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

Latest revision as of 22:44, 6 October 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

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

@@ Line 3: / Line 3: @@
 __TOC__
-== Solution ==
+== Diagram ==
 <asy>
 defaultpen(fontsize(12)+0.8); size(300);
@@ Line 18: / Line 18: @@
 </asy>
-=== Solution 1 (Homothety)===
+== Solution 1 (Homothety)==
 <asy>
@@ Line 31: / Line 31: @@
 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,right+down,p); dot("$I$",P,left+up,p);
+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>.
@@ Line 42: / Line 41: @@
 https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png
-=== Solution 2 ===
+== Solution 2 ==
 [[Image:2007 AIME II-15b.gif]]
-Consider a 13-14-15 triangle.  <math>A=84.</math>  [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
+Consider a 13-14-15 triangle.  <math>A=84.</math>  (By [[Heron's Formula]] or by 5-12-13 and 9-12-15 right triangles.)
 The inradius is <math>r=\frac{A}{s}=\frac{84}{21}=4</math>, where <math>s</math> is the semiperimeter.  Scale the triangle with the inradius by a linear scale factor, <math>u.</math>
@@ Line 66: / Line 65: @@
 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.
@@ Line 82: / Line 81: @@
 <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. ~bluesoul
-== Diagram for Solution 1 ==
+==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
-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>.
+~pinkpig
-<asy>
-unitsize(0.75cm);
-pair A, B, C, Oa, Ob, Oc, Od, O, I;
-path circ1, circ2;
-// Homotethy factor - backplugged from solution
+==Solution 5==
-real k = 64/129;
-real r = 260/129;
-B = (0, 0);
+<asy>
-C = (14, 0);
+defaultpen(fontsize(12)+0.8); size(300);
-circ1 = circle(B, 13);
+pair A,B,C,X,Y,Z,P,Q,R;
-circ2 = circle(C, 15);
+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>
-A = intersectionpoints(circ1, circ2)[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
-I = incenter(A, B, C);
-Oa = (65*A + 64*I)/129;
+<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>
-Ob = (65*B + 64*I)/129;
-Oc = (65*C + 64*I)/129;
-Od = circumcenter(Oa, Ob, Oc);
+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>).
-O = circumcenter(A, B, C);
-draw(circle(Oa, r));
+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:
-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);
+<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>
-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);
+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).
-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>
+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.
-==Sidenote (Generalization)==
+The homethety turned out to be <math>\mathcal{H} \left(I, \dfrac{64}{129}\right)</math>
-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
+~sml1809
 == See also ==

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