Difference between revisions of "2005 AIME I Problems/Problem 7"
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Revision as of 20:03, 4 July 2013
Problem
In quadrilateral and
Given that
where
and
are positive integers, find
Solution
Solution 1
![AIME 2005I Solution 7 1.png](https://wiki-images.artofproblemsolving.com//4/41/AIME_2005I_Solution_7_1.png)
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 2
![AIME 2005I Solution 7 2.png](https://wiki-images.artofproblemsolving.com//2/2f/AIME_2005I_Solution_7_2.png)
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:
This simplifies to
; the quadratic formula yields the (discard the negative result) same result of
.
Solution 3
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
.
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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