Difference between revisions of "2009 AMC 10B Problems/Problem 20"
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BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\ | BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\ | ||
BD(\sqrt5+1)=2\\ | BD(\sqrt5+1)=2\\ | ||
| − | BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2} \Longrightarrow | + | BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2} \Longrightarrow B}.</math> |
== See Also == | == See Also == | ||
Revision as of 22:51, 17 January 2016
Problem
Triangle 
has a right angle at 
, 
, and 
. The bisector of 
meets 
at 
. What is 
?
![[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]](http://latex.artofproblemsolving.com/f/0/b/f0b6594577eb2fc6916bc732b5f99843acda5c60.png)
Solution
By the Pythagorean Theorem, 
. Then, from the Angle Bisector Theorem, we have:
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. 
