Difference between revisions of "2015 AIME I Problems/Problem 7"
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This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> | This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> | ||
− | and <math>ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{ | + | and <math>ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}</math> |
Since <math>AE</math> = <math>AM + ME</math>, we get | Since <math>AE</math> = <math>AM + ME</math>, we get | ||
− | <math>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{ | + | <math>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a</math> |
− | <math>\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{ | + | <math>\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a</math> |
− | <math>\Rightarrow \frac{ | + | <math>\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a</math> |
+ | |||
+ | <math>\Rightarrow -21\sqrt{11} = \sqrt5a\frac{14 - 20}{7}</math> | ||
+ | |||
+ | <math>\Rightarrow \frac{49\sqrt{11}}{2} = \sqrt5a</math> | ||
+ | |||
+ | <math>\Rightarrow \frac{7\sqrt{11}} = \frac{4a\sqrt{5}}{7}</math> | ||
Revision as of 18:20, 20 March 2015
Problem
7. In the diagram below, is a square. Point is the midpoint of . Points and lie on , and and lie on and , respectively, so that is a square. Points and lie on , and and lie on and , respectively, so that is a square. The area of is 99. Find the area of .
Solution
We begin by denoting the length , giving us and . Since angles and are complimentary, we have that (and similarly the rest of the triangles are triangles). We let the sidelength of be , giving us:
and .
Since ,
,
Solving for in terms of yields .
We now use the given that , implying that . We also draw the perpendicular from E to ML and label the point of intersection P.
This gives that and
Since = , we get
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.