[/quote]
,
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?
interesting if 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
and circumcenter 
.
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
and 
be points on segment 
denote the circle passing through 
.
denote the circle passing through 
.
)
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
airline companies in a country. Between any two cities, there is exactly one 
and circumscribed about the incircle 
,
.
,
,
,
,
(distinct from 
intersects 
(distinct from 
be the midpoint of the arc 
of 
at 
and 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
Y by BuiBaAnh, Davi-8191, a_friendwr_a, megarnie, Adventure10, Mango247, Rounak_iitr, Tastymooncake2, Funcshun840, cubres, MS_asdfgzxcvb, and 3 other users
Given a triangle 
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on the arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the meeting point of the lines 
and 
lies on .
Proposed by Tai Wai Ming and Wang Chongli, Hong Kong
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on the arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the meeting point of the lines 
and 
lies on .Proposed by Tai Wai Ming and Wang Chongli, Hong Kong
.
and 
are similar. But this is easy because it is enough to show that 
.
.
which is clearly true.
at K. Let the parallel from I to BC cut AF to P. Then 
is cyclic (because AK is antiparallel to IP or by simple angle chase).
.
.
since then, 
will be parallelogram and G the intersection of DK and IE.
and we want to show 
it is enough 
which is obvious from similar triangles 
since arcs BD and DC are equal.
it is not necessary.
and 
are collinear
and 
.
,
.
.
.
bisects 
because 
is similar to 
.
is similar to 
.
.
.
and 
are similar. Then 
and 
are parallel. But 
.
.
.
.
(*)
.
because 
and 
because 
the intersection of 
by Menelaus
by (*)
,
are collinear.
.
so 
so 
and since 
by the similarity, it is sufficient to show 
.
the problem reduces to showing
and 
and we're done. To show that, inverse the plane with regard to 
where 
.
are mapped to 
resp. Hence 
,
be a point on 
.
.
.
,
.
.
and 
,
,
.
,
.
,
and 
,
,
and 
are congruent. To do this, we need a spiral similarity between 
and 
.
,
.
,
similar to 
.
.
,
must be a parallelogram. Reversely, if 
,
is similar to 
.
and 
.
.
is isosceles, so that 
.
,
.
is parallel to 
.
.
,
.
is fixed. It is then sufficient to show that the result is true for some choice of 
and 
)
and 
,
,
.
,
,
trivially meets 
meets 
is cyclic, because 
.
is a parallelogram.
is parallel to 
.
.
is equal to the angle between 
.
.
and 
.
.
,
is parallel to 
.
.
.
is cyclic, thus 
,
cases.
through 
e
be the 
.
by homothety reasons (then it passes through 
are spirally similar as 
.
lies on 
.
implying the result.
;
,
,
.
is similar. Let 
be the intersection of 
and 
and 
,
.
,
is cyclic. This means that 
since 
,
.
.
intersect 
and 
at points 
and 
be 
.
and 
,
.
to 
maps 
.
.
since 
.
,
so it is sufficient that 
,
.
,
,
.
,
.
,
intersect 
at 
and 
,
is parallel to 
,
,
,
,
,
and 
are all similar
are concyclic, and 
is parallel to 
and 
,
.
,
,
is parallel to 
,
.
,
,
of 
,
.
,
,
.
This concludes the proof. 
condition. We prove a stronger version of the problem statement with MMP.
and 
Fix 
and animate 
Then 
are fixed and 
vary as a function of 
is a projective map since rotations preserve cross ratios. Furthermore, 
is a projective map, so 
is projective. On the other hand, by homothety about 
we know 
the map 
is projective. It suffices to show these maps coincide for 
and 
and denote the 
.
.
and 
so the conclusion follows immediately. 
