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
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
Solution 4 (THINK OUTSIDE THE BOX)
Extend and so they intersect at a point . Then note that is the incenter of , implying that is on the angle bisector of . Now because is both an angle bisector and a median of , is isosceles. Then we can start angle chasing:
Let and . Then , implying that , implying that , or that . Substituting this into the rest of the diagram, we find that .
Then , or . Moreover, , or . Similarly, , or . Then using Law of Cosines on , to get that , or .
We finish by using the formula , as follows:
.
.
-dragoon
Solution 5 (incircle)
As shown in paragraph one of solution 4, extending and to , we realize that is isosceles, thus . Let . And, midpoint is the incenter of . Construct perpendiculars to sides respectively (constructing the radii of the incircle). Let . The semiperimeter . Since is the tangent off the incircle, . So, . Because ,
.
By Heron's formula and the inradius area formula,
Then, . Finally,
Thus, our answer is
Remark
and are the tangent points of the X-mixtilinear incircle (of ). This may be useful, but I haven't looked into it.
Solution 6 (bash)
Let the midpoint of be . Angle-chase and observe that . Let and and . As a result of this similarity, we write
which gives . Similarly, we write
and
to get and .
We now have all required side lengths; we can find the area of with Heron's formula. Doing so yields . We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between and is , and between and is . Thus, the area ratios are and , respectively, so adding together we have . Multiplying this by our , we have as their total area. Adding this to our original area, we have .
The square of this is .
~~Technodoggo
Solution 7 (similar to 4)
As in solution 4, let , so is the incenter of and . Let . Then the normalized barycentric coordinates of , , and with respect to are , , and . So we have giving . The sidelengths of are thus , , and giving . Also, we have so that . The area squared is thus . ~~ golue3120
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 |
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