Just curious...
by aops29, Apr 3, 2020, 1:22 PM
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How much time (per week) do you spend on olympiad math?
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Properties Of The Sharky-Devil Point
by aops29, Sep 7, 2019, 2:23 PM
This blog post is going to be about the Sharky-Devil Point! The reason I'm making this is because I am starting to see this configuration quite frequently. Hopefully this is helpful!
A quick note about the naming:
A friend of mine - whose Discord name happens to be Sharky Kesa - told me about this in April of 2019. My name and profile picture seemed to resemble Satan, and hence the name. Please don't deprecate this name
First, definitions:
, the 
-SD point is defined to be the intersection of ray 
with 
. Here, 
is the incenter, and 
is the foot of the perpendicular from the -incircle touch point to the -intouch chord.
To the untrained eye, this configuration might seem obscure; what properties could such a point possess? Actually, the situtation is quite the opposite...
In the following,
is the contact triangle of 
.
![\[\textrm{(Property 1) } R \textrm{ lies on the circumcircle of } AEF\]](//latex.artofproblemsolving.com/f/a/3/fa3588e4feca14c19d36ffb63357ead516a4a278.png)
![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
draw(D--P);
draw(I--R);
draw(circumcircle(A,E,F),dotted+red);
draw(rightanglemark(I,R,A,2));
pair Ma = foot(I,E,F);
dot(Ma);
draw(I--A);
draw(rightanglemark(I,Ma,F,2));
[/asy]](//latex.artofproblemsolving.com/c/5/b/c5b5f9ffcb3a4367e2083433e84e620f891be928.png)
We show that
. To do so, invert around the incircle 
. As this swaps with the nine-point circle of , the image of is , the foot of the perpendicular from 
to 
. Thus, 
, where we let 
be the midpoint of .
![\[\textrm{(Property 2) } \textrm{The second intersection of } RI \textrm{ with the circumcircle is the } A \textrm{-antipode} \]](//latex.artofproblemsolving.com/d/f/d/dfd918b9492e1b7fb64a3f75de2ad8ae44f0bcd0.png)
![[asy]
import olympiad;
import cse5;
// config
unitsize(100);
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
pair A1 = OP(Line(R,I,10),w);
dot(A1);
label("$A'$",A1,S);
draw(R--A1);
draw(A--I);
draw(R--A);
draw(rightanglemark(A,R,I,2));
draw(circumcircle(A,E,F),dashed+red);
draw(A--A1,dashed+blue);
[/asy]](//latex.artofproblemsolving.com/4/2/f/42fc7829ae92bacabb54ea005d7de714365e5ecc.png)
This is clearly true, since
![\[\textrm{(Property 3) } R \textrm{ is the Miquel Point of } BCEF \]](//latex.artofproblemsolving.com/9/7/2/9727d4bf8c6e514a3ef0e69efc4876625eba1156.png)
![[asy]
import olympiad;
import cse5;
// config
unitsize(100);
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
draw(R--F--E--cycle,green);
draw(R--B--C--cycle,green);
draw(circumcircle(A,E,F));
draw(I--R);
[/asy]](//latex.artofproblemsolving.com/a/9/9/a99fb415f8a3a46b8830a3510d361746e397b755.png)
To demonstrate this fact, simply note that is the intersection of 
and .
![\[\textrm{(Property 4) The second intersection of line } RD \textrm{ with the circumcircle is the midpoint of the arc } BC \textrm{ not containing } A \]](//latex.artofproblemsolving.com/d/e/5/de55a1eb2782f01478b85dad0f90e9d10b68ebf1.png)
![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
pair M = OP(Line(R,D,10),w);
dot(M);
label("$M$",M,S);
draw(R--M,dashed+red);
draw(circumcircle(A,E,F),dotted+blue);
pair K = extension(A,I,B,C);
dot(K);
label("$K$",K,SE);
draw(arc(M,abs(M-I),30,150),dashed);
draw(circumcircle(I,D,K),dotted+green);
draw(A--M);
draw(R--I);
draw(R--A);
draw(rightanglemark(A,R,I,2));
draw(I--D);
draw(rightanglemark(I,D,K,2));
[/asy]](//latex.artofproblemsolving.com/5/8/e/58ee243ea7b38d57a3bd9a52f27efd34bd287422.png)
Now we shall prove this. Let
, and invert around 
. We are required to show that the image of is . Under this inversion, is swapped with 
. Thus, is swapped with the circle with diameter 
. As was the intersection of and , 
must be the intersection of 
and the circle with diameter , which is clearly . Hence we are done.