and that 
and with 
we get that 
and 
![[asy]
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pen dotstyle = black; /* point style */
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draw((11.478869149674523,-2.745873896612408)--(4.101180235347286,5.637284029866867), linewidth(1.));
draw((4.101180235347286,5.637284029866867)--(1.52,-3.), linewidth(1.));
draw((4.101180235347286,5.637284029866867)--(1.52,-3.), linewidth(1.) + qqffff);
draw((1.52,-3.)--(11.478869149674523,-2.745873896612408), linewidth(1.) + qqffff);
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draw((6.569849493777312,-5.632405395509976)--(4.101180235347286,5.637284029866867), linewidth(1.) + wwccff);
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label("$C$", (11.571713944281683,-2.5255200216579543), NE * labelscalefactor);
dot((4.101180235347286,5.637284029866867),dotstyle);
label("$A$", (4.194732135062415,5.856375718228946), NE * labelscalefactor);
dot((6.569849493777312,-5.632405395509976),linewidth(4.pt) + dotstyle);
label("$D$", (6.6613390557920775,-5.448906001781996), NE * labelscalefactor);
dot((10.490979088141643,-3.9949414711319724),dotstyle);
label("$E$", (10.589638966583761,-3.7588234820227844), NE * labelscalefactor);
dot((2.5703092250260915,-4.197057689176556),linewidth(4.pt) + dotstyle);
label("$F_{1}$", (2.6645222860912354,-4.010051964689694), NE * labelscalefactor);
dot((2.7615820377563653,-2.9683178480860586),linewidth(4.pt) + dotstyle);
label("$F$", (2.8472339098489883,-2.776748504324864), NE * labelscalefactor);
dot((5.3512791992080295,-0.06952623869681132),linewidth(4.pt) + dotstyle);
label("$I$", (5.450874548396965,0.12379852282945857), NE * labelscalefactor);
dot((4.056430618482198,-1.518922043391435),linewidth(4.pt) + dotstyle);
label("$G$", (4.149054229122977,-1.3378944672325623), NE * labelscalefactor);
dot((1.2629522122759738,3.0529091907011026),linewidth(4.pt) + dotstyle);
label("$X$", (0.9972787193017416,3.0471845029535003), NE * labelscalefactor);
dot((3.2042965798356873,-0.12431200994271625),linewidth(4.pt) + dotstyle);
label("$S$", (3.30401296924337,0.05528166392030134), NE * labelscalefactor);
dot((4.908564657128707,-2.9135320768401525),linewidth(4.pt) + dotstyle);
label("$Q$", (4.994095489002583,-2.731070598385426), NE * labelscalefactor);
dot((5.968345846524453,-2.8864890401201238),linewidth(4.pt) + dotstyle);
label("$P$", (6.067526278579381,-2.7082316454157067), NE * labelscalefactor);
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[/asy]](http://latex.artofproblemsolving.com/6/a/a/6aa4ca40d40e93d5f91fa453d51f65be281dd4d9.png)
are collinear.
cyclic.
is a parallelogram, this implies that 
bisects segment 
.
.
bisects 
.
,
is a cyclic quadrilateral Since 
it is 
.
,
.
.![$[IF]$](http://latex.artofproblemsolving.com/7/8/2/782899635cf77445262ce0b52aa0ace7d5e33cc1.png)
is a parallelogram". We have to prove that 
.
.
(the rule that 
.
from 
.
.
,
.
.
we just said, the expression goes into 
,
,
,
,
are colinear.
we have 
and 
we have to show that
.
.
or 
or 
.
and 
we have 
or 
or 
be the 
i
.
.
by Fact 
,
bisects 
Pascal on 
![[asy]
import olympiad;
import cse5;
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(110);
dot("$A$", A, dir(A));
pair B = dir(210);
dot("$B$", B, dir(B));
pair C = dir(330);
dot("$C$", C, dir(C));
pair I = incenter(A, B, C);
dot("$I$", I, dir(45));
pair O = origin;
dot("$O$", O, dir(45));
pair D = extension(A, I, O, B+C);
dot("$D$", D, dir(225));
pair E = dir(310);
dot("$E$", E, dir(E));
pair F_1 = B*C/E;
dot("$F_1$", F_1, dir(F_1));
pair F = extension(B, C, A, F_1);
dot("$F$", F, dir(F));
pair G = midpoint(I--F);
dot("$G$", G, dir(-15));
pair P = extension(E, I, D, G);
dot("$P$", P, dir(P));
pair I_A = 2*D-I;
dot("$I_A$", I_A, dir(I_A));
pair Q = extension(D, P, A, F);
dot("$Q$", Q, dir(230));
pair Z = extension(I, Q, I_A, F);
dot(Z);
draw(I--Z--I_A);
dot(extension(A, I, B, C));
draw(A--B--C--cycle, blue);
draw(unitcircle, red);
draw(A--F_1--E--P--D--cycle, magenta);
draw(A--E, dotted+green);
draw(I--F);
draw(D--I_A, dotted);
/* Source generated by TSQ:
A = dir 110
B = dir 210
C = dir 330
I = incenter A B C 45
O = origin 45
D = extension A I O B+C 225
E = dir 310
F_1 = B*C/E
F = extension B C A F_1
G = midpoint I--F -15
P = extension E I D G
I_A = 2*D-I
Q = extension D P A F 230
Z := extension I Q I_A F
:= Z
I--Z--I_A
:= extension A I B C
A--B--C--cycle blue
unitcircle red
A--F_1--E--P--D--cycle magenta
A--E dotted green
I--F
D--I_A dotted
*/
[/asy]](http://latex.artofproblemsolving.com/c/0/5/c05a9385fefdeccf1d0b0d9c152966783cc53164.png)
,
,
.
are tangent to each other. Since 
,
.
