AoPSWiki

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

Personal tools

Search

Toolbox

2001 AMC 12 Problems/Problem 24

From AoPSWiki

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

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 $\angle ACE=90^\circ$ , 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 • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2001_AMC_12_Problems/Problem_24"

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us