Difference between revisions of "2015 AIME I Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math> | + | We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complimentary, we have that <math>\triangle CDE ~ \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us <math>JC = \sqrt5 * FC = \sqrt5 * FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} * HJ = \frac{a}{\sqrt5}</math>. |
Revision as of 17:48, 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
.
Problem
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
.