be a quadrilateral and let 
be a line. Let 
at points 
respectively. Given that these six points on 
,
are coaxal.
Y by anantmudgal09, artsolver, hwl0304, agimog, doxuanlong15052000, acegikmoqsuwy2000, anhtaitran, Davi-8191, tenplusten, nguyendangkhoa17112003, lahmacun, Purple_Planet, super.shamik, MathsLion, adityaguharoy, MatBoy-123, megarnie, HWenslawski, rayfish, Mogmog8, OronSH, Rounak_iitr, MathLuis, Adventure10, hairtail, cubres
Let 
be a triangle with altitude 
. The 
-excircle touches 
at 
, and intersects the circumcircle at two points 
and 
. Prove that one can select points 
and 
on lines 
and 
such that quadrilateral 
is a rhombus.
Danielle Wang and Evan Chen
be a triangle with altitude 
. The 
-excircle touches 
at 
, and intersects the circumcircle at two points 
and 
. Prove that one can select points 
and 
on lines 
and 
such that quadrilateral 
is a rhombus.Danielle Wang and Evan Chen
This post has been edited 2 times. Last edited by 62861, May 18, 2018, 11:58 PM
perpendicular bisector of
and we see 
perpendicular bisector of 
denote the incenter, 
the 
the midpoint of 
,
the tangents from 
,
then concur at 
(unless 
,
,![[asy]
size(12cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M = dir(270);
pair J = 2*M-I;
pair D = foot(J, B, C);
pair E = foot(A, B, C);
pair F = IP(CP(J, D), unitcircle);
pair G = OP(CP(J, D), unitcircle);
draw(unitcircle, lightblue);
draw(arc(J,abs(D-J),-30,210), lightblue);
pair L = midpoint(A--E);
pair V = extension(G, D, L, L+B-C);
pair N = extension(F, D, L, L+B-C);
draw(A--B--C--cycle, lightblue);
draw(A--E, lightblue);
draw(G--V, lightblue);
draw(N--F, lightblue);
draw(E--V--A--N--cycle, heavygreen);
draw(V--N, heavygreen);
pair H = extension(G, F, B, C);
draw(C--H--G, lightblue);
draw(circumcircle(B, I, C), lightred);
pair Y = IP(circumcircle(B, I, C), circumcircle(D, F, G));
pair Z = OP(circumcircle(B, I, C), circumcircle(D, F, G));
pair T = -D+2*foot(J, D, I);
draw(T--H, blue);
draw(Z--H, blue);
draw(L--T, blue);
draw(Y--I--Z, blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M$", M, dir(M));
dot("$J$", J, dir(J));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(270));
dot("$G$", G, dir(270));
dot("$L$", L, dir(L));
dot("$V$", V, dir(V));
dot("$N$", N, dir(N));
dot("$H$", H, dir(H));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$T$", T, dir(T));
[/asy]](http://latex.artofproblemsolving.com/3/f/3/3f3fe5274607b20df93fcc9b3b47df0168facb8a.png)
and 
be the tangents from 
is the symmedian of 
,
.
passes through 
,
from the circle, which then becomes the bundle 
which is the desired result. 
.
;
is fixed, the excircle maps to a line parallel to 
,
is the inradius. To prove this, let 
denote the 
denote the area of 
;
which is true, as desired.
and 
,
is parallel to 
.
also lies on the perpendicular bisector of 
.
where M is the midpoint of 
,
is harmonic or what is equivalent, tangent at X to the excircle, BC, FG concur. By extraversion of ISL 2002 G7, 
is tangent to the excircle and thus by an application of POP, we are done.
be the midpoint of 
and 
is bisected by 
.
hit the excircle again at 
,
is harmonic.
.
be the 
e
be the 
to be on the perpendicular bisector of 
the midpoint of 
.
and 
.
.
,
is cyclic.
denote the power of 
with respect to a circle 
.
and 
.
and 
are linear in 
.
,
and similarly for 
onto 
and 
and 
.
it follows 
and 
and similarly for 
.