These results can be consolidated into one lemma:
![\[\textbf{The Sharky-Devil Lemma}\]](//latex.artofproblemsolving.com/3/9/9/399887fb7279520308658fc59f42f4bd7569f739.png)
In triangle, let be the contact triangle, and let 
be the midpoint of the arc 
not containing in . Suppose ray 
meets again at 
. If is the incenter of and ray 
intersects again at 
, then is the antipode of . If 
, then 
.
![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,S);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair T = OP(w,circumcircle(A,E,F));
dot(T);
label("$T$",T,NW);
pair M = OP(Line(T,D,10),w);
dot(M);
label("$M$",M,S);
draw(D--P);
draw(rightanglemark(D,P,E,2),green);
pair S = OP(Line(T,I,10),w);
dot(S);
label("$S$",S,SE);
draw(T--S);
draw(A--S,dashed+red);
draw(A--T);
draw(rightanglemark(A,T,S,2),red);
draw(M--T,blue);
[/asy]](//latex.artofproblemsolving.com/9/c/6/9c62c38c69c310b596e7a0acee9df6779e85b30c.png)
Hope you enjoyed! I might edit this later to add some problems!
![\[\textbf{DO NOT FORGET THIS CONFIGURATION!}\]](//latex.artofproblemsolving.com/1/9/2/1926a66468df6e02bdd48aeadbeb8eb105466546.png)
A quick note about the naming:
A friend of mine - whose Discord name happens to be Sharky Kesa - told me about this in April of 2019. My name and profile picture seemed to resemble Satan, and hence the name. Please don't deprecate this name
First, definitions:
, the 
-SD point is defined to be the intersection of ray 
with 
. Here, 
is the incenter, and 
is the foot of the perpendicular from the -incircle touch point to the -intouch chord.To the untrained eye, this configuration might seem obscure; what properties could such a point possess? Actually, the situtation is quite the opposite...
In the following,
is the contact triangle of 
.![\[\textrm{(Property 1) } R \textrm{ lies on the circumcircle of } AEF\]](http://latex.artofproblemsolving.com/f/a/3/fa3588e4feca14c19d36ffb63357ead516a4a278.png)
![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
draw(D--P);
draw(I--R);
draw(circumcircle(A,E,F),dotted+red);
draw(rightanglemark(I,R,A,2));
pair Ma = foot(I,E,F);
dot(Ma);
draw(I--A);
draw(rightanglemark(I,Ma,F,2));
[/asy]](http://latex.artofproblemsolving.com/c/5/b/c5b5f9ffcb3a4367e2083433e84e620f891be928.png)
We show that
. To do so, invert around the incircle 
. As this swaps with the nine-point circle of , the image of is , the foot of the perpendicular from 
to 
. Thus, 
, where we let 
be the midpoint of .![\[\textrm{(Property 2) } \textrm{The second intersection of } RI \textrm{ with the circumcircle is the } A \textrm{-antipode} \]](http://latex.artofproblemsolving.com/d/f/d/dfd918b9492e1b7fb64a3f75de2ad8ae44f0bcd0.png)
![[asy]
import olympiad;
import cse5;
// config
unitsize(100);
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
pair A1 = OP(Line(R,I,10),w);
dot(A1);
label("$A'$",A1,S);
draw(R--A1);
draw(A--I);
draw(R--A);
draw(rightanglemark(A,R,I,2));
draw(circumcircle(A,E,F),dashed+red);
draw(A--A1,dashed+blue);
[/asy]](http://latex.artofproblemsolving.com/4/2/f/42fc7829ae92bacabb54ea005d7de714365e5ecc.png)
This is clearly true, since
![\[\textrm{(Property 3) } R \textrm{ is the Miquel Point of } BCEF \]](http://latex.artofproblemsolving.com/9/7/2/9727d4bf8c6e514a3ef0e69efc4876625eba1156.png)
![[asy]
import olympiad;
import cse5;
// config
unitsize(100);
