2009 AMC 10B Problem 20

Attempt Problem 20 of the 2009 AMC 10B 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 2009 AMC 10B solutions, or check the answer key.

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20.

Triangle

ABC

has a right angle at

B,

AB=1,

and

BC=2.

The bisector of

\angle BAC

meets

\overline{BC}

at

D.

What is

BD?

\dfrac{\sqrt3-1}{2}

\dfrac{\sqrt5-1}{2}

\dfrac{\sqrt5+1}{2}

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

2\sqrt3-1

Answer: B

Concepts:angle bisector theorem Pythagorean Theorem

Difficulty rating: 1600

Small hint:

By the Angle Bisector Theorem,

\dfrac{BD}{DC}=\dfrac{AB}{AC}

Big hint:

AC=\sqrt5

and

BD+DC=2

Solution:

By the Pythagorean Theorem,

AC=\sqrt{1^2+2^2}=\sqrt5.

The Angle Bisector Theorem gives

\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5},

so

DC=\sqrt5\,BD.

Since

BD+DC=2,

we have

BD(1+\sqrt5)=2,

so

BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.

Thus, the correct answer is B.