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

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=== Solution 3 (Basically 1 but less complicated) ===
 
=== Solution 3 (Basically 1 but less complicated) ===
 
As in solution 1, in triangle <math>ABC</math> we have <math>AB = 1+4 = 5</math> and <math>BC=4-1 = 3</math>, thus by the Pythagorean theorem or pythagorean triples in general, we have <math>AC=4</math>.
 
As in solution 1, in triangle <math>ABC</math> we have <math>AB = 1+4 = 5</math> and <math>BC=4-1 = 3</math>, thus by the Pythagorean theorem or pythagorean triples in general, we have <math>AC=4</math>.
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Let <math>r</math> be the radius. Let <math>s</math> be the perpendicular intersecting point <math>S</math> and line <math>BC</math>. <math>AC=s</math> because <math>s,</math> both perpendicular radii, and <math>AC</math> form a rectangle. We just have to find <math>AC</math> in terms of <math>r</math> and solve for <math>r</math> now.
+
Let <math>r</math> be the radius. Let <math>s</math> be the perpendicular intersecting point <math>S</math> and line <math>BC</math>. <math>AC=s</math> because <math>s,</math> both perpendicular radii, and <math>AC</math> form a rectangle. We just have to find <math>AC</math> in terms of <math>r</math> and solve for <math>r</math> now. From the Pythagorean theorem and subtracting to get lengths, we get <math>AC=4=\sqrt{(r+1)^2 - (1-r)^2} + \sqrt{(r+4)^2 - (4-r)^2}</math>, which is simply $4=\sqrt{4r}+\sqrt{16r} \implies \sqrt{r}=\frac{2}{3} \implies r= \boxed{
  
 
=== Video Solution ===
 
=== Video Solution ===

Revision as of 01:24, 9 January 2024

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. What is the radius of the third circle?

[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), S=(12/9,4/9); 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); label("$S$",S,N); filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); [/asy]

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


Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $A$ to $\overline{BC}$. 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 $s<0$, which puts the center on the wrong side of $A$, so this is not the correct case.

(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$, this circle turns out to have the same radius as circle $B$, with center directly left of center $B$, and tangent to $B$ directly above center $A$.)

The second case solves to $s=\frac 43$. We then have $4r = s^2 = \frac {16}9$, hence $r = \boxed{\frac 49}$.


More generally, for two large circles of radius $a$ and $b$, the radius $c$ of the small circle is $c = \frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2} = \frac{1}{\left(1/\sqrt{a}+1/\sqrt{b}\right)^2}$.

Equivalently, we have that $1/\sqrt{c} = 1/\sqrt{a} + 1/\sqrt{b}$.

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 $r = \boxed{\frac 49}$.

Solution 3 (Basically 1 but less complicated)

As in solution 1, in triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem or pythagorean triples in general, we have $AC=4$. Let $r$ be the radius. Let $s$ be the perpendicular intersecting point $S$ and line $BC$. $AC=s$ because $s,$ both perpendicular radii, and $AC$ form a rectangle. We just have to find $AC$ in terms of $r$ and solve for $r$ now. From the Pythagorean theorem and subtracting to get lengths, we get $AC=4=\sqrt{(r+1)^2 - (1-r)^2} + \sqrt{(r+4)^2 - (4-r)^2}$, which is simply $4=\sqrt{4r}+\sqrt{16r} \implies \sqrt{r}=\frac{2}{3} \implies r= \boxed{

Video Solution

https://youtu.be/zOwYoFOUg2U

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

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