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
draw(R--F--E--cycle,green);
draw(R--B--C--cycle,green);
draw(circumcircle(A,E,F));
draw(I--R);
[/asy]](http://latex.artofproblemsolving.com/a/9/9/a99fb415f8a3a46b8830a3510d361746e397b755.png)
To demonstrate this fact, simply note that
is the intersection of 
and .![\[\textrm{(Property 4) The second intersection of line } RD \textrm{ with the circumcircle is the midpoint of the arc } BC \textrm{ not containing } A \]](http://latex.artofproblemsolving.com/d/e/5/de55a1eb2782f01478b85dad0f90e9d10b68ebf1.png)
![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,SE);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair R = OP(w,circumcircle(A,E,F));
dot(R);
label("$R$",R,NW);
// config over
pair M = OP(Line(R,D,10),w);
dot(M);
label("$M$",M,S);
draw(R--M,dashed+red);
draw(circumcircle(A,E,F),dotted+blue);
pair K = extension(A,I,B,C);
dot(K);
label("$K$",K,SE);
draw(arc(M,abs(M-I),30,150),dashed);
draw(circumcircle(I,D,K),dotted+green);
draw(A--M);
draw(R--I);
draw(R--A);
draw(rightanglemark(A,R,I,2));
draw(I--D);
draw(rightanglemark(I,D,K,2));
[/asy]](http://latex.artofproblemsolving.com/5/8/e/58ee243ea7b38d57a3bd9a52f27efd34bd287422.png)
Now we shall prove this. Let
, and invert around 
. We are required to show that the image of is . Under this inversion, is swapped with 
. Thus, is swapped with the circle with diameter 
. As was the intersection of and , 
must be the intersection of 
and the circle with diameter , which is clearly . Hence we are done.These results can be consolidated into one lemma:
![\[\textbf{The Sharky-Devil Lemma}\]](http://latex.artofproblemsolving.com/3/9/9/399887fb7279520308658fc59f42f4bd7569f739.png)
In triangle
, let be the contact triangle, and let 
be the midpoint of the arc 
not containing in . Suppose ray 
meets again at 
. If is the incenter of and ray 
intersects again at 
, then is the antipode of . If 
, then 
.![[asy]
unitsize(100);
import olympiad;
import cse5;
//the config
pair A,B,C,I;
B=origin;
C=2*right;
A=1.2*dir(65);
I=incenter(A,B,C);
dot(A);
dot(B);
dot(C);
dot(I);
label("$I$",I,S);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,E);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
pair D,E,F;
D=foot(I,B,C);
E=foot(I,C,A);
F=foot(I,A,B);
dot(D);
dot(E);
dot(F);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
draw(D--E--F--cycle);
pair P=foot(D,E,F);
dot(P);
label("$P$",P,N);
path w = circumcircle(A,B,C);
pair T = OP(w,circumcircle(A,E,F));
dot(T);
label("$T$",T,NW);
pair M = OP(Line(T,D,10),w);
dot(M);
label("$M$",M,S);
draw(D--P);
draw(rightanglemark(D,P,E,2),green);
pair S = OP(Line(T,I,10),w);
dot(S);
label("$S$",S,SE);
draw(T--S);
draw(A--S,dashed+red);
draw(A--T);
draw(rightanglemark(A,T,S,2),red);
draw(M--T,blue);
[/asy]](http://latex.artofproblemsolving.com/9/c/6/9c62c38c69c310b596e7a0acee9df6779e85b30c.png)
Hope you enjoyed! I might edit this later to add some problems!
![\[\textbf{DO NOT FORGET THIS CONFIGURATION!}\]](http://latex.artofproblemsolving.com/1/9/2/1926a66468df6e02bdd48aeadbeb8eb105466546.png)
This post has been edited 6 times. Last edited by aops29, Sep 7, 2019, 4:02 PM
Reason: typo + warning
Reason: typo + warning
Standard Pythagorean rules apply. No beans. All numbers are ratios of two postive integers. Also, no beans.
August 2020
April 2020
March 2020
September 2019