Difference between revisions of "1991 AIME Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | <center> | + | <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>. | 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>. | [[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=\boxed{384}</math>. |
== See also == | == See also == | ||
Revision as of 21:50, 11 April 2008
Problem
A hexagon is inscribed in a circle. Five of the sides have length 
and the sixth, denoted by 
, has length 
. Find the sum of the lengths of the three diagonals that can be drawn from 
.
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]](http://latex.artofproblemsolving.com/a/5/b/a5b56caddf0c74876a022cbb3946a4eda30b38d0.png)
Let 
, 
, and 
.
Ptolemy's Theorem on 
gives 
, and Ptolemy on 
gives 
.
Subtracting these equations give 
, and from this 
. Ptolemy on 
gives 
, and from this 
. Finally, plugging back into the first equation gives 
, so 
.
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 | ||
