2022 AIME II Problems/Problem 11
Contents
Problem
Let be a convex quadrilateral with
,
, and
such that the bisectors of acute angles
and
intersect at the midpoint of
. Find the square of the area of
.
Solution 1
According to the problem, we have ,
,
,
, and
Because is the midpoint of
, we have
, so:
Then, we can see that is an isosceles triangle with
Therefore, we could start our angle chasing: .
This is when we found that points ,
,
, and
are on a circle. Thus,
. This is the time we found that
.
Thus,
Point is the midpoint of
, and
.
.
The area of this quadrilateral is the sum of areas of triangles:
Finally, the square of the area is
~DSAERF-CALMIT (https://binaryphi.site)
Solution 2
Denote by the midpoint of segment
.
Let points
and
be on segment
, such that
and
.
Denote ,
,
,
.
Denote . Because
is the midpoint of
,
.
Because is the angle bisector of
and
,
.
Hence,
and
.
Hence,
.
Because is the angle bisector of
and
,
.
Hence,
and
.
Hence,
.
Because is the midpoint of segment
,
.
Because
and
,
.
Thus,
.
Thus,
\[
\alpha + \theta = \beta + \phi . \hspace{1cm} (1)
\]
In ,
.
In addition,
.
Thus,
\[
\alpha + \beta = \theta + \phi . \hspace{1cm} (2)
\]
Taking , we get
.
Taking
, we get
.
Therefore, .
Hence, and
.
Thus,
and
.
In , by applying the law of cosines,
.
Hence,
.
Hence,
.
Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}
Therefore, the square of is
.
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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