Difference between revisions of "2015 AIME I Problems/Problem 7"
Line 89: | Line 89: | ||
label("$M$",M,dir(90)); | label("$M$",M,dir(90)); | ||
label("$N$",N,dir(180)); | label("$N$",N,dir(180)); | ||
− | label("$P$",P,dir( | + | label("$P$",P,dir(235)); </asy> |
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> |
Revision as of 20:15, 21 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
So our final answer is
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.