Difference between revisions of "2023 AIME I Problems/Problem 12"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 3 ( | + | ==Solution 3 (Synthetic)== |
Drop the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{AC}</math>, <math>\overline{BC}</math>, and call them <math>Q,R,</math> and <math>S</math> respectively. This gives us three similar right triangles <math>FQP</math>, <math>ERP</math>, and <math>DSP.</math> | Drop the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{AC}</math>, <math>\overline{BC}</math>, and call them <math>Q,R,</math> and <math>S</math> respectively. This gives us three similar right triangles <math>FQP</math>, <math>ERP</math>, and <math>DSP.</math> | ||
Revision as of 00:28, 14 February 2023
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
Solution (Coord 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 (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 , 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
~Danielzh
Solution 5
By the law of cosines, Similarly we get and . implies that , , and are three cyclic quadrilaterals. 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.