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

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==Solution 4==
 
==Solution 4==
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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
+
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
 
<math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. ~bluesoul
  

Revision as of 22:57, 24 September 2021

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

2007 AIME II-15.png

Solution 1 (homothety)

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

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

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]

See also

2007 AIME II (Problems • Answer Key • Resources)
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