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Difference between revisions of "2009 AIME II Problems/Problem 5"

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
  
 
Let <math>X</math> be the intersection of the circles with centers <math>B</math> and <math>E</math>, and <math>Y</math> be the intersection of the circles with centers <math>C</math> and <math>E</math>. Since the radius of <math>B</math> is <math>3</math>, <math>AX =4</math>. Assume <math>AE</math> = <math>p</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+p</math>. <math>AC = 8</math>, and angle <math>CAE = 60</math> degrees because we are given that triangle <math>T</math> is equilateral. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain
 
Let <math>X</math> be the intersection of the circles with centers <math>B</math> and <math>E</math>, and <math>Y</math> be the intersection of the circles with centers <math>C</math> and <math>E</math>. Since the radius of <math>B</math> is <math>3</math>, <math>AX =4</math>. Assume <math>AE</math> = <math>p</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+p</math>. <math>AC = 8</math>, and angle <math>CAE = 60</math> degrees because we are given that triangle <math>T</math> is equilateral. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain
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<math>p =\frac {28}{20} = \frac {7}{5}</math>. The radius of circle <math>E</math> is <math>4 + \frac {7}{5} = \frac {27}{5}</math>, so the answer is <math>27 + 5 = \boxed{032}</math>.
 
<math>p =\frac {28}{20} = \frac {7}{5}</math>. The radius of circle <math>E</math> is <math>4 + \frac {7}{5} = \frac {27}{5}</math>, so the answer is <math>27 + 5 = \boxed{032}</math>.
  
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=== Solution 2 ===
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==Video Solution==
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https://www.youtube.com/watch?v=KKVxQV4hszo&t=7s
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== See Also ==
 
== See Also ==
  
 
{{AIME box|year=2009|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2009|n=II|num-b=4|num-a=6}}
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[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:54, 9 August 2020

Problem 5

Equilateral triangle $T$ is inscribed in circle $A$, which has radius $10$. Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$. Circles $C$ and $D$, both with radius $2$, are internally tangent to circle $A$ at the other two vertices of $T$. Circles $B$, $C$, and $D$ are all externally tangent to circle $E$, which has radius $\dfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep}; draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5)); dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy]


Solution

Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC = 8$, and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$, we obtain

$(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60$.

The $2$ and the $\cos 60$ terms cancel out:

$p^2 + 12p +36 = p^2 + 64 - 8p$

$12p+ 36 = 64 - 8p$

$p =\frac {28}{20} = \frac {7}{5}$. The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$, so the answer is $27 + 5 = \boxed{032}$.

Video Solution

https://www.youtube.com/watch?v=KKVxQV4hszo&t=7s

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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