2009 AMC 10B Problem 20

By the Pythagorean theorem, $AC = \sqrt{1^2 + 2^2} = \sqrt5$.

The bisector from $A$ divides the opposite side in the ratio of the adjacent sides:
$$\frac{BD}{DC} = \frac{AB}{AC} = \frac{1}{\sqrt5}.$$
Since $BD + DC = 2$, we get
$$BD = 2 \cdot \frac{1}{1 + \sqrt5} = \frac{2}{\sqrt5 + 1}.$$
Multiply numerator and denominator by $\sqrt5 - 1$:
$$BD = \frac{2(\sqrt5 - 1)}{5 - 1} = \frac{\sqrt5 - 1}{2}.$$

Sanity check: $\frac{\sqrt5 - 1}{2} \approx 0.618$, which is less than $DC \approx 1.382$, consistent with $AB < AC$.

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt5 - 1}{2}}$.

The angle bisector theorem converts an angle condition into a length ratio, and the ratio uses the two sides that form the bisected angle. Combined with the known total of the split side, it is a one-line computation; the only care needed is rationalizing to match the answer format.

Let $\angle BAD = \theta$, so $\tan 2\theta = \frac{BC}{AB} = 2$. From $\tan 2\theta = \frac{2t}{1 - t^2}$ with $t = \tan\theta$, $t^2 + t - 1 = 0$ and $t = \frac{\sqrt5 - 1}{2}$; then $BD = AB \tan\theta = \frac{\sqrt5 - 1}{2}$.

• Splitting BC in the ratio of the legs AB : BC = 1 : 2 instead of AB : AC, which gives BD = 2/3.
• Writing the ratio upside down and obtaining $DC$ instead of $BD$, i.e. choice (C) after rationalizing.