Difference between revisions of "2001 AMC 12 Problems/Problem 18"
(→Solution) |
m (→Solution 1) |
||
(24 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> 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. | + | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> 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> | <asy> | ||
Line 32: | Line 32: | ||
\text{(E) }\frac {1}{2} | \text{(E) }\frac {1}{2} | ||
</math> | </math> | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
Line 39: | Line 41: | ||
<asy> | <asy> | ||
unitsize(1cm); | unitsize(1cm); | ||
− | pair A=(0,1), B=(4,4), C=(4,1); | + | pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); |
dot(A); dot(B); | dot(A); dot(B); | ||
draw( circle(A,1) ); | draw( circle(A,1) ); | ||
Line 49: | Line 51: | ||
label("$B$",B,N); | label("$B$",B,N); | ||
label("$C$",C,E); | label("$C$",C,E); | ||
+ | label("$S$",S,N); | ||
− | filldraw( circle( | + | filldraw( circle(S,4/9), lightgray, black ); |
− | dot( | + | dot(S); |
draw( rightanglemark(A,C,B) ); | draw( rightanglemark(A,C,B) ); | ||
+ | draw( S -- A ); | ||
+ | draw( S -- B ); | ||
</asy> | </asy> | ||
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>. | ||
− | |||
− | |||
− | Let <math>r</math> be the radius of the small circle, and let <math>s</math> be the | + | 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: | ||
Line 84: | Line 87: | ||
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>. |
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | More generally, for two large circles of radius <math>a</math> and <math>b</math>, the radius <math>c</math> of the small circle is <math>c = \frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2} = \frac{1}{\left(1/\sqrt{a}+1/\sqrt{b}\right)^2}</math>. | ||
+ | |||
+ | Equivalently, we have that <math>1/\sqrt{c} = 1/\sqrt{a} + 1/\sqrt{b}</math>. | ||
=== Solution 2 === | === Solution 2 === | ||
Line 92: | Line 106: | ||
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> | ||
− | < | + | <cmath>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</cmath> |
+ | <cmath>\frac{16}{r^2}-\frac{40}{r}+9=0</cmath> | ||
+ | <cmath>\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0</cmath> | ||
+ | |||
+ | Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>. | ||
− | <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> | ||
− | + | ~Wezzerwez7254 | |
− | + | === Video Solution === | |
+ | https://youtu.be/zOwYoFOUg2U | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=17|num-a=19}} | {{AMC12 box|year=2001|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.