2009 AMC 10B Problems/Problem 20

Problem

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 $BD$ ?

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,1), B=(0,0), C=(2,0); pair D=extension(A,bisectorpoint(B,A,C),B,C); pair[] ds={A,B,C,D}; dot(ds); draw(A--B--C--A--D); label("$1$",midpoint(A--B),W); label("$B$",B,SW); label("$D$",D,S); label("$C$",C,SE); label("$A$",A,NW); draw(rightanglemark(C,B,A,2)); [/asy]$

$\text{(A) } \frac {\sqrt3 - 1}{2} \qquad \text{(B) } \frac {\sqrt5 - 1}{2} \qquad \text{(C) } \frac {\sqrt5 + 1}{2} \qquad \text{(D) } \frac {\sqrt6 + \sqrt2}{2} \qquad \text{(E) } 2\sqrt 3 - 1$

Solution

Let $\angle BAD = \alpha$ , then $\angle BAC = 2\alpha$ .

Let $BD = x$ , we then have $\tan \alpha = \frac x1 = x$ and $\tan (2\alpha) = \frac 21 = 2$ .

We can now use the formula $\tan (2\alpha) = \frac{2 \tan\alpha}{1 - \tan^2 \alpha}$ . Substituting the values for $\tan\alpha$ and $\tan(2\alpha)$ , we get the equation $x^2 + x - 1 = 0$ .

This quadratic equation has two roots. However, one of them is negative, hence our $x$ is the positive root $\boxed{ \frac{\sqrt 5 - 1}2 }$ .

Note

The formula for $\tan (2\alpha)$ can easily be derived using the better-known formulas $\sin (2\alpha)=2\sin\alpha\cos\alpha$ and $\cos (2\alpha)=\cos^2\alpha - \sin^2\alpha$ as follows:

$\tan (2\alpha) = \dfrac{ \sin (2\alpha) }{ \cos (2\alpha) } = \dfrac{ 2\sin\alpha\cos\alpha }{ \cos^2\alpha - \sin^2\alpha } = \dfrac{ \frac{ 2\sin\alpha\cos\alpha }{ \cos^2\alpha } }{ \frac{ \cos^2\alpha - \sin^2\alpha }{ \cos^2\alpha } } = \frac{2 \tan\alpha}{1 - \tan^2 \alpha}$

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2009 AMC 10B Problems/Problem 20

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Problem

Solution

Note

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