A Simple Lemma
by liberator, Aug 14, 2014, 2:16 PM
The configurations arising from the following lemma and its corollaries, as well as the statement itself, has been used to solve many Olympiad geometry problems: for example, RMM 2013 Problem 3, Concurrency problem, and Concyclic points. Such is this humble lemma's ubiquity that I have thought it fit to give it a place on this blog.
Lemma: Let
be the antipode of 
w.r.t the circumcircle of 
. If 
is the orthocenter of 
, then 
is a parallelogram.
![[asy]
/* Free script by liberator, 14 August 2014 */
unitsize(2.5cm);
defaultpen(fontsize(10pt));
pointpen=black;
/* Initialize objects */
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair D = foot(A,B,C);
pair E = foot(B,C,A);
pair F = foot(C,A,B);
pair H = orthocenter(A,B,C);
pair M = midpoint(B--C);
pair Ap = rotate(180, M)*H;
/* Draw objects */
draw(H--B--Ap--C--cycle, rgb(0.4, 0.6, 0.8)+linewidth(1));
draw(A--B--C--cycle, rgb(0.4, 0.6, 0.9));
draw(A--D, rgb(0.4, 0.6, 0.9));
draw(B--E, rgb(0.4, 0.6, 0.9));
draw(C--F, rgb(0.4, 0.6, 0.9));
draw(H--Ap, rgb(0.4, 0.6, 0.9)+dashed+linewidth(1));
draw(unitcircle, rgb(0.9,0,0));
markscalefactor=0.01;
/* Draw angle marks */
draw(rightanglemark(B,F,C), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(A,C,Ap), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(B,E,C), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(Ap,B,F), rgb(0.4, 0.6, 0.9));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(A));
dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(H); label("$H$", H, dir(-90));
dot(M); label("$M$", M, dir(M));
dot(Ap); label("$A'$", Ap, dir(Ap));
[/asy]](//latex.artofproblemsolving.com/4/2/7/427406602ecd400450e7fad7ec0a866a48f30ccf.png)
Proof:
, so 
; similarly, 
and the result follows.
Corollary 1: If
is the midpoint of 
, then is the reflection of in .
Corollary 2: If
is the antipode of w.r.t the circumcircle of 
, then 
is a parallelogram.
Corollary 3: The reflection of in is .
Of course, there are similar results for the antipodes of
and the antipodes of w.r.t the circumcircles of 
.
Lemma: Let
be the antipode of 
w.r.t the circumcircle of 
. If 
is the orthocenter of 
, then 
is a parallelogram.![[asy]
/* Free script by liberator, 14 August 2014 */
unitsize(2.5cm);
defaultpen(fontsize(10pt));
pointpen=black;
/* Initialize objects */
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair D = foot(A,B,C);
pair E = foot(B,C,A);
pair F = foot(C,A,B);
pair H = orthocenter(A,B,C);
pair M = midpoint(B--C);
pair Ap = rotate(180, M)*H;
/* Draw objects */
draw(H--B--Ap--C--cycle, rgb(0.4, 0.6, 0.8)+linewidth(1));
draw(A--B--C--cycle, rgb(0.4, 0.6, 0.9));
draw(A--D, rgb(0.4, 0.6, 0.9));
draw(B--E, rgb(0.4, 0.6, 0.9));
draw(C--F, rgb(0.4, 0.6, 0.9));
draw(H--Ap, rgb(0.4, 0.6, 0.9)+dashed+linewidth(1));
draw(unitcircle, rgb(0.9,0,0));
markscalefactor=0.01;
/* Draw angle marks */
draw(rightanglemark(B,F,C), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(A,C,Ap), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(B,E,C), rgb(0.4, 0.6, 0.9));
draw(rightanglemark(Ap,B,F), rgb(0.4, 0.6, 0.9));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(A));
dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(H); label("$H$", H, dir(-90));
dot(M); label("$M$", M, dir(M));
dot(Ap); label("$A'$", Ap, dir(Ap));
[/asy]](http://latex.artofproblemsolving.com/4/2/7/427406602ecd400450e7fad7ec0a866a48f30ccf.png)
Proof:
, so 
; similarly, 
and the result follows.Corollary 1: If
is the midpoint of 
, then is the reflection of in .Corollary 2: If
is the antipode of w.r.t the circumcircle of 
, then 
is a parallelogram.Corollary 3: The reflection of
in is .Of course, there are similar results for the antipodes of
and the antipodes of w.r.t the circumcircles of 
.This post has been edited 2 times. Last edited by liberator, Aug 18, 2014, 9:45 AM
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