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2007 AIME II Problems/Problem 15

Revision as of 21:58, 31 July 2022 by Shihan (talk | contribs) (→‎Diagram for Solution 1)

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

[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

See also

2007 AIME II (Problems • Answer Key • Resources)
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