)]], " solves some problem ", Cell[BoxData[ \(TraditionalForm\`p\)]], " that is solved by at least ", Cell[BoxData[ \(TraditionalForm\`\[LeftCeiling]21/6\[RightCeiling] = 4\)]], " boys. By assumption, ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(4\)\)\)\)]], " implies that ", Cell[BoxData[ \(TraditionalForm\`p \[Element] \(P\_-\)\)]], ", so every girl solves at least one problem in ", Cell[BoxData[ \(TraditionalForm\`\(P\_-\)\)]], ". Thus " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->False], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[GreaterEqual]", "\[VerticalSeparator]", "G", "\[VerticalSeparator]", "."}], TraditionalForm]], "NumberedEquation"], Cell["In view of (2) and (3) we have ", "Text"], Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->False], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "=", UnderscriptBox["\[Sum]", \(p \[Element] P\), LimitsPositioning->False], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "-", UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->False], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[LessEqual]", "5", "\[VerticalSeparator]", "G", "\[VerticalSeparator]", "."}], TraditionalForm]], "DisplayFormula"], Cell[TextData[{ "Also, each boy solves a problem that is solved by at least four girls, so \ each boy solves a problem ", Cell[BoxData[ \(TraditionalForm\`p \[Element] \(P\_+\)\)]], ". Thus ", Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->True], "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]", "\[GreaterEqual]", "\[VerticalSeparator]", "B", "\[VerticalSeparator]"}], TraditionalForm]]], ", and the calculation proceeds as before using (2). ", Cell[BoxData[ \(TraditionalForm\`\[Placeholder]\)]], " \nUsing the claim just established, we find " }], "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(T\)\(\ \[VerticalSeparator]\)\)\)]], " " }]], Cell[TextData[Cell[BoxData[ \(TraditionalForm\` = \)]]]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] P\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[CenterDot]", "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]"}], TraditionalForm]]], " " }]]}, {Cell[""], Cell[TextData[Cell[BoxData[ \(TraditionalForm\` = \)]]]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[CenterDot]", "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]", "+", UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", "\[CenterDot]", "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]"}], TraditionalForm]]], " " }]]}, {Cell[""], Cell[TextData[Cell[BoxData[ \(TraditionalForm\` \[LessEqual] \)]]]], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{"2", UnderscriptBox["\[Sum]", \(p \[Element] \(P\_+\)\), LimitsPositioning->True], "\[VerticalSeparator]", "G", \((p)\), "\[VerticalSeparator]", \(+2\), UnderscriptBox["\[Sum]", \(p \[Element] \(P\_-\)\), LimitsPositioning->True], "\[VerticalSeparator]", "B", \((p)\), "\[VerticalSeparator]"}], TraditionalForm]]], " " }]]}, {Cell[""], Cell[TextData[Cell[BoxData[ \(TraditionalForm\` \[LessEqual] \)]]]], Cell[TextData[Cell[ BoxData[ \(TraditionalForm\`10 \[VerticalSeparator] G \[VerticalSeparator] \(+10\) \[VerticalSeparator] B \[VerticalSeparator] = 20\[CenterDot]21. \)]]]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula"], Cell["This contradicts (1), so the proof is complete. ", "Text"], Cell["Solution 2", "MathCaption"], Cell[TextData[{ "Let us use some of the notation given in the first solution. Suppose that \ for every ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], " either ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(2\)\)\)\)]], " or ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(2\)\)\)\)]], ". For each ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], ", color ", Cell[BoxData[ \(TraditionalForm\`p\)]], " red if ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(2\)\)\)\)]], " and otherwise color it black. In this way, if ", Cell[BoxData[ \(TraditionalForm\`p\)]], " is red then ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(2\)\)\)\)]], " and if ", Cell[BoxData[ \(TraditionalForm\`p\)]], " is black then ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[LessEqual]\)\(2\)\)\)\)]], ". Consider a chessboard with 21 rows, each representing one of the girls, \ and 21 columns, each representing one of the boys. For each ", Cell[BoxData[ \(TraditionalForm\`g \[Element] G\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b \[Element] B\)]], ", color the square corresponding to ", Cell[BoxData[ \(TraditionalForm\`\((g, b)\)\)]], " as follows: pick ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P(g) \[Intersection] P(b)\)]], " and assign ", Cell[BoxData[ \(TraditionalForm\`p\)]], "'s color to that square. (By condition (b), there is always an available \ choice.) By the Pigeonhole Principle, one of the two colors is assigned to at \ least ", Cell[BoxData[ \(TraditionalForm\`\[LeftCeiling]441/2\[RightCeiling] = 221\)]], " squares, and thus some row has at least ", Cell[BoxData[ \(TraditionalForm\`\[LeftCeiling]221/21\[RightCeiling] = 11\)]], " black squares or some column has at least 11 red squares. \nSuppose the \ row corresponding to ", Cell[BoxData[ \(TraditionalForm\`g \[Element] G\)]], " has at least 11 black squares. Then for each of 11 squares, the black \ problem that was chosen in assigning the color was solved by at most 2 boys. \ Thus we account for at least ", Cell[BoxData[ \(TraditionalForm\`\[LeftCeiling]11/2\[RightCeiling] = 6\)]], " distinct problems solved by ", Cell[BoxData[ \(TraditionalForm\`g\)]], ". In view of condition (a), ", Cell[BoxData[ \(TraditionalForm\`g\)]], " solves only these problems. But then at most 12 boys solve a problem also \ solved by ", Cell[BoxData[ \(TraditionalForm\`g\)]], ", in violation of condition (b). \nIn exactly the same way, a \ contradiction is reached if we suppose that some column has at least 11 red \ squares. Hence some ", Cell[BoxData[ \(TraditionalForm\`p \[Element] P\)]], " satisfies ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(G( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(\[VerticalSeparator]\)\(B( p)\)\(\[VerticalSeparator]\)\(\(\[GreaterEqual]\)\(3\)\)\)\)]], ". " }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problem 4", "SubsectionIcon"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`n\)]], " be an odd integer greater than 1, and let ", Cell[BoxData[ \(TraditionalForm\`k\_1, k\_2, \[Ellipsis], k\_n\)]], " be given integers. For each of the ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " permutations ", Cell[BoxData[ \(TraditionalForm\`a = \((a\_1, a\_2, \[Ellipsis], a\_n)\)\)]], " of ", Cell[BoxData[ \(TraditionalForm\`1, 2, \[Ellipsis], n\)]], ", let " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{"S", \((a)\), "=", UnderoverscriptBox["\[Sum]", \(i = 1\), "n", LimitsPositioning->False], \(k\_i\), \(a\_i\), "."