Difference between revisions of "2022 AIME II Problems/Problem 11"
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The area of <math>ABCD</math> is <math>(1 + 2 + 3) \sqrt{5} = 6\sqrt{5},</math> its square is <math>\boxed{180}.</math> | The area of <math>ABCD</math> is <math>(1 + 2 + 3) \sqrt{5} = 6\sqrt{5},</math> its square is <math>\boxed{180}.</math> | ||
− | + | '''Shelomovskii, vvsss, www.deoma-cmd.ru''' | |
==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 13:02, 18 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 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 hence coincides with
The inscribed angles is symmetric to with respect to
Solution
Let and on Then
Quadrilateral is inscribed. Let Then
Circle has center 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 areas of triangles and as well as and are equal.
The area of is its square is
Shelomovskii, vvsss, www.deoma-cmd.ru
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.