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Mock AIME 1 2010 Problems/Problem 9

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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