Difference between revisions of "2022 AIME II Problems/Problem 11"
(→Solution 1) |
(→Solution 3 (Visual)) |
||
Line 137: | Line 137: | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 4 (THINK OUTSIDE THE BOX)== | ||
+ | |||
+ | Extend <math>AB</math> and <math>CD</math> so they intersect at a point <math>X</math>. Then note that <math>M</math> is the incenter of <math>\triangle{XAD}</math>, implying that <math>M</math> is on the angle bisector of <math>X</math>. Now because <math>XM</math> is both an angle bisector and a median of <math>\triangle{XBC}</math>, <math>\triangle{XBC}</math> is isosceles. Then we can start angle chasing: | ||
+ | |||
+ | Let <math>\angle{BAM}=a, \angle{CDM}=b,</math> and <math>\angle{XBC}=c</math>. Then <math>\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c</math>, implying that <math>\angle{BMA}+\angle{CMD}=a+b</math>, implying that <math>2c-(a+b)=(a+b)</math>, or that <math>c=a+b</math>. Substituting this into the rest of the diagram, we find that <math>\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}</math>. | ||
+ | |||
+ | Then <math>\frac{AB}{BM}=\frac{MC}{CD}</math>, or <math>BM=CM=\sqrt{6}</math>. Moreover, <math>\frac{AB}{AM}=\frac{AM}{AD}</math>, or <math>AM=\sqrt{14}</math>. Similarly, <math>\frac{CD}{MD}=\frac{MD}{AD}</math>, or <math>DM=\sqrt{21}</math>. Then using Law of Cosines on <math>\triangle{AMD}</math>, to get that <math>cos\angle{AMD}=-\frac{\sqrt{6}}{6}</math>, or <math>sin\angle{AMD}=\frac{\sqrt{30}}{6}</math>. | ||
+ | |||
+ | We finish by using the formula <math>K=\frac{1}{2}absinC</math>, as follows: | ||
+ | |||
+ | <math>[ABCD}=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}</math>. | ||
+ | |||
+ | <math>6\sqrt{5}^2=\boxed{180</math>. | ||
+ | |||
+ | -dragoon | ||
==Video Solution by The Power of Logic== | ==Video Solution by The Power of Logic== |
Revision as of 17:29, 7 July 2023
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:
$[ABCD}=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}$ (Error compiling LaTeX. Unknown error_msg).
$6\sqrt{5}^2=\boxed{180$ (Error compiling LaTeX. Unknown error_msg).
-dragoon
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.