2007 AIME II Problems/Problem 15
Problem
Four circles and with the same radius are drawn in the interior of triangle such that is tangent to sides and , to and , to and , and is externally tangent to and . If the sides of triangle are and the radius of can be represented in the form , where and are relatively prime positive integers. Find
Solution
Solution 1
First, apply Heron's formula to find that . The semiperimeter is , so the inradius is .
Now consider the incenter of . Let the radius of one of the small circles be . Let the centers of the three little circles tangent to the sides of be , , and . Let the center of the circle tangent to those three circles be . The homothety maps to ; since , is the circumcenter of and therefore maps the circumcenter of to . Thus, , where is the circumradius of . Substituting , and the answer is .
Solution 2
Consider a 13-14-15 triangle. [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is , where is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,
The circumradius is where and are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, .
Cut and combine the triangles, as shown. Then solve for :
The solution is .
Diagram for Solution 1
Here is a diagram illustrating solution 1. Note that unlike in the solution refers to the circumcenter of . Instead, is used for the center of the third circle, .
[asy] unitsize(0.75cm); pair A, B, C, Oa, Ob, Oc, Od, O, I; path circ1, circ2;
// Homotethy factor - backplugged from solution real k = 64/129; real r = 260/129;
B = (0, 0); C = (14, 0);
circ1 = circle(B, 13); circ2 = circle(C, 15);
A = intersectionpoints(circ1, circ2)[0]; I = incenter(A, B, C);
Oa = (65*A + 64*I)/129; Ob = (65*B + 64*I)/129; Oc = (65*C + 64*I)/129;
Od = circumcenter(Oa, Ob, Oc); O = circumcenter(A, B, C);
draw(circle(Oa, r)); draw(circle(Ob, r)); draw(circle(Oc, r)); draw(circle(Od, r));
draw(incircle(Oa, Ob, Oc)^^incircle(A, B, C)^^I--foot(I, A, C), green); draw(A--B--C--cycle); draw(Oa--Ob--Oc--cycle, blue); draw(A--I--B^^I--C, blue); draw(Oa--foot(Oa, A, C)^^Oc--foot(Oc, A, C), blue); draw(rightanglemark(Oa, foot(Oa, A, C), C)^^rightanglemark(Oc, foot(Oc, A, C), A)); dot(I); dot(Oa); dot(Ob); dot(Oc); dot(Od); dot(O, red);
label("", A, N); label("", B, S); label("", C, S); label("", I, S); label("", Oa, NW); label("", Ob, SW); label("", Oc, SE); label("", Od, N); label("", O, SE, red);
[/asy]
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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