2012 AIME I Problems/Problem 12

Problem 12

Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solutions

Solution 1

Let $CD = 2a$ . Using angle bisector theorem on $\triangle CDB$ , $\frac{2a}{8}=\frac{CB}{15}$ , so $CB = \frac{15a}{4}$ . Then, drop the altitude from $D$ to $CB$ and call the foot $F$ . Thus, $CF = a$ , $FD = a\sqrt{3}$ , and $FB = \frac{11a}{4}$ . Finally, $\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}$ . Our answer is $\boxed{018}$ .

Solution 2

Without loss of generality, set $CB = 1$ . Then, by the Angle Bisector Theorem on triangle $DCB$ , we have $CD = \frac{8}{15}$ . We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$ , which we can simplify to get $BD = \frac{13}{15}$ .

Now, we have $\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$ , so $\cos \angle B = \frac{11}{13}$ . In addition, $\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$ , so $\tan \angle B = \frac{4\sqrt{3}}{11}$ .

Our final answer is $4+3+11 = \boxed{018}$ .

Solution 3

(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)

Find values for all angles in terms of $\angle B$ . $\angle CEB = 150-B$ , $\angle CED = 30+B$ , $\angle CDE = 120-B$ , $\angle CDA = 60+B$ , and $\angle A = 90-B$ .

Use the law of sines on $\triangle CED$ and $\triangle CEB$ :

In $\triangle CED$ , $\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}$ . This simplifies to $16 = \frac{CE}{\sin (120-B)}$ .

In $\triangle CEB$ , $\frac{15}{\sin 30} = \frac{CE}{\sin B}$ . This simplifies to $30 = \frac{CE}{\sin B}$ .

Solve for $CE$ and equate them so that you get $16\sin (120-B) = 30\sin B$ .

From this, $\frac{8}{15} = \frac{\sin B}{\sin (120-B)}$ .

Use a trig identity on the denominator on the right to obtain: $\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}$

This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$

This gives $8\sqrt{3}\cos B+8\sin B=30\sin B$ Dividing by $\cos B$ , we have ${8\sqrt{3}}= 22\tan B$

$\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ . Our final answer is $4 + 3 + 11 = \boxed{018}$ .

Solution 4

(This solution avoids advanced trigonometry)

Let $X$ be the foot of the perpendicular from $D$ to $\overline{BC}$ , and let $Y$ be the foot of the perpendicular from $E$ to $\overline{BC}$ .

Now let $EY=x$ . Clearly, triangles $EYB$ and $DXB$ are similar with $\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}$ , so $DX=\frac{23}{15}x$ .

Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$ . For example, we see that $CY=x\sqrt{3}$ and $CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x$ . Therefore $XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x$ .

Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}$ , so $BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}$ .

Thus $\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}$ , and our answer is $4+3+11=\boxed{018}$ .

Solution 5

(Another solution without trigonometry)

Extend $CD$ to point $F$ such that $\overline{AF} \parallel \overline{CB}$ . It is then clear that $\triangle AFD$ is similar to $\triangle BCD$ .

Let $AC=p$ , $BC=q$ . Then $\tan \angle B = p/q$ .

With the Angle Bisector Theorem, we get that $CD=\frac{8}{15}q$ . From 30-60-90 $\triangle CAF$ , we get that $AF=\frac{1}{\sqrt{3}}p$ and $FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q$ .

From $\triangle AFD \sim \triangle BCD$ , we have that $\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}$ . Simplifying yields $\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1$ , and $\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}$ , so our answer is $4+3+11=\boxed{018}$ .

Solution 6

Let $CB = 1$ , and let the feet of the altitudes from $D$ and $E$ to $\overline{CB}$ be $D'$ and $E'$ , respectively. Also, let $DE = 8k$ and $EB = 15k$ . We see that $BD' = 15k\cos B$ and $BE' = 23k\cos B$ by right triangles $\triangle{BDD'}$ and $\triangle{BEE'}$ . From this we have that $D'E' = 8k\cos B$ . With the same triangles we have $DD' = 23k\sin B$ and $EE' = 15k\sin B$ . From 30-60-90 triangles $\triangle{CDD'}$ and $\triangle{CEE'}$ , we see that $CD' = \frac{23k\sqrt{3}\sin B}{3}$ and $CE' = 15k\sqrt{3}\sin B$ , so $D'E' = \frac{22k\sqrt{3}\sin B}{3}$ . From our two values of $D'E'$ we get: $8k\cos B = \frac{22k\sqrt{3}\sin B}{3}$

$\frac{\sin B}{\cos B} = \frac{8k}{\frac{22k\sqrt{3}}{3}} = \tan B$

$\tan B = \frac{8}{\frac{22\sqrt{3}}{3}} = \frac{24}{22\sqrt{3}} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$ Our answer is then $4+3+11 = \boxed{018}$ .

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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