}], TraditionalForm]], "DisplayFormula"], Cell[TextData[{ "Prove that there are two permutations ", Cell[BoxData[ \(TraditionalForm\`b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b \[NotEqual] c\)]], ", such that ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " is a divisor of ", Cell[BoxData[ \(TraditionalForm\`S(b) - S(c)\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon", CounterAssignments->{{"NumberedEquation", 0}}], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`\[Sum]S(a)\)]], " be the sum of ", Cell[BoxData[ \(TraditionalForm\`S(a)\)]], " over all ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " permutations ", Cell[BoxData[ \(TraditionalForm\`a = \((a\_1, a\_2, \[Ellipsis], a\_n)\)\)]], ". We compute ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\(S(a)\)\ mod\ \(n!\)\)]], " two ways, one of which depends on the desired conclusion being false, and \ reach a contradiction when ", Cell[BoxData[ \(TraditionalForm\`n\)]], " is odd. \n", StyleBox["First way.", "TI"], " In ", Cell[BoxData[ \(TraditionalForm\`\[Sum]S(a)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`k\_1\)]], " is multiplied by each ", Cell[BoxData[ \(TraditionalForm\`i \[Element] {1, \[Ellipsis], n}\)]], " a total of ", Cell[BoxData[ \(TraditionalForm\`\(\((n - 1)\)!\)\)]], " times, once for each permutation of ", Cell[BoxData[ \(TraditionalForm\`{1, \[Ellipsis], n}\)]], " in which ", Cell[BoxData[ \(TraditionalForm\`a\_1 = i\)]], ". Thus the coefficient of ", Cell[BoxData[ \(TraditionalForm\`k\_1\)]], " in ", Cell[BoxData[ \(TraditionalForm\`\[Sum]S(a)\)]], " is " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(\((n - 1)\)!\) \((1 + 2 + \[CenterEllipsis] + n)\) = \(\((n + 1)\)!\)/2. \)], "DisplayFormula"], Cell[TextData[{ "The same is true for all ", Cell[BoxData[ \(TraditionalForm\`k\_i\)]], ", so " }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]S(a)\), "=", RowBox[{\(\(\((n + 1)\)!\)\/2\), RowBox[{ UnderoverscriptBox["\[Sum]", \(i = 1\), "n", LimitsPositioning->False], \(\(k\_i\)\(.\)\)}]}]}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ StyleBox["Second way.", "TI"], " If ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " is not a divisor of ", Cell[BoxData[ \(TraditionalForm\`S(b) - S(c)\)]], " for any ", Cell[BoxData[ \(TraditionalForm\`b \[NotEqual] c\)]], ", then each ", Cell[BoxData[ \(TraditionalForm\`S(a)\)]], " must have a different remainder mod ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], ". Since there are ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], " permutations, these remainders must be precisely the numbers ", Cell[BoxData[ \(TraditionalForm\`0, 1, 2, \[Ellipsis], \(n!\) - 1\)]], ". Thus " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\[Sum]S( a) \[Congruent] \(\(\((\(n!\) - 1)\) \(n!\)\)\/2\) mod\ \(\(n!\)\(.\)\)\)], "NumberedEquation"], Cell["Combining (1) and (2), we get ", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\(\((n + 1)\)!\)\/2\), RowBox[{ UnderoverscriptBox["\[Sum]", \(i = 1\), "n", LimitsPositioning->False], \(k\_i\)}]}], "\[Congruent]", \(\(\(\((\(n!\) - 1)\) \(n!\)\)\/2\) mod\ \(\(n!\)\(.\)\)\)}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ "Now, for ", Cell[BoxData[ \(TraditionalForm\`n\)]], " odd, the left side of (3) is congruent to 0 modulo ", Cell[BoxData[ \(TraditionalForm\`\(n!\)\)]], ", while for ", Cell[BoxData[ \(TraditionalForm\`n > 1\)]], " the right side is not congruent to 0 (", Cell[BoxData[ \(TraditionalForm\`\(n!\) - 1\)]], " is odd). For ", Cell[BoxData[ \(TraditionalForm\`n > 1\)]], " and odd, we have a contradiction. " }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problem 5", "SubsectionIcon"], Cell[TextData[{ "In a triangle ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], ", let ", Cell[BoxData[ \(TraditionalForm\`A P\)]], " bisect ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B A C\)]], ", with ", Cell[BoxData[ \(TraditionalForm\`P\)]], " on ", Cell[BoxData[ \(TraditionalForm\`B C\)]], ", and let ", Cell[BoxData[ \(TraditionalForm\`B Q\)]], " bisect ", Cell[BoxData[ \(TraditionalForm\`\[Angle] A B C\)]], ", with ", Cell[BoxData[ \(TraditionalForm\`Q\)]], " on ", Cell[BoxData[ \(TraditionalForm\`C A\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "It is known that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B A C = 60\^\[EmptySmallCircle]\)]], " and that ", Cell[BoxData[ \(TraditionalForm\`A B + B P = A Q + Q B\)]], ". " }], "Text", ZeroWidthTimes->True], Cell[TextData[{ "What are the possible angles of triangle ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], "? " }], "Text", ZeroWidthTimes->True], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon"], Cell[TextData[{ "Denote the angles of ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], " by ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 60\^\[EmptySmallCircle]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Beta]\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\[Gamma]\)]], ". Extend ", Cell[BoxData[ \(TraditionalForm\`A B\)]], " to ", Cell[BoxData[ \(TraditionalForm\`P\^\[Prime]\)]], " so that ", Cell[BoxData[ \(TraditionalForm\`B P\^\[Prime] = B P\)]], ", and construct ", Cell[BoxData[ \(TraditionalForm\`P\^\[Prime]\[Prime]\)]], " on ", Cell[BoxData[ \(TraditionalForm\`A Q\)]], " so that ", Cell[BoxData[ \(TraditionalForm\`A P\^\[Prime]\[Prime] = A P\^\[Prime]\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`B \( P\^\[Prime]\) P\)]], " is an isosceles triangle with base angle ", Cell[BoxData[ \(TraditionalForm\`\[Beta]/2\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`A Q + Q P\^\[Prime]\[Prime] = \(A B + B P\^\[Prime] = \(A B + B P = A Q + Q B\)\)\)]], ", it follows that ", Cell[BoxData[ \(TraditionalForm\`Q P\^\[Prime]\[Prime] = Q B\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`A \( P\^\[Prime]\) P\^\[Prime]\[Prime]\)]], " is equilateral and ", Cell[BoxData[ \(TraditionalForm\`A P\)]], " bisects the angle at ", Cell[BoxData[ \(TraditionalForm\`A\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`P P\^\[Prime] = P P\^\[Prime]\[Prime]\)]], ". 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" }], "Text"], Cell[TextData[{ StyleBox["Proof.", "TI"], " Suppose to the contrary that ", Cell[BoxData[ \(TraditionalForm\`B P P\^\[Prime]\[Prime]\)]], " is a nondegenerate triangle. We have that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] P B Q = \(\[Angle] P \( P\^\[Prime]\) B = \(\[Angle] P \( P\^\[Prime]\[Prime]\) Q = \[Beta]/2\)\)\)]], ". Thus the diagram appears as below, or else with ", Cell[BoxData[ \(TraditionalForm\`P\)]], " is on the other side of ", Cell[BoxData[ \(TraditionalForm\`B P\^\[Prime]\[Prime]\)]], ". In either case, the assumption that ", Cell[BoxData[ \(TraditionalForm\`B P P\^\[Prime]\[Prime]\)]], " is nondegenerate leads to ", Cell[BoxData[ \(TraditionalForm\`B P = \(P P\^\[Prime]\[Prime] = P P\^\[Prime]\)\)]], ", thus to the conclusion that ", Cell[BoxData[ \(TraditionalForm\`B P P\^\[Prime]\)]], " is equilateral, and finally to the absurdity ", Cell[BoxData[ \(TraditionalForm\`\[Beta]/2 = 60\^\[EmptySmallCircle]\)]], " so ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] + \[Beta] = \(60\^\[EmptySmallCircle] + 120\^\[EmptySmallCircle] = 180\^\[EmptySmallCircle]\)\)]], ". 