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

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Now we can note that <math>\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ</math>, hence also the triangle <math>EBC</math> is isosceles and we have <math>BE=CE</math>.
 
Now we can note that <math>\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ</math>, hence also the triangle <math>EBC</math> is isosceles and we have <math>BE=CE</math>.
  
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Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle ACE=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>.
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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>.
  
 
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 = \boxed{75^\circ}</math>.

Revision as of 18:14, 13 February 2010

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

See Also

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
All AMC 12 Problems and Solutions
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