Difference between revisions of "2004 AMC 10A Problems/Problem 23"
m (→Solution) |
(→Solution) |
||
Line 19: | Line 19: | ||
<math> \mathrm{(A) \ } \frac{2}{3} \qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2} \qquad \mathrm{(C) \ } \frac{7}{8} \qquad \mathrm{(D) \ } \frac{8}{9} \qquad \mathrm{(E) \ } \frac{1+\sqrt{3}}{3} </math> | <math> \mathrm{(A) \ } \frac{2}{3} \qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2} \qquad \mathrm{(C) \ } \frac{7}{8} \qquad \mathrm{(D) \ } \frac{8}{9} \qquad \mathrm{(E) \ } \frac{1+\sqrt{3}}{3} </math> | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
Line 72: | Line 72: | ||
So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | ||
+ | ==Solution 2== | ||
+ | Using [[Descartes' Circle Formula]], <math>\left(1-\frac{1}{2}(\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)</math>. Solving this gives us linear equation with <math>r=\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | ||
== See also == | == See also == |
Revision as of 13:58, 8 February 2016
Contents
Problem
Circles ,
, and
are externally tangent to each other and internally tangent to circle
. Circles
and
are congruent. Circle
has radius
and passes through the center of
. What is the radius of circle
?
Solution 1
Let
be the center of circle
for all
and let
be the tangent point of
. Since the radius of
is the diameter of
, the radius of
is
. Let the radius of
be
and let
. If we connect
, we get an isosceles triangle with lengths
. Then right triangle
has legs
and hypotenuse
. Solving for
, we get
.
Also, right triangle has legs
, and hypotenuse
. Solving,
So the answer is .
Solution 2
Using Descartes' Circle Formula, . Solving this gives us linear equation with
.
See also
- <url>viewtopic.php?t=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.