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

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__TOC__
 
__TOC__
== Solution ==
+
== Diagram ==
[[Image:2007 AIME II-15.png]]
+
<asy>
=== Solution 1 ===
+
defaultpen(fontsize(12)+0.8); size(300);
 +
 
 +
pair A,B,C,X,Y,Z,P,Q,R;
 +
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
 +
real r=260/129;
 +
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
 +
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);
 +
draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray);
 +
label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
 +
pen p=fontsize(10)+linewidth(3);
 +
dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p);
 +
</asy>
 +
 
 +
== Solution 1 (Homothety)==
 +
 
 +
<asy>
 +
defaultpen(fontsize(12)+0.8); size(350);
 +
 
 +
pair A,B,C,X,Y,Z,P,Q,R,Zp;
 +
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
 +
real r=260/129;
 +
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
 +
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z);
 +
draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6));
 +
label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
 +
pen p=fontsize(10)+linewidth(3);
 +
dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p);
 +
</asy>
 +
 
 
First, apply [[Heron's formula]] to find that <math>[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84</math>. The semiperimeter is <math>21</math>, so the [[inradius]] is <math>\frac{A}{s} = \frac{84}{21} = 4</math>.  
 
First, apply [[Heron's formula]] to find that <math>[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84</math>. The semiperimeter is <math>21</math>, so the [[inradius]] is <math>\frac{A}{s} = \frac{84}{21} = 4</math>.  
  
 
Now consider the [[incenter]] <math>I</math> of <math>\triangle ABC</math>. Let the [[radius]] of one of the small circles be <math>r</math>. Let the centers of the three little circles tangent to the sides of <math>\triangle ABC</math> be <math>O_A</math>, <math>O_B</math>, and <math>O_C</math>. Let the center of the circle tangent to those three circles be <math>O</math>. The [[homothety]] <math>\mathcal{H}\left(I, \frac{4-r}{4}\right)</math> maps <math>\triangle ABC</math> to <math>\triangle XYZ</math>; since <math>OO_A = OO_B = OO_C = 2r</math>, <math>O</math> is the circumcenter of <math>\triangle XYZ</math> and <math>\mathcal{H}</math> therefore maps the circumcenter of <math>\triangle ABC</math> to <math>O</math>. Thus, <math>2r = R \cdot \frac{4 - r}{4}</math>, where <math>R</math> is the [[circumradius]] of <math>\triangle ABC</math>. Substituting <math>R = \frac{abc}{4[ABC]} = \frac{65}{8}</math>, <math>r = \frac{260}{129}</math> and the answer is <math>\boxed{389}</math>.
 
Now consider the [[incenter]] <math>I</math> of <math>\triangle ABC</math>. Let the [[radius]] of one of the small circles be <math>r</math>. Let the centers of the three little circles tangent to the sides of <math>\triangle ABC</math> be <math>O_A</math>, <math>O_B</math>, and <math>O_C</math>. Let the center of the circle tangent to those three circles be <math>O</math>. The [[homothety]] <math>\mathcal{H}\left(I, \frac{4-r}{4}\right)</math> maps <math>\triangle ABC</math> to <math>\triangle XYZ</math>; since <math>OO_A = OO_B = OO_C = 2r</math>, <math>O</math> is the circumcenter of <math>\triangle XYZ</math> and <math>\mathcal{H}</math> therefore maps the circumcenter of <math>\triangle ABC</math> to <math>O</math>. Thus, <math>2r = R \cdot \frac{4 - r}{4}</math>, where <math>R</math> is the [[circumradius]] of <math>\triangle ABC</math>. Substituting <math>R = \frac{abc}{4[ABC]} = \frac{65}{8}</math>, <math>r = \frac{260}{129}</math> and the answer is <math>\boxed{389}</math>.
  
=== Solution 2 ===
+
 
 +
https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png
 +
 
 +
== Solution 2 ==
 
[[Image:2007 AIME II-15b.gif]]
 
[[Image:2007 AIME II-15b.gif]]
  
Line 34: Line 65:
 
The solution is <math>260+129=\boxed{389}</math>.
 
The solution is <math>260+129=\boxed{389}</math>.
  
=== Solution 3 (elementary)===
+
== Solution 3 (elementary)==
  
 
Let <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>O</math> be the centers of circles <math>\omega_{A}</math>, <math>\omega_{B}</math>, <math>\omega_{C}</math>, <math>\omega</math>, respectively, and let <math>x</math> be their radius.
 
Let <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>O</math> be the centers of circles <math>\omega_{A}</math>, <math>\omega_{B}</math>, <math>\omega_{C}</math>, <math>\omega</math>, respectively, and let <math>x</math> be their radius.
Line 46: Line 77:
 
Solving, we get <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>.
 
Solving, we get <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>.
  
