Difference between revisions of "2015 AIME I Problems/Problem 7"
(→Solution 1: Pythagorean Bash) |
|||
Line 18: | Line 18: | ||
This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> and <math>ME = \sqrt5 \cdot MP = \sqrt5 \cdot (ML - EG) = \sqrt5 \cdot (ML - (EC - (GF + FC))</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 (ML - EG) = \sqrt5 \cdot (ML - (EC - (GF + FC))</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2015|n=I|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 17:59, 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
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.