2007 AMC 10B Problem 4

P1-10~2 minPrint

Circles Angles & Polygons Set up the equation/formula and compute; no special trick needed

The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$

Show solution without answering

Key insight

The three center angles add to 360, so angle AOC is 100. The angle at B looks at the same arc, so it is half of that.

Solution

We want the angle at $B$ . An angle whose corner is on the circle is half of the center angle that opens onto the same arc. Angle $ABC$ opens onto arc $AC$ , so we need the center angle $AOC$ .

1. The three angles around the center $O$ fill a full turn of $360^\circ$ :
$\angle AOB + \angle BOC + \angle AOC = 360^\circ.$
2. Two of them are given, so
$\angle AOC = 360^\circ - 140^\circ - 120^\circ = 100^\circ.$
3. Angle $ABC$ has its corner $B$ on the circle and looks at the same arc $AC$ , so it is half as big:
$\angle ABC = \tfrac{1}{2} \times 100^\circ = 50^\circ.$

The answer is $\boxed{\textbf{(D)}\ 50}$ .

Why this works

Each angle of a triangle drawn in a circle is half of the center angle across from it. Here that gives $\angle A = 60^\circ$ , $\angle B = 50^\circ$ , $\angle C = 70^\circ$ , and these add to $180^\circ$ as they should. When two center angles are given, the third always comes from $360^\circ$ .

Alternative approach

Use only isosceles triangles. The segments $OA$ , $OB$ , $OC$ are all radii, so they are equal.

Triangle $OAB$ has two equal sides and top angle $140^\circ$ . Its two bottom angles share $180^\circ - 140^\circ = 40^\circ$ , so each is $20^\circ$ .
Triangle $OBC$ has top angle $120^\circ$ , so each of its bottom angles is $(180^\circ - 120^\circ) \div 2 = 30^\circ$ .
The angle at $B$ in triangle $ABC$ is made of one bottom angle from each: $20^\circ + 30^\circ = 50^\circ$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The angle $\angle ABC$ has its vertex on the circle and opens onto arc $AC$ (the arc not containing $B$ ). Its measure is half the central angle $\angle AOC$ that subtends the same arc.

The three central angles at $O$ fill the full turn:
$\angle AOC = 360^\circ - 120^\circ - 140^\circ = 100^\circ.$
Hence
$\angle ABC = \tfrac{1}{2}\cdot 100^\circ = 50^\circ.$

The answer is $\boxed{\textbf{(D)}\ 50}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 12: Circles and Angles
Introduction to Geometry, Chapter 9: Polygons (§9.1, §9.2)