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
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Animated Video Solution
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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 | ||
All AIME Problems and Solutions |
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