Difference between revisions of "2009 AMC 10B Problems/Problem 20"
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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 has a right angle at , , and . The bisector of meets at . What is ?
Solution 1
By the Pythagorean Theorem, . Then, from the Angle Bisector Theorem, we have:
Solution 2
Let . Notice and . By the double angle identity,
Remarks: You could also use tangent half angle formula
Solution 3
Let .
Make a line so that it is perpendicular to . Since is an angle bisector, is congruent to . Using the Pythagorean Theorem:
We know that by the congruent triangles, so . We know that , , and . We now have right triangle and its 3 sides. Using the Pythagorean Thereom, we get:
So,
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.