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

(Diagram for Solution 1)
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__TOC__
 
__TOC__
 
== Solution ==
 
== Solution ==
[[Image:2007 AIME II-15.png]]
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<asy>
=== Solution 1 (homothety)===
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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>.
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 +
 +
https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png
  
 
=== Solution 2 ===
 
=== Solution 2 ===
Line 47: Line 78:
  
 
==Solution 4==
 
==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}}. After solving, we get
+
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
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<math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. ~bluesoul
=== 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$.
 
<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);
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==Sidenote (Generalization)==
circ2 = circle(C, 15);
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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
  
A = intersectionpoints(circ1, circ2)[0];
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~pinkpig
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>
 
  
 
== See also ==
 
== See also ==

Revision as of 21:58, 31 July 2022

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 (ProblemsAnswer KeyResources)
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