Difference between revisions of "2023 AIME I Problems/Problem 12"

m (Solution 5)
(Solution 1 (Coordinates Bash))
 
(36 intermediate revisions by 11 users not shown)
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Let <math>\triangle ABC</math> be an equilateral triangle with side length <math>55.</math> Points <math>D,</math> <math>E,</math> and <math>F</math> lie on <math>\overline{BC},</math> <math>\overline{CA},</math> and <math>\overline{AB},</math> respectively, with <math>BD = 7,</math> <math>CE=30,</math> and <math>AF=40.</math> Point <math>P</math> inside <math>\triangle ABC</math> has the property that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math>
 
Let <math>\triangle ABC</math> be an equilateral triangle with side length <math>55.</math> Points <math>D,</math> <math>E,</math> and <math>F</math> lie on <math>\overline{BC},</math> <math>\overline{CA},</math> and <math>\overline{AB},</math> respectively, with <math>BD = 7,</math> <math>CE=30,</math> and <math>AF=40.</math> Point <math>P</math> inside <math>\triangle ABC</math> has the property that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math>
  
==Solution==
+
==Diagram==
By Miquel's theorem, <math>P=(AEF)\cap(BFD)\cap(CDE)</math> (intersection of circles). 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>.
+
<asy>
 +
/* Made by MRENTHUSIASM */
 +
 
 +
size(300);
 +
 
 +
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];
 +
 
 +
draw(anglemark(A,E,P,20),red);
 +
draw(anglemark(B,F,P,20),red);
 +
draw(anglemark(C,D,P,20),red);
 +
add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red));
 +
add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red));
 +
add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red));
 +
draw(A--B--C--cycle^^P--E^^P--F^^P--D);
 +
 
 +
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(150),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>
 +
~MRENTHUSIASM
 +
 
 +
==Solution 1 (Coordinates Bash)==
 +
By Miquel's theorem, <math>P=(AEF)\cap(BFD)\cap(CDE)</math> (intersection of circles)<math>\text{*}</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>.
  
 
By the extended law of sines, the radius of circle <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>.
 
By the extended law of sines, the radius of circle <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>.
Line 9: Line 46:
 
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);
 +
draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);
 +
 
 +
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(150),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>
  
[[File:AIME 2023-I12 Geogebra Diagram.png]]
+
<math>\text{*}</math>
 +
Basic angle chasing gives <math>\angle BDP=180-\angle CDP.</math> Because <math>\angle BFP+\angle BDP=180+\angle BFP-\angle CDP=180,</math> which means that <math>BFPD</math> is cyclic, and that <math>P</math> passes through the circumcircle of triangle <math>BFD.</math> Similar reasoning leads us to the fact that <math>P</math> also passes through the circumcircles of triangles <math>AEF</math> and <math>CDE,</math> which means that <math>P=(AEF)\cap(BFD)\cap(CDE).</math> Continue as above.
  
==Solution 2==
+
==Solution 2 (Vectors/Complex)==
  
 
Denote <math>\theta = \angle AEP</math>.
 
Denote <math>\theta = \angle AEP</math>.
Line 75: Line 143:
 
</cmath>
 
</cmath>
  
Therefore, <math>\tan^2 \theta = \boxed{\textbf{(075) }}</math>.
+
Therefore, <math>\tan^2 \theta = \boxed{075}</math>.
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
==Solution 3 (no trig)==
+
==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>  
  
Line 88: Line 156:
 
Finally, <math>\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.</math>  
 
Finally, <math>\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.</math>  
  
Thus, <math>\tan^2 AEP = \boxed{075.}</math>
+
Thus, <math>\tan^2 AEP = \boxed{075}.</math>
  
 
~anon
 
~anon
  
==Solution 4 (LOC)==
+
<i><b>Claim</b></i>
This solution is heavily inspired by AIME 1999 Problem 14 Solution 2 (a similar question)
+
[[File:Carnot theorem new.png|400px|right]]
 +
a) Carnot's theorem. Given triangle <math>\triangle ABC</math> and point <math>P.</math>  Let <math>PS \perp BC,</math>
 +
<math>PR \perp AC, PQ \perp AB.</math> <math>P</math> doesn't have to be inside <math>\triangle ABC.</math>
 +
 
