2001 AMC 12 Problems/Problem 24

Revision as of 19:05, 27 May 2017 by Dyoder13 (talk | contribs) (Solution)

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); \ 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.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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