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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 9"

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</asy>
 
</asy>
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Let <math>\omega_1</math> have center <math>A</math>, <math>\omega_2</math> have center <math>B</math>, and <math>\omega_3</math> have center <math>C</math>. Further, let <math>\overline{AB}</math> intersect <math>\omega_1</math> at <math>D</math>, <math>\omega_2</math> at <math>E</math>, and <math>\omega_3</math> at <math>T</math>, as in the diagram. Let <math>r</math> be the radius of <math>\omega_3</math> and let <math>TE=x</math>.
 +
 +
Because <math>\overline{AB}</math> is tangent to <math>\omega_3</math>, <math>\overline{CT} \perp \overline{AB}</math>. Because <math>\omega_1</math> and <math>\omega_3</math> are tangent, we know that the line joining their centers goes through their point of tangency. Thus, because <math>\omega_1</math> has radius <math>5</math>, <math>AC=5-r</math>. Similarly, <math>BC=7-r</math>. Because <math>AB=10</math> with <math>AE=5</math> and <math>BD=7</math>, <math>AB=2</math>. Thus, <math>DT=2-x</math>, <math>AD=5-2=3</math>, and <math>BE=7-2=5</math>.
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 +
By the [[Pythagorean Theorem]] in <math>\triangle BTC</math>, we have the following equation that we can solve for <math>r</math>:
 +
\begin{align*}
 +
BT^2+TC^2 &= BC^2 \\
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(5+x)^2+r^2 &= (7-r)^2 \\
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25+10x+x^2+r^2 &= 49-14r+r^2 \\
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x^2+10x-24 &= -14r \\
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r &= -\frac{x^2+10x-24}{14}
 +
\end{align*}
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By using Pythagoras again in <math>\triangle ATC</math>, we have the following equation:
 +
\begin{align*}
 +
AT^2+TC^2 &= AC^2 \\
 +
(3+(2-x))^2+r^2 &= (5-r)^2 \\
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(5-x)^2+r^2 &= r^2-10r+25 \\
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x^2-10x+25 &= -10r+25 \\
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x^2-10x &= -10r
 +
\end{align*}
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Substituting the expression for <math>r</math> we found earlier, we see the following:
 +
\begin{align*}
 +
x^2-10x &= -10(-\frac{x^2+10x-24}{14}) \\
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x^2-10x &= \frac{5x^2+50x-120}{7} \\
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7x^2-70x &= 5x^2+50x-120 \\
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2x^2-120x+120 &= 0 \\
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x^2-60x+60 &= 0 \\
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x &= \frac{60 \pm \sqrt{3600-240}}2 \\
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x &= 30 \pm \sqrt{900-60} \\
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x &= 30 \pm 2\sqrt{210}
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\end{align*}
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Because <math>x<2</math>, <math>x=30 - 2\sqrt{210}</math>. Now, we can plug this value for <math>x</math> into our expression for <math>r</math> to get our answer:
 +
\begin{align*}
 +
r &= -\frac{x^2+10x-24}{14} \\
 +
&= -\frac{(30-2\sqrt{210})^2+10(30-2\sqrt{210})-24}{14} \\
 +
&= -\frac{900+840-120\sqrt{210}+300-20\sqrt{210}-24}{14} \\
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&= -\frac{1740+276-140\sqrt{210}}{14} \\
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&= 10\sqrt{210}-\frac{2016}{14} \\
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&= 10\sqrt{210}-144
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\end{align*}
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Thus, our answer is <math>10+210+144=\boxed{364}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{Mock AIME box|year=2010|n=1|num-b=8|num-a=10}}
 
{{Mock AIME box|year=2010|n=1|num-b=8|num-a=10}}
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 15:52, 2 August 2024

Problem

Let $\omega_1$ and $\omega_2$ be circles of radii 5 and 7, respectively, and suppose that the distance between their centers is 10. There exists a circle $\omega_3$ that is internally tangent to both $\omega_1$ and $\omega_2$, and tangent to the line joining the centers of $\omega_1$ and $\omega_2$. If the radius of $\omega_3$ can be expressed in the form $a \sqrt{b} - c$, where $a$, $b$, and $c$ are integers, and $b$ is not divisible by the square if any prime, find the value of $a + b + c$.

