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

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m (Solution 1)
 
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== Problem ==
 
== Problem ==
  
In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB</math>.
+
In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB.</math>
  
 
<math>
 
<math>
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</math>
 
</math>
  
== Solution ==
+
== Solution 1 ==
  
 
<asy>
 
<asy>
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defaultpen(0.8);
 
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 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 ortho=rotate(-50)*(D-A);
pair E=intersectionpoint(A--D, C--(C+10*ortho));
+
pair E=extension(C, ortho, A, D);
draw(A--B--C--cycle);  
+
draw(A--B);
 +
draw(B--C);
 +
draw(A--C);
 +
draw(C--E);
 +
draw(E--D);
 
draw(A--D);
 
draw(A--D);
draw(C--E, dashed);
+
draw(B--E, dotted);
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("$A$",A,SW);
 
label("$B$",B,SE);
 
label("$B$",B,SE);
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Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>.
 
Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>.
  
Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>.
+
Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = 75^\circ \boxed{D}</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Draw a good diagram! Now, let's call <math>BD=t</math>, so <math>DC=2t</math>. Given the rather nice angles of <math>\angle ABD = 45^\circ</math> and <math>\angle ADC = 60^\circ</math> as you can see, let's do trig. Drop an altitude from <math>A</math> to <math>BC</math>; call this point <math>H</math>. We realize that there is no specific factor of <math>t</math> we can call this just yet, so let <math>AH=kt</math>. Notice that in <math>\triangle{ABH}</math> we get <math>BH=kt</math>. Using the 60-degree angle in <math>\triangle{ADH}</math>, we obtain <math>DH=\frac{\sqrt{3}}{3}kt</math>. The comparable ratio is that <math>BH-DH=t</math>. If we involve our <math>k</math>, we get:
  
 +
<math>kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t</math>. Eliminating <math>t</math> and removing radicals from the denominator, we get <math>k=\frac{3+\sqrt{3}}{2}</math>. From there, one can easily obtain <math>HC=3t-kt=\frac{3-\sqrt{3}}{2}t</math>. Now we finally have a desired ratio. Since <math>\tan\angle ACH = 2+\sqrt{3}</math> upon calculation, we know that <math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and cosine using <math>\sin(x)^2+\cos(x)^2=1</math>, and perhaps the half- or double-angle formulas, you get <math>\boxed{75^\circ}</math>.
  
==Trig Bash==
+
== Solution 3==
WLOG, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADB = 60^\circ</math> and <math>\angle ADC = 120^\circ</math>.
+
Without loss of generality, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADC = 60^\circ</math> and <math>\angle ADB = 120^\circ</math>.
  
Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>.
+
Using Law of Sines on triangle <math>ADB</math>, we find that <cmath>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.</cmath> Since we know that <cmath>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},</cmath> <cmath>\sin 45^\circ = \frac{\sqrt{2}}{2},</cmath> <cmath>\sin 120^\circ = \frac{\sqrt{3}}{2},</cmath> we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>.
  
Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the RHS, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>.
+
Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <cmath>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).</cmath> Simplifying the right side, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>.
  
Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>.
+
Now, we apply Law of Sines to triangle <math>ABC</math> to see that <cmath>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.</cmath> After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <cmath>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.</cmath>
  
Dividing the RHS through by <math>\sqrt{3}</math>, we see that <math>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4}</math>, so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>.
+
Dividing the right side by <math>\sqrt{3}</math>, we see that <cmath>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},</cmath> so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>.
  
 
Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines.
 
Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines.

Latest revision as of 11:56, 27 December 2020

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(-50)*(D-A); pair E=extension(C, ortho, A, D); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(E--D); draw(A--D); draw(B--E, dotted);  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 $\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = 75^\circ \boxed{D}$.

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 $\tan(75)=2+\sqrt{3}$ 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

Without loss of generality, we can assume that $BD = 1$ and $CD = 2$. As above, we are able to find that $\angle ADC = 60^\circ$ and $\angle ADB = 120^\circ$.

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 \[\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},\] \[\sin 45^\circ = \frac{\sqrt{2}}{2},\] \[\sin 120^\circ = \frac{\sqrt{3}}{2},\] 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 right side, 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 right side 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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