Difference between revisions of "2022 AIME II Problems/Problem 11"
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Finally, the square of the area is <math>(6\sqrt{5})^2=\boxed{180}</math> | Finally, the square of the area is <math>(6\sqrt{5})^2=\boxed{180}</math> | ||
− | ~DSAERF-CALMIT | + | ~DSAERF-CALMIT (https://binaryphi.site) |
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=10|num-a=12}} | {{AIME box|year=2022|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:29, 18 February 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
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)
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.