.
and 
.
onto 
,
,
than 
tangents to 
touch AB,AC,BC in E,F,D.Let the E,F-midlines of triangles 
and 
cut in 
.
so 
but also 
where 
is the length of a tangent from 
,
tangent to 
and so the conclusion follows.
in 
and let 
is on the radical axis of 
and so 
are concurrent 
is harmonic and hence 
so the rest is proving that 
are anti-parallel,:
Hence the conclusion.
intersect the 
denote the incenter and 
is harmonic, whence projecting through 
and let 
.
,
.
,
,
or equivalently that 
is harmonic. It then suffices to show 
.
passes through 
,
,
,
and 
concur. Then it would suffice to show that 
.
imply that 
lie on the perpendicular bisector of 
Define 
It suffices to prove that 
,
be the other extouch points of the 
Define 
Then 
,
so by the Ceva configuration, 
be the intersection of the 
Since 
and 
we have 
by LaHire. We can see that the polars form an angle harmonic bundle. So consider their projections onto 
means 
,
so 
,
Since the projection is a harmonic bundle, 
,
and by LaHire, 
Define 
are collinear.
excircle, 
are collinear. 
is simply the median of 
.
,
since 
Then 
by PoP. This means that 
are collinear, 
are tangents to the excircle, and 
are collinear. So,
and 
and 
intersect 
be the intersections of lines 
is an isosceles trapezoid. In fact, we will show, via length calculations, that line 
is exactly halfway between parallel lines 
as desired. 
,
is cyclic; let 
be the reflection of 
is cyclic with perpendicular diagonals; the equation we're trying to prove follows by power of a point on the diagonals.
centered at 
is sent to the perpendicular bisector of 
![\[-1 = (VN;L \infty) \overset{D}{=} (GF;KD),\]](http://latex.artofproblemsolving.com/3/c/b/3cbeba1000b1f0e424824cc23d591d9e05a38a14.png)
.
to be tangent to the excircle, where 
.
.
meets the excircle at two points 
and 
are tangents to the excircle, 
is harmonic, which implies the desired. 
are concurrent. Call this concurrency point 
is tangent to the 
,
,
and so![\[
-1 = (DP;FG) \stackrel D= (P_\infty M;NV)
\]](http://latex.artofproblemsolving.com/9/b/2/9b2e9da0316b5daa36044947d10e304a7357dd5a.png)
where 
is the point of infinity along line 
,
![[asy]
import olympiad; import cse5;
defaultpen(fontsize(10pt));
usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp");
size(8cm);
pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) {
r=r/2;
pair Bp=unit(B-A)*r+A;
pair Cp=unit(C-A)*r+A;
pair P=Bp+Cp-A;
D(Bp--P--Cp,q);
return A;
}
pointpen=black+linewidth(2);
pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0);
pen polyfill=polyline+opacity(0.1);
pen angleline=linewidth(pathpen)+rgb(0,0.4,0);
pen anglefill=angleline+opacity(0.4);
markscalefactor=0.01;
size(15cm);
pair A=dir(120),B=dir(210),C=dir(330);
// filldraw(A--B--C--cycle,polyfill,polyline);
D(A--B--C--cycle,polyline);
pair I=incenter(A,B,C);
pair I_A=2*circumcenter(B,I,C)-I;
path exc=CP(I_A,foot(I_A,B,C));
path cir=circumcircle(A,B,C);
pair F=IP(exc,cir,1);
pair G=IP(exc,cir,0);
D(exc); D(cir);
pair D=foot(I_A,B,C),E=foot(A,B,C);
pair M=(A+E)/2;
D(B--foot(I_A,A,B)); D(C--foot(I_A,A,C));
D(A--E);
pair V=extension(D,G,M,M+B-E),N=extension(D,F,M,M+C-E);
D(V--N); D(F--N); D(G--V); D(D--M);
pair L=2*foot(I_A,M,D)-D;
pair X=(D+L)/2;
pair Y=extension(B,C,F,G);
D(D--L); D(C--Y--L); D(Y--I_A); D(circumcircle(B,I,C));
D("A",D(A),A);
D("B",D(B),W);
D("C",D(C),dir(0));
D("I",D(I),W);
D("I_A",D(I_A),S);
D("F",D(F),dir(181));
D("G",D(G),dir(-80));
D("D",D(D),dir(-98));
D("E",D(E),S);
D("M",D(M),dir(48));
D("V",D(V),W);
D("N",D(N),dir(0));
D("L",D(L),dir(0));
D("X",D(X),dir(3));
D("Y",D(Y),dir(45));
[/asy]](http://latex.artofproblemsolving.com/a/7/6/a762a3d0f70642f820deab3d350eaedc1b7c5f3f.png)
,
.
