Difference between revisions of "2013 AIME I Problems/Problem 3"
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Revision as of 14:00, 4 July 2013
Contents
Problem 3
Let 
be a square, and let 
and 
be points on 
and 
respectively. The line through parallel to 
and the line through parallel to divide into two squares and two nonsquare rectangles. The sum of the areas of the two squares is 
of the area of square 
Find 
Solution
It's important to note that 
is equivalent to 
We define 
as the length of the side of larger inner square, which is also 
, 
as the length of the side of the smaller inner square which is also 
, and 
as the side length of . Since we are given that the sum of the areas of the two squares is of the the area of ABCD, we can represent that as 
. The sum of the two nonsquare rectangles can then be represented as 
.
Looking back at what we need to find, we can represent as 
. We have the numerator, and dividing
by two gives us the denominator 
. Dividing 
gives us an answer of 
.
Solution 2
Let the side of the square be 
. Therefore the area of the square is also .
We label as and as . Notice that what we need to find is equivalent to: 
.
Since the sum of the two squares (
) is (as stated in the problem) the area of the whole square, it is clear that the
sum of the two rectangles is 
. Since these two rectangles are congruent, they
each have area: 
. Also note that the area of this is 
. Plugging this into our equation we get:
See also
| 2013 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
