2008 AIME II Problems/Problem 11
Problem
In triangle , , and . Circle has radius and is tangent to and . Circle is externally tangent to and is tangent to and . No point of circle lies outside of . The radius of circle can be expressed in the form , where , , and are positive integers and is the product of distinct primes. Find .
Solution
Let and be the feet of the perpendiculars from and to , respectively. Let the radius of be . We know that . From draw segment such that is on . Clearly, and . Also, we know is a right triangle.
To find , consider the right triangle . Since is tangent to , then bisects . Let ; then . Dropping the altitude from to , we recognize the right triangle, except scaled by .
So we get that . From the half-angle identity, we find that . Therefore, . By similar reasoning in triangle , we see that .
We conclude that .
So our right triangle has sides , , and .
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get , for a final answer of .
Solution
first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r
Solution 2(pure synthetic)
Refer to the above diagram. Let the larger circle have center , the smaller have center , and the incenter be . We can easily calculate that the area of , and and , where is the inradius.
Now, Line is the perpendicular bisector of , as is isosceles. Letting the point of intersection be , we get that and , and are collinear as is equidistant from and . By Pythagoras, , and we notice that is a 3-4-5 right triangle.
Letting be the desired radius and letting be the projection of onto , we find that . Similarly, we find that the distance between the projection from onto , , and , is . From there, we let the projection of onto be , and we have , , and . We finish with Pythagoras on , whence we get the desired answer of . - Spacesam
Solution 3
Let the incenter be O and the altitude from A to be T. Note that by AA, and Note that from , the inradius of the big triangle is Using ravi substitution(or noticing that is an altitude), we then have that From similar triangles, we can now find We have Now, note that as in solution 1, drawing the perpendicular from Q to (call it Z) yields Then, from this, Using similar similarity as was done to find we have . Now adding all these up and equating them to yields
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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