Difference between revisions of "2022 AIME II Problems/Problem 11"
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(→Solution 3 (Visual)) |
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==Solution 3 (Visual)== | ==Solution 3 (Visual)== | ||
− | + | [[File:AIME 2022 11a.png|300px|right]] | |
<b><i>Lemma</b></i> | <b><i>Lemma</b></i> | ||
Line 94: | Line 94: | ||
<b><i> Proof</b></i> | <b><i> Proof</b></i> | ||
+ | |||
Let <math>A = 2\alpha.</math> Then <math>\angle DBC = \angle DCB = \alpha.</math> | Let <math>A = 2\alpha.</math> Then <math>\angle DBC = \angle DCB = \alpha.</math> | ||
− | Let <math>E</math> be the intersection point of the perpendicular dropped from <math>D</math> to <math>AB</ | + | Let <math>E</math> be the intersection point of the perpendicular dropped from <math>D</math> to <math>AB</math> with the circle. |
− | Then the sum of arcs < | + | |
− | Let < | + | Then the sum of arcs <math>\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = \pi.</math> |
+ | <cmath>\overset{\Large\frown} {BE} = \pi – 2\alpha – \overset{\Large\frown}{AC}.</cmath> | ||
+ | |||
+ | Let <math>E'</math> be the point of intersection of the line <math>CM</math> with the circle. | ||
+ | <math>CM</math> is perpendicular to <math>AD, \angle AMC = \frac {\pi}{2} – \alpha,</math> the sum of arcs <math>\overset{\Large\frown}{A}C + \overset{\Large\frown}{BE'} = \pi – 2\alpha,</math> hence <math>E'</math> coincides with <math>E.</math> | ||
+ | |||
+ | The inscribed angles <math>\angle DEM = \angle DEB, M</math> is symmetric to <math>B</math> with respect to <math>DE, DM = DB.</math> | ||
+ | |||
+ | <b><i> Solution</b></i> | ||
+ | |||
+ | Let <math>AB' = AB, DC' = DC, B'</math> and <math>C'</math> on <math>AD.</math> | ||
+ | Then <math>AB' = 2, DC' = 3, B'C' = 2 = AB'.</math> | ||
+ | Quadrilateral <math>ABMC'</math> is inscribed. | ||
==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 15:34, 1 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.
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.