2010 AMC 10B Problem 6

P1-10~1 minPrint

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

A circle is centered at $O$ , $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$ ?

Show solution without answering

Key insight

Angles CAB and COB open onto the same arc CB. The one with its corner on the circle is half the one at the center: 25 degrees.

Solution

Draw the picture: $A$ and $B$ at the two ends of a diameter, $O$ in the middle, and $C$ on the circle so that $\angle COB = 50^\circ$ .

1. Look at which part of the circle each angle "sees". Both $\angle COB$ and $\angle CAB$ open onto the same arc, the arc from $C$ to $B$ .
2. $\angle COB$ has its corner at the center, so the arc $CB$ measures $50^\circ$ .
3. $\angle CAB$ has its corner $A$ on the circle. An angle with its corner on the circle is always half of the arc it opens onto. So
$\angle CAB = \frac{50^\circ}{2} = 25^\circ.$

The answer is $\boxed{\textbf{(B)}\ 25}$ .

Why this works

This is the inscribed angle rule: an angle with its corner on the circle is half the angle at the center that opens onto the same arc. Whenever a problem gives you an angle at the center and asks about an angle at a point on the circle (or the other way round), check whether they share an arc.

Alternative approach

If you do not remember the rule, use an isosceles triangle.

$OA$ and $OC$ are both radii, so triangle $AOC$ has two equal sides, and its two base angles at $A$ and $C$ are equal.
$A$ , $O$ , $B$ lie on one straight line, so $\angle AOC = 180^\circ - 50^\circ = 130^\circ$ .
The two base angles share what is left: $\frac{180^\circ - 130^\circ}{2} = 25^\circ$ . The base angle at $A$ is exactly $\angle CAB$ .

Concise solution

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

Both $\angle COB$ and $\angle CAB$ open onto the same arc $CB$ . The first has its vertex at the center, so it is a central angle equal to the arc, $50^\circ$ . The second has its vertex $A$ on the circle, so it is an inscribed angle, and an inscribed angle is half the arc it cuts:
$\angle CAB = \frac{1}{2}\cdot 50^\circ = 25^\circ.$

The answer is $\boxed{\textbf{(B)}\ 25}$ .

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)