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

Latest revision as of 15:34, 15 October 2023

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:

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

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

@@ Line 24: / Line 24: @@
 <math>
 \text{(A) } \frac {\sqrt3 - 1}{2}
 \qquad
@@ Line 35: / Line 36: @@
 </math>
-== Solution ==
+== Solution 1 ==
+By the Pythagorean Theorem, <math>AC=\sqrt5</math>. Then, from the Angle Bisector Theorem, we have:
+<math>\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 Answer B}.</math>
+== 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>
-Let <math>\angle BAD = \alpha</math>, then <math>\angle BAC = 2\alpha</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:
-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>y^2+(\sqrt{5}-1)^2=(2-y)^2</math>
-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>.
+<math>-4y=2-2\sqrt{5}</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>.
+So, <math>y=BD=\boxed{\frac{\sqrt5-1}{2} \implies B}.</math>
-=== Note ===
+~HelloWorld21
-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:
+== Video Solution by OmegaLearn ==
+https://youtu.be/4_x1sgcQCp4?t=4816
-<cmath>
+~ pi_is_3.14
-\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}
-</cmath>
 == See Also ==
 {{AMC10 box|year=2009|ab=B|num-b=19|num-a=21}}
+{{MAA Notice}}

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

Page

Toolbox

Search