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2005 AIME II Problems/Problem 14

Revision as of 20:04, 11 December 2016 by Ychen (talk | contribs) (→‎Solution 2 (Similar Triangles))

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 $\boxed{463}$.

See also

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
All AIME Problems and Solutions

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