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\ \[Beta]/2\)\)]], ", so ", Cell[BoxData[ \(TraditionalForm\`\[Beta] = 80\^\[EmptySmallCircle]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Gamma] = 40\^\[EmptySmallCircle]\)]], ". Thus ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], " is a 60\[Hyphen]80\[Hyphen]40 degree triangle. " }], "Text", ZeroWidthTimes->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problem 6", "SubsectionIcon"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`a, b, c, d\)]], " be integers with ", Cell[BoxData[ \(TraditionalForm\`a > b > c > d > 0\)]], ". Suppose that " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a c + b d = \((b + d + a - c)\) \(\((b + d - a + c)\)\(.\)\)\)], "DisplayFormula"], Cell[TextData[{ "Prove that ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " is not prime. " }], "Text"], Cell[CellGroupData[{ Cell["Solution", "SubsubsectionIcon"], Cell["Solution 1", "MathCaption"], Cell[TextData[{ "Suppose to the contrary that ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " is prime. Note that " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a b + c d = \(\((a + d)\) c + \((b - c)\) a = m\[CenterDot]gcd \((a + d, b - c)\)\)\)], "DisplayFormula"], Cell[TextData[{ "for some positive integer ", Cell[BoxData[ \(TraditionalForm\`m\)]], ". By assumption, either ", Cell[BoxData[ \(TraditionalForm\`m = 1\)]], " or gcd", Cell[BoxData[ \(TraditionalForm\`\((a + d, b - c)\) = 1\)]], ". We consider these alternatives in turn. \n", StyleBox["Case (i):", "TI"], " ", Cell[BoxData[ \(TraditionalForm\`m = 1\)]], ". Then " }], "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`gcd(a + d, b - c)\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a b + c d > a b + c d - \((a - b + c + d)\)\)]], " " }]]}, {Cell[""], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((a + d)\) \((c - 1)\) + \((b - c)\) \((a + 1)\)\)]], " " }]]}, {Cell[""], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` \[GreaterEqual] \)]], " " }]], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\(\(gcd(a + d, b - c)\)\(,\)\)\)]]]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", CellTags->"\@@FnameCK"], Cell[TextData[{ "which is false. \n", StyleBox["Case (ii):", "TI"], " ", Cell[BoxData[ \(TraditionalForm\`gcd(a + d, b - c) = 1\)]], ". Substituting ", Cell[BoxData[ \(TraditionalForm\`a c + b d = \((a + d)\) b - \((b - c)\) a\)]], " for the left\[Hyphen]hand side of ", Cell[BoxData[ \(TraditionalForm\`a c + b d = \((b + d + a - c)\) \((b + d - a + c)\)\)]], ", we obtain " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\((a + d)\) \((a - c - d)\) = \((b - c)\) \(\((b + c + d)\)\(.\)\)\)], "DisplayFormula"], Cell[TextData[{ "In view of this, there exists a positive integer ", Cell[BoxData[ \(TraditionalForm\`k\)]], " such that " }], "Text", CellTags->"\@@FnameCK"], Cell[BoxData[ FormBox[GridBox[{ {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a - c - d\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(\(k(b - c)\)\(,\)\)\)]], " " }]]}, {Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`b + c + d\)]], " " }]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\` = \)]], " " }]], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\(\(k(a + d)\)\(.\)\)\)]]]]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", CellTags->"\@@FnameCK"], Cell[TextData[{ "Adding these equations, we obtain ", Cell[BoxData[ \(TraditionalForm\`a + b = k(a + b - c + d)\)]], " and thus ", Cell[BoxData[ \(TraditionalForm\`k(c - d) = \((k - 1)\) \((a + b)\)\)]], ". Recall that ", Cell[BoxData[ \(TraditionalForm\`a > b > c > d\)]], ". If ", Cell[BoxData[ \(TraditionalForm\`k = 1\)]], " then ", Cell[BoxData[ \(TraditionalForm\`c = d\)]], ", a contradiction. If ", Cell[BoxData[ \(TraditionalForm\`k \[GreaterEqual] 2\)]], " then " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(2 \[GreaterEqual] k\/\(k - 1\) = \(a + b\)\/\(c - d\) > 2\)\(,\)\)\)], "DisplayFormula"], Cell[TextData[{ "a contradiction. \nSince a contradiction is reached in both (i) and (ii), \ ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " is not prime. " }], "Text"], Cell["Solution 2", "MathCaption", CounterAssignments->{{"NumberedEquation", 0}}], Cell[TextData[{ "The equality ", Cell[BoxData[ \(TraditionalForm\`a c + b d = \((b + d + a - c)\) \((b + d - a + c)\)\)]], " is equivalent to " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a\^2 - a c + c\^2 = b\^2 + b d + \(\(d\^2\)\(.\)\)\)], "NumberedEquation"], Cell[TextData[{ "Let ", Cell[BoxData[ \(TraditionalForm\`A B C D\)]], " be the quadrilateral with ", Cell[BoxData[ \(TraditionalForm\`A B = a\)]], ", ", Cell[BoxData[ \(TraditionalForm\`B C = d\)]], ", ", Cell[BoxData[ \(TraditionalForm\`C D = b\)]], ", ", Cell[BoxData[ \(TraditionalForm\`A D = c\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B A D = 60\^\[EmptySmallCircle]\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\[Angle] B C D = 120\^\[EmptySmallCircle]\)]], ". Such a quadrilateral exists in view of (1) and the Law of Cosines; the \ common value in (1) is ", Cell[BoxData[ \(TraditionalForm\`B D\^2\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`\[Angle] A B C = \[Alpha]\)]], ", so that ", Cell[BoxData[ \(TraditionalForm\`\[Angle] C D A = 180\^\[EmptySmallCircle] - \[Alpha]\)]], ". Applying the Law of Cosines to triangles ", Cell[BoxData[ \(TraditionalForm\`A B C\)]], " and ", Cell[BoxData[ \(TraditionalForm\`A C D\)]], " gives " }], "Text", ZeroWidthTimes->True], Cell[BoxData[ \(TraditionalForm\`a\^2 + d\^2 - 2 a d cos \[Alpha] = \(A C\^2 = b\^2 + c\^2 + 2 b c cos \( \(\[Alpha]\)\(.\)\)\)\)], "DisplayFormula"], Cell[TextData[{ "Hence ", Cell[BoxData[ \(TraditionalForm\`2 cos \[Alpha] = \((a\^2 + d\^2 - b\^2 - c\^2)\)/\((a d + b c)\)\)]], ", and " }], "Text"], Cell[BoxData[ \(TraditionalForm\`A C\^2 = \(a\^2 + d\^2 - a \( \(d\)\(\ \)\) \(a\^2 + d\^2 - b\^2 - c\^2\)\/\(a d + b c\) = \ \(\((a b + c d)\) \((a c + b d)\)\)\/\(\(a d + b c\)\(.\)\)\)\)], \ "DisplayFormula"], Cell[TextData[{ "Because ", Cell[BoxData[ \(TraditionalForm\`A B C D\)]], " is cyclic, Ptolemy's Theorem gives " }], "Text", ZeroWidthTimes->True], Cell[BoxData[ \(TraditionalForm\`\((A C\[CenterDot]B D)\)\^2 = \((a b + c \ d)\)\^2\)], "DisplayFormula"], Cell["It follows that ", "Text"], Cell[BoxData[ \(TraditionalForm\`\((a c + b d)\) \((a\^2 - a c + c\^2)\) = \((a b + c d)\) \(\((a d + b c)\)\(.\)\)\)], "NumberedEquation"], Cell[TextData[{ "(", StyleBox["Note.", "TI"], " Straightforward algebra can also be used obtain (2) from (1).) Next \ observe that " }], "Text"], Cell[BoxData[ \(TraditionalForm\`a b + c d > a c + b d > a d + b \( \(c\)\(.