Mock AIME 1 2010 Problems/Problem 9

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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 $r$: BT2+TC2=BC2(5+x)2+r2=(7r)225+10x+x2+r2=4914r+r2x2+10x24=14rr=x2+10x2414 By using Pythagoras again in $\triangle ATC$, we have the following equation: AT2+TC2=AC2(3+(2x))2+r2=(5r)2(5x)2+r2=r210r+25x210x+25=10r+25x210x=10r Substituting the expression for $r$ we found earlier, we see the following: x210x=10(x2+10x2414)x210x=5x2+50x12077x270x=5x2+50x1202x2120x+120=0x260x+60=0x=60±36002402x=30±90060x=30±2210 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: r=x2+10x2414=(302210)2+10(302210)2414=900+840120210+300202102414=1740+27614021014=10210201614=10210144 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