Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(wik)
(→‎Solution)
 
(5 intermediate revisions by 5 users not shown)
Line 2: Line 2:
 
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>.
  
−
[[Image:1991_AIME-14.png]]
+
== Solution ==
 +
<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);
 +
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></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>.
 +
 
 +
[[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=\boxed{384}</math>.
  
−
== Solution ==
+
== Video Solution by OmegaLearn ==
−
[[Image:1991_AIME-14a.png]]
+
https://youtu.be/DVuf-uXjfzY?t=522
  
−
Let <math>x=AC</math>, <math>y=AD</math>, and <math>z=AE</math>.
+
~ pi_is_3.14
−
[[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 ==
Line 15: 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_14&oldid=187353"