\)\)\)], "NumberedEquation"], Cell[TextData[{ "The first inequality follows from ", Cell[BoxData[ \(TraditionalForm\`\((a - d)\) \((b - c)\) > 0\)]], ", and the second from ", Cell[BoxData[ \(TraditionalForm\`\((a - b)\) \((c - d)\) > 0\)]], ". " }], "Text"], Cell[TextData[{ "Now assume that ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " is prime. It then follows from (3) that ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a c + b d\)]], " are relatively prime. Hence, from (2), it must be true that ", Cell[BoxData[ \(TraditionalForm\`a c + b d\)]], " divides ", Cell[BoxData[ \(TraditionalForm\`a d + b c\)]], ". However, this is impossible by (3). Thus ", Cell[BoxData[ \(TraditionalForm\`a b + c d\)]], " must not be prime. " }], "Text"], Cell[TextData[{ StyleBox["Note.", "TI"], " Examples of 4\[Hyphen]tuples ", Cell[BoxData[ \(TraditionalForm\`\((a, b, c, d)\)\)]], " that satisfy the given conditions are ", Cell[BoxData[ \(TraditionalForm\`\((21, 18, 14, 1)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((65, 50, 34, 11)\)\)]], ". " }], "Text"] }, Closed]] }, Closed]] }, Closed]] }, FrontEndVersion->"4.1 for Macintosh", ScreenRectangle->{{0, 1152}, {0, 850}}, ScreenStyleEnvironment->"Working", WindowSize->{649, 599}, WindowMargins->{{2, Automatic}, {Automatic, 0}}, PrintingCopies->1, PrintingPageRange->{1, Automatic}, Magnification->1, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions for On-line Help", "Subtitle"], Cell["\<\ These styles are carefully tuned for all aspects of the online \ help. 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The \"Hyperlink\" style is for links within the same Notebook, \ or between Notebooks. The others point to the various sections of the help \ browser. 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FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLink", "Printout"], FontSize->11, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["MainBookLink", "EnhancedPrintout"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["MainBookLink", "EnhancedPrintoutNonGray"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MainBookLinkMR"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLinkMR", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["MainBookLinkMR", "EnhancedPrintout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["MainBookLinkMR", "EnhancedPrintoutNonGray"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnsLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["AddOnsLink", "EnhancedPrintout"], FontFamily->"WRICourier", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["AddOnsLink", "EnhancedPrintoutNonGray"], FontFamily->"WRICourier", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLinkText"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Times", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnsLinkText", "Printout"], FontSize->11, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["AddOnsLinkText", "EnhancedPrintout"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["AddOnsLinkText", "EnhancedPrintoutNonGray"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuide", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["RefGuideLink", "EnhancedPrintout"], FontFamily->"WRICourier", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["RefGuideLink", "EnhancedPrintoutNonGray"], FontFamily->"WRICourier", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Printout"], FontSize->11, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["GettingStartedLink", "EnhancedPrintout"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["GettingStartedLink", "EnhancedPrintoutNonGray"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["OtherInformationLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "OtherInformation", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["OtherInformationLink", "Printout"], FontSize->11, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["OtherInformationLink", "EnhancedPrintout"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["OtherInformationLink", "EnhancedPrintoutNonGray"], FontFamily->"Palatino", FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["OtherInformationLinkMR"], StyleMenuListing->None, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "OtherInformation", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["OtherInformationLinkMR", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["OtherInformationLinkMR", "EnhancedPrintout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}], Cell[StyleData["OtherInformationLinkMR", "EnhancedPrintoutNonGray"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Palette Styles", "Section"], Cell["\<\ The cells below define styles that define standard \ ButtonFunctions, for use in palette buttons.\ \>", "Text"], Cell[StyleData["Paste"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, After]}]&)}], Cell[StyleData["Evaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["EvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionMove[ FrontEnd`InputNotebook[ ], All, Cell, 1], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}] }, Closed]], Cell[CellGroupData[{ Cell["Inline Formatting", "Section"], Cell["\<\ These styles are for modifying individual words or letters in a \ cell exclusive of the cell tag.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["JT"], StyleMenuListing->None, FontFamily->"JFontText"], Cell[StyleData["JT", "Printout"]], Cell[StyleData["JT", "EnhancedPrintout"]], Cell[StyleData["JT", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["JS"], StyleMenuListing->None, FontFamily->"JFontSans"], Cell[StyleData["JS", "Printout"]], Cell[StyleData["JS", "EnhancedPrintout"]], Cell[StyleData["JS", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["JSB"], StyleMenuListing->None, FontFamily->"JFontSansBold"], Cell[StyleData["JSB", "Printout"]], Cell[StyleData["JSB", "EnhancedPrintout"]], Cell[StyleData["JSB", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RM"], StyleMenuListing->None, FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["RM", "Printout"]], Cell[StyleData["RM", "EnhancedPrintout"]], Cell[StyleData["RM", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["BF"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontWeight->"Bold"], Cell[StyleData["BF", "Printout"]], Cell[StyleData["BF", "EnhancedPrintout"]], Cell[StyleData["BF", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["IT"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontSlant->"Italic"], Cell[StyleData["IT", "Printout"]], Cell[StyleData["IT", "EnhancedPrintout"]], Cell[StyleData["IT", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TR"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["TR", "Printout"]], Cell[StyleData["TR", "EnhancedPrintout"]], Cell[StyleData["TR", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TI"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Times", FontSlant->"Italic"], Cell[StyleData["TI", "Printout"]], Cell[StyleData["TI", "EnhancedPrintout"]], Cell[StyleData["TI", "EnhancedPrintoutNonGray"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TB"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["TB", "Printout"]], Cell[StyleData["TB", "EnhancedPrintout"], FontFamily->"Palatino"], Cell[StyleData["TB", "EnhancedPrintoutNonGray"], FontFamily->"Palatino"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["TBI"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["TBI", "Printout"]], Cell[StyleData["TBI", "EnhancedPrintout"], FontFamily->"Palatino"], Cell[StyleData["TBI", "EnhancedPrintoutNonGray"], FontFamily->"Palatino"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MR"], ConversionRules:>{"HTML" -> {"", ""}}, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->" "}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