Difference between revisions of "2023 AIME I Problems/Problem 12"
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==Solution== | ==Solution== | ||
+ | By Miquel's theorem, <math>P=(AEF)\cap(BFD)\cap(CDE)</math>. The law of cosines can be used to compute <math>DE=42</math>, <math>EF=35</math>, and <math>FD=13</math>. Toss the points on the coordinate plane; let <math>B=(-7, 0)</math>, <math>D=(0, 0)</math>, and <math>C=(48, 0)</math>, where we will find <math>\tan^{2}\left(\measuredangle CDP\right)</math> with <math>P=(BFD)\cap(CDE)</math>. | ||
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+ | By the extended law of sines, the radius of <math>(BFD)</math> is <math>\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}</math>. Its center lies on the line <math>x=-\frac{7}{2}</math>, and the origin is a point on it, so <math>y=\frac{23}{6}\sqrt{3}</math>. | ||
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+ | The radius of <math>(CDE)</math> is <math>\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}</math>. The origin is also a point on it, and its center is on the line <math>x=24</math>, so <math>y=2\sqrt{3}</math>. | ||
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+ | The equations of the two circles are <cmath>\begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*}</cmath> These equations simplify to <cmath>\begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*}</cmath> Subtracting these two equations gives that both their points of intersection, <math>D</math> and <math>P</math>, lie on the line <math>55x-\tfrac{11}{3}\sqrt{3}y=0</math>. Hence, <math>\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}</math>. To scale, the configuration looks like the figure below. | ||
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+ | [[File:AIME 2023-I12 Geogebra Diagram.png]] | ||
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+ | ==Solution 2== | ||
Denote <math>\theta = \angle AEP</math>. | Denote <math>\theta = \angle AEP</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution | + | ==Solution 3 (no trig)== |
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> | ||
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~anon | ~anon | ||
− | ==Solution | + | ==Solution 4 (LOC)== |
This solution is heavily inspired by AIME 1999 Problem 14 Solution 2 (a similar question) | This solution is heavily inspired by AIME 1999 Problem 14 Solution 2 (a similar question) | ||
Revision as of 13:19, 9 February 2023
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
By Miquel's theorem, . 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 is . Its center lies on the line , and the origin is a point on it, so .
The radius of 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
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 (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 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
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.