Difference between revisions of "2004 AMC 10B Problems/Problem 16"
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<math> \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} </math> | <math> \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} </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | The situation | + | 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{\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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label("$C$",C,ENE); | label("$C$",C,ENE); | ||
</asy> | </asy> | ||
+ | |||
+ | ==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 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>r = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}</math> | ||
== See also == | == See also == | ||
{{AMC10 box|year=2004|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2004|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:05, 5 January 2022
Contents
Problem
Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?
Solution 1
The situation is shown in the picture below. The radius we seek is . Clearly
. The point
is the center of the equilateral triangle
, thus
is
of the altitude of this triangle. We get that
. Therefore the radius we seek is
.
Solution 2
Using Descartes' Circle Formula, we can assign curvatures to all the circles: ,
,
, and
(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:
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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