Difference between revisions of "2009 AMC 10B Problems/Problem 20"
(→Solution 1) |
(→Poop) |
||
(4 intermediate revisions by 2 users not shown) | |||
Line 40: | Line 40: | ||
By the Pythagorean Theorem, <math>AC=\sqrt5</math>. Then, from the Angle Bisector Theorem, we have: | 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>\frac{BD}{1}=\frac{2-BD}{\sqrt5}</cmath> |
− | BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5} | + | <cmath>BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}</cmath> |
− | BD(\sqrt5+1)=2 | + | <cmath>BD(\sqrt5+1)=2</cmath> |
− | BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2} \implies {B}</cmath> | + | <cmath>BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2}} \implies {B}</cmath> |
== Solution 2 == | == Solution 2 == |
Latest revision as of 12:50, 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.