Difference between revisions of "2023 AIME I Problems/Problem 12"
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The radius of circle <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>. | The radius of circle <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>. | ||
− | 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 | + | 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: |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
− | [ | + | size(400); |
+ | |||
+ | pair A, B, C, D, E, F, P; | ||
+ | A = 55*sqrt(3)/3 * dir(90); | ||
+ | B = 55*sqrt(3)/3 * dir(210); | ||
+ | C = 55*sqrt(3)/3 * dir(330); | ||
+ | D = B + 7*dir(0); | ||
+ | E = A + 25*dir(C-A); | ||
+ | F = A + 40*dir(B-A); | ||
+ | P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; | ||
+ | |||
+ | filldraw(D--E--F--cycle,yellow); | ||
+ | draw(A--B--C--cycle^^circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E)); | ||
+ | |||
+ | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(B),linewidth(4)); | ||
+ | dot("$C$",C,1.5*dir(C),linewidth(4)); | ||
+ | dot("$D$",D,1.5*S,linewidth(4)); | ||
+ | dot("$E$",E,1.5*dir(30),linewidth(4)); | ||
+ | dot("$F$",F,1.5*dir(95),linewidth(4)); | ||
+ | dot("$P$",P,1.5*dir(-30),linewidth(4)); | ||
+ | |||
+ | label("$7$",midpoint(B--D),1.5*S,red); | ||
+ | label("$30$",midpoint(C--E),1.5*dir(30),red); | ||
+ | label("$40$",midpoint(A--F),1.5*dir(150),red); | ||
+ | </asy> | ||
==Solution 2 (Vectors/Complex)== | ==Solution 2 (Vectors/Complex)== |
Revision as of 16:20, 28 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
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 (LOC)
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: (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
Animated Video Solution
~Star League (https://starleague.us)
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.