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
Claim
a) Carnot's theorem. Given triangle and point Let doesn't have to be inside
Prove that
b) Let be the equilateral triangle. Prove that (The sum of the lengths of the alternating segments split by the perpendiculars from a point within an equilateral triangle is equal to half the perimeter.)
Proof
a) b)
vladimir.shelomovskii@gmail.com, vvsss
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 (Combining Solutions 3 and 4)
We begin by using the fact stated in Solution 3 that, for any point in an equilateral triangle, the lengths of the three perpendicular lines dropped to the sides of the triangle add up to the altitude of that triangle. To make things simple, let's assign . We can label these three perpendiculars as: Simplifying, we get Now, as stated and quoting Solution 4, "Draw line segments from to points , , and . [We know that] the angle measure of , , and is
Using Law of Cosines (note that )
We can perform this operation :
Leaving us with (after combining and simplifying) ".
Now, we can use our previous equation along with this one to get: .
This equation becomes: As so, our answer is ~Solution by armang32324 (Mathemagics Club)
Solution 6
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
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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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