2005 AIME I Problems/Problem 7
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Problem
In quadrilateral 
and 
Given that 
where 
and 
are positive integers, find 
Contents |
Solution
Solution 1
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
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 
.
See also
| 2005 AIME I (Problems • Resources) | ||
| Preceded by Problem 6 | Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
