Difference between revisions of "2001 AMC 12 Problems/Problem 18"
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− | Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the perpendicular distance from <math> | + | Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the perpendicular distance from <math>S</math> to <math>\overline{AC}</math>. 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}</math> and <math>SB=\sqrt{(4-s)^2 + (4-r)^2}</math>. Hence we get the following two equations: | 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: | ||
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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>. | 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</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, | + | Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. |
− | (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 <math>4:1</math> ratio between radii of <math>A</math> and <math>B</math>, this circle turns out to have the same radius as circle <math>B</math>, with center directly left of center <math>B</math>, and tangent to <math>B</math> directly above center <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>. | + | |
+ | In the first case clearly <math>s<0</math>, which puts the center on the wrong side of <math>A</math>, 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 <math>4:1</math> ratio between radii of <math>A</math> and <math>B</math>, this circle turns out to have the same radius as circle <math>B</math>, with center directly left of center <math>B</math>, and tangent to <math>B</math> directly above center <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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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>. | ||
+ | |||
+ | === 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>. | ||
+ | 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=s=4=\sqrt{(r+1)^2 - (1-r)^2} + \sqrt{(r+4)^2 - (4-r)^2}</math>, which is simply <math>4=\sqrt{4r}+\sqrt{16r} \implies \sqrt{r}=\frac{2}{3} \implies r= \boxed{\textbf{(D) } \frac{4}{9}}.</math> | ||
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+ | ~Wezzerwez7254 | ||
=== Video Solution === | === Video Solution === |
Latest revision as of 00:35, 7 February 2024
Contents
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 . Moreover, the small circle is tangent to both other circles, hence we have and .
We have and . Hence we get the following two equations:
Simplifying both, we get
As in our case both and are positive, we can divide the second one by the first one to get .
Now there are two possibilities: either , or .
In the first case clearly , which puts the center on the wrong side of , 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 ratio between radii of and , this circle turns out to have the same radius as circle , with center directly left of center , and tangent to directly above center .)
The second case solves to . We then have , hence .
More generally, for two large circles of radius and , the radius of the small circle is .
Equivalently, we have that .
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 .
Solution 3 (Basically 1 but less complicated)
As in solution 1, in triangle we have and , thus by the Pythagorean theorem or pythagorean triples in general, we have . Let be the radius. Let be the perpendicular intersecting point and line . because both perpendicular radii, and form a rectangle. We just have to find in terms of and solve for now. From the Pythagorean theorem and subtracting to get lengths, we get , which is simply
~Wezzerwez7254
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 |
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