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2009 AMC 10B Problems/Problem 20: Difference between revisions

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== Problem ==
 
Triangle <math>ABC</math> has a right angle at <math>B</math>, <math>AB=1</math>, and <math>BC=2</math>. The bisector of <math>\angle BAC</math> meets <math>\overline{BC}</math> at <math>D</math>. What is <math>BD</math>?
 
<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>
 
<math>
 
\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
</math>
 
== Solution 1 ==
 
By the Pythagorean Theorem, <math>AC=\sqrt5</math>. Then, from the Angle Bisector Theorem, we have:
 
<cmath>\frac{BD}{1}=\frac{2-BD}{\sqrt5}</cmath>
<cmath>BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}</cmath>
<cmath>BD(\sqrt5+1)=2</cmath>
<cmath>BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2}} \implies {B}</cmath>
 
== Solution 2 ==
 
Let <math>\theta = \angle BAD = \angle DAC</math>. Notice <math>\tan \theta = BD</math> and <math>\tan 2 \theta = 2</math>. By the double angle identity, <cmath>2 = \frac{2 BD}{1 - BD^2} \implies BD = \boxed{\frac{\sqrt5 - 1}{2} \implies B}.</cmath>
 
Remarks: You could also use tangent half angle formula
 
== Solution 3 ==
Let <math>BD=y</math>.
 
Make <math>DE</math> a line so that it is perpendicular to <math>AC</math>. Since <math>AD</math> is an angle bisector, <math>\triangle AED</math> is congruent to <math>\triangle ABD</math>.
Using the Pythagorean Theorem:
 
<math>AC^2=1^2+2^2</math>
 
<math>AC^2=5</math>
 
<math>AC=\sqrt{5}</math>
 
We know that <math>AE=1</math> by the congruent triangles, so <math>EC=\sqrt{5}-1</math>. We know that <math>DE=y</math>, <math>EC=\sqrt{5}-1</math>, and <math>DC=2-y</math>. We now have right triangle <math>DEC</math> and its 3 sides. Using the Pythagorean Thereom, we get:
 
<math>y^2+(\sqrt{5}-1)^2=(2-y)^2</math>
 
<math>-4y=2-2\sqrt{5}</math>
 
So, <math>y=BD=\boxed{\frac{\sqrt5-1}{2} \implies B}.</math>
 
~HelloWorld21
 
== Video Solution by OmegaLearn ==
https://youtu.be/4_x1sgcQCp4?t=4816
 
~ pi_is_3.14
 
== See Also ==
 
{{AMC10 box|year=2009|ab=B|num-b=19|num-a=21}}
{{MAA Notice}}

Latest revision as of 11:50, 10 August 2024

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 1

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

\[\frac{BD}{1}=\frac{2-BD}{\sqrt5}\] \[BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\] \[BD(\sqrt5+1)=2\] \[BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2}} \implies {B}\]

Solution 2

Let $\theta = \angle BAD = \angle DAC$. Notice $\tan \theta = BD$ and $\tan 2 \theta = 2$. By the double angle identity, \[2 = \frac{2 BD}{1 - BD^2} \implies BD = \boxed{\frac{\sqrt5 - 1}{2} \implies B}.\]

Remarks: You could also use tangent half angle formula

Solution 3

Let $BD=y$.

Make $DE$ a line so that it is perpendicular to $AC$. Since $AD$ is an angle bisector, $\triangle AED$ is congruent to $\triangle ABD$. Using the Pythagorean Theorem:

$AC^2=1^2+2^2$

$AC^2=5$

$AC=\sqrt{5}$

We know that $AE=1$ by the congruent triangles, so $EC=\sqrt{5}-1$. We know that $DE=y$, $EC=\sqrt{5}-1$, and $DC=2-y$. We now have right triangle $DEC$ and its 3 sides. Using the Pythagorean Thereom, we get:

$y^2+(\sqrt{5}-1)^2=(2-y)^2$

$-4y=2-2\sqrt{5}$

So, $y=BD=\boxed{\frac{\sqrt5-1}{2} \implies B}.$

~HelloWorld21

Video Solution by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=4816

~ pi_is_3.14

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

These problems are copyrighted © by the Mathematical Association of America.

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