Difference between revisions of "2018 AIME II Problems/Problem 12"
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But <math>\sin{APB}=\sin{APD}</math>, so <cmath>(AP-CP)(BP-DP)=0</cmath> | But <math>\sin{APB}=\sin{APD}</math>, so <cmath>(AP-CP)(BP-DP)=0</cmath> | ||
Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here). | Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here). | ||
− | Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for triangles <math>APB</math> and <math>BPC</math>, it is clear that <math>x^2+y^2-100=-(x^2+y^2-196)</math>, or <cmath>x^2+y^2=148...(1)</cmath> Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>x^2+z^2=180...(2)</cmath>. It follows that <cmath>z^2-y^2=32...(3)</cmath>. Now, denote angle <math>APB</math> by <math>\alpha</math>. Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2}</cmath>, giving <cmath>5y=3z</cmath>. Plugging this back to equations (1), (2), and (3), it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath> | + | Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for triangles <math>APB</math> and <math>BPC</math>, it is clear that |
+ | |||
+ | <math>x^2+y^2-100=2 \cdot x \cdot y \cdot \cos{APB}=-(2 \cdot x \cdot y \cdot \cos{\pi-CPB})=-(x^2+y^2-196)</math>, or <cmath>x^2+y^2=148...(1)</cmath> Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>x^2+z^2=180...(2)</cmath>. It follows that <cmath>z^2-y^2=32...(3)</cmath>. Now, denote angle <math>APB</math> by <math>\alpha</math>. Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2}</cmath>, giving <cmath>5y=3z</cmath>. Plugging this back to equations (1), (2), and (3), it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath> | ||
--Solution by MicGu | --Solution by MicGu | ||
+ | |||
==Solution 4 == | ==Solution 4 == | ||
As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). | As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). |
Revision as of 00:32, 18 November 2020
Contents
[hide]Problem
Let be a convex quadrilateral with
,
, and
. Assume that the diagonals of
intersect at point
, and that the sum of the areas of triangles
and
equals the sum of the areas of triangles
and
. Find the area of quadrilateral
.
Solution 1
For reference, , so
is the longest of the four sides of
. Let
be the length of the altitude from
to
, and let
be the length of the altitude from
to
. Then, the triangle area equation becomes
.
What an important finding! Note that the opposite sides and
have equal length, and note that diagonal
bisects diagonal
. This is very similar to what happens if
were a parallelogram with
, so let's extend
to point
, such that
is a parallelogram. In other words,
and
. Now, let's examine
. Since
, the triangle is isosceles, and
. Note that in parallelogram
,
and
are congruent, so
and thus
. Define
, so
. We use the Law of Cosines on
and
:
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot
on
gives
and therefore
. Seeing that
, we conclude that
is a 3-4-5 right triangle, so
. Then, the area of
is
. Since
, points
and
are equidistant from
, so
and hence
. -kgator
Just to be complete -- and
can actually be equal. In this case,
, but
must be equal to
. We get the same result. -Mathdummy.
Solution 2 (Another way to get the middle point)
So, let the area of triangles
,
,
,
. Suppose
and
, then it is easy to show that
Also, because
we will have
So
So
So
So
As a result,
Then, we have
Combine the condition
we can find out that
so
is the midpoint of
~Solution by (Frank FYC)
Solution 3 (With yet another way to get the middle point)
Using the formula for the area of a triangle,
But
, so
Hence
(note that
makes no difference here).
Now, assume that
,
, and
. Using the cosine rule for triangles
and
, it is clear that
, or
Likewise, using the cosine rule for triangles
and
,
. It follows that
. Now, denote angle
by
. Since
,
which simplifies to
, giving
. Plugging this back to equations (1), (2), and (3), it can be solved that
. Then, the area of the quadrilateral is
--Solution by MicGu
Solution 4
As in all other solutions, we can first find that either or
, but it's an AIME problem, we can take
, and assume the other choice will lead to the same result (which is true).
From , we have
,
=>
, therefore,
By Law of Cosine,
Square (1) and (2), add them, we get
Solve,
=>
,
-Mathdummy
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.