 +
Prove that <math>AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.</math>
 +
 
 +
b) Let <math>\triangle ABC</math> be the equilateral triangle. Prove that <math>AQ + BS + CR = \frac {3}{2} AB.</math> (The sum of the lengths of the alternating segments split by the perpendiculars from a point <math>P</math> within an equilateral triangle is equal to half the perimeter.)
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
a) <math>AP^2 = AQ^2 + PQ^2 = AR^2 + PR^2,</math>
 +
<cmath>BP^2 = BQ^2 + PQ^2 = BS^2 + PS^2,</cmath>
 +
<cmath>CP^2 = CR^2 + PR^2 = CS^2 + PS^2 \implies</cmath>
 +
<math>AQ^2 + PQ^2 + BS^2 + PS^2 + CR^2 + PR^2 = AR^2 + PR^2 + BQ^2 + PQ^2 + CS^2 + PS^2,</math>
 +
<cmath>AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.</cmath>
 +
b) <math>AQ + BQ = BS + CS = CR + AR = AB = AC = BC,</math>
 +
<cmath>AQ^2 + BQ^2 + 2 AQ \cdot BQ + BS^2 + CS^2 + 2 BS \cdot CS + AR^2 + CR^2+ 2 AR \cdot CR = 3 AB^2.</cmath>
 +
[[File:2023 AIME I 11.png|350px|right]]
 +
<cmath>2AQ^2 + 2 AQ \cdot BQ + 2 BS^2 + 2 BS \cdot CS + 2 CR^2 + 2 AR \cdot CR = 3 AB^2.</cmath>
 +
<cmath>AQ (AQ + BQ) + BS (BS + CS) + CR (AR + CR) = \frac {3}{2} AB^2.</cmath>
 +
<cmath>AQ \cdot AB + BS \cdot BC + CR \cdot AC = \frac {3}{2} AB^2.</cmath>
 +
<cmath>AQ + BS + CR = \frac {3}{2} AB.</cmath>
 +
 
 +
'''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 <math>P</math> to points <math>A</math>, <math>B</math>, and <math>C</math>. And label the angle measure of <math>\angle{BFP}</math>, <math>\angle{CDP}</math>, and <math>\angle{AEP}</math> to be <math>\alpha</math>
 
Draw line segments from <math>P</math> to points <math>A</math>, <math>B</math>, and <math>C</math>. And label the angle measure of <math>\angle{BFP}</math>, <math>\angle{CDP}</math>, and <math>\angle{AEP}</math> to be <math>\alpha</math>
  
Using Law of Cosines (note that <math>cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180^\circ-\alpha}=-cos{\alpha}</math>)
+
Using Law of Cosines (note that <math>\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}</math>)
  
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
(1) BP^2 &= FP^2+15^2-2*FP*15*cos(\alpha)\\
+
(1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\
(2) BP^2 &= DP^2+7^2+2*DP*7*cos(\alpha)\\
+
(2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\
(3) CP^2 &= DP^2+48^2-2*DP*48*cos(\alpha)\\
+
(3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\
(4) CP^2 &= EP^2+30^2+2*EP*30*cos(\alpha)\\
+
(4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\
(5) AP^2 &= EP^2+25^2-2*EP*25*cos(\alpha)\\
+
(5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\
(6) AP^2 &= FP^2+40^2+2*FP*40*cos(\alpha)\\
+
(6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
We can perform this operation: (1) - (2) + (3) - (4) + (5) - (6)
+
We can perform this operation <math>(1) - (2) + (3) - (4) + (5) - (6)</math>:
  
 
Leaving us with (after combining and simplifying)
 
Leaving us with (after combining and simplifying)
 
+
<cmath>\cos{\alpha}=\frac{-11}{2\cdot(DP+EP+FP)}</cmath>
<cmath>
 
\begin{align*}
 
\cos{\alpha}=\frac{-11}{2*(DP+EP+FP)}
 
\end{align*}
 
</cmath>
 
  
 
Therefore, we want to solve for <math>DP+EP+FP</math>
 
Therefore, we want to solve for <math>DP+EP+FP</math>
  
Notice that <math>\angle{APC}=\angle{APC}=\angle{APC}=120^\circ</math>
+
Notice that <math>\angle{DPE}=\angle{EPF}=\angle{FPD}=120^\circ</math>
  
 
We can use Law of Cosines again to solve for the sides of <math>\triangle{DEF}</math>, which have side lengths of <math>13</math>, <math>42</math>, and <math>35</math>, and area <math>120\sqrt{3}</math>.
 