and 
.
and thus 
.
and 
,![\[YL^2 = YX \cdot YI_A = YB \cdot YC = YF \cdot YG\text{.}\]](http://latex.artofproblemsolving.com/e/f/a/efaf7e6ecfcc6ffa5a666d17bf38a42cd93b720c.png)
Thus 
.
we have 
.
is a harmonic quadrilateral, or that the tangents at 
.
,
and 
.
is harmonic, so we are done.
through 
where 
.
.
to be harmonic, the following are equivalent: 
is tangent to 
is the polar of 
;
and 
at points 
satisfies 
by Ceva-Menelaus. The midpoint 
of 
satisfies 
so 
.
lies on the polar of 
.
)
.
,
.
must satisfy 
,
and 
and let 
respectively. If 
which will evidently show that 
are collinear. Let the line 
intersect the 
is tangent to 
We first show that 
Indeed, if 
is the midpoint of 
then Power of a Point implies 
since 
is cyclic 
)
we see that 
implying that 
Then the Shooting Lemma implies the result.
Radical center on 
implies that 
are collinear. Noting that 
is tangent to 
is harmonic. Then 
and we are done.
We would like to prove 
and the 
We want to prove 
Let tangent at 
or 
(points 
are collinear follows from homothethy at 
the polar of 
The polar is the line between the intersection between the circle with diameter 
and the excircle; let these intersections be 
.
is cyclic. Thus, by Radical Axis Theorem, 
concur, at 
,
.
,
be the circumcenter, 
intersects 
,![\[-1 = (V,N;M,\infty) \overset{D}{=} (F,G;K,D).\]](http://latex.artofproblemsolving.com/0/0/4/0041e5f78175410c5b86223d1d2d490c8a3eb774.png)
is harmonic. Hence, it suffices to show that 
is tangent to 
is harmonic
,
,
concur, and we also know that the concurrency point is 
be the midpoint of 
and let 
it tangent to 
and 
are tangent to 
.
are tangent to 
is harmonic, hence 
through 
is cyclic. Let 
denote the circumcircle of 
and 
with 
.![\[ TP \cdot TI_A = TB \cdot TC = TF \cdot TG = TD^2 \]](http://latex.artofproblemsolving.com/5/d/f/5dfb15ae1831373dab737e1b3546a7d2489851de.png)
then it follows 
,![\[ \angle TPD = \angle TDI_A = 90^\circ \]](http://latex.artofproblemsolving.com/8/0/d/80dc2a053fa743b211705e55ba5a0bf3f2c11ce3.png)
however, we also have ![\[ \angle TPI = 180^\circ - \angle I_API = 180^\circ - 90^\circ = 90^\circ \]](http://latex.artofproblemsolving.com/d/b/f/dbf8c3007673dc5448a3990bb90be3b24438278a.png)
hence 
as needed. 
and 
,
is cyclic.
is tangent to 
;
,![\[ \measuredangle VFG = \measuredangle DFG = \measuredangle BDG = \measuredangle VNG \]](http://latex.artofproblemsolving.com/3/5/b/35b987ba77ae3bfe1ee3c8282f506d73a77412b1.png)
as needed. 
and since 
.
by definition as well. Thus 
Let 
.
.
bisector of 
and 
then 
.
or we just need to prove that 
is tangent to 
,
