Mock AIME 1 2010 Problems/Problem 9
Problem
Let and be circles of radii 5 and 7, respectively, and suppose that the distance between their centers is 10. There exists a circle that is internally tangent to both and , and tangent to the line joining the centers of and . If the radius of can be expressed in the form , where , , and are integers, and is not divisible by the square if any prime, find the value of .
Solution
Let have center , have center , and have center . Further, let intersect at , at , and at , as in the diagram. Let be the radius of and let .
Because is tangent to , . Because and are tangent, we know that the line joining their centers goes through their point of tangency. Thus, because has radius , . Similarly, . Because with and , . Thus, , , and .
By the Pythagorean Theorem in , we have the following equation that we can solve for : \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 , 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 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 , . Now, we can plug this value for into our expression for 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 .
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 |