Difference between revisions of "2001 AMC 12 Problems/Problem 18"
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In the triangle <math>ABC</math> we have <math>AB = 1+4 = 5</math> and <math>BC=4-1 = 3</math>, thus by the [[Pythagorean theorem]] we have <math>AC=4</math>. | In the triangle <math>ABC</math> we have <math>AB = 1+4 = 5</math> and <math>BC=4-1 = 3</math>, thus by the [[Pythagorean theorem]] we have <math>AC=4</math>. | ||
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− | Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the <math> | + | Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the perpendicular distance from <math>A</math> to <math>\overline{BC}. Moreover, the small circle is tangent to both other circles, hence we have </math>SA=1+r<math> and </math>SB=4+r<math>. |
− | We have <math>SA = \sqrt{s^2 + (1-r)^2}< | + | We have </math>SA = \sqrt{s^2 + (1-r)^2}<math> and </math>SB=\sqrt{(4-s)^2 + (4-r)^2}<math>. Hence we get the following two equations: |
<cmath> | <cmath> | ||
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</cmath> | </cmath> | ||
− | As in our case both <math>r< | + | As in our case both </math>r<math> and </math>s<math> are positive, we can divide the second one by the first one to get </math>\left( \frac{4-s}s \right)^2 = 4<math>. |
− | Now there are two possibilities: either <math>\frac{4-s}s=-2< | + | Now there are two possibilities: either </math>\frac{4-s}s=-2<math>, or </math>\frac{4-s}s=2<math>. In the first case clearly </math>s<0<math>, hence 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 - a large circle whose center is somewhere to the left of </math>A<math>.) The second case solves to </math>s=\frac 43<math>. We then have </math>4r = s^2 = \frac {16}9<math>, hence </math>r = \boxed{\frac 49}<math>. |
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+ | More generally, for two large circles of radius </math>a<math> and </math>b<math>, the radius of the small circle is </math>\frac{ab}{(a+b)^2}$ | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 23:56, 19 August 2023
Problem
A circle centered at with a radius of 1 and a circle centered at
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?
Solution
Solution 1
In the triangle we have
and
, thus by the Pythagorean theorem we have
.
Let be the radius of the small circle, and let
be the perpendicular distance from
to
SA=1+r
SB=4+r$.
We have$ (Error compiling LaTeX. Unknown error_msg)SA = \sqrt{s^2 + (1-r)^2}SB=\sqrt{(4-s)^2 + (4-r)^2}$. Hence we get the following two equations:
<cmath> \begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*} </cmath>
Simplifying both, we get
<cmath> \begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*} </cmath>
As in our case both$ (Error compiling LaTeX. Unknown error_msg)rs
\left( \frac{4-s}s \right)^2 = 4$.
Now there are two possibilities: either$ (Error compiling LaTeX. Unknown error_msg)\frac{4-s}s=-2\frac{4-s}s=2
s<0
A
s=\frac 43
4r = s^2 = \frac {16}9
r = \boxed{\frac 49}$.
More generally, for two large circles of radius$ (Error compiling LaTeX. Unknown error_msg)ab
\frac{ab}{(a+b)^2}$
Solution 2
The horizontal line is the equivalent of a circle of curvature , thus we can apply Descartes' Circle Formula.
The four circles have curvatures , and
.
We have
Simplifying, we get
Obviously cannot equal
, therefore
.
Video Solution
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.