=== Diagram for Solution 1 ===
+
==Solution 4==
 +
According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is<math>2r</math>. Now denoting <math>AB=13;BC=14;AC=15</math>, and centers of circles tangent to <math>AB,AC;AC,BC;AB,BC</math> are relatively <math>M,N,O</math> with <math>OJ,NK</math> both perpendicular to <math>BC</math>. It is easy to know that <math>tanB=\frac{12}{5}</math>, so <math>tan\angle OBJ=\frac{2}{3}</math> according to half angle formula. Similarly, we can find <math>tan\angle NCK=\frac{1}{2}</math>. So we can see that <math>JK=ON=14-\frac{7x}{2}</math>. Obviously, <math>\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}</math> . After solving, we get
 +
<math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. ~bluesoul
 +
 
 +
==Sidenote (Generalization)==
 +
If four circles <math>\omega,</math> <math>\omega_{A},</math> <math>\omega_{B},</math> and <math>\omega_{C}</math> with the same radius are drawn in the interior of triangle <math>ABC</math> such that <math>\omega_{A}</math> is tangent to sides <math>AB</math> and <math>AC</math>, <math>\omega_{B}</math> to <math>BC</math> and <math>BA</math>, <math>\omega_{C}</math> to <math>CA</math> and <math>CB</math>, and <math>\omega</math> is externally tangent to <math>\omega_{A},</math> <math>\omega_{B},</math> and <math>\omega_{C}</math>. If <math>ABC</math> has side lengths <math>a,b,</math> and <math>c</math>, then the radius of <math>\omega</math> can be written as <cmath>\frac{abc\sqrt{2b^{2}c^{2}+2a^{2}b^{2}+2a^{2}c^{2}-a^{4}-b^{4}-c^{4}}}{4b^{2}c^{2}+4a^{2}b^{2}+4a^{2}c^{2}-2a^{4}-2b^{4}-2c^{4}+2a^{2}bc+2ab^{2}c+2abc^{2}},</cmath> or, more simply as, <cmath>\frac{abc\cdot K}{8K^{2}+sabc},</cmath> where <math>K</math> is the area of the triangle and <math>s</math> is the semiperimeter
 +
 
 +
~pinkpig
 +
 
 +
==Solution 5==
  
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>
 
<asy>
unitsize(0.75cm);
+
defaultpen(fontsize(12)+0.8); size(300);
pair A, B, C, Oa, Ob, Oc, Od, O, I;
 
path circ1, circ2;
 
  
// Homotethy factor - backplugged from solution
+
pair A,B,C,X,Y,Z,P,Q,R;
real k = 64/129;
+
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
real r = 260/129;
+
real r=260/129;
 +
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
 +
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);
 +
draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray);
 +
label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
 +
draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue);
 +
pen p=fontsize(10)+linewidth(3);
 +
dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p);  dot("$H$",(7,0),down,p);
 +
</asy>
  
B = (0, 0);
+
Let <math>O_A, O_B, O_C, O</math> be the centers of <math>w_A, w_B, w_C, w</math>, respectively. Also, let <math>I</math> be the [[incenter]] of <math>ABC</math> and <math>r</math> be the radius of circle <math>w</math>. Since <math>AB||O_AO_B</math>, <math>BC||O_BO_C</math>, and <math>CA||O_CO_A</math>, we know that
C = (14, 0);
 
  
circ1 = circle(B, 13);
+
<cmath>\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.</cmath>
circ2 = circle(C, 15);
 
  
A = intersectionpoints(circ1, circ2)[0];
+
That means <math>\angle ABC = \angle O_AO_BO_C</math>, <math>\angle BAC = \angle O_BO_AO_C</math>, and <math>\angle ACB = \angle O_AO_CO_B</math>. Thus, <math>\triangle ABC \sim \triangle O_AO_BO_C</math>. We also know that we are scaling each side of <math>\triangle ABC</math> (from <math>AB</math> to <math>O_AO_B</math> for instance), about <math>I</math> (since A,O_A,I are [[collinear]]; same apply with <math>B</math> and <math>C</math>).
I = incenter(A, B, C);
 
  
Oa = (65*A + 64*I)/129;
+
Now, let the [[homothety]] <math>\mathcal{H} (I, x)</math> map <math>\triangle ABC</math> to <math>\triangle O_AO_BO_C</math>. To start off, we know the [[circumradius]] of <math>O_AO_BO_C</math> is <math>O</math>, since <math>OO_A = OO_B = OO_C = 2r</math>. Since <math>O_AO_B = 13x</math>, <math>O_BO_C = 14x</math>, <math>O_CO_A = 15x</math>, we can get an relationship involving <math>x</math> and <math>r</math> via another way to find the circumradius:
Ob = (65*B + 64*I)/129;
 
Oc = (65*C + 64*I)/129;
 
