2023 AIME I Problems/Problem 12
Problem
Let be an equilateral triangle with side length Points and lie on and respectively, with and Point inside has the property that Find
Solution
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 2 (no trig)
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 3 (LOC)
This solution is heavily 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: (1) - (2) + (3) - (4) + (5) - (6)
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 DEF, which have side lengths of 13, 42, and 35, 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
~Danielzh
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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