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Difference between revisions of "1991 AIME Problems/Problem 14"

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A [[hexagon]] is inscribed in a [[circle]]. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A_{}^{}</math>.
 
A [[hexagon]] is inscribed in a [[circle]]. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A_{}^{}</math>.
  
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[[Image:1991_AIME-14.png]]
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== Solution ==
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<center>[[Image:AIME_1991_Solution_14.png]]</center>
  
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== Solution ==
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Let <math>x=AC=BF</math>, <math>y=AD=BE</math>, and <math>z=AE=BD</math>.
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[[Image:1991_AIME-14a.png]]
 
  
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Let <math>x=AC</math>, <math>y=AD</math>, and <math>z=AE</math>.
 
 
[[Ptolemy's Theorem]] on <math>ABCD</math> gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on <math>ACDE</math> gives <math>x\cdot z+81^2=y^2</math>.
 
[[Ptolemy's Theorem]] on <math>ABCD</math> gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on <math>ACDE</math> gives <math>x\cdot z+81^2=y^2</math>.
 
Subtracting these equations give <math>y^2-81y-112\cdot 81=0</math>, and from this <math>y=144</math>. Ptolemy on <math>ADEF</math> gives <math>81y+81^2=z^2</math>, and from this <math>z=135</math>. Finally, plugging back into the first equation gives <math>x=105</math>, so <math>x+y+z=105+144+135=384</math>.
 
Subtracting these equations give <math>y^2-81y-112\cdot 81=0</math>, and from this <math>y=144</math>. Ptolemy on <math>ADEF</math> gives <math>81y+81^2=z^2</math>, and from this <math>z=135</math>. Finally, plugging back into the first equation gives <math>x=105</math>, so <math>x+y+z=105+144+135=384</math>.

Revision as of 19:15, 16 November 2007

Problem

A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$.

Solution

AIME 1991 Solution 14.png

Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$.

Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDE$ gives $x\cdot z+81^2=y^2$. Subtracting these equations give $y^2-81y-112\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into the first equation gives $x=105$, so $x+y+z=105+144+135=384$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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