Solution

[asy] import geometry; size(10cm); point A = origin; point B = (10,0); point C = (5-(30-2*sqrt(210)),10*sqrt(210)-144); point D,E,T; circle a = circle(A,5); circle b = circle(B,7); circle c = circle(C,10*sqrt(210)-144); // Defining D, E, and T pair[] d = intersectionpoints(b,A--B); D = d[0]; pair[] e = intersectionpoints(a,A--B); E = e[0]; pair[] t = intersectionpoints(c,A--B); T = t[0]; // Circles draw(a); draw(b); draw(c); // Segments AB, BC, AC, and CT draw(A--B); draw(B--C); draw(A--C); draw(C--T); // Point Labels dot(A); label("A",A,WSW); dot(B); label("B",B,ESE); dot(C); label("C",C,(0,0.5)); dot(D); label("D",D,SW); dot(E); label("E",E,SE); dot(T); label("T",T,S); // Right angle mark markscalefactor=0.05; draw(rightanglemark(C,T,A)); [/asy]

Let $\omega_1$ have center $A$, $\omega_2$ have center $B$, and $\omega_3$ have center $C$. Further, let $\overline{AB}$ intersect $\omega_1$ at $D$, $\omega_2$ at $E$, and $\omega_3$ at $T$, as in the diagram. Let $r$ be the radius of $\omega_3$ and let $TE=x$.

Because $\overline{AB}$ is tangent to $\omega_3$, $\overline{CT} \perp \overline{AB}$. Because $\omega_1$ and $\omega_3$ are tangent, we know that the line joining their centers goes through their point of tangency. Thus, because $\omega_1$ has radius $5$, $AC=5-r$. Similarly, $BC=7-r$. Because $AB=10$ with $AE=5$ and $BD=7$, $AB=2$. Thus, $DT=2-x$, $AD=5-2=3$, and $BE=7-2=5$.

By the Pythagorean Theorem in $\triangle BTC$, we have the following equation that we can solve for $r$: \begin{align*} BT^2+TC^2 &= BC^2 \\ (5+x)^2+r^2 &= (7-r)^2 \\ 25+10x+x^2+r^2 &= 49-14r+r^2 \\ x^2+10x-24 &= -14r \\ r &= -\frac{x^2+10x-24}{14} \end{align*} By using Pythagoras again in $\triangle ATC$, we have the following equation: \begin{align*} AT^2+TC^2 &= AC^2 \\ (3+(2-x))^2+r^2 &= (5-r)^2 \\ (5-x)^2+r^2 &= r^2-10r+25 \\ x^2-10x+25 &= -10r+25 \\ x^2-10x &= -10r \end{align*} Substituting the expression for $r$ we found earlier, we see the following: \begin{align*} x^2-10x &= -10(-\frac{x^2+10x-24}{14}) \\ x^2-10x &= \frac{5x^2+50x-120}{7} \\ 7x^2-70x &= 5x^2+50x-120 \\ 2x^2-120x+120 &= 0 \\ x^2-60x+60 &= 0 \\ x &= \frac{60 \pm \sqrt{3600-240}}2 \\ x &= 30 \pm \sqrt{900-60} \\ x &= 30 \pm 2\sqrt{210} \end{align*} Because $x<2$, $x=30 - 2\sqrt{210}$. Now, we can plug this value for $x$ into our expression for $r$ to get our answer: \begin{align*} r &= -\frac{x^2+10x-24}{14} \\ &= -\frac{(30-2\sqrt{210})^2+10(30-2\sqrt{210})-24}{14} \\ &= -\frac{900+840-120\sqrt{210}+300-20\sqrt{210}-24}{14} \\ &= -\frac{1740+276-140\sqrt{210}}{14} \\ &= 10\sqrt{210}-\frac{2016}{14} \\ &= 10\sqrt{210}-144 \end{align*} Thus, our answer is $10+210+144=\boxed{364}$.

See Also

Mock AIME 1 2010 (Problems, Source)
Preceded by
Problem 8 		Followed by
Problem 10
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