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Difference between revisions of "2009 AMC 10B Problems/Problem 20"
(→‎Solution: much better solution.)
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== Solution ==
 
== Solution ==
  
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Let <math>\angle BAD = \alpha</math>, then <math>\angle BAC = 2\alpha</math>.
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By the Pythagorean Theorem, <math>AC=\sqrt5</math>. The Angle Bisector Theorem now yields that
  
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Let <math>BD = x</math>, we then have <math>\tan \alpha = \frac x1 = x</math> and <math>\tan (2\alpha) = \frac 21 = 2</math>.
+
<math>\frac{BC}{1}=\frac{2-BC}{\sqrt5}\\
−
 
+
BC\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\
−
We can now use the formula <math>\tan (2\alpha) = \frac{2 \tan\alpha}{1 - \tan^2 \alpha}</math>. Substituting the values for <math>\tan\alpha</math> and <math>\tan(2\alpha)</math>, we get the equation <math>x^2 + x - 1 = 0</math>.
+
BC(\sqrt5+1)=2\\
−
 
+
BC=\frac{2}{\sqrt5+1}=\boxed{\sqrt5-1}.</math>
−
This quadratic equation has two roots. However, one of them is negative, hence our <math>x</math> is the positive root <math>\boxed{ \frac{\sqrt 5 - 1}2 }</math>.
 
−
 
 
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=== Note ===
 
−
 
 
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The formula for <math>\tan (2\alpha)</math> can easily be derived using the better-known formulas <math>\sin (2\alpha)=2\sin\alpha\cos\alpha</math> and <math>\cos (2\alpha)=\cos^2\alpha - \sin^2\alpha</math> as follows:
 
−
 
 
−
<cmath>
 
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\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}
 
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</cmath>
 
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2009|ab=B|num-b=19|num-a=21}}

Revision as of 22:18, 13 March 2009

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, $AC=\sqrt5$. The Angle Bisector Theorem now yields that

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

See Also

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
All AMC 10 Problems and Solutions
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