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

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== 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 (NO TRIGONOMETRY!!!)==
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Draw <math>\overline{CD}</math> from circle center <math>C</math> to center <math>D</math>. Let its midpoint be point <math>F</math>.
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Draw <math>\overline{BF}</math> perpendicular to line segment <math>CD</math> and intersecting <math>CD</math> at point <math>F</math>.
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Let <math>G</math> be the common external tangent point of circle <math>B</math> and circle <math>E</math>. <math>\overline{GA} = 4</math>
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Circle centers <math>A</math> and <math>E</math> lie on <math>\overline{BF}</math>.
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Draw <math>\overline{AC}</math> from circle center <math>A</math> to center <math>C</math>. <math>\overline{AC}</math> has length <math>8</math>.
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Triangle <math>ACF</math> is a right triangle with <math>\angle ACF = 30^{\circ}</math>.
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(If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle <math>ACF</math> is similar to one of the triangles).
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<math>\overline{AF} = 4</math> and <math>\overline{CF} = 4\sqrt{3}</math>.
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Let the radius of circle <math>E</math> have length <math>r</math>. Draw <math>\overline{EC}</math> from circle center <math>E</math> to center <math>C</math>.
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<math>\overline{EC}</math> has length <math>r + 2</math>. <math>\overline{EF}</math> has length <math>GA + AF - GE = 8 - r</math>.
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Since <math>CEF</math> is a right triangle, by the Pythagorean theorem <math>EF^2 + CF^2 = EC^2</math>.
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<math>(8 - r)^2 + (4\sqrt{3})^2 = (r + 2)^2</math>. Solving, <math>r = \frac {27}{5}</math> and the answer is <math>27 + 5 = \boxed{032}</math>.
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-unhappyfarmer
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==Video Solution==
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https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh
  
 
== 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 18:45, 22 March 2025

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 1

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 $27 + 5 = \boxed{032}$.

Solution 2 (NO TRIGONOMETRY!!!)

Draw $\overline{CD}$ from circle center $C$ to center $D$. Let its midpoint be point $F$.

Draw $\overline{BF}$ perpendicular to line segment $CD$ and intersecting $CD$ at point $F$.

Let $G$ be the common external tangent point of circle $B$ and circle $E$. $\overline{GA} = 4$

Circle centers $A$ and $E$ lie on $\overline{BF}$.

Draw $\overline{AC}$ from circle center $A$ to center $C$. $\overline{AC}$ has length $8$.

Triangle $ACF$ is a right triangle with $\angle ACF = 30^{\circ}$.

(If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle $ACF$ is similar to one of the triangles).

$\overline{AF} = 4$ and $\overline{CF} = 4\sqrt{3}$.

Let the radius of circle $E$ have length $r$. Draw $\overline{EC}$ from circle center $E$ to center $C$.

$\overline{EC}$ has length $r + 2$. $\overline{EF}$ has length $GA + AF - GE = 8 - r$.

Since $CEF$ is a right triangle, by the Pythagorean theorem $EF^2 + CF^2 = EC^2$.

$(8 - r)^2 + (4\sqrt{3})^2 = (r + 2)^2$. Solving, $r = \frac {27}{5}$ and the answer is $27 + 5 = \boxed{032}$.

-unhappyfarmer

Video Solution

https://youtu.be/fZAChuJDlSw?si=wJUPmgVRlYwazauh

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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