Difference between revisions of "2015 AIME I Problems/Problem 7"
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<cmath>AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x</cmath> | <cmath>AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x</cmath> | ||
− | Now, it is trivial to see that <math>[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2 | + | Now, it is trivial to see that <math>[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.\Box</math> |
==Solution 2== | ==Solution 2== |
Revision as of 21:52, 5 March 2019
Contents
[hide]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 1
Let us find the proportion of the side length of and
. Let the side length of
and the side length of
.
Now, examine . We know
, and triangles
and
are similar to
since they are
triangles. Thus, we can rewrite
in terms of the side length of
.
Now examine . We can express this length in terms of
since
. By using similar triangles as in the first part, we have
Now, it is trivial to see that
Solution 2
We begin by denoting the length
, giving us
and
. Since angles
and
are complementary, 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
to
and label the point of intersection
:
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.