Difference between revisions of "Mock AIME II 2012 Problems/Problem 9"
(Created page with "==Problem== In <math>\triangle ABC</math>, <math>AB=12</math>, <math>AC=20</math>, and <math>\angle ABC=120^\circ</math>. <math>D, E,</math> and <math>F</math> lie on <math>\ove...") |
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Start out by finding <math>BC</math>. Remark that by the law of sines, <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle ACB)}{12}</math>. Therefore <math>\sin(\angle ACB)=\frac{3}{5}*\frac{\sqrt{3}}{2}=\frac{3\sqrt3}{10}</math>. We know <math>90^\circ>\angle ACB>0^\circ</math> because <math>\angle ABC>90^\circ</math> in <math>\triangle ABC</math>, therefore <math>\angle ACB</math> is in the first quadrant. Use the Pythagorean Identity to give us <math>\cos^2(\angle ACB)+\sin^2(\angle ACB)=1\implies \cos(\angle ACB)=\frac{\sqrt{73}}{10}</math>. | Start out by finding <math>BC</math>. Remark that by the law of sines, <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle ACB)}{12}</math>. Therefore <math>\sin(\angle ACB)=\frac{3}{5}*\frac{\sqrt{3}}{2}=\frac{3\sqrt3}{10}</math>. We know <math>90^\circ>\angle ACB>0^\circ</math> because <math>\angle ABC>90^\circ</math> in <math>\triangle ABC</math>, therefore <math>\angle ACB</math> is in the first quadrant. Use the Pythagorean Identity to give us <math>\cos^2(\angle ACB)+\sin^2(\angle ACB)=1\implies \cos(\angle ACB)=\frac{\sqrt{73}}{10}</math>. | ||
− | Now, note that <math>\angle BAC</math> is the same thing as <math>60^\circ-\angle ACB</math>, from <math>\triangle ABC</math>, therefore we have <math>\sin(\angle BAC)=\sin(60^\circ-\angle ACB)=\sin(60^\circ)\cos(\angle ACB)-\sin(\angle ACB)\cos(60^\circ) | + | Now, note that <math>\angle BAC</math> is the same thing as <math>60^\circ-\angle ACB</math>, from <math>\triangle ABC</math>, therefore we have <math>\sin(\angle BAC)=\sin(60^\circ-\angle ACB)=\sin(60^\circ)\cos(\angle ACB)-\sin(\angle ACB)\cos(60^\circ)</math> |
+ | <math>\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}</math>. | ||
Next, use the law of sines to give us <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle BAC)}{BC}</math>. This gives us <math>\frac{\sqrt3}{40}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20BC}</math>. This gives us <math>2\sqrt{73}-6=BC</math>. | Next, use the law of sines to give us <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle BAC)}{BC}</math>. This gives us <math>\frac{\sqrt3}{40}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20BC}</math>. This gives us <math>2\sqrt{73}-6=BC</math>. |
Revision as of 03:13, 5 April 2012
Problem
In ,
,
, and
.
and
lie on
, and
, respectively. If
, and
, the area of
can be expressed in the form
where
are all positive integers, and
and
do not have any perfect squares greater than
as divisors. Find
.
Solution
Here is a diagram (note that D should be on AC and F should be on BC):
.
Start out by finding
. Remark that by the law of sines,
. Therefore
. We know
because
in
, therefore
is in the first quadrant. Use the Pythagorean Identity to give us
.
Now, note that is the same thing as
, from
, therefore we have
$\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}$ (Error compiling LaTeX. Unknown error_msg).
Next, use the law of sines to give us . This gives us
. This gives us
.
Now, we use coordinates to find that the coordinates of ,
,
, and
.
is going to be the point
from adding point
to create
triangle
as shown in this diagram (where
is the right angle,
is the
angle, and
is the
angle):
Now, note that we have to find the coordinates of and
. Assume WLOG that
is
and
be
.
is obviously
,
is going to be
and the coordinates of
is going to be
. Since
and
have the same
value, we can find
by using
. The base is going to be equal to the distance from
to
, which is the same thing as
, and the height is the change in the
coordinate from
to
. This is the same thing as
. Hence, plugging these into
give us
. The answer is thus
.