["MR", "Printout"]], Cell[StyleData["MR", "EnhancedPrintout"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"], Cell[StyleData["MR", "EnhancedPrintoutNonGray"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MO"], ConversionRules:>{"HTML" -> {"", ""}}, HyphenationOptions->{"HyphenationCharacter"->""}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["MO", "Printout"]], Cell[StyleData["MO", "EnhancedPrintout"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"], Cell[StyleData["MO", "EnhancedPrintoutNonGray"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MB"], ConversionRules:>{"HTML" -> {"", ""}}, HyphenationOptions->{"HyphenationCharacter"->""}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["MB", "Printout"]], Cell[StyleData["MB", "EnhancedPrintout"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"], Cell[StyleData["MB", "EnhancedPrintoutNonGray"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MBO"], ConversionRules:>{"HTML" -> {"", ""}}, HyphenationOptions->{"HyphenationCharacter"->""}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["MBO", "Printout"]], Cell[StyleData["MBO", "EnhancedPrintout"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"], Cell[StyleData["MBO", "EnhancedPrintoutNonGray"], HyphenationOptions->{"HyphenationCharacter"->""}, FontFamily->"WriCourier"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SR"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["SR", "Printout"]], Cell[StyleData["SR", "EnhancedPrintout"], FontFamily->"Futura"], Cell[StyleData["SR", "EnhancedPrintoutNonGray"], FontFamily->"Futura"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SO"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO", "Printout"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO", "EnhancedPrintout"], FontFamily->"Futura"], Cell[StyleData["SO", "EnhancedPrintoutNonGray"], FontFamily->"Futura"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SB"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["SB", "Printout"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["SB", "EnhancedPrintout"], FontFamily->"Futura"], Cell[StyleData["SB", "EnhancedPrintoutNonGray"], FontFamily->"Futura"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SBO"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["SBO", "Printout"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["SBO", "EnhancedPrintout"], FontFamily->"Futura"], Cell[StyleData["SBO", "EnhancedPrintoutNonGray"], FontFamily->"Futura"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["M4"], ShowCellBracket->True, StyleMenuListing->None, FontFamily->"Math4", CharacterEncoding->"Math4"], Cell[StyleData["M4", "Printout"], ShowCellBracket->True, StyleMenuListing->None, FontFamily->"Math4", CharacterEncoding->"Math4"], Cell[StyleData["M4", "EnhancedPrintout"], ShowCellBracket->True, StyleMenuListing->None, FontFamily->"Math4", CharacterEncoding->"Math4"], Cell[StyleData["M4", "EnhancedPrintoutNonGray"], ShowCellBracket->True, StyleMenuListing->None, FontFamily->"Math4", CharacterEncoding->"Math4"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SO10"], ConversionRules:>{"HTML" -> {"", ""}}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "Printout"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->8, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "EnhancedPrintout"], StyleMenuListing->None, FontFamily->"Futura", FontSize->8, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "EnhancedPrintoutNonGray"], StyleMenuListing->None, FontFamily->"Futura", FontSize->8, FontWeight->"Plain", FontSlant->"Italic"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["C1"], StyleMenuListing->None, FontColor->RGBColor[0.4, 0, 1]], Cell[StyleData["C1", "Printout"], FontColor->GrayLevel[0]], Cell[StyleData["C1", "EnhancedPrintout"], FontColor->GrayLevel[0]], Cell[StyleData["C1", "EnhancedPrintoutNonGray"], FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["C1MR"], StyleMenuListing->None, FontFamily->"Courier", FontColor->RGBColor[0.4, 0, 1]], Cell[StyleData["C1MR", "Printout"], FontColor->GrayLevel[0]], Cell[StyleData["C1MR", "EnhancedPrintout"], FontColor->GrayLevel[0]], Cell[StyleData["C1MR", "EnhancedPrintoutNonGray"], FontColor->GrayLevel[0]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Emphasis Boxes and Pictures", "Section"], Cell[CellGroupData[{ Cell[StyleData["Box"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnSpacings->1}], Cell[StyleData["Box", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["Box", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["Box", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["DoubleBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnSpacings->2, RowAlignments->Top}], Cell[StyleData["DoubleBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["DoubleBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["DoubleBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["1ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnSpacings->1}], Cell[StyleData["1ColumnBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["1ColumnBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["1ColumnBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["2ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->{0.31, 0.67}}], Cell[StyleData["2ColumnBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["2ColumnEvenBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->0.46}], Cell[StyleData["2ColumnEvenBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnEvenBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnEvenBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["2ColumnSmallBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnSpacings->1.5, ColumnWidths->0.35, ColumnAlignments->{Right, Left}}], Cell[StyleData["2ColumnSmallBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnSmallBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["2ColumnSmallBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["3ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->0.32}], Cell[StyleData["3ColumnBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["3ColumnBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["3ColumnBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["3ColumnSmallBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnSpacings->1.5, ColumnWidths->0.24, ColumnAlignments->{Right, Center, Left}}], Cell[StyleData["3ColumnSmallBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["3ColumnSmallBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["3ColumnSmallBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["4ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->{0.13, 0.35, 0.13, 0.35}}], Cell[StyleData["4ColumnBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["4ColumnBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[0.900008]], Cell[StyleData["4ColumnBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->10, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["5ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->0.202}], Cell[StyleData["5ColumnBox", "Printout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[0.900008]], Cell[StyleData["5ColumnBox", "EnhancedPrintout"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[0.900008]], Cell[StyleData["5ColumnBox", "EnhancedPrintoutNonGray"], CellMargins->{{2, 0}, {0, 8}}, FontSize->9, Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["6ColumnBox"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], FrameBoxOptions->{BoxFrame->0.5, BoxMargins->True}, GridBoxOptions->{ColumnWidths->{0.12, 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Cell[StyleData["NumberedEquation", "Printout"], CellMargins->{{39, 55}, {0, 10}}, FontSize->10], Cell[StyleData["NumberedEquation", "EnhancedPrintout"], CellMargins->{{39, 55}, {0, 10}}, FontFamily->"Palatino", FontSize->10], Cell[StyleData["NumberedEquation", "EnhancedPrintoutNonGray"], CellMargins->{{39, 55}, {0, 10}}, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["EquationNumber"], LineSpacing->{1.83, 0}, LanguageCategory->"Formula"], Cell[StyleData["EquationNumber", "Printout"], LineSpacing->{1.7, 0}], Cell[StyleData["EquationNumber", "EnhancedPrintout"], LineSpacing->{1.7, 0}, FontFamily->"Palatino"], Cell[StyleData["EquationNumber", "EnhancedPrintoutNonGray"], LineSpacing->{1.7, 0}, FontFamily->"Palatino"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Program"], CellMargins->{{16, Inherited}, {Inherited, 10}}, CellHorizontalScrolling->True, LanguageCategory->"Formula", FontFamily->"Courier"], Cell[StyleData["Program", "Printout"], 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"Brackets"], CellMargins->{{72, 4}, {1, 14}}, CellGroupingRules->{"SectionGrouping", 50}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], Cell[StyleData["ObjectName", "Printout"], CellMargins->{{2, 0}, {1, 10}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SubObjectNameClosed"], ShowGroupOpenCloseIcon->True, CellMargins->{{10, 4}, {1, 10}}, PrivateCellOptions->{"DefaultCellGroupOpen"->False}, CellGroupingRules->{"SectionGrouping", 90}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->14, FontWeight->"Bold"], Cell[StyleData["SubObjectNameClosed", "Brackets"], CellMargins->{{72, 4}, {1, 14}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], Cell[StyleData["SubObjectNameClosed", "Printout"], CellMargins->{{2, 0}, {1, 10}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SubObjectName"], ShowGroupOpenCloseIcon->True, CellMargins->{{24, 4}, {1, 10}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->16, FontWeight->"Bold"], Cell[StyleData["SubObjectName", "Brackets"], ShowGroupOpenCloseIcon->True, CellMargins->{{90, 4}, {1, 14}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontFamily->"Courier", FontSize->12], Cell[StyleData["SubObjectName", "Printout"], CellMargins->{{2, 0}, {1, 10}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontFamily->"Courier", FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GroupObjectName"], ShowGroupOpenCloseIcon->True, CellMargins->{{10, 4}, {0, 14}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->16, FontWeight->"Bold"], Cell[StyleData["GroupObjectName", "Brackets"], ShowGroupOpenCloseIcon->True, CellMargins->{{72, 10}, {2, 12}}, CellGroupingRules->{"SectionGrouping", 70}, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphIndent->-27, CounterIncrements->"Outline2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], Cell[StyleData["GroupObjectName", "Printout"], HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GroupObjectNameClosed"], ShowGroupOpenCloseIcon->True, CellMargins->{{10, 4}, {0, 14}}, PrivateCellOptions->{"DefaultCellGroupOpen"->False}, CellGroupingRules->{"SectionGrouping", 80}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->16, FontWeight->"Bold"], Cell[StyleData["GroupObjectNameClosed", "Brackets"], ShowGroupOpenCloseIcon->True, CellMargins->{{72, 10}, {2, 12}}, CellGroupingRules->{"SectionGrouping", 70}, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphIndent->-27, CounterIncrements->"Outline2", FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], Cell[StyleData["GroupObjectNameClosed", "Printout"], HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ObjectNameWarningMessages"], CellMargins->{{10, 4}, {1, 10}}, CellGroupingRules->{"SectionGrouping", 65}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->16, FontWeight->"Bold"], Cell[StyleData["ObjectNameWarningMessages", "Brackets"], CellMargins->{{10, 4}, {1, 10}}, CellGroupingRules->{"SectionGrouping", 65}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontFamily->"Courier", FontSize->16, FontWeight->"Bold"], Cell[StyleData["ObjectNameWarningMessages", "Printout"], CellMargins->{{2, 0}, {1, 10}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Command"], CellMargins->{{10, 4}, {0, 14}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, CounterIncrements->"Section", StyleMenuListing->None, FontFamily->"Helvetica", FontSize->16, FontWeight->"Bold"], Cell[StyleData["Command", "Printout"], CellMargins->{{2, 0}, {0, 14}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["CharacterName"], CellFrame->{{0, 0}, {0.5, 0.5}}, CellMargins->{{10, 4}, {0, 4}}, CellFrameMargins->4, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None], Cell[StyleData["CharacterName", "Brackets"], CellMargins->{{72, 4}, {1, 14}}, CellGroupingRules->{"SectionGrouping", 90}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Courier", FontSize->12, FontWeight->"Bold"], Cell[StyleData["CharacterName", "Printout"], CellMargins->{{2, 0}, {0, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Usage"], CellFrame->{{0, 0}, {0.5, 0.5}}, CellMargins->{{10, 4}, {0, 4}}, CellFrameMargins->5, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 6}, ParagraphIndent->-10, StyleMenuListing->None], Cell[StyleData["Usage", "Brackets"], CellFrame->{{0, 0}, {0.5, 0.5}}, CellMargins->{{72, 4}, {0, 4}}, CellFrameMargins->5, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 6}, ParagraphIndent->-10, StyleMenuListing->None], Cell[StyleData["Usage", "Printout"], CellMargins->{{2, 0}, {0, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontSize->10, FontWeight->"Plain"] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ErrorMessage"], CellFrame->{{0, 0}, {0.5, 0.5}}, ShowCellBracket->True, CellMargins->{{10, 4}, {0, 4}}, CellGroupingRules->"OutputGrouping", PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{14, Inherited}, {Inherited, Inherited}}, CellFrameMargins->5, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"-"}, FormatType->InputForm, StyleMenuListing->None, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["ErrorMessage", "Brackets"], CellMargins->{{72, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None], Cell[StyleData["ErrorMessage", "Printout"], HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->9, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Notes"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 6}, ParagraphIndent->-10, StyleMenuListing->None, FontFamily->"Times"], Cell[StyleData["Notes", "Brackets"], CellMargins->{{72, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 6}, ParagraphIndent->-10, StyleMenuListing->None, FontFamily->"Times"], Cell[StyleData["Notes", "Printout"], CellMargins->{{2, 0}, {0, 