2005 AIME I Problems/Problem 7
Problem
In quadrilateral 
and 
Given that 
where 
and 
are positive integers, find 
Contents
Solution
Solution 1
![[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); [/asy]](http://latex.artofproblemsolving.com/1/b/b/1bbf437694cfea9d7049c523e552fb18592a1f17.png)
Solution 2
Draw the perpendiculars from 
and 
to 
, labeling the intersection points as 
and 
. This forms 2 
right triangles, so 
and 
. Also, if we draw the horizontal line extending from to a point 
on the line 
, we find another right triangle 
. 
. The Pythagorean Theorem yields that 
, so 
. Therefore, 
, and 
.
Solution 3
Extend 
and 
to an intersection at point . We get an equilateral triangle 
. We denote the length of a side of 
as 
and solve for it using the Law of Cosines: ![\[12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos{60}\]](http://latex.artofproblemsolving.com/c/2/d/c2d267f767804bf3bd39941692aeba42fd97a999.png)
![\[144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)\]](http://latex.artofproblemsolving.com/7/2/c/72c8eaac07f971c4b517db5481326b8765930c76.png)
This simplifies to 
; the quadratic formula yields the (discard the negative result) same result of 
.
Solution 4
Extend and to meet at point , forming an equilateral triangle . Draw a line from parallel to so that it intersects at point . Then, apply Stewart's Theorem on 
. Let 
. ![\[2x(x-2) + 12^2x = 2x^2 + x^2(x-2)\]](http://latex.artofproblemsolving.com/d/0/2/d02a599e57533e7b5ef3ecf315b88dc5262e2e80.png)
![\[x^3 - 2x^2 - 140x = 0\]](http://latex.artofproblemsolving.com/2/6/1/26188645723b907de148b80ec31fed400b3ab0cf.png)
By the quadratic formula (discarding the negative result), 
, giving 
for a final answer of 
.
See also
| 2005 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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