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2009 AMC 10B Problem 20

Below is the professionally curated solution for Problem 20 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Concepts:angle bisector theoremPythagorean Theorem

Difficulty rating: 1600

20.

Triangle ABCABC has a right angle at B,B, AB=1,AB=1, and BC=2.BC=2. The bisector of ∠BAC\angle BAC meets BC‾\overline{BC} at D.D. What is BD?BD?

3−12\dfrac{\sqrt3-1}{2}

5−12\dfrac{\sqrt5-1}{2}

5+12\dfrac{\sqrt5+1}{2}

6+22\dfrac{\sqrt6+\sqrt2}{2}

23−12\sqrt3-1

Solution:

By the Pythagorean Theorem, AC=12+22=5.AC=\sqrt{1^2+2^2}=\sqrt5. The Angle Bisector Theorem gives BDDC=ABAC=15, \dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5}, so DC=5 BD.DC=\sqrt5\,BD.

Since BD+DC=2,BD+DC=2, we have BD(1+5)=2,BD(1+\sqrt5)=2, so BD=21+5=5−12. BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.

Thus, the correct answer is B.

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