2010 AMC 10A Problem 16

Attempt Problem 16 of the 2010 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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16.

Nondegenerate ABC\triangle ABC has integer side lengths, BD\overline{BD} is an angle bisector, AD=3,AD = 3, and DC=8.DC = 8. What is the smallest possible value of the perimeter?

3030

3333

3535

3636

3737

Answer: B
Concepts:angle bisector theoremtriangle inequalityoptimization
Difficulty rating: 1600
Solution:

Using the Angle Bisector Theorem, we have that AB3=BC8 \dfrac{AB}{3} = \dfrac{BC}{8} AB=38BC. AB = \dfrac{3}{8} BC.

For ABAB and BCBC to be integers, we must have that BCBC is a multiple of 8.8.

To minimize the perimeter, we can set BC=8BC = 8 and AB=3.AB = 3. This, however, makes the triangle degenerate.

BCBC must then be 1616 and AB=6.AB = 6. Since AC=AD+DC=11,AC = AD + DC = 11, the perimeter is 16+6+11=33. 16 + 6 + 11 = 33.

Thus, B is the correct answer.

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