We can use Law of Cosines again to solve for the sides of <math>\triangle{DEF}</math>, which have side lengths of <math>13</math>, <math>42</math>, and <math>35</math>, and area <math>120\sqrt{3}</math>.
Line 132: Line 221:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
[\triangle{DEF}] &= \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3}
+
[\triangle{DEF}] &= \frac{1}{2}\cdot\sin{120°}\cdot(xy+yz+zx) = 120\sqrt{3}
 
\\
 
\\
xy+yz+zx &= 2^5*3*5
+
xy+yz+zx &= 2^5\cdot3\cdot5
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
Line 142: Line 231:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
x^2+y^2+xy=42^2\\
+
x^2+y^2+xy&=42^2\\
y^2+z^2+yz=35^2\\
+
y^2+z^2+yz&=35^2\\
z^2+x^2+zx=13^2\\
+
z^2+x^2+zx&=13^2\\
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
Line 156: Line 245:
 
(x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2}
 
(x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2}
 
\\
 
\\
(x+y+z)^2 &= \frac{42^2+35^2+13^2+3*2^5*3*5}{2}
+
(x+y+z)^2 &= \frac{42^2+35^2+13^2+3\cdot2^5\cdot3\cdot5}{2}
 
\\
 
\\
 
(x+y+z)^2 &= 2299
 
(x+y+z)^2 &= 2299
Line 168: Line 257:
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
\cos{\alpha} = \frac{-1}{2\sqrt{19}}\\
+
\cos{\alpha} &= \frac{-1}{2\sqrt{19}}\\
\sin{\alpha} = \frac{5\sqrt{3}}{2\sqrt{19}}
+
\sin{\alpha} &= \frac{5\sqrt{3}}{2\sqrt{19}}
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
Giving us
+
Giving us <cmath>\tan^2{\alpha}=\boxed{075}</cmath>
 +
 
 +
~[[Daniel Zhou's Profile|Danielzh]]
 +
 
 +
==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 <math>\angle AEP = \angle BFP = \angle CDP = \alpha</math>. We can label these three perpendiculars as:
 +
<cmath>PD\cdot\sin{\alpha} + PE\cdot\sin{\alpha} + PF\cdot\sin{\alpha} = \dfrac{55 \sqrt{3}}{2}</cmath>
 +
Simplifying, we get
 +
<cmath>PD + PE + PF = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}</cmath>
 +
Now, as stated and quoting Solution 4,
 +
"Draw line segments from <math>P</math> to points <math>A</math>, <math>B</math>, and <math>C</math>. [We know that] the angle measure of <math>\angle{AEP}</math>, <math>\angle{BFP}</math>, and <math>\angle{CDP}</math> is <math>\alpha</math>
 +
 
 +
Using Law of Cosines (note that <math>\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}</math>)
  
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
\tan^2{\alpha}=\boxed{075}
+
(1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\
 +
(2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\
 +
(3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\
 +
(4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\
 +
(5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\
 +
(6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
~Danielzh
+
We can perform this operation <math>(1) - (2) + (3) - (4) + (5) - (6)</math>:
 +
 
 +
Leaving us with (after combining and simplifying)
 +
<cmath>PD + PE + PF=\frac{11}{2\cdot\cos{\alpha}}</cmath>".
  
==Solution 5==
+
Now, we can use our previous equation along with this one to get:
[[File:752F4325-0FE4-4963-A7A3-77870D2C5B81.png|200px|thumb|left|This diagram isn't to scale]]
+
<cmath>\frac{11}{2\cdot\cos{\alpha}} = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}</cmath>.
 +
 
 +
This equation becomes:
 +
<cmath>\tan{\alpha} = 5\sqrt{3}</cmath>
 +
As so, our answer is
 +
<cmath>
 +
\left(5\sqrt3\right)^2=\boxed{075}.
 +
</cmath>
 +
~Solution by armang32324 (Mathemagics Club)
 +
 
 +
==Solution 6==
 
By the law of cosines,
 
By the law of cosines,
 
<cmath>
 
<cmath>
 
FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.
 
FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.
 
</cmath>
 
</cmath>
Similarly we get <math>FD=13</math> and <math>DE=42</math>. <math>\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta</math> implies that <math>AFPE</math>, <math>BDPF</math>, and <math>CDPE</math> are three cyclic quadrilaterals. Using the law of sines in each,
+
Similarly we get <math>FD=13</math> and <math>DE=42</math>. <math>\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta</math> implies that <math>AFPE</math>, <math>BDPF</math>, and <math>CDPE</math> are three cyclic quadrilaterals, as shown 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];
 +
 
 +
draw(anglemark(A,E,P,20),red);
 +
draw(anglemark(B,F,P,20),red);
 +
draw(anglemark(C,D,P,20),red);
 +
add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red));
 +
add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red));
 +
add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red));
 +
draw(A--B--C--cycle^^P--E^^P--F^^P--D);
 +
draw(P--A^^P--B^^P--C,dashed);
 +
draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);
 +
 
 +
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(150),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>
 +
Using the law of sines in each,
 
<cmath>
 
<cmath>
 
\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.
 
\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.
Line 204: Line 360:
 
Lastly, <math>\sin\angle AEP=\frac{5\sqrt{57}}{38}</math>, then <math>\tan\angle AEP=5\sqrt3</math>. The answer is
 
Lastly, <math>\sin\angle AEP=\frac{5\sqrt{57}}{38}</math>, then <math>\tan\angle AEP=5\sqrt3</math>. The answer is
 
<cmath>
 
<cmath>
\left(5\sqrt3\right)^2=\boxed{75}.
+
\left(5\sqrt3\right)^2=\boxed{075}.
 
</cmath>
 
</cmath>
 +
 +
==Video Solution==
 +
 +
https://www.youtube.com/watch?v=EdwM8GpY_yc
 +
 +
~MathProblemSolvingSkills.com
 +
 +
==Animated Video Solution==
 +
 +
https://youtu.be/5d98iXeyu4E
 +
 +
~Star League (https://starleague.us)
 +
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/BYR3DollZeA
  
 
==See also==
 
==See also==

Latest revision as of 14:40, 13 April 2024

Problem

Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$

Diagram

[asy] /* Made by MRENTHUSIASM */  size(300);  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];  draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D);  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(150),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] ~MRENTHUSIASM

Solution 1 (Coordinates Bash)

By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles)$\text{*}$. The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$.

By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$. Its center lies on the line $x=-\frac{7}{2}$, and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$.

The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$. The origin is also a point on it, and its center is on the line $x=24$, so $y=2\sqrt{3}$.

The equations of the two circles are \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*} These equations simplify to \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*} Subtracting these two equations gives that both their points of intersection, $D$ and $P$, lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$. Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$. 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); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);  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(150),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]

$\text{*}$ Basic angle chasing gives $\angle BDP=180-\angle CDP.$ Because $\angle BFP+\angle BDP=180+\angle BFP-\angle CDP=180,$ which means that $BFPD$ is cyclic, and that $P$ passes through the circumcircle of triangle $BFD.$ Similar reasoning leads us to the fact that $P$ also passes through the circumcircles of triangles $AEF$ and $CDE,$ which means that $P=(AEF)\cap(BFD)\cap(CDE).$ Continue as above.

Solution 2 (Vectors/Complex)

Denote $\theta = \angle AEP$.

In $AFPE$, we have $\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0$. Thus, \[ AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)} + EA e^{- i 120^\circ} = 0. \]

Taking the real and imaginary parts, we get \begin{align*} AF + FP \cos \theta + PE \cos \left( \theta + 60^\circ \right) + EA \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (1) \\ FP \sin \theta + PE \sin \left( \theta + 60^\circ \right) + EA \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (2) \end{align*}

In $BDPF$, analogous to the analysis of $AFPE$ above, we get \begin{align*} BD + DP \cos \theta + PF \cos \left( \theta + 60^\circ \right) + FB \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (3) \\ DP \sin \theta + PF \sin \left( \theta + 60^\circ \right) + FB \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (4) \end{align*}

