An Amazing Geometry Problem
by djmathman, Sep 8, 2020, 7:47 PM
This problem statement is so cool 
KöMaL A.774 (March 2020), Lajos Fonyó. Let
be the circumcenter of triangle 
, and 
be an arbitrary point on the circumcircle of . Let points 
, 
and 
be the orthogonal projections of point onto lines 
, 
and 
, respectively. Prove that the incenter of triangle 
is on the Simson-Wallace line of triangle corresponding to point .
I know I say this all the time, but I'm very happy to have solved this
.
KöMaL A.774 (March 2020), Lajos Fonyó. Let
be the circumcenter of triangle 
, and 
be an arbitrary point on the circumcircle of . Let points 
, 
and 
be the orthogonal projections of point onto lines 
, 
and 
, respectively. Prove that the incenter of triangle 
is on the Simson-Wallace line of triangle corresponding to point .
not containing 
.
the orthocenter of 
;
,
,
be the feet of the altitudes from 
,
,
,
,
be the reflections of 
be the Simson line of 
,
,
be the reflections of 
the incenter of 
.![[asy]
size(300);
defaultpen(linewidth(0.6)+fontsize(10));
pair A = (0,6), B = (-2,0), C = (4,0), H = orthocenter(A,B,C), O = circumcenter(A,B,C), D = rotate(114,O)*C;
pair Ap = foot(D,B,C), Bp = foot(D,A,C), Cp = foot(D,A,B), X = foot(D,O,A), Y = foot(D,O,B), Z = foot(D,O,C);
pair P = foot(A,B,C), Q = foot(B,A,C), R = foot(C,A,B);
pair Pp = 2* P - H, Qp = 2*Q-H, Rp = 2*R-H, Xp = 2*X-D, Yp = 2*Y-D, Zp = 2*Z-D;
pair I = incenter(Xp,Yp,Zp);
draw(A--B--C--A,lightblue);
draw(circumcircle(A,B,C)^^O--A^^O--B^^O--C,heavygreen);
draw(Xp--Yp--Zp--cycle,red);
draw(Ap--Cp,orange);
draw(H--I,orange+linetype("3 3"));
draw(A--Pp^^B--Qp^^C--Rp,purple);
draw(A--Cp--D^^Bp--D^^Ap--D,magenta+linetype("3 3"));
draw(Pp--Qp--Rp--cycle,red+linetype("3 3"));
dot("$A$",A,NW,linewidth(3.3));
dot("$B$",B,dir(O--B),linewidth(3.3));
dot("$C$",C,dir(O--C),linewidth(3.3));
dot("$D$",D,dir(O--D),linewidth(3.3));
dot("$H$",H,2.3*dir(65),linewidth(3.3));
dot("$I$",I,dir(H--I),linewidth(3.3));
dot("$O$",O,S,linewidth(3.3));
dot("$X'$",Xp,dir(O--Xp),linewidth(3.3));
dot("$Y'$",Yp,dir(O--Yp),linewidth(3.3));
dot("$Z'$",Zp,dir(O--Zp),linewidth(3.3));
dot("$P'$",Pp,dir(O--Pp),linewidth(3.3));
dot("$Q'$",Qp,dir(O--Qp),linewidth(3.3));
dot("$R'$",Rp,dir(O--Rp),linewidth(3.3));
dot("$A'$",Ap,dir(D--Ap),linewidth(3.3));
dot("$B'$",Bp,dir(D--Bp),linewidth(3.3));
dot("$C'$",Cp,dir(D--Cp),linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/0/9/1/091686c2bbfa34b3a267336e0eda65a84efedeaf.png)
and 
are congruent.
.
Analogous (though not entirely congruent) angle chases hold for the other two pairs of angles. This means that 
;
mapping 
to 
and is hence also the incenter of 
;
,
That is, the angle line 
makes with the line 
is equal to the angle it makes with 
.
.
sends 
.
,
also sends 
I know I say this all the time, but I'm very happy to have solved this
.This post has been edited 5 times. Last edited by djmathman, Sep 9, 2020, 4:38 PM
is the incenter of 
.
,
,
,
.
April 2025
November 2024