Difference between revisions of "2007 AIME II Problems/Problem 15"

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Problem

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ , $\omega_{B}$ to $BC$ and $BA$ , $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Solution 1

First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . The semiperimeter is $21$ , so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$ .

Now consider the incenter $I$ of $\triangle ABC$ . Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $O_A$ , $O_B$ , and $O_C$ . Let the center of the circle tangent to those three circles be $O$ . The homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ maps $\triangle ABC$ to $\triangle XYZ$ ; since $OO_A = OO_B = OO_C = 2r$ , $O$ is the circumcenter of $\triangle XYZ$ and $\mathcal{H}$ therefore maps the circumcenter of $\triangle ABC$ to $O$ . Thus, $2r = R \cdot \frac{4 - r}{4}$ , where $R$ is the circumradius of $\triangle ABC$ . Substituting $R = \frac{abc}{4[ABC]} = \frac{65}{8}$ , $r = \frac{260}{129}$ and the answer is $\boxed{389}$ .

Solution 2

Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]

The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$

The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$ .

Cut and combine the triangles, as shown. Then solve for $4u$ :

$\frac{65}{8}v=8u$

$v=\frac{64}{65}u$

$u+v=1$

$u+\frac{64}{65}u=1$

$\frac{129}{65}u=1$

$4u=\frac{260}{129}$

The solution is $260+129=\boxed{389}$ .

Diagram for Solution 1

Here is a diagram illustrating solution 1. Note that unlike in the solution $O$ refers to the circumcenter of $\triangle ABC$ . Instead, $O_\omega$ is used for the center of the third circle, $\omega$ . $[asy] unitsize(0.75cm); pair A, B, C, Oa, Ob, Oc, Od, O, I; path circ1, circ2; // Homotethy factor - backplugged from solution real k = 64/129; real r = 260/129; B = (0, 0); C = (14, 0); circ1 = circle(B, 13); circ2 = circle(C, 15); A = intersectionpoints(circ1, circ2)[0]; I = incenter(A, B, C); Oa = (65*A + 64*I)/129; Ob = (65*B + 64*I)/129; Oc = (65*C + 64*I)/129; Od = circumcenter(Oa, Ob, Oc); O = circumcenter(A, B, C); draw(circle(Oa, r)); draw(circle(Ob, r)); draw(circle(Oc, r)); draw(circle(Od, r)); draw(incircle(Oa, Ob, Oc)^^incircle(A, B, C)^^I--foot(I, A, C), green); draw(A--B--C--cycle); draw(Oa--Ob--Oc--cycle, blue); draw(A--I--B^^I--C, blue); draw(Oa--foot(Oa, A, C)^^Oc--foot(Oc, A, C), blue); draw(rightanglemark(Oa, foot(Oa, A, C), C)^^rightanglemark(Oc, foot(Oc, A, C), A)); dot(I); dot(Oa); dot(Ob); dot(Oc); dot(Od); dot(O, red); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); label("$I$", I, S); label("$O_A$", Oa, NW); label("$O_B$", Ob, SW); label("$O_C$", Oc, SE); label("$O_\omega$", Od, N); label("$O$", O, SE, red); [/asy]$

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