8}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["NotesList"], CellMargins->{{20, 4}, {2, 8}}, CellHorizontalScrolling->True, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ScriptLevel->1, SingleLetterItalics->True, StyleMenuListing->None, FontFamily->"Times"], Cell[StyleData["NotesList", "Brackets"], CellMargins->{{82, 4}, {2, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ScriptLevel->1, SingleLetterItalics->True, StyleMenuListing->None, FontFamily->"Times"], Cell[StyleData["NotesList", "Printout"], CellMargins->{{12, 0}, {2, 8}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ExampleSection"], CellFrame->{{0, 0}, {0, 0.5}}, ShowGroupOpenCloseIcon->True, CellMargins->{{11, 4}, {0, 10}}, CellGroupingRules->{"SectionGrouping", 70}, CellFrameMargins->12, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontSize->13, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 0.4]], Cell[StyleData["ExampleSection", "Printout"], CellMargins->{{3, 0}, {0, 10}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->9, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ExampleText"], CellMargins->{{20, 40}, {0, 18}}, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 6}, StyleMenuListing->None, FontColor->RGBColor[0, 0, 0.4]], Cell[StyleData["ExampleText", "Printout"], CellMargins->{{2, 80}, {0, 8}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->8, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ExampleSubsection"], ShowCellBracket->True, CellMargins->{{20, 4}, {0, 16}}, CellGroupingRules->{"SectionGrouping", 80}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Bold"], Cell[StyleData["ExampleSubsection", "Printout"], CellMargins->{{12, 0}, {0, 16}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->7] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["LineHead"], CellFrame->{{0, 0}, {0.5, 0}}, CellMargins->{{10, 4}, {8, 18}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->12, FontWeight->"Bold"], Cell[StyleData["LineHead", "Printout"], CellMargins->{{2, 0}, {8, 18}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FlushInput"], CellMargins->{{10, 4}, {1, 4}}, Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, CellLabelAutoDelete->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultInputFormatType, ShowSpecialCharacters->Automatic, HyphenationOptions->{"HyphenationCharacter"->"-"}, FormatType->StandardForm, ShowStringCharacters->True, NumberMarks->True, StyleMenuListing->None, FontWeight->"Bold"], Cell[StyleData["FlushInput", "Printout"], CellMargins->{{2, 0}, {1, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FlushOutput"], CellMargins->{{10, Inherited}, {Inherited, Inherited}}, CellEditDuplicate->True, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, CellLabelAutoDelete->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"-"}, FormatType->StandardForm, StyleMenuListing->None], Cell[StyleData["FlushOutput", "Printout"], CellMargins->{{2, Inherited}, {Inherited, Inherited}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["DialogItem"], CellDingbat->"\[EmptyDiamond]", CellMargins->{{20, 4}, {0, 12}}, CellGroupingRules->{"SectionGrouping", 70}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Bold"], Cell[StyleData["DialogItem", "Printout"], CellMargins->{{12, 0}, {0, 12}}, PageBreakBelow->False, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->8] }, Closed]], Cell[StyleData["GroupDivider"], CellMargins->{{10, Inherited}, {Inherited, Inherited}}, CellEditDuplicate->True, CellGroupingRules->{"TitleGrouping", 0}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None], Cell[CellGroupData[{ Cell[StyleData["Highlight"], CellMargins->{{10, 4}, {0, 18}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->12, Background->GrayLevel[0.900008]], Cell[StyleData["Highlight", "Printout"], CellMargins->{{2, 0}, {0, 18}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, LineSpacing->{1, 3}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["UserNotes"], CellFrame->{{2, 0}, {0, 0.25}}, CellMargins->{{10, 4}, {0, 4}}, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphIndent->-7, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10], Cell[StyleData["UserNotes", "Printout"], CellMargins->{{2, 0}, {0, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, LineSpacing->{1, 3}, FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["MoreInfo"], CellMargins->{{10, 4}, {0, 4}}, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphIndent->-7, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontColor->GrayLevel[1], Background->RGBColor[0.700008, 0.4, 0.100008]], Cell[StyleData["MoreInfo", "Printout"], CellMargins->{{2, 0}, {0, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, LineSpacing->{1, 3}, FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["UserTips"], CellFrame->{{2, 0}, {0, 0.25}}, CellMargins->{{10, 4}, {0, 4}}, Hyphenation->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphIndent->-7, StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontColor->GrayLevel[1], Background->RGBColor[0.2, 0.500008, 0.700008]], Cell[StyleData["UserTips", "Printout"], CellMargins->{{2, 0}, {0, 4}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, LineSpacing->{1, 3}, FontSize->8, Background->GrayLevel[0.700008]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Theorem"], CellFrame->{{4, 0}, {0, 0}}, CellMargins->{{12, 62}, {0, 14}}, CellFrameMargins->4, CellFrameColor->RGBColor[0, 0, 0.8], HyphenationOptions->{"HyphenationCharacter"->"-"}, LineSpacing->{1, 1}, ParagraphSpacing->{0, 8}], Cell[StyleData["Theorem", "Printout"], CellMargins->{{2, 0}, {0, 14}}, CellFrameMargins->3, CellFrameColor->GrayLevel[0.8], HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->9, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Commentary"], CellFrame->{{2, 0}, {0, 0}}, CellMargins->{{36, 10}, {7, 7}}, PageBreakBelow->False, CellFrameMargins->8, CellFrameColor->RGBColor[0, 0.2, 1], LineSpacing->{1, 3}, FontSlant->"Italic"], Cell[StyleData["Commentary", "Printout"], CellMargins->{{36, 0}, {6, 6}}, CellFrameColor->GrayLevel[0.8], FontSize->10], Cell[StyleData["Commentary", "EnhancedPrintout"], CellMargins->{{36, 0}, {6, 6}}, CellFrameColor->GrayLevel[0.8], FontFamily->"Palatino", FontSize->10], Cell[StyleData["Commentary", "EnhancedPrintoutNonGray"], CellMargins->{{36, 0}, {6, 6}}, CellFrameColor->GrayLevel[0.8], FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Level1Icon"], ShowGroupOpenCloseIcon->True, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 8}, StyleMenuListing->None], Cell[StyleData["Level1Icon", "Printout"], CellMargins->{{2, 0}, {0, 8}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Level2Icon"], ShowGroupOpenCloseIcon->True, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"-"}, ParagraphSpacing->{0, 8}, StyleMenuListing->None], Cell[StyleData["Level2Icon", "Printout"], CellMargins->{{2, 0}, {0, 8}}, HyphenationOptions->{"HyphenationCharacter"->"-"}, FontSize->10] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Manual.nb Merge", "Subsection"], Cell[CellGroupData[{ Cell["Function List", "Subsubsection"], Cell[CellGroupData[{ Cell[StyleData["FunctionName"], CellDingbat->"\[FilledSquare]", CellMargins->{{24, 