and circumcenter 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
Y by Me.021, mathematicsy, zunaid_cnd, megarnie, ImSh95, Adventure10, Mango247, Rounak_iitr, deplasmanyollari, ohhh, Tastymooncake2, ehuseyinyigit
Let 
be a triangle with 
, and let 
be the midpoint of 
. Let 
be a point such that 
and 
is parallel to . Let 
and 
be points on the lines 
and 
, respectively, so that 
lies on the segment 
, 
lies on the segment 
, and 
. Prove that the quadrilateral 
is cyclic.
be a triangle with 
, and let 
be the midpoint of 
. Let 
be a point such that 
and 
is parallel to . Let 
and 
be points on the lines 
and 
, respectively, so that 
lies on the segment 
, 
lies on the segment 
, and 
. Prove that the quadrilateral 
is cyclic.This post has been edited 1 time. Last edited by bluecarneal, Aug 14, 2019, 2:05 PM
Reason: fix typo
Reason: fix typo
be a triangle and 
.
again at 
.
.
are collinear.![[asy]
size(12cm);
pair P=dir(125); pair B=dir(200); pair C=dir(-20);
pair A= 2*extension(B, C, P, P+dir(90)*P)-P;
pair M=2*foot(origin, A, -P)+P;
pair X=P+0.6*(B-P);
pair Y=extension(A, X, P, C);
pair Z=extension(A, X, P, M);
draw(anglemark(P, M, X, t=10),green);
draw(anglemark(Y, M, P, t=10),green);
draw(unitcircle);
draw(P--B--C--cycle);
draw(M--A);
draw(circumcircle(P, A, M), dotted);
draw(A--P--M);
draw(X--M--Y, red);
draw(A--Y, dashed+blue);
string[] names = {"$P$", "$B$", "$C$", "$A$", "$M$", "$X$", "$Y$", "$Z$"};
pair[] pts = {P, B, C, A, M, X, Y, Z};
pair[] labels = {dir(90), B, C, A, M, dir(120), dir(45), Z};
for(int i=0; i<names.length; ++i){
dot(names[i], pts[i], dir(labels[i]));
}
[/asy]](http://latex.artofproblemsolving.com/2/5/a/25a4b37bb1976584abeaee27a534dfa675aedf87.png)
and 
.
by the inversion. Since 
bisects 
,![\[(MA, MZ; MX, MY)=-1=(P, M; B, C)\stackrel{P}{=}(A, Z; X, Y').\]](http://latex.artofproblemsolving.com/e/3/b/e3b2d431002e9fa046279ad9431c16011576b3d1.png)
Therefore 
,
,
,
is cyclic by the given angles. We denote by ![[asy]size(250); defaultpen(fontsize(11pt)); pair D = dir(130); pair E = -conj(D); pair M = midpoint(D--E); pair X = dir(230); pair Y = dir(25); pair S = -X+2*foot(origin, X, M); pair T = -Y+2*foot(origin, Y, M); pair B = extension(X, T, D, E); pair C = extension(Y, S, D, E); pair P = extension(X, T, S, Y); pair A = 1/conj(M); pair O = origin;
markscalefactor *= 0.6; draw(anglemark(M, X, P), lightblue); draw(anglemark(P, Y, M), lightblue); filldraw(A--B--C--cycle, invisible, red); draw(P--X--S, lightblue); draw(P--Y--T, lightblue); draw(D--E, red); filldraw(unitcircle, invisible, heavycyan); draw(O--A, red); draw(P--A, dotted+red);
dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$M$", M, dir(295)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$O$", O, dir(-90));
/* TSQ Source:
D = dir 130 E = -conj(D) M = midpoint D--E R295 X = dir 230 Y = dir 25 S = -X+2*foot origin X M T = -Y+2*foot origin Y M B = extension X T D E C = extension Y S D E P = extension X T S Y A = 1/conj(M) O = origin R-90
!markscalefactor *= 0.6; anglemark M X P lightblue anglemark P Y M lightblue A--B--C--cycle 0.1 lightred / red P--X--S lightblue P--Y--T lightblue D--E red unitcircle 0.1 lightcyan / heavycyan O--A red P--A dotted red
*/ [/asy]](http://latex.artofproblemsolving.com/5/4/d/54d62439a0f1123a1c96903ea9312f0b2b9c6986.png)
.
.
to a chord of the circle, since then 
and 
.
is cyclic.
passing through 
.
be the antipodal of 
and 
,
and 
are cyclic. Therefore, 
The conclusion follows.
be reflections of 
across 
respectively. The angle condition gives
Let 
be the reflection of 
.
,
as desired.
We obtain the following equivalent problem: 
and 
.
be the antipode of 
.
such that 
and 
such that 
.
