Difference between revisions of "2009 AMC 10B Problems/Problem 20"

Revision as of 22:49, 17 January 2016

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

By the Pythagorean Theorem, $AC=\sqrt5$ . Then, from the Angle Bisector Theorem, we have:

$\frac{BD}{1}=\frac{2-BD}{\sqrt5}\\ BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\ BD(\sqrt5+1)=2\\ BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2}}.$

2009 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 39: / Line 39: @@
 == Solution ==
-By the Pythagorean Theorem, <math>AC=\sqrt5</math>. The Angle Bisector Theorem now yields that
+By the Pythagorean Theorem, <math>AC=\sqrt5</math>. Then, from the Angle Bisector Theorem, we have:
 <math>\frac{BD}{1}=\frac{2-BD}{\sqrt5}\\

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Difference between revisions of "2009 AMC 10B Problems/Problem 20"

Revision as of 22:49, 17 January 2016

Problem

Solution

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