Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 18"

(→‎Problem)
(LaTeX correctly, people!)
Line 94: Line 94:
 
The four circles have curvatures <math>0, 1, \frac 14</math>, and <math>\frac 1r</math>.
 
The four circles have curvatures <math>0, 1, \frac 14</math>, and <math>\frac 1r</math>.
  
−
We have <math>2(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2})=(0+1+\frac 14+\frac 1r)^2</math>
+
We have <math>2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2</math>
  
 
Simplifying, we get <math>\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}</math>
 
Simplifying, we get <math>\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}</math>
  
−
<math>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</math>
+
<cmath>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</cmath>
−
 
+
<cmath>\frac{16}{r^2}-\frac{40}{r}+9=0</cmath>
−
<math>\frac{16}{r^2}-\frac{40}{r}+9=0</math>
+
<cmath>\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0</cmath>
−
 
 
−
<math>(\frac{4}{r}-9)(\frac{4}{r}-1)=0</math>
 
  
 
Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>.
 
Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>.

Revision as of 13:33, 25 September 2016

Problem

A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is

[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]

$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$

Solution

Solution 1

[asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); draw( rightanglemark(A,C,B) ); [/asy]

In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem we have $AC=4$.

We can now pick a coordinate system where the common tangent is the $x$ axis and $A$ lies on the $y$ axis. In this coordinate system we have $A=(0,1)$ and $B=(4,4)$.

Let $r$ be the radius of the small circle, and let $s$ be the $x$-coordinate of its center $S$. We then know that $S=(s,r)$, as the circle is tangent to the $x$ axis. Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$.

We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$. Hence we get the following two equations:

\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}

Simplifying both, we get

\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}

As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$.

Now there are two possibilities: either $\frac{4-s}s=-2$, or $\frac{4-s}s=2$. In the first case clearly $s<0$, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the $x$ axis - a large circle whose center is somewhere to the left of $A$.) The second case solves to $s=\frac 43$. We then have $4r = s^2 = \frac {16}9$, hence $r = \boxed{\frac 49}$.

Solution 2

The horizontal line is the equivalent of a circle of curvature $0$, thus we can apply Descartes' Circle Formula.

The four circles have curvatures $0, 1, \frac 14$, and $\frac 1r$.

We have $2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2$

Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$

\[\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0\] \[\frac{16}{r^2}-\frac{40}{r}+9=0\] \[\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0\]

Obviously $r$ cannot equal $4$, therefore $r = \boxed{\frac 49}$.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_18&oldid=80393"