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

Revision as of 22:42, 24 December 2020

1 Problem
2 Solution 1
3 Solution 2 (Similar Triangles)
4 Solution 3 (LoC and LoS bash)
5 Solution 4 (Ratio Lemma and Angle Bisector Theorem)
6 Solution 5 (Isogonal lines with respect to A angle bisesector)
7 Solution 6 (Tangent subtraction formulas)
8 Solution 8(literally 2 minute solution)
9 Solution 9 (Short and no IQ Required Altogether-Bash)
10 See also

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 $HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}$ .

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)

Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.

$[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label("$G$",G,N); label("$F$",F,N); [/asy]$

Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\angle DAG = \alpha$ . Now we know that $\overline{DG} = \frac{24}{5}$ and $\overline{GC} = \frac{18}{5}$ . Therefore, $\overline{AG} = \frac{52}{5}$ , so $\tan{(\alpha)} = \frac{6}{13}$ . Our goal now is to use tangent $\angle EAG$ in triangle $AEG$ . We set $\overline{BE}$ to $x$ , so $\overline{ED} = 9 - x$ and $\overline{EC} = 15 - x$ , so $\overline{EG} = \frac{4}{5}(15-x)$ and $\overline{GC} = \frac{3}{5}(15-x)$ so $\overline{AG} = \frac{3x+25}{5}$ . Now we just need tangent of $\angle EAG$ .

We find this using $\tan{(EAG)} = \tan{(A - \alpha)} = \frac{\tan{A} - \tan{\alpha}}{1 + \tan{A}\tan{\alpha}}$ , which is $\frac{\frac{12}{5} - \frac{6}{13}}{1 + \frac{12}{5}\frac{6}{13}}$ or $\frac{126}{137}$ . Now we solve the equation $\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}$ , so $x = \frac{2535}{463}$

Solution 8(literally 2 minute solution)

Let $ED = x$ , such that $BE = 9-x$ . Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$ , and we can solve to get $x = \frac{1632}{463}$ (and $BE = \frac{1146}{463}$ ). Hence, our answer is $\boxed{463}$ . - Spacesam

Solution 9 (Short and no IQ Required Altogether-Bash)

Diagram borrowed from 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$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$

Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have $AB^2=AC^2+BC^2-2(AC)(BC)\cos C$ Solving gets $\cos C=\frac{3}{5}$ , which implies that $\sin C=\sqrt{1-\cos C}=\frac{4}{5}$ Applying Stewart's Theorem with cevian $AD$ we have $(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2$ Solving gets $AD=\frac{4\sqrt{205}}{5}$ .

Applying Law of Sines on $\bigtriangleup ACD$ to solve for $\sin CAD$ we have $\frac{AD}{\sin C}=\frac{CD}{\sin CAD}$ Solving gets $\sin CAD=\frac{6\sqrt{205}}{205}$ . Thus $\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}$ .

Applying Law of Sines on $\bigtriangleup ABC$ we have $\frac{AC}{\sin B}=\frac{AB}{\sin C}$ Solving gets $\sin B=\frac{56}{65}$ .

Applying Stewart's Theorem with cevian $AE$ we have $(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2$ $(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2$ Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}$

Finally, applying Law of Sines on $\bigtriangleup BAE$ we have $\frac{AE}{\sin B}=\frac{BE}{\sin BAE}$ $\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}$ $7605BE^2-32342BE+2535=0$ Solving the easy quadratic equation gets $BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}$

~ Nafer

@@ Line 57: / Line 57: @@
 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
 <cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}</cmath>
-It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}</math>.
+It follows that <math>AH = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}</math>.
 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

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