2001 AMC 12 Problems/Problem 24

Problem

In $\triangle ABC$ , $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB$ .

$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$

Solution

$[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); pair E=intersectionpoint(A--D, C--(C+10*ortho)); draw(A--B--C--cycle); draw(A--D); draw(C--E, dashed); draw(B--E, dashed); draw( rightanglemark(C,E,D) ); draw( scale(2)*anglemark(B,A,D) ); draw( anglemark(D,B,A) ); label("$15^\circ$",(0.6,0),5*ENE); label("$45^\circ$",B,5*WNW,UnFill); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,S); [/asy]$

We start with the observation that $\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ$ , and $\angle ADC = 15^\circ + 45^\circ = 60^\circ$ .

We can draw the height $CE$ from $C$ onto $AD$ . In the triangle $CED$ , we have $\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12$ . Hence $ED = CD/2$ .

By the definition of $D$ , we also have $BD=CD/2$ , therefore $BD=DE$ . This means that the triangle $BDE$ is isosceles, and as $\angle BDE=120^\circ$ , we must have $\angle BED = \angle EBD = 30^\circ$ .

Then we compute $\angle ABE = 45^\circ - 30^\circ = 15^\circ$ , thus $\angle ABE = \angle BAE$ and the triangle $ABE$ is isosceles as well. Hence $AE=BE$ .

Now we can note that $\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ$ , hence also the triangle $EBC$ is isosceles and we have $BE=CE$ .

Combining the previous two observations we get that $AE=EC$ , and as $\angle AEC=90^\circ$ , this means that $\angle CAE = \angle ACE = 45^\circ$ .

Finally, we get $\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}$ .

Trig Bash

WLOG, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $\angle ADB = 60^\circ$ and $\angle ADC = 120^\circ$ .

Using Law of Sines on triangle $ADB$ , we find that $\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}$ . Since we know that $\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$ , $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , and $\sin 120^\circ = \frac{\sqrt{3}}{2}$ , we can compute $AD$ to equal $1+\sqrt{3}$ and $AB$ to be $\frac{3\sqrt{2}+\sqrt{6}}{2}$ .

Next, we apply Law of Cosines to triangle $ADC$ to see that $AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)$ . Simplifying the RHS, we get $AC^2 = 6$ , so $AC = \sqrt{6}$ .

Now, we apply Law of Sines to triangle $ABC$ to see that $\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}$ . After rearranging and noting that $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , we get $\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}$ .

Dividing the RHS through by $\sqrt{3}$ , we see that $\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4}$ , so $\angle ACB$ is either $75^\circ$ or $105^\circ$ . Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75^\circ}$ .

Note that we can also confirm that $\angle ACB \neq 105^\circ$ by computing $\angle CAB$ with Law of Sines.

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2001 AMC 12 Problems/Problem 24

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Solution

Trig Bash

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