Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 12"
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== Solution == | == Solution == | ||
+ | Let <math>\angle{AB'C'} = \theta</math>. By some angle chasing in <math>\triangle{AB'E}</math>, we find that <math>\angle{EAB'} = 90^{\circ} - 2 \theta</math>. Before we apply the law of sines, we're going to want to get everything in terms of <math>\sin \theta</math>, so note that <math>\sin \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the following: | ||
+ | |||
+ | <math>\frac{\sin \theta}{5}=\frac{1 - 2 \sin^2 \theta}{23} \implies \sin \theta = \frac{-23 \pm 27}{20}</math>, | ||
+ | but since <math>\theta < 180^{\circ}</math>, we go with the positive solution. Thus, <math>\sin \theta = \frac15</math>. | ||
+ | |||
+ | Denote the intersection of <math>B'C'</math> and <math>AE</math> with <math>G</math>. By another application of the law of sines, <math>B'G = \frac{23}{\sqrt{24}}</math> and <math>AE = 10\sqrt{6}</math>. Since <math>\sin \theta = \frac15, GE = \frac{115}{\sqrt{24}}</math>, and <math>AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}</math>. Note that <math>\triangle{EB'G} \sim \triangle{C'AG}</math>, so <math>\frac{EG}{B'G}=\frac{C'G}{AG} \implies C'G = \frac{25}{\sqrt{24}}</math>. | ||
+ | |||
+ | Now we have that <math>AB = AE + EB = 10\sqrt{6} + 23</math>, and <math>B'C' = BC = B'G + C'G = \frac{23}{\sqrt{24}} + \frac{25}{\sqrt{24}} = \frac{48}{\sqrt{24}}=4\sqrt{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 + 240 = \boxed{338}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{Mock AIME box|year=Pre 2005|n=1|num-b=11|num-a=13|source=14769}} | ||
+ | |||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 22:41, 26 December 2016
Problem
is a rectangular sheet of paper.
and
are points on
and
respectively such that
. If
is folded over
,
maps to
on
and
maps to
such that
. If
and
, then the area of
can be expressed as
square units, where
and
are integers and
is not divisible by the square of any prime. Compute
.
Solution
Let . By some angle chasing in
, we find that
. Before we apply the law of sines, we're going to want to get everything in terms of
, so note that
. Now, we use law of sines, which gives us the following:
,
but since
, we go with the positive solution. Thus,
.
Denote the intersection of and
with
. By another application of the law of sines,
and
. Since
, and
. Note that
, so
.
Now we have that , and
. Thus, the area of
is
, and our final answer is
.
See also
Mock AIME 1 Pre 2005 (Problems, Source) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |