2023 AIME I Problems/Problem 12
Contents
Problem
Let be an equilateral triangle with side length
Points
and
lie on
and
respectively, with
and
Point
inside
has the property that
Find
Diagram
~MRENTHUSIASM
Solution 1 (Coordinates Bash)
By Miquel's theorem, (intersection of circles). The law of cosines can be used to compute
,
, and
. Toss the points on the coordinate plane; let
,
, and
, where we will find
with
.
By the extended law of sines, the radius of circle is
. Its center lies on the line
, and the origin is a point on it, so
.
The radius of circle is
. The origin is also a point on it, and its center is on the line
, so
.
The equations of the two circles are These equations simplify to
Subtracting these two equations gives that both their points of intersection,
and
, lie on the line
. Hence,
. To scale, the configuration looks like the figure below:
Solution 2 (Vectors/Complex)
Denote .
In , we have
.
Thus,
Taking the real and imaginary parts, we get
In , analogous to the analysis of
above, we get
Taking , we get
Taking , we get
Taking , we get
Therefore,
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Synthetic)
Drop the perpendiculars from to
,
,
, and call them
and
respectively. This gives us three similar right triangles
,
, and
The sum of the perpendiculars to a point within an equilateral triangle is always constant, so we have that
The sum of the lengths of the alternating segments split by the perpendiculars from a point within an equilateral triangle is always equal to half the perimeter, so
which means that
Finally,
Thus,
~anon
Solution 4 (Law of Cosines)
This solution is inspired by AIME 1999 Problem 14 Solution 2 (a similar question)
Draw line segments from to points
,
, and
. And label the angle measure of
,
, and
to be
Using Law of Cosines (note that )
We can perform this operation :
Leaving us with (after combining and simplifying)
Therefore, we want to solve for
Notice that
We can use Law of Cosines again to solve for the sides of , which have side lengths of
,
, and
, and area
.
Label the lengths of ,
, and
to be
,
, and
.
Therefore, using the area formula,
In addition, we know that
By using Law of Cosines for ,
, and
respectively
Because we want , which is
, we see that
So plugging the results back into the equation before, we get
Giving us
Solution 5
By the law of cosines,
Similarly we get
and
.
implies that
,
, and
are three cyclic quadrilaterals, as shown below:
Using the law of sines in each,
So we can set
,
, and
. Let
,
, and
. Applying Ptolemy theorem in the cyclic quadrilaterals,
We can solve out
,
,
. By the law of cosines in
,
. The law of sines yield
.
Lastly,
, then
. The answer is
Video Solution
https://www.youtube.com/watch?v=EdwM8GpY_yc
~MathProblemSolvingSkills.com
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by MOP 2024
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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