Taking $(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)$, we get \[ AF \sin \left( \theta + 60^\circ \right) + \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5) \]

Taking $(3) \cdot \sin \theta - (4) \cdot \cos \theta$, we get \[ BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6) \]

Taking $(5) + (6)$, we get \[ AF \sin \left( \theta + 60^\circ \right)  - EA \sin \theta + BD \sin \theta  + FB \sin \left( \theta + 120^\circ \right) . \]

Therefore, \begin{align*} \tan \theta & = \frac{\frac{\sqrt{3}}{2} \left( AF + FB \right)} {\frac{FB}{2} + EA - \frac{AF}{2} - BD} \\ & = 5 \sqrt{3} . \end{align*}

Therefore, $\tan^2 \theta = \boxed{075}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Synthetic)

Drop the perpendiculars from $P$ to $\overline{AB}$, $\overline{AC}$, $\overline{BC}$, and call them $Q,R,$ and $S$ respectively. This gives us three similar right triangles $FQP$, $ERP$, and $DSP.$

The sum of the perpendiculars to a point $P$ within an equilateral triangle is always constant, so we have that $PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.$

The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is always equal to half the perimeter, so $QA+RC+SB = \dfrac{165}{2},$ which means that $FQ+ER+DS = QA+RC+SB - CE - AF - BD = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.$

Finally, $\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.$

Thus, $\tan^2 AEP = \boxed{075}.$

~anon

Claim

Carnot theorem new.png

a) Carnot's theorem. Given triangle $\triangle ABC$ and point $P.$ Let $PS \perp BC,$ $PR \perp AC, PQ \perp AB.$ $P$ doesn't have to be inside $\triangle ABC.$

Prove that $AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.$

b) Let $\triangle ABC$ be the equilateral triangle. Prove that $AQ + BS + CR = \frac {3}{2} AB.$ (The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is equal to half the perimeter.)

Proof

a) $AP^2 = AQ^2 + PQ^2 = AR^2 + PR^2,$ \[BP^2 = BQ^2 + PQ^2 = BS^2 + PS^2,\] \[CP^2 = CR^2 + PR^2 = CS^2 + PS^2 \implies\] $AQ^2 + PQ^2 + BS^2 + PS^2 + CR^2 + PR^2 = AR^2 + PR^2 + BQ^2 + PQ^2 + CS^2 + PS^2,$ \[AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.\] b) $AQ + BQ = BS + CS = CR + AR = AB = AC = BC,$ \[AQ^2 + BQ^2 + 2 AQ \cdot BQ + BS^2 + CS^2 + 2 BS \cdot CS + AR^2 + CR^2+ 2 AR \cdot CR = 3 AB^2.\]

2023 AIME I 11.png

\[2AQ^2 + 2 AQ \cdot BQ + 2 BS^2 + 2 BS \cdot CS + 2 CR^2 + 2 AR \cdot CR = 3 AB^2.\] \[AQ (AQ + BQ) + BS (BS + CS) + CR (AR + CR) = \frac {3}{2} AB^2.\] \[AQ \cdot AB + BS \cdot BC + CR \cdot AC = \frac {3}{2} AB^2.\] \[AQ + BS + CR = \frac {3}{2} AB.\]

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 $P$ to points $A$, $B$, and $C$. And label the angle measure of $\angle{BFP}$, $\angle{CDP}$, and $\angle{AEP}$ to be $\alpha$

Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$)

\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}

We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$:

Leaving us with (after combining and simplifying) \[\cos{\alpha}=\frac{-11}{2\cdot(DP+EP+FP)}\]

Therefore, we want to solve for $DP+EP+FP$

Notice that $\angle{DPE}=\angle{EPF}=\angle{FPD}=120^\circ$

We can use Law of Cosines again to solve for the sides of $\triangle{DEF}$, which have side lengths of $13$, $42$, and $35$, and area $120\sqrt{3}$.

Label the lengths of $PD$, $PE$, and $PF$ to be $x$, $y$, and $z$.

