Difference between revisions of "2013 AIME I Problems/Problem 3"
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We define <math>a</math> as the length of the side of larger inner square, which is also <math>EB</math>, <math>b</math> as the length of the side of the smaller inner square which is also <math>AE</math>, and <math>s</math> as the side length of <math>ABCD</math>. Since we are given that the sum of the areas of the two squares is<math>\frac{9}{10}</math> of the the area of ABCD, we can represent that as <math>a^2 + b^2 = \frac{9s^2}{10}</math>. The sum of the two nonsquare rectangles can then be represented as <math>2ab = \frac{s^2}{10}</math>. | We define <math>a</math> as the length of the side of larger inner square, which is also <math>EB</math>, <math>b</math> as the length of the side of the smaller inner square which is also <math>AE</math>, and <math>s</math> as the side length of <math>ABCD</math>. Since we are given that the sum of the areas of the two squares is<math>\frac{9}{10}</math> of the the area of ABCD, we can represent that as <math>a^2 + b^2 = \frac{9s^2}{10}</math>. The sum of the two nonsquare rectangles can then be represented as <math>2ab = \frac{s^2}{10}</math>. | ||
− | Looking back at what we need to find, we can represent <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> as <math>\dfrac{a^2 + b^2}{ab}</math>. We have the numerator, and dividing<math>\frac{s^2}{10}</math> by two gives us the denominator <math>\frac{s^2}{20}</math>. Dividing <math>\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}</math> gives us an answer of <math>018</math>. | + | Looking back at what we need to find, we can represent <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> as <math>\dfrac{a^2 + b^2}{ab}</math>. We have the numerator, and dividing<math>\frac{s^2}{10}</math> by two gives us the denominator <math>\frac{s^2}{20}</math>. Dividing <math>\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}</math> gives us an answer of <math>\boxed{018}</math>. |
Revision as of 16:00, 16 March 2013
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
.