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Difference between revisions of "2004 AMC 10B Problems/Problem 16"

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==Solution 1==
 
==Solution 1==
  
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The situation is shown in the picture below. The radius we seek is <math>SD = AD + AS</math>. Clearly <math>AD=1</math>. The point <math>S</math> is clearly the center of the equilateral triangle <math>ABC</math>, thus <math>AS</math> is <math>2/3</math> of the altitude of this triangle. We get that <math>AS = \frac23 \cdot \sqrt 3</math>. Therefore the radius we seek is
+
The situation is shown in the picture below. The radius we seek is <math>SD = AD + AS</math>. Clearly <math>AD=1</math>. The point <math>S</math> is the center of the equilateral triangle <math>ABC</math>, thus <math>AS</math> is <math>2/3</math> of the altitude of this triangle. We get that <math>AS = \frac23 \cdot \sqrt 3</math>. Therefore the radius we seek is
 
<math>1 + \frac23 \cdot \sqrt 3 = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}</math>.
 
<math>1 + \frac23 \cdot \sqrt 3 = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}</math>.
  
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==Solution 2==
 
==Solution 2==
  
−
Using [[Descartes' Circle Formula]], we can assign curvatures to all the circles: <math>1</math>, <math>1</math>, <math>1</math>, and <math>-\frac{1}{r}</math> (b/c it is the circle internally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve:
+
Using [[Descartes' Circle Formula]], we can assign curvatures to all the circles: <math>1</math>, <math>1</math>, <math>1</math>, and <math>-\frac{1}{r}</math> (b/c the bigger circle is externally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve:
  
 
<math>2(1^2+1^2+1^2+(-\frac{1}{r})^2) = (1+1+1-\frac{1}{r})^2</math>
 
<math>2(1^2+1^2+1^2+(-\frac{1}{r})^2) = (1+1+1-\frac{1}{r})^2</math>

Latest revision as of 13:05, 5 January 2022

Problem

Three circles of radius $1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?

$\mathrm{(A) \ } \frac{2 + \sqrt{6}}{3} \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } \frac{2 + 3\sqrt{2}}{2} \qquad \mathrm{(D) \ } \frac{3 + 2\sqrt{3}}{3} \qquad \mathrm{(E) \ } \frac{3 + \sqrt{3}}{2}$

Solution 1

The situation is shown in the picture below. The radius we seek is $SD = AD + AS$. Clearly $AD=1$. The point $S$ is the center of the equilateral triangle $ABC$, thus $AS$ is $2/3$ of the altitude of this triangle. We get that $AS = \frac23 \cdot \sqrt 3$. Therefore the radius we seek is $1 + \frac23 \cdot \sqrt 3 = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}$.


[asy] unitsize(2cm); pair A=(0,0), B=dir(0)*2, C=dir(60)*2; draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); dot(A); dot(B); dot(C); draw(A--B--C--cycle); pair D=A+dir(210), E=B+dir(-30), F=C+dir(90); draw(circumcircle(D,E,F)); dot(D); dot(E); dot(F); pair S=(A+B+C)/3; dot(S); draw(S--D); draw(S--E); draw(S--F); label("$S$",S,S); label("$A$",A,SE); label("$D$",D,SW); label("$B$",B,NE); label("$C$",C,ENE); [/asy]

Solution 2

Using Descartes' Circle Formula, we can assign curvatures to all the circles: $1$, $1$, $1$, and $-\frac{1}{r}$ (b/c the bigger circle is externally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve:

$2(1^2+1^2+1^2+(-\frac{1}{r})^2) = (1+1+1-\frac{1}{r})^2$

$r = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}$

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions

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