Difference between revisions of "2022 AIME II Problems/Problem 11"
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Because <math>M</math> is the midpoint of segment <math>BC</math>, <math>MB = MC</math>. | Because <math>M</math> is the midpoint of segment <math>BC</math>, <math>MB = MC</math>. | ||
Because <math>MP = MB</math> and <math>MQ = MC</math>, <math>MP = MQ</math>. | Because <math>MP = MB</math> and <math>MQ = MC</math>, <math>MP = MQ</math>. | ||
+ | |||
Thus, <math>\angle MPD = \angle MQA</math>. | Thus, <math>\angle MPD = \angle MQA</math>. | ||
+ | |||
Thus, | Thus, | ||
− | \[ | + | <cmath>\[ |
\alpha + \theta = \beta + \phi . \hspace{1cm} (1) | \alpha + \theta = \beta + \phi . \hspace{1cm} (1) | ||
− | \] | + | \]</cmath> |
In <math>\triangle AMD</math>, <math>\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta</math>. | In <math>\triangle AMD</math>, <math>\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta</math>. | ||
In addition, <math>\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi</math>. | In addition, <math>\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi</math>. | ||
Thus, | Thus, | ||
− | \[ | + | <cmath>\[ |
\alpha + \beta = \theta + \phi . \hspace{1cm} (2) | \alpha + \beta = \theta + \phi . \hspace{1cm} (2) | ||
− | \] | + | \]</cmath> |
Taking <math>(1) + (2)</math>, we get <math>\alpha = \phi</math>. | Taking <math>(1) + (2)</math>, we get <math>\alpha = \phi</math>. | ||
Line 75: | Line 77: | ||
Therefore, | Therefore, | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
{\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ | {\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) \\ | & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ | ||
& = 6 \sqrt{5} . | & = 6 \sqrt{5} . | ||
− | \end{align*} | + | \end{align*}</cmath> |
Therefore, the square of <math>{\rm Area} \ ABCD</math> is <math>\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}</math>. | Therefore, the square of <math>{\rm Area} \ ABCD</math> is <math>\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}</math>. |
Revision as of 14:40, 1 June 2022
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,
In ,
.
In addition,
.
Thus,
Taking , we get
.
Taking
, we get
.
Therefore, .
Hence, and
.
Thus,
and
.
In , by applying the law of cosines,
.
Hence,
.
Hence,
.
Therefore,
Therefore, the square of is
.
~Steven Chen (www.professorchenedu.com)
Solution 3 (Visual)
Lemma
In the triangle is the midpoint of
is the point of intersection of the circumscribed circle and the bisector of angle
Then
Proof
Let Then
Let be the intersection point of the perpendicular dropped from
to
BE + AC + CD = π. BE = π – 2α – AC.
E'
CM$ with the circle.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.