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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>.  
  
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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>.
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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 15:00, 16 March 2013

Problem 3

Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$


Solution

It's important to note that $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$

We define $a$ as the length of the side of larger inner square, which is also $EB$, $b$ as the length of the side of the smaller inner square which is also $AE$, and $s$ as the side length of $ABCD$. Since we are given that the sum of the areas of the two squares is$\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \frac{9s^2}{10}$. The sum of the two nonsquare rectangles can then be represented as $2ab = \frac{s^2}{10}$.

Looking back at what we need to find, we can represent $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\dfrac{a^2 + b^2}{ab}$. We have the numerator, and dividing$\frac{s^2}{10}$ by two gives us the denominator $\frac{s^2}{20}$. Dividing $\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}$ gives us an answer of $\boxed{018}$.

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