Difference between revisions of "2013 AIME I Problems/Problem 3"
(→Solution) |
m |
||
Line 1: | Line 1: | ||
− | |||
== Problem 3 == | == Problem 3 == | ||
Let <math>ABCD</math> be a square, and let <math>E</math> and <math>F</math> be points on <math>\overline{AB}</math> and <math>\overline{BC},</math> respectively. The line through <math>E</math> parallel to <math>\overline{BC}</math> and the line through <math>F</math> parallel to <math>\overline{AB}</math> divide <math>ABCD</math> into two squares and two nonsquare rectangles. The sum of the areas of the two squares is <math>\frac{9}{10}</math> of the area of square <math>ABCD.</math> Find <math>\frac{AE}{EB} + \frac{EB}{AE}.</math> | Let <math>ABCD</math> be a square, and let <math>E</math> and <math>F</math> be points on <math>\overline{AB}</math> and <math>\overline{BC},</math> respectively. The line through <math>E</math> parallel to <math>\overline{BC}</math> and the line through <math>F</math> parallel to <math>\overline{AB}</math> divide <math>ABCD</math> into two squares and two nonsquare rectangles. The sum of the areas of the two squares is <math>\frac{9}{10}</math> of the area of square <math>ABCD.</math> Find <math>\frac{AE}{EB} + \frac{EB}{AE}.</math> | ||
Line 11: | Line 10: | ||
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>. | 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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2013|n=I|num-b=2|num-a=4}} |
Revision as of 20:40, 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 .
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 |