Difference between revisions of "2022 AIME II Problems/Problem 11"
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+ | ==Problem 11== | ||
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+ | Let <math>ABCD</math> be a convex quadrilateral with <math>AB=2, AD=7,</math> and <math>CD=3</math> such that the bisectors of acute angles <math>\angle{DAB}</math> and <math>\angle{ADC}</math> intersect at the midpoint of <math>\overline{BC}.</math> Find the square of the area of <math>ABCD.</math> | ||
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+ | [[2022 AIME II Problems/Problem 11|Solution]] | ||
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==Solution 1== | ==Solution 1== | ||
Revision as of 19:47, 25 March 2023
Contents
[hide]Problem 11
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)
Claim
In the triangle is the midpoint of
is the point of intersection of the circumcircle and the bisector of angle
Then
Proof
Let Then
Let be the intersection point of the perpendicular dropped from
to
with the circle.
Then the sum of arcs
Let be the point of intersection of the line
with the circle.
is perpendicular to
the sum of arcs
coincides with
The inscribed angles is symmetric to
with respect to
Solution
Let and
on
Then
Quadrilateral is cyclic.
Let
Then
Circle centered at
is its diameter,
since they both complete
to
since they are the exterior angles of an isosceles
by two angles.
The height dropped from to
is
The areas of triangles and
are equal to
area of
is
The area of
is
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by The Power of Logic
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.