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

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== Problem ==
 
== Problem ==
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A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by <math>\overline{AB}</math>, has length 31. Find the sum of the lengths of the three diagonals that can be drawn from <math>A_{}^{}</math>.
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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>.
 +
 
 +
[[Image:1991_AIME-14.png]]
  
 
== Solution ==
 
== Solution ==
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Let x=AC, y=AD, and z=AE.
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[[Image:1991_AIME-14a.png]]
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Ptolemy's Theorem on ABCD gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on ACDE gives <math>x\cdot z+81^2=y^2</math>.
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Subtracting these equations give <math>y^2-81y-112\cdot 81=0</math>, and from this y=144. Ptolemy on ADEF gives <math>81y+81^2=z^2</math>, and from this z=135. Finally, plugging back into the first equation gives x=105, so x+y+z=105+144+135=<math>384</math>.
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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>ACDE4 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$.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=13|num-a=15}}
 
{{AIME box|year=1991|num-b=13|num-a=15}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 11:55, 21 October 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 $A_{}^{}$.

File:1991 AIME-14.png

Solution

File:1991 AIME-14a.png

Let $x=AC$, $y=AD$, and $z=AE$. Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDE4 gives$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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