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Difference between revisions of "2005 AIME II Problems/Problem 14"
(→‎Solution 3 (LoC and LoS bash))
Line 53: Line 53:
 
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have
 
Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have
 
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{BEA}}</cmath>
 
<cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{BEA}}</cmath>
−
As a result, our goal is to find <math>\sin{BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>).  
+
As a result, our goal is to find <math>\sin{\angle BEA}</math> and <math>\sin{\theta}</math> (we already know <math>AB</math>).  
  
 
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>H</math>. By law of cosines on <math>\triangle ABC</math> we have
 
Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>H</math>. By law of cosines on <math>\triangle ABC</math> we have
Line 61: Line 61:
 
Note that by PT on <math>\triangle AHD</math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC</math> (where we square everything to avoid taking the square root) we see
 
Note that by PT on <math>\triangle AHD</math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC</math> (where we square everything to avoid taking the square root) we see
 
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}</cmath>
 
<cmath>\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}</cmath>
 +
How are we going to find <math>\sin{\angle BEA}</math> though? <math>\angle BEA</math> and <math>\theta</math> are in the same triangle. Applying Law of Sines on <math>\triangle ABC</math> we see that
 +
<cmath>\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}</cmath>
 +
<math>\theta</math>, <math>\angle B</math>, and <math>\angle BEA</math> are all in the same triangle. We know they add up to <math>180^{\circ}</math>. There's a good chance we can exploit this using the identity <math>\sin{p} = \sin{180^{\circ}-p}</math>.
  
−
 
+
We have that <math>\sin{180^{\circ} - (\theta + \angle B)} = \sin{\angle BEA} = \sin{\theta + \angle B}</math>. Success! We know <math>\sin{\theta}</math> and <math>\sin{\angle B}</math> already. Applying the <math>\sin</math> addition formula we see
 +
<cmath>\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}}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 00:24, 28 July 2017

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 \[\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{BEA}}\] 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 \[\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}\]

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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