# 2013 AIME I Problems/Problem 9

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## Problem

A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.

$[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]$

## Solution 1

Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it. $[asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); [/asy]$ We have $AF=6\sqrt{3}$ and $FD=3$, so $AD=3\sqrt{13}$. Denote $\angle DAF = \theta$; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$.

In triangle $AXY$, $AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}$, and $AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}$.

In triangle $AMX$, we get $\angle AMX=60^\circ-\theta$ and then use sine-law to get $MX=\tfrac 12 AX\csc(60^\circ-\theta)$; similarly, from triangle $ANX$ we get $NX=\tfrac 12 AX\csc(60^\circ+\theta)$. Thus $$MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).$$ Since $\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)$, we get \begin{align*} \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} \end{align*} Then $$MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}$$

The answer is $39 + 39 + 35 = \boxed{113}$.

## Solution 2

Let $P$ and $Q$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded.

Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.

Let $a$, $b$, and $x$ be the lengths $AP$, $AQ$, and $PQ$, respectively.

We have $PD = a$, $QD = b$, $BP = 12 - a$, $CQ = 12 - b$, $BD = 9$, and $CD = 3$.

Using the Law of Cosines on $BPD$:

$a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}$

$a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$

$a = \frac{39}{5}$

Using the Law of Cosines on $CQD$:

$b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}$

$b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$

$b = \frac{39}{7}$

Using the Law of Cosines on $DPQ$:

$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$

$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$

$x = \frac{39 \sqrt{39}}{35}$

The solution is $39 + 39 + 35 = \boxed{113}$.

## Solution 3

Proceed with the same labeling as in Solution 1.

$\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$

$\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ$

Therefore, $\angle PDB = \angle CQD$.

Similarly, $\angle BPD = \angle QDC$.

Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ are similar triangles, so

$\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}$.

Solving this system of equations yields $a = \frac{39}{5}$ and $b = \frac{39}{7}$.

Using the Law of Cosines on $APQ$:

$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$

$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$

$x = \frac{39 \sqrt{39}}{35}$

The solution is $39 + 39 + 35 = \boxed{113}$.

### Note

Once you find $DP$ and $DQ$, you can scale down the triangle by a factor of $\frac{39}{35}$ so that all sides are integers. Applying Law of cosines becomes easier, you just need to remember to scale back up.

## Solution 4 (Coordinate Bash)

We let the original position of $A$ be $A$, and the position of $A$ after folding be $D$. Also, we put the triangle on the coordinate plane such that $A=(0,0)$, $B=(-6,-6\sqrt3)$, $C=(6,-6\sqrt3)$, and $D=(3,-6\sqrt3)$.

$[asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = (9,0); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label("D", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label("A",X,dir(90)); draw(A--X); [/asy]$

Note that since $A$ is reflected over the fold line to $D$, the fold line is the perpendicular bisector of $AD$. We know $A=(0,0)$ and $D=(3,-6\sqrt3)$. The midpoint of $AD$ (which is a point on the fold line) is $(\tfrac32, -3\sqrt3)$. Also, the slope of $AD$ is $\frac{-6\sqrt3}{3}=-2\sqrt3$, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$, or $\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}$. Then, using point slope form, the equation of the fold line is $$y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)$$$$y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}$$ Note that the equations of lines $AB$ and $AC$ are $y=\sqrt3x$ and $y=-\sqrt3x$, respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$: $$\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}$$$$\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}$$ Therefore, the point of intersection is $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$. Now, lets find the intersection with $AC$. Substituting for $y$ yields $$-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}$$$$\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}$$ Therefore, the point of intersection is $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$. Now, we just need to use the distance formula to find the distance between $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ and $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$. $$\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}$$ The number 39 is in all of the terms, so let's factor it out: $$39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}$$$$\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}$$ Therefore, our answer is $39+39+35=\boxed{113}$, and we are done.

Solution by nosaj.