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

Revision as of 18:45, 26 December 2020

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 $260+129=\boxed{389}$ .

@@ Line 33: / Line 33: @@
 The solution is <math>260+129=\boxed{389}</math>.
-=== Diagram for Solution 1 ===
-Here is a diagram illustrating solution 1. Note that unlike in the solution <math>O</math> refers to the circumcenter of <math>\triangle ABC</math>. Instead, <math>O_\omega</math> is used for the center of the third circle, <math>\omega</math>.
-[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("<math>A</math>", A, N);
-label("<math>B</math>", B, S);
-label("<math>C</math>", C, S);
-label("<math>I</math>", I, S);
-label("<math>O_A</math>", Oa, NW);
-label("<math>O_B</math>", Ob, SW);
-label("<math>O_C</math>", Oc, SE);
-label("<math>O_\omega</math>", Od, N);
-label("<math>O</math>", 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

Page

Toolbox

Search