2012 AMC 10A Problem 4 — Solution | LIVE by Po-Shen Loh

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2012 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2012 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10A solutions, or check the answer key.

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Concepts:angle chasing optimization

Difficulty rating: 770

4.

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

$0$

$2$

$4$

$6$

$12$

Solution:

Note that both of the angles share the ray $AB.$ To minimize the desired degree, we want $D$ to be between $A$ and $C.$

This would make $\angle CBD = \angle ABC - \angle ABD = 4^{\circ}.$

Thus, C is the correct answer.

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Problem 4 in Other Years

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