Difference between revisions of "2018 AIME II Problems/Problem 12"
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==Solution 3 (With yet another way to get the middle point)== | ==Solution 3 (With yet another way to get the middle point)== | ||
− | Using the formula for the area of a triangle, <cmath>(\frac{1}{2}AP | + | Using the formula for the area of a triangle, <cmath>(\frac{1}{2}AP\cdotBP+\frac{1}{2}CP\cdotDP)\sin{APB}=(\frac{1}{2}AP\cdotDP+\frac{1}{2}CP\cdotBP)\sin{APD}</cmath> |
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). |
Revision as of 17:05, 26 February 2021
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,
\[(\frac{1}{2}AP\cdotBP+\frac{1}{2}CP\cdotDP)\sin{APB}=(\frac{1}{2}AP\cdotDP+\frac{1}{2}CP\cdotBP)\sin{APD}\] (Error compiling LaTeX. Unknown error_msg)
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.