,
are concurrent at 
is cyclic. Symmetrically, 
is also cyclic. This implies that 
![[asy]
unitsize(2.7cm);
defaultpen(fontsize(10pt));
pair P = dir(110);
pair B = dir(230);
pair C = dir(310);
pair P1 = dir(70);
pair A = (P+P1)/2;
pair M = (B+C)/2;
pair Q = 1.4*M-0.4*A;
pair X = foot(Q,P,B);
pair Y = foot(Q,P,C);
dot(P);dot(B);dot(C);dot(A);dot(Q);dot(X);dot(Y);dot(M);
draw(X--P--Y);
draw(P--B--C--cycle,linewidth(1.5));
draw(X--Q--A--P);
draw(Y--Q,dashed);
draw(X--M--Y);
draw(B--Q--C);
draw(circumcircle(A,X,Y),dotted);
markscalefactor = 0.02;
draw(anglemark(M,X,P));
draw(anglemark(P,Y,M));
draw(anglemark(M,Q,B));
draw(anglemark(C,Q,M));
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,NE);
label("$P$",P,dir(P));
label("$Q$",Q,S);
label("$X$",X,SW);
label("$Y$",Y,SE);
label("$M$",M,2*NE);
[/asy]](http://latex.artofproblemsolving.com/0/b/d/0bdff2904d300b2684ee505b0b8faa10657ccf1f.png)
.
,
Thus 
or 
lies on circle with diameter 
.
,![\[\angle BXM=\angle CYM=\angle KBC=\angle KCB.\]](http://latex.artofproblemsolving.com/8/f/f/8ff148ce639983451768b2854ec7d60a44ffa3e2.png)
Line 
meets the parallel from 
![[asy]
size(8cm);
pathpen=rgb(0.4,0.6,0.8); pointpen=black; pointfontpen=fontsize(10);
real r = 2.2;
pair P=dir(150), B=dir(230), C=dir(-50), A=extension(P,P+dir(0),(B+C)/2,origin), X=r*B+(1-r)*P, Y=IP(circumcircle(A,P,X),L(P,C,-0.1,10)), M=(B+C)/2, Xp=extension(M,X,P,C), Yp=extension(M,Y,P,B), R=extension(A,P,X,Y);
DPA(R--X--P--Y^^X--Xp^^Y--Yp);
DPA(M--A^^B--C,dashed+pathpen);
D(P--R--Yp,red);
D(circumcircle(X,Y,Xp),heavygreen+dashed);
D(rightanglemark(M,A,P,3));
D(anglemark(M,X,P));
D(anglemark(P,Y,M));
D("A",A,N);
D("B",B,SW);
D("C",C,ENE);
D("P",P,NW);
D("X",X,SW);
D("Y",Y,SE);
D("M",M,SSE);
D("X'",Xp,NNE);
D("Y'",Yp,NW);
D(R);
[/asy]](http://latex.artofproblemsolving.com/6/8/f/68f587bfb33ae271881d5175d44971bdcec02e5e.png)
,
.
,
,
,
concur. Then 
,
![[asy]
size(6cm);
pathpen=rgb(0.4,0.6,0.8); pointpen=black; pointfontpen=fontsize(10);
real r = 1.8;
pair P=dir(125), B=dir(230), C=dir(-50), A=extension(P,P+dir(0),(B+C)/2,origin), X=r*B+(1-r)*P, Y=IP(circumcircle(A,P,X),L(P,C,-0.1,10)), M=(B+C)/2;
D(P--X--Y--cycle);
D(X--M--Y,red);
D(X--A--Y,heavygreen);
D(circumcircle(A,P,X),heavygreen+dashed);
DPA(B--C,dashed+pathpen);
D(anglemark(M,X,P));
D(anglemark(P,Y,M));
D("A",A,N);
D("B",B,SW);
D("C",C,SSW);
D("P",P,NW);
D("X",X,SW);
D("Y",Y,SE);
D("M",M,N);
[/asy]](http://latex.artofproblemsolving.com/c/0/a/c0aa0b527067a23b2d3d9e44241e878713cb9e3e.png)
,
meet 
.
is a homography.
,
is a reflection with respect to a fixed line through 
is a homography.
is fixed, so 
is a rotation about 
,
and 
as the unit circle, so that
We now consider point 
,
(Clearly, none of these factors are zero since 
are all necessarily distinct.)
This equation is a little painful to work with, but we'll come back to it.
and 
.
and 
,
We also have
This gives us a system of two equations in two unknowns. We manipulate to get
Now, we define point 
as follows:
We see that 
,
.
and 
.
:![$$\frac{a^*-p}{a^*-m} = \frac{m-p^2\overline{m}}{p(1-m\overline{m})} = \frac1{1-|m|^2}\left[\frac{m}p - \overline{m}p\right] \in i\mathbb{R}$$](http://latex.artofproblemsolving.com/a/a/1/aa1f2db02e8e57ae1a48b834206776f44bcfaa43.png)
(Since 
,
and there's no danger of dividing by zero here.)
;
.
