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...") |
|||
(One intermediate revision by one other user not shown) | |||
Line 44: | Line 44: | ||
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>. | ||
Line 84: | Line 85: | ||
Now, note that we have to find the coordinates of <math>E,F</math> and <math>D</math>. Assume WLOG that <math>A</math> is <math>(0,0)</math> and <math>B</math> be <math>(12,0)</math>. <math>E</math> is obviously <math>(3,0)</math>, <math>F</math> is going to be <math>(12+\frac{9+\sqrt{73}-12}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math> and the coordinates of <math>D</math> is going to be <math>(\frac{9+\sqrt{73}}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math>. Since <math>D</math> and <math>F</math> have the same <math>y</math> value, we can find <math>[EDF]</math> by using <math>\frac12*b*h</math>. The base is going to be equal to the distance from <math>D</math> to <math>F</math>, which is the same thing as <math>12+\frac{9+\sqrt{73}-12}{4}-\frac{9+\sqrt{73}}{4}=9=b</math>, and the height is the change in the <math>y</math> coordinate from <math>E</math> to <math>D</math>. This is the same thing as <math>\frac{\sqrt{219}-3\sqrt3}{4}-0=\frac{\sqrt{219}-3\sqrt3}{4}=h</math>. Hence, plugging these into <math>[EDF]=\frac12*b*h</math> give us <math>\frac12*9*\frac{\sqrt{219}-3\sqrt3}{4}=\frac{9\sqrt{219}-27\sqrt3}{8}</math>. The answer is thus <math>9+219+27+3+8=\boxed{266}</math>. | Now, note that we have to find the coordinates of <math>E,F</math> and <math>D</math>. Assume WLOG that <math>A</math> is <math>(0,0)</math> and <math>B</math> be <math>(12,0)</math>. <math>E</math> is obviously <math>(3,0)</math>, <math>F</math> is going to be <math>(12+\frac{9+\sqrt{73}-12}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math> and the coordinates of <math>D</math> is going to be <math>(\frac{9+\sqrt{73}}{4}, \frac{\sqrt{219}-3\sqrt3}{4})</math>. Since <math>D</math> and <math>F</math> have the same <math>y</math> value, we can find <math>[EDF]</math> by using <math>\frac12*b*h</math>. The base is going to be equal to the distance from <math>D</math> to <math>F</math>, which is the same thing as <math>12+\frac{9+\sqrt{73}-12}{4}-\frac{9+\sqrt{73}}{4}=9=b</math>, and the height is the change in the <math>y</math> coordinate from <math>E</math> to <math>D</math>. This is the same thing as <math>\frac{\sqrt{219}-3\sqrt3}{4}-0=\frac{\sqrt{219}-3\sqrt3}{4}=h</math>. Hence, plugging these into <math>[EDF]=\frac12*b*h</math> give us <math>\frac12*9*\frac{\sqrt{219}-3\sqrt3}{4}=\frac{9\sqrt{219}-27\sqrt3}{8}</math>. The answer is thus <math>9+219+27+3+8=\boxed{266}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>S</math> be the area of the large triangle and let <math>x</math> be the area we are trying to find. We have that <math>\Delta ADE</math> is similar to <math>\Delta ABC</math> with a ratio of <math>1:4</math> while <math>\Delta CDF</math> is similar to <math>\Delta ABC</math> with a ratio of <math>3:4</math>. Then the area of <math>\Delta ADE = \frac{S}{16}</math> and the area of <math>\Delta CDE = \frac{9S}{16}</math>. Lastly, the area of <math>\Delta EBF = \frac{1}{2}EB*BF*\sin B = \frac{1}{2}* \frac{3AB}{4}*\frac{BC}{4}*\sin B = \frac{3}{16}S</math>. The area of <math>\Delta ABC</math> is equal to the sum of the areas of the four smaller triangles so <math>S= \frac{S}{16}+\frac{9S}{16}+\frac{3S}{16} + x</math>. Thus <math>x = \frac{3S}{16}</math>. | ||
+ | |||
+ | It remains to find <math>S</math>. First we can use the Law of Cosines on <math>\Delta ABC</math> to find <math>12^2+BC^2 + 12BC = 400</math> which solves to <math>BC = -6 + 2\sqrt{73}</math>. Then <math>S = \frac{1}{2} AB* BC*\sin{120} = -18\sqrt3 + 6\sqrt{219}</math>. Then substituting yields <math>x =\frac{-27\sqrt 3 + 9 \sqrt{219}}{8}</math> so the answer is <math>\boxed{266}</math> |
Latest revision as of 14:04, 27 December 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
.
Solution 2
Let be the area of the large triangle and let
be the area we are trying to find. We have that
is similar to
with a ratio of
while
is similar to
with a ratio of
. Then the area of
and the area of
. Lastly, the area of
. The area of
is equal to the sum of the areas of the four smaller triangles so
. Thus
.
It remains to find . First we can use the Law of Cosines on
to find
which solves to
. Then
. Then substituting yields
so the answer is