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Difference between revisions of "1991 AIME Problems/Problem 14"
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== Solution ==
 
== Solution ==
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<center>[[Image:AIME_1991_Solution_14.png]]</center>
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<center><asy>defaultpen(fontsize(9));
 +
pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));
 +
draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--E);
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dot(A^^B^^C^^D^^E^^F);
 +
label("\(A\)",A,(-1,-1));label("\(B\)",B,(1,-1));label("\(C\)",C,(1,0));
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label("\(D\)",D,(1,1));label("\(E\)",E,(-1,1));label("\(F\)",F,(-1,0));
 +
label("31",A/2+B/2,(0.7,1));label("81",B/2+C/2,(0.45,-0.2));
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label("81",C/2+D/2,(-1,-1));label("81",D/2+E/2,(0,-1));
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label("81",E/2+F/2,(1,-1));label("81",F/2+A/2,(1,1));
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label("\(x\)",A/2+C/2,(-1,1));label("\(y\)",A/2+D/2,(1,-1.5));
 +
label("\(z\)",A/2+E/2,(1,0));
 +
</asy></center><!-- asy replaced Image:AIME 1991 Solution 14.png by minsoens -->
  
 
Let <math>x=AC=BF</math>, <math>y=AD=BE</math>, and <math>z=AE=BD</math>.
 
Let <math>x=AC=BF</math>, <math>y=AD=BE</math>, and <math>z=AE=BD</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>.
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[[Ptolemy's Theorem]] on <math>ABCD</math> gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on <math>ACDF</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=\boxed{384}</math>.
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/DVuf-uXjfzY?t=522
 +
 
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
Line 14: Line 30:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 02:56, 23 January 2023

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

[asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--E); dot(A^^B^^C^^D^^E^^F); label("\(A\)",A,(-1,-1));label("\(B\)",B,(1,-1));label("\(C\)",C,(1,0)); label("\(D\)",D,(1,1));label("\(E\)",E,(-1,1));label("\(F\)",F,(-1,0)); label("31",A/2+B/2,(0.7,1));label("81",B/2+C/2,(0.45,-0.2)); label("81",C/2+D/2,(-1,-1));label("81",D/2+E/2,(0,-1)); label("81",E/2+F/2,(1,-1));label("81",F/2+A/2,(1,1)); label("\(x\)",A/2+C/2,(-1,1));label("\(y\)",A/2+D/2,(1,-1.5)); label("\(z\)",A/2+E/2,(1,0)); [/asy]

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 $ACDF$ 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 $x+y+z=105+144+135=\boxed{384}$.

Video Solution by OmegaLearn

https://youtu.be/DVuf-uXjfzY?t=522

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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