We will show that the denominators are equal, i.e., that
which will imply the desired result. Indeed, subtracting the RHS from the LHS and expanding, the result is
![$$=2\left[p^2(\overline{m}x-1)(\overline{m}y-1) - (m-x)(m-y)\right]$$](http://latex.artofproblemsolving.com/9/e/6/9e61c0954100ff62db061da82426c5c48838f26d.png)
which is zero due to the equation at the top of the solution satisfied by 
.
![$\text{[D]DIT}$](http://latex.artofproblemsolving.com/2/9/e/29e83d4aebf4be88b17265f0d2322a4434b0cfb8.png)
mean?
.
,
,
.
and 
.
in terms of the other three angles.
,
.
,
.
.
.
.
,
from 
to get 
(where 
,
,
)
is the same when we scale down 
,
.
.
and 
variables, it’s the same when we scale down 
.
and 
and 
.
.
are concyclic. Let their circumcentre be 
and 
and 
.
is perpendicular bisector of 
.
lie on circle with diameter 
and 
congruent by LOS, so if we suppose they intersect for a second time at 
,
.![\[\angle PAN = \angle PYN = \angle PXN = 90 \implies APXYN \text{ cyclic.} \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/1/a/c1ab73d5697af486c9fab25ab72433ab26feffcb.png)
.
since 
.
since 
.
we have 
and it follows that 
as desired.![[asy]
import geometry;
size(10cm);
defaultpen(fontsize(9pt));
pen pri; pri=RGB(24, 105, 174);
pen sec; sec=RGB(217, 165, 179);
pen tri; tri=RGB(126, 123, 235);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pair A,B,C,X,Y,P,Z,M;
pair A = dir(90);
pair B = dir(210);
pair C = dir(330);
pair M = midpoint(B--C);
P=A-1;
Z=(0, -1);
pair X = intersectionpoints(circle(B,M,Z),line(P,B))[0];
pair Y = intersectionpoints(circle(C,M,Z),line(P,C))[1];
filldraw((path)(A--B--C--cycle), white+0.1*pri, pri);
draw(A--P);
draw(circumcircle(B,M,Z), sec+dashed);
draw(circumcircle(Z,M,C), sec+dashed);
draw(P--X);
draw(P--C);
draw(A--Z);
filldraw(circumcircle(A,X,Y), tfil, tri);
dot("$A$",A,dir(90));
dot("$B$",B,dir(210));
dot("$C$",C,dir(260));
dot("$M$",M,dir(210));
dot("$P$",P,dir(180));
dot("$X$",X,dir(260));
dot("$Y$",Y,dir(260));
dot("$Z$",Z,dir(260));
[/asy]](http://latex.artofproblemsolving.com/b/f/d/bfd084f609db0ebed50d3c3cfc9d907757b89759.png)
which 
Note that 
and similarly 
We also know that 
since 
Thus 
which implies the result.
.
,
is cyclic. We have that 
,
is cyclic. Similarly, 
is cyclic. Therefore, 
,
and 
intersect on 
.
be the second intersection of these circles. We have to show that 
,
where 
(equal circumradii) and 
,
b
as desired.
denote directed angles modulo 
.
,
are concyclic.
are concyclic.
are concyclic as desired.
then since 
we get that the two circle are equal.
then we have that 
so 
are collinear.
so 
lie on the same circle.
and 
and 
since cyclic quadrilaterals exist and 
and 
are cyclic
and 
are cyclic, and 
Then 
by butterfly. Therefore 
is cyclic. Observe that by reim's we have 
cyclic.
then we have![\[ \measuredangle MYC = \measuredangle BXM = \measuredangle BZM = \measuredangle MZC\]](http://latex.artofproblemsolving.com/b/f/c/bfceb40f3d74bcb1353c476da22aff404b751331.png)
Where the last equality is due to the isosceles condition. So 
is cyclic hence![\[ \measuredangle ZYP = \measuredangle ZYC = \measuredangle ZMC = 90^{\circ} = \measuredangle ZAP\]](http://latex.artofproblemsolving.com/d/e/e/deeb52d5a2c71edde620978c7b94313b1791d6e5.png)
Hence we get that 
is cyclic.
is cyclic proving the required result.
then reflect about the angle bisector of 
.
is the intersection of the symmedian and the circumcircle of 
,
and 
are reflections over the symmedian and 
and 
.
be the line through 
perpendicular to 
.
is the same as 
.
(harmonic quadrilateral) and 
(angle bisector). 
with 
.
is cyclic and 
which implies 
to be cyclic. Similarly, 
is also cyclic. Therefore, 
is cyclic and we are done.
and 
.
of circle 
and 
and 
.
meet at 
contains 
,
implies 
implies 
,
and thus 
,
,
,
,
.
.
,
must be cyclic. Also, 
,
is also cyclic. Now finally note that 
,
,
.
are collinear. Since 
are concylic points , since 
.
Since 
are concylic points which 
that 
Let the intersection be 
Thus, 
and similarly, 
However, since 
we get that 
is cyclic, which finishes.
