2005 AIME II Problems/Problem 14

Problem

In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Solution 1

$[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$

By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$ , we have

$\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align*}$

Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$ . The answer is $q = \boxed{463}$ .

Solution 2 (Similar Triangles)

Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.

From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21*6*7*8}$ = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS= $\frac{24}{5}$ . Consequently, from Pythagorean theorem, SC = $\frac{18}{5}$ and AS = 14-SC = $\frac{52}{5}$ . We can also use pythagorean triangle on triangle AQB to determine that BQ = $\frac{33}{5}$ .

Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles.

Firstly: $\triangle ARE \sim \triangle ASD$ . From there, we have $\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}$ .

Next: $\triangle BRE \sim \triangle BQA$ . From there, we have $\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}$ .

Solve the system to get $x = \frac{2184}{463}$ and $y = \frac{4732}{463}$ . Notice that 463 is prime, so even though we use pythagorean theorem on x and 13-y, the denominator won't change. The answer we desire is $\boxed{463}$ .

Solution 3 (LoC and LoS bash)

Let $\angle CAD = \angle BAE = \theta$ . Note by Law of Sines on $\triangle BEA$ we have $\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}$ As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$ ).

Let the foot of the altitude from $A$ to $BC$ be $H$ . By law of cosines on $\triangle ABC$ we have $169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}$ It follows that $AH = \frac{56}{5}$ and $HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}$ .

Note that by PT on $\triangle AHD$ we have that $AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}$ . By Law of Sines on $\triangle ADC$ (where we square everything to avoid taking the square root) we see $\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.$ How are we going to find $\sin{\angle BEA}$ though? $\angle BEA$ and $\theta$ are in the same triangle. Applying Law of Sines on $\triangle ABC$ we see that $\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.$ $\theta$ , $\angle B$ , and $\angle BEA$ are all in the same triangle. We know they add up to $180^{\circ}$ . There's a good chance we can exploit this using the identity $\sin{p} = \sin{180^{\circ}-p}$ .

We have that $\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}$ . Success! We know $\sin{\theta}$ and $\sin{\angle B}$ already. Applying the $\sin$ addition formula we see $\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.$ This is the last stretch! Applying Law of Sines a final time on $\triangle BEA$ we see $\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.$ It follows that the answer is $\boxed{463}$ .

Solution 4 (Ratio Lemma and Angle Bisector Theorem)

Let $AK$ be the angle bisector of $\angle A$ such that $K$ is on $BC$ .

Then $\angle KAB = \angle KAC$ , and thus $\angle KAE = \angle KAD$ .

By the Ratio Lemma, $\frac{BE}{KE} = \frac{BA}{KA} * \frac{\sin{BAE}}{\sin{KAE}}$ and $\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}$ .

This implies that $\frac{BE}{KE*BA} = \frac{CD}{KD*CA}$ .

Thus, $\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}$ .

$DK = CK - 6 = 14*15/27 - 6 = 16/9$ . Thus, $\frac{BE}{KE} = \frac{13*54}{14*16}$ .

Additionally, $BE + KE = 9$ . Solving gives that $q = 463.$

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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