Difference between revisions of "2015 AIME I Problems/Problem 7"
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<math>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{3\sqrt{11}}{2}}{2} = a</math> | <math>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{3\sqrt{11}}{2}}{2} = a</math> | ||
+ | |||
+ | <math>\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{3\sqrt{11}}{2}) = 2\sqrt5a</math> | ||
== See also == | == See also == |
Revision as of 18:06, 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.