Therefore, using the $\sin$ area formula,

\begin{align*} [\triangle{DEF}] &= \frac{1}{2}\cdot\sin{120°}\cdot(xy+yz+zx) = 120\sqrt{3} \\ xy+yz+zx &= 2^5\cdot3\cdot5 \end{align*}

In addition, we know that

\begin{align*} x^2+y^2+xy&=42^2\\ y^2+z^2+yz&=35^2\\ z^2+x^2+zx&=13^2\\ \end{align*}

By using Law of Cosines for $\triangle{DPE}$, $\triangle{EPF}$, and $\triangle{FPD}$ respectively

Because we want $DP+EP+FP$, which is $x+y+z$, we see that

\begin{align*} (x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2} \\ (x+y+z)^2 &= \frac{42^2+35^2+13^2+3\cdot2^5\cdot3\cdot5}{2} \\ (x+y+z)^2 &= 2299 \\ x+y+z &= 11\sqrt{19} \end{align*}

So plugging the results back into the equation before, we get

\begin{align*} \cos{\alpha} &= \frac{-1}{2\sqrt{19}}\\ \sin{\alpha} &= \frac{5\sqrt{3}}{2\sqrt{19}} \end{align*}

Giving us \[\tan^2{\alpha}=\boxed{075}\]

~Danielzh

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 $\angle AEP = \angle BFP = \angle CDP = \alpha$. We can label these three perpendiculars as: \[PD\cdot\sin{\alpha} + PE\cdot\sin{\alpha} + PF\cdot\sin{\alpha} = \dfrac{55 \sqrt{3}}{2}\] Simplifying, we get \[PD + PE + PF = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\] Now, as stated and quoting Solution 4, "Draw line segments from $P$ to points $A$, $B$, and $C$. [We know that] the angle measure of $\angle{AEP}$, $\angle{BFP}$, and $\angle{CDP}$ is $\alpha$

Using Law of Cosines (note that $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$)

\begin{align*} (1) \ BP^2 &= FP^2+15^2-2\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}

We can perform this operation $(1) - (2) + (3) - (4) + (5) - (6)$:

Leaving us with (after combining and simplifying) \[PD + PE + PF=\frac{11}{2\cdot\cos{\alpha}}\]".

Now, we can use our previous equation along with this one to get: \[\frac{11}{2\cdot\cos{\alpha}} = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}\].

This equation becomes: \[\tan{\alpha} = 5\sqrt{3}\] As so, our answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\] ~Solution by armang32324 (Mathemagics Club)

Solution 6

By the law of cosines, \[FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.\] Similarly we get $FD=13$ and $DE=42$. $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$, $BDPF$, and $CDPE$ are three cyclic quadrilaterals, as shown 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];  draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D); draw(P--A^^P--B^^P--C,dashed); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue);  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(150),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] Using the law of sines in each, \[\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.\] So we can set $AP=35k$, $BP=13k$, and $CP=42k$. Let $PD=d$, $PE=e$, and $PF=f$. Applying Ptolemy theorem in the cyclic quadrilaterals, \[\begin{cases}AP\cdot FE=AF\cdot PE+AE\cdot PF,\\CP\cdot ED=CE\cdot PD+CD\cdot PE,\\BP\cdot DF=BD\cdot PF+BF\cdot PD.\end{cases} \implies \begin{cases} 1225k=40e+25f,\\1764k=30d+48e,\\169k=15d+7f, \end{cases}\] We can solve out $d=\frac{54k}5$, $e=30k$, $f=k$. By the law of cosines in $\triangle PEF$, $FE=\sqrt{900k^2+k^2-60k\cdot\left(\frac{-1}2\right)}=\sqrt{931}k$. The law of sines yield $\frac{\sin\angle AEP}{\sin\angle FAE}=\frac{AP}{FE}=\frac{35k}{\sqrt{931}k}=\frac{35}{\sqrt{931}}$. Lastly, $\sin\angle AEP=\frac{5\sqrt{57}}{38}$, then $\tan\angle AEP=5\sqrt3$. The answer is \[\left(5\sqrt3\right)^2=\boxed{075}.\]

Video Solution

https://www.youtube.com/watch?v=EdwM8GpY_yc

~MathProblemSolvingSkills.com

Animated Video Solution

https://youtu.be/5d98iXeyu4E

~Star League (https://starleague.us)

Video Solution by MOP 2024

https://youtu.be/BYR3DollZeA

See also

2023 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png