Difference between revisions of "2004 AMC 10A Problems/Problem 23"
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− | Let <math>O_{i}</math> be the center of circle <math>i</math> for all <math>i \in \{A,B,C,D\}</math> and let <math>E</math> be the tangent point of <math>B,C</math>. Since the radius of <math>D</math> is the diameter of <math>A</math>, the radius of <math>D</math> is <math>2</math>. Let the radius of <math>B,C</math> be <math>r</math> and let <math>O_{D}E = x</math>. If we connect <math>O_{A},O_{B},O_{C}</math>, we get an [[isosceles triangle]] with lengths <math>1 + r, 2r</math>. Then right triangle <math>O_{D}O_{B} | + | Let <math>O_{i}</math> be the center of circle <math>i</math> for all <math>i \in \{A,B,C,D\}</math> and let <math>E</math> be the tangent point of <math>B,C</math>. Since the radius of <math>D</math> is the diameter of <math>A</math>, the radius of <math>D</math> is <math>2</math>. Let the radius of <math>B,C</math> be <math>r</math> and let <math>O_{D}E = x</math>. If we connect <math>O_{A},O_{B},O_{C}</math>, we get an [[isosceles triangle]] with lengths <math>1 + r, 2r</math>. Then right triangle <math>O_{D}O_{B}E</math> has legs <math>r, x</math> and [[hypotenuse]] <math>2-r</math>. Solving for <math>x</math>, we get <math>x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}</math>. |
− | Also, right triangle <math>O_{A}O_{B} | + | Also, right triangle <math>O_{A}O_{B}E</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving, |
<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} |
Revision as of 10:59, 20 July 2019
Contents
Problem
Circles , , and are externally tangent to each other and internally tangent to circle . Circles and are congruent. Circle has radius and passes through the center of . What is the radius of circle ?
Solution 1
Let be the center of circle for all and let be the tangent point of . Since the radius of is the diameter of , the radius of is . Let the radius of be and let . If we connect , we get an isosceles triangle with lengths . Then right triangle has legs and hypotenuse . Solving for , we get .
Also, right triangle has legs , and hypotenuse . Solving,
So the answer is .
Solution 2
Using Descartes' Circle Formula, . Solving this gives us linear equation with .
See also
<url>viewtopic.php?=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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