10}, {10, 12}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, StyleMenuListing->None, FontFamily->"Courier", FontSize->14, FontWeight->"Bold", CellTags->"FunctionName"], Cell[StyleData["FunctionName", "Printout"], CellMargins->{{9, 0}, {10, 10}}, FontSize->12], Cell[StyleData["FunctionName", "EnhancedPrintout"], CellMargins->{{9, 0}, {10, 10}}, FontFamily->"Courier", FontSize->12], Cell[StyleData["FunctionName", "EnhancedPrintoutNonGray"], CellMargins->{{9, 0}, {10, 10}}, FontFamily->"Courier", FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SubsectionIcon"], ShowGroupOpenCloseIcon->True, CellMargins->{{24, 10}, {0, 30}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, StyleMenuListing->None, FontSize->14, FontWeight->"Bold", CellTags->"FunctionSection"], Cell[StyleData["SubsectionIcon", "Printout"], CellMargins->{{20, 0}, {0, 20}}, FontSize->13], Cell[StyleData["SubsectionIcon", "EnhancedPrintout"], CellMargins->{{20, 0}, {0, 20}}, FontFamily->"Palatino", FontSize->10], Cell[StyleData["SubsectionIcon", "EnhancedPrintoutNonGray"], CellMargins->{{20, 0}, {0, 20}}, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionSubsection"], CellDingbat->"\[FilledSquare]", CellMargins->{{36, 10}, {0, 6}}, CellGroupingRules->{"SectionGrouping", 60}, PageBreakBelow->False, StyleMenuListing->None, FontWeight->"Bold", CellTags->"FunctionSubsection"], Cell[StyleData["FunctionSubsection", "Printout"], CellMargins->{{30, 0}, {0, 6}}], Cell[StyleData["FunctionSubsection", "EnhancedPrintout"], CellMargins->{{30, 0}, {0, 6}}, FontFamily->"Palatino", FontSize->11], Cell[StyleData["FunctionSubsection", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {0, 6}}, FontFamily->"Palatino", FontSize->11] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SubsubsectionIcon"], ShowGroupOpenCloseIcon->True, CellMargins->{{36, 10}, {0, 6}}, CellGroupingRules->{"SectionGrouping", 60}, PageBreakBelow->False, StyleMenuListing->None, FontWeight->"Bold", CellTags->"SubsubsectionIcon"], Cell[StyleData["SubsubsectionIcon", "Printout"], CellMargins->{{30, 0}, {0, 6}}, FontSize->11], Cell[StyleData["SubsubsectionIcon", "EnhancedPrintout"], CellMargins->{{30, 0}, {0, 6}}, FontFamily->"Palatino", FontSize->9], Cell[StyleData["SubsubsectionIcon", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {0, 6}}, FontFamily->"Palatino", FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionDescription"], CellMargins->{{36, 10}, {4, 0}}, LineSpacing->{1, 3}, ParagraphSpacing->{0.7, 0}, CounterIncrements->"Text", StyleMenuListing->None, FontWeight->"Plain"], Cell[StyleData["FunctionDescription", "Printout"], CellMargins->{{30, 0}, {4, 0}}, FontSize->11], Cell[StyleData["FunctionDescription", "EnhancedPrintout"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10], Cell[StyleData["FunctionDescription", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionDescriptionRuled"], CellFrame->{{0, 0}, {0.25, 0.25}}, CellMargins->{{36, 10}, {4, 0}}, LineSpacing->{1, 3}, ParagraphSpacing->{0.7, 0}, CounterIncrements->"Text", StyleMenuListing->None], Cell[StyleData["FunctionDescriptionRuled", "Printout"], CellMargins->{{30, 0}, {4, 0}}, FontSize->11], Cell[StyleData["FunctionDescriptionRuled", "EnhancedPrintout"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10], Cell[StyleData["FunctionDescriptionRuled", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionDescriptionRuleTop"], CellFrame->{{0, 0}, {0, 0.25}}, CellMargins->{{36, 10}, {4, 0}}, LineSpacing->{1, 3}, ParagraphSpacing->{0.7, 0}, CounterIncrements->"Text", StyleMenuListing->None], Cell[StyleData["FunctionDescriptionRuleTop", "Printout"], CellMargins->{{30, 0}, {4, 0}}, FontSize->11], Cell[StyleData["FunctionDescriptionRuleTop", "EnhancedPrintout"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10], Cell[StyleData["FunctionDescriptionRuleTop", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionDescriptionRuleBottom"], CellFrame->{{0, 0}, {0.25, 0}}, CellMargins->{{36, 10}, {4, 0}}, LineSpacing->{1, 3}, ParagraphSpacing->{0.7, 0}, CounterIncrements->"Text", StyleMenuListing->None], Cell[StyleData["FunctionDescriptionRuleBottom", "Printout"], CellMargins->{{30, 0}, {4, 0}}, FontSize->11], Cell[StyleData["FunctionDescriptionRuleBottom", "EnhancedPrintout"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10], Cell[StyleData["FunctionDescriptionRuleBottom", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {4, 0}}, TextJustification->0, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionExample"], CellMargins->{{36, 10}, {4, 0}}, StyleMenuListing->None, FontSize->10], Cell[StyleData["FunctionExample", "Printout"], CellMargins->{{30, 0}, {4, 0}}, FontSize->9], Cell[StyleData["FunctionExample", "EnhancedPrintout"], CellMargins->{{30, 0}, {4, 0}}, FontFamily->"Palatino", FontSize->9], Cell[StyleData["FunctionExample", "EnhancedPrintoutNonGray"], CellMargins->{{30, 0}, {4, 0}}, FontFamily->"Palatino", FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["FunctionTextNote"], CellMargins->{{36, 4}, {0, 4}}, LineSpacing->{1, 3}, ParagraphSpacing->{0, 4}, LineIndent->0, StyleMenuListing->None], Cell[StyleData["FunctionTextNote", "Printout"], CellMargins->{{36, 4}, {0, 4}}, FontSize->9], Cell[StyleData["FunctionTextNote", "EnhancedPrintout"], CellMargins->{{36, 4}, {0, 4}}, FontFamily->"Palatino", FontSize->8], Cell[StyleData["FunctionTextNote", "EnhancedPrintoutNonGray"], CellMargins->{{36, 4}, {0, 4}}, FontFamily->"Palatino", FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["3ColumnFunctionTable"], CellMargins->{{36, 4}, {0, 8}}, CellHorizontalScrolling->True, PageBreakWithin->False, AutoIndent->False, AutoSpacing->False, LineIndent->0, StyleMenuListing->None, GridBoxOptions->{RowSpacings->1.5, ColumnSpacings->2, ColumnWidths->{0.28, 0.28, 0.43}, ColumnAlignments->{Left}}], Cell[StyleData["3ColumnFunctionTable", "Printout"], CellMargins->{{30, 0}, {0, 8}}, FontSize->10], Cell[StyleData["3ColumnFunctionTable", "EnhancedPrintout"], CellMargins->{{30, 0}, {0, 8}}, FontSize->10], Cell[StyleData["3ColumnFunctionTable", 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Cell[CellGroupData[{ Cell[StyleData["ItemizedTextNote2"], CellMargins->{{41, 4}, {0, 4}}, Hyphenation->True, LineSpacing->{1, 2}, CounterIncrements->"Text", StyleMenuListing->None], Cell[StyleData["ItemizedTextNote2", "Printout"], CellMargins->{{38, 4}, {0, 2}}, FontSize->11], Cell[StyleData["ItemizedTextNote2", "EnhancedPrintout"], CellMargins->{{38, 4}, {0, 2}}, FontFamily->"Palatino", FontSize->10], Cell[StyleData["ItemizedTextNote2", "EnhancedPrintoutNonGray"], CellMargins->{{38, 4}, {0, 2}}, FontFamily->"Palatino", FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["ItemizedTopBox"], CellFrame->{{0.5, 0.5}, {0, 0.5}}, CellMargins->{{37, 4}, {0, 8}}, CellHorizontalScrolling->True, LineIndent->0, StyleMenuListing->None, Background->RGBColor[1, 0.6, 0.6], GridBoxOptions->{RowSpacings->1.5, ColumnSpacings->1, ColumnWidths->{0.31, 0.62}, ColumnAlignments->{Left}}], Cell[StyleData["ItemizedTopBox", "Printout"], FontSize->10, Background->GrayLevel[0.900008]], 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