Difference between revisions of "2001 AMC 12 Problems/Problem 24"

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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 1

$[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); \ label("$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 $\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}$ .

Solution 2

Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:

$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $arctan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $sin(x)^2+cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75^\circ}$ .

Solution 3

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.

@@ Line 32: / Line 32: @@
 We start with the observation that <math>\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ</math>, and <math>\angle ADC = 15^\circ + 45^\circ = 60^\circ</math>.
-We can draw the height <math>CE</math> from <math>C</math> onto <math>AB</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>.
+We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>.
 By the definition of <math>D</math>, we also have <math>BD=CD/2</math>, therefore <math>BD=DE</math>. This means that the triangle <math>BDE</math> is isosceles, and as <math>\angle BDE=120^\circ</math>, we must have <math>\angle BED = \angle EBD = 30^\circ</math>.

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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