  
Od = circumcenter(Oa, Ob, Oc);
+
<cmath>[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}</cmath>
O = circumcenter(A, B, C);
 
  
draw(circle(Oa, r));
+
Take notice of the [[inradius]] of <math>ABC</math>. We get the inradius to be <math>[\triangle ABC ] = sr_0 \Longrightarrow r_0=4</math>. Let the tangency point of the [[incircle]] and side <math>BC</math> be <math>H</math>. We know <math>IH = 4</math>. We also know that we can cut off the part of <math>IH</math> that is outside of <math>\triangle O_AO_BO_C</math> to get the inradius of <math>\triangle O_AO_BO_C</math>. To part that is outside <math>\triangle O_AO_BO_C</math> turns out just to be the radius of circle <math>w_B</math> (as seen in the picture). That means the inradius of <math>\triangle O_AO_BO_C</math> is just <math>4-r</math>. We can calculate that incradius in another way, though. We know that the inradius of <math>\triangle ABC</math> is <math>4</math>, which means the inradius of <math>\triangle O_AO_BO_C</math> is just <math>4x</math> (by our homethety ratio).
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);
+
Thus, we have <math>4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}</math>. That gives <math>\boxed{389}</math> as our final answer.
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);
+
The homethety turned out to be <math>\mathcal{H} \left(I, \dfrac{64}{129}\right)</math>
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>
+
~sml1809
  
 
== See also ==
 
== See also ==

Latest revision as of 17:51, 4 January 2024

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 $m+n.$

Diagram

[asy] defaultpen(fontsize(12)+0.8); size(300);  pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p); [/asy]

Solution 1 (Homothety)

[asy] defaultpen(fontsize(12)+0.8); size(350);  pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p); [/asy]

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 $\boxed{389}$.


https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png

Solution 2

2007 AIME II-15b.gif

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

Solution 3 (elementary)

Let $A'$, $B'$, $C'$, and $O$ be the centers of circles $\omega_{A}$, $\omega_{B}$, $\omega_{C}$, $\omega$, respectively, and let $x$ be their radius.

Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$.

Since $OA'=OB'=OC'=2x$, $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$. Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$. Then the inradius of $A'B'C'=r-x$, so now we compute r. Computing the inradius by $A=rs$, we find that the inradius of $ABC$ is $4$. Additionally, using the circumradius formula $R=\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\frac{65}{8}$. Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$.

\[\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}\]

Solving, we get $x=\frac{260}{129}$, so our answer is $260+129=\boxed{389}$.

Solution 4

According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is$2r$. Now denoting $AB=13;BC=14;AC=15$, and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular to $BC$. It is easy to know that $tanB=\frac{12}{5}$, so $tan\angle OBJ=\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$. So we can see that $JK=ON=14-\frac{7x}{2}$. Obviously, $\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}$ . After solving, we get $x=\frac{260}{129}$, so our answer is $260+129=\boxed{389}$. ~bluesoul

Sidenote (Generalization)

If 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 $ABC$ has side lengths $a,b,$ and $c$, then the radius of $\omega$ can be written as \[\frac{abc\sqrt{2b^{2}c^{2}+2a^{2}b^{2}+2a^{2}c^{2}-a^{4}-b^{4}-c^{4}}}{4b^{2}c^{2}+4a^{2}b^{2}+4a^{2}c^{2}-2a^{4}-2b^{4}-2c^{4}+2a^{2}bc+2ab^{2}c+2abc^{2}},\] or, more simply as, \[\frac{abc\cdot K}{8K^{2}+sabc},\] where $K$ is the area of the triangle and $s$ is the semiperimeter

~pinkpig

Solution 5

[asy] defaultpen(fontsize(12)+0.8); size(300);  pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p);  dot("$H$",(7,0),down,p); [/asy]

Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$, respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$. Since $AB||O_AO_B$, $BC||O_BO_C$, and $CA||O_CO_A$, we know that

\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]

That means $\angle ABC = \angle O_AO_BO_C$, $\angle BAC = \angle O_BO_AO_C$, and $\angle ACB = \angle O_AO_CO_B$. Thus, $\triangle ABC \sim \triangle O_AO_BO_C$. We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear; same apply with $B$ and $C$).

Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$. To start off, we know the circumradius of $O_AO_BO_C$ is $O$, since $OO_A = OO_B = OO_C = 2r$. Since $O_AO_B = 13x$, $O_BO_C = 14x$, $O_CO_A = 15x$, we can get an relationship involving $x$ and $r$ via another way to find the circumradius:

\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]

Take notice of the inradius of $ABC$. We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$. Let the tangency point of the incircle and side $BC$ be $H$. We know $IH = 4$. We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$. To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$. We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$, which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).

Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$. That gives $\boxed{389}$ as our final answer.

The homethety turned out to be $\mathcal{H} \left(I, \dfrac{64}{129}\right)$

~sml1809

See also

2007 AIME II (ProblemsAnswer KeyResources)
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