Difference between revisions of "2007 AIME II Problems/Problem 15"
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− | == | + | == Diagram == |
− | + | <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>. | ||
− | + | ||
+ | https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png | ||
+ | |||
+ | == Solution 2 == | ||
[[Image:2007 AIME II-15b.gif]] | [[Image:2007 AIME II-15b.gif]] | ||
Line 33: | Line 64: | ||
The solution is <math>260+129=\boxed{389}</math>. | The solution is <math>260+129=\boxed{389}</math>. | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | Now, triangles <math>ABC</math> and <math>A'B'C'</math> are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for <math>x</math>. | ||
+ | |||
+ | Since <math>OA'=OB'=OC'=2x</math>, <math>O</math> is the circumcenter of triangle <math>A'B'C'</math> and its circumradius is <math>2x</math>. Let <math>I</math> denote the incenter of triangle <math>ABC</math> and <math>r</math> the inradius of <math>ABC</math>. Then the inradius of <math>A'B'C'=r-x</math>, so now we compute r. Computing the inradius by <math>A=rs</math>, we find that the inradius of <math>ABC</math> is <math>4</math>. Additionally, using the circumradius formula <math>R=\frac{abc}{4K}</math> where <math>K</math> is the area of <math>ABC</math> and <math>R</math> is the circumradius, we find <math>R=\frac{65}{8}</math>. Now we can equate the ratio of circumradius to inradius in triangles <math>ABC</math> and <math>A'B'C'</math>. | ||
+ | |||
+ | <cmath>\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}</cmath> | ||
+ | |||
+ | Solving, we get <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. | ||
+ | |||
+ | ==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== | ||
+ | |||
+ | <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 <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 | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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>). | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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). | ||
+ | |||
+ | 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. | ||
+ | |||
+ | The homethety turned out to be <math>\mathcal{H} \left(I, \dfrac{64}{129}\right)</math> | ||
+ | |||
+ | ~sml1809 | ||
== See also == | == See also == |
Latest revision as of 17:51, 4 January 2024
Problem
Four circles
and
with the same radius are drawn in the interior of triangle
such that
is tangent to sides
and
,
to
and
,
to
and
, and
is externally tangent to
and
. If the sides of triangle
are
and
the radius of
can be represented in the form
, where
and
are relatively prime positive integers. Find
Contents
Diagram
Solution 1 (Homothety)
First, apply Heron's formula to find that . The semiperimeter is
, so the inradius is
.
Now consider the incenter of
. Let the radius of one of the small circles be
. Let the centers of the three little circles tangent to the sides of
be
,
, and
. Let the center of the circle tangent to those three circles be
. The homothety
maps
to
; since
,
is the circumcenter of
and
therefore maps the circumcenter of
to
. Thus,
, where
is the circumradius of
. Substituting
,
and the answer is
.
https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png
Solution 2
Consider a 13-14-15 triangle. [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is , where
is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,
The circumradius is where
and
are the side-lengths. Scale the triangle with the circumradius by a linear scale factor,
.
Cut and combine the triangles, as shown. Then solve for :
The solution is .
Solution 3 (elementary)
Let ,
,
, and
be the centers of circles
,
,
,
, respectively, and let
be their radius.
Now, triangles and
are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for
.
Since ,
is the circumcenter of triangle
and its circumradius is
. Let
denote the incenter of triangle
and
the inradius of
. Then the inradius of
, so now we compute r. Computing the inradius by
, we find that the inradius of
is
. Additionally, using the circumradius formula
where
is the area of
and
is the circumradius, we find
. Now we can equate the ratio of circumradius to inradius in triangles
and
.
Solving, we get , so our answer is
.
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. Now denoting
, and centers of circles tangent to
are relatively
with
both perpendicular to
. It is easy to know that
, so
according to half angle formula. Similarly, we can find
. So we can see that
. Obviously,
. After solving, we get
, so our answer is
. ~bluesoul
Sidenote (Generalization)
If four circles
and
with the same radius are drawn in the interior of triangle
such that
is tangent to sides
and
,
to
and
,
to
and
, and
is externally tangent to
and
. If
has side lengths
and
, then the radius of
can be written as
or, more simply as,
where
is the area of the triangle and
is the semiperimeter
~pinkpig
Solution 5
Let be the centers of
, respectively. Also, let
be the incenter of
and
be the radius of circle
. Since
,
, and
, we know that
That means ,
, and
. Thus,
. We also know that we are scaling each side of
(from
to
for instance), about
(since A,O_A,I are collinear; same apply with
and
).
Now, let the homothety map
to
. To start off, we know the circumradius of
is
, since
. Since
,
,
, we can get an relationship involving
and
via another way to find the circumradius:
Take notice of the inradius of . We get the inradius to be
. Let the tangency point of the incircle and side
be
. We know
. We also know that we can cut off the part of
that is outside of
to get the inradius of
. To part that is outside
turns out just to be the radius of circle
(as seen in the picture). That means the inradius of
is just
. We can calculate that incradius in another way, though. We know that the inradius of
is
, which means the inradius of
is just
(by our homethety ratio).
Thus, we have . That gives
as our final answer.
The homethety turned out to be
~sml1809
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 | ||
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