2012 AMC 10A Problem 4

P1-10~2 minPrint

Angles & Polygons Split into exhaustive cases and handle each

Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$ . What is the smallest possible degree measure for angle $CBD$ ?

Show solution without answering

Key insight

Rays BC and BD can be on the same side of BA or on opposite sides. The same side gives 24 - 20 = 4, the smaller choice.

Solution

Both angles have their corner at $B$ and both use the ray $BA$ . The problem does not show a picture, so try the two ways the rays can be placed.

Ray $BC$ is $24^\circ$ away from ray $BA$ , and ray $BD$ is $20^\circ$ away from ray $BA$ .
Opposite sides of $BA$ . The two angles sit next to each other, so $\angle CBD = 24^\circ + 20^\circ = 44^\circ$ .
Same side of $BA$ . Now ray $BD$ lies inside the angle $ABC$ , and the angle between $BD$ and $BC$ is what is left over: $\angle CBD = 24^\circ - 20^\circ = 4^\circ$ .
It cannot be $0^\circ$ : that would need $BC$ and $BD$ to be the same ray, but they are different distances ( $24^\circ$ and $20^\circ$ ) from $BA$ .

The smallest possible value is $4^\circ$ .

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

When two angles share a corner and one side, their other sides can land on the same side or on opposite sides, so the angle between them is either the difference or the sum. The words "smallest possible" tell you the picture is not fixed. Draw both pictures and compare.

Concise solution

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

Both angles have vertex $B$ and share ray $BA$ . Ray $BC$ makes a $24^\circ$ angle with ray $BA$ , and ray $BD$ makes a $20^\circ$ angle with it. There are two possible pictures.

If $C$ and $D$ are on opposite sides of line $BA$ , the angles sit next to each other and $\angle CBD = 24^\circ + 20^\circ = 44^\circ$ .

If $C$ and $D$ are on the same side of ray $BA$ , the angle $ABD$ lies inside angle $ABC$ , and $\angle CBD = 24^\circ - 20^\circ = 4^\circ$ .

The smaller of the two is $4^\circ$ ; it cannot be $0$ because the two rays make different angles with $BA$ and so cannot coincide.

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 2: Angles
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Geometry, Chapter 9: Polygons (§9.1, §9.2)
Prealgebra, Chapter 14: Counting (§14.3)