Difference between revisions of "2005 AIME II Problems/Problem 14"

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==Solution 6 (Tangent subtraction formulas)==
 
==Solution 6 (Tangent subtraction formulas)==
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We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude
  
 
== See also ==
 
== See also ==

Revision as of 22:46, 19 February 2019

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 the Pythagorean theorem 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 the 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 $q = 463.$

Alternate: By the ratio lemma, $BD/DC = (13/14)*(\sin BAD/\sin DAC)$ $EC/EB = (14/13)*(\sin EAC/\sin BAE)$

Combining these, we get $(BD/DC)(14/13) = (EC/EB)(13/14)$ $(3/2)(14/13)(14/13) = (15-x)(x)$

$x = 2535/463$ Thus, $q = 463$

Solution 5 (Isogonal lines with respect to A angle bisesector)

Since $AE$ and $AD$ are isogonal with respect to the $A$ angle bisector, we have \[\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.\] To prove this, let $\angle BAE=\angle DAC=x$ and $\angle BAD=\angle CAE=y.$ Then, by the Ratio Lemma, we have \[\frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}\] \[\frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}\] and multiplying these together proves the formula for isogonal lines. Hence, we have \[\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}\] so our desired answer is $\boxed{463}.$


Solution 6 (Tangent subtraction formulas)

We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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