[/quote]
,
be an acute triangle with orthocenter 
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: 
,
.
interesting if 
of 
and 
is also a divisor of 
and 
be points on segment 
denote the circle passing through 
.
denote the circle passing through 
at 
.
)
.
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.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
such that 
Y by Davi-8191, Mathuzb, sa2001, Gumnaami_1945, itslumi, Adventure10, mathematicsy, and 1 other user
Denote by 
midpoint of side 
in an isosceles triangle 
with 
. Take a point 
on a smaller arc 
of circumcircle of triangle 
. Denote by 
point inside of angle 
such that 
and 
.
Prove that
does not depend on choice of .
Author: Farzan Barekat, Canada
midpoint of side 
in an isosceles triangle 
with 
. Take a point 
on a smaller arc 
of circumcircle of triangle 
. Denote by 
point inside of angle 
such that 
and 
.Prove that
does not depend on choice of .Author: Farzan Barekat, Canada
and 
to 
and 
respectively intersect at 
.
.
.
is isosceles
is the median as well as altitude of the triangle
is cyclic with 
being the diameter
.
.
.
is isosceles, so 
and 
.
,
is cyclic. Thus, 
and 
.
,![\[ \angle{BAM}=\angle{BXM}=\angle{MXN}-\angle{BXN}=\angle{MTB}-\angle{CTM}.\]](http://latex.artofproblemsolving.com/2/3/6/236a6f9e0fcb050cc99ec7abd89f7aa542b24233.png)
is independent of 
is also not dependent on the choice of 
be the symetrical of 
and 
and 
is cyclic to conclude that 
is cyclic which leads to![\[ \angle{MTB}-\angle{CTM}= \angle LTB = \angle MAB \]](http://latex.artofproblemsolving.com/d/2/e/d2e8ac9730e11bcd0e288358f9048c4de47de4ab.png)
Define 
and 
Let 
be the reflection of 
meet 
for a second time at 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/9/9/c/99c9ae474aa5186b025d73ab4a4bed7713d2cc76.png)
is a parallelogram because its diagonals bisect one another. Hence, 
Meanwhile, 
because 
is the perpendicular bisector of 
Then since 
,
is tangent to 
Thus, the Alternate Segment Theorem yields 
Hence,
But clearly 
is fixed, as desired. 
to be the unit circle. Furthermore, let 
.
.
,![\[\frac{t-m}{x-m}=-\frac{\bar t - \frac{1}{m}}{\frac{1}{x}-\frac{1}{m}} \implies t-m=mx \bar t - x \implies \bar t = \frac{t-m+x}{mx}\]](http://latex.artofproblemsolving.com/e/6/8/e6834d009d457d940fdeb52be1a0140bab873bc2.png)
Now since 
,![\[(t-x)(\bar t - \tfrac{1}{x})=(1-x)(1-\tfrac{1}{x}) \implies t \bar t - \frac{t}{x}- \bar t x + x +\frac{1}{x} -1=0\]](http://latex.artofproblemsolving.com/1/5/c/15ce441b1cb89c3430a3f7495eed5bc6be27f5b1.png)
Plugging in our first equation into our second, we get ![\[t\left(\frac{t-m+x}{mx}\right)-\frac{t}{x}-\left(\frac{t-m+x}{m}\right) + x + \frac{1}{x} -1=0 \implies t^2 - 2mt + (m-1)x^2+m=0\]](http://latex.artofproblemsolving.com/1/5/4/15478e60b12437d3ff900d60f1f71a8655de7487.png)
Now let 
and call the other root 
.
are collinear), so now consider the following complex number which has argument 
![\[\frac{\frac{s-r}{1-r}}{\frac{2m-1-r}{s-r}}=-\frac{(s-r)^2}{(1-r)(1-s)}=\frac{4m^2-4((m-1)x^2+m)}
{(m-1)(1-x^2)}=\frac{4(m-x^2)}{1-x^2}\]](http://latex.artofproblemsolving.com/3/c/8/3c842376a702578546102091cc8b209f3bd2c982.png)
Now notice that since 
,
lies on the unit circle, and so we see that the argument of this complex number is 
which is independent of 
,
,
.
is a midline in 
,![\[ \angle MTC=\angle BTC - \angle MTC=\angle BNM-\angle MXN=\angle MXT-\angle MXN=\angle NXT=\angle NXB \]](http://latex.artofproblemsolving.com/4/f/f/4ff229e5ff3d044f6a1e862cb0ebdfa90d0de0de.png)
Hence, 
which is fixed.
.
.
,![\[\begin{aligned} \angle MTB - \angle CTM &= \angle MTB - (\angle CTB - \angle MTB) \\ &= 2\angle MTB - \angle CTB \\ &= 2\angle MTN - \angle MNB \\ &= 2\angle MXN - (\pi - \angle MNT) \\ &= 2\angle MXN - \angle MXT \\ &= 2\angle MXN - (\angle MXN + \angle NXT) \\ &= \angle MXN - \angle NXT \\ &= (\angle MXB + \angle BXN) - \angle NXT \\ &= (\angle MXB + \angle NXT) - \angle NXT \\ &= \angle MXB \\ &= \angle MAB \end{aligned}\]](http://latex.artofproblemsolving.com/b/e/1/be1cee4099c3c8ae5ff4fe4c74b492b39e56e6c1.png)
which does not depend on 
be the midpoint of 
then 
also 
is cyclic.
is the midline of 
As desired.
is (showed in black) a cute shape, which surprisingly is independent of 
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draw(circle((9.924343193909214,-0.8781989574561639), 6.288863996049319), linewidth(2.) + linetype("2 2") + blue); /* locus construction */
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linewidth(2.));
/* dots and labels */
dot((6.38,1.),dotstyle);
dot((9.02,4.02),dotstyle);
dot((9.074301470817678,-1.9716560339900866),dotstyle);
dot((3.74,-2.02),dotstyle);
dot((14.408602941635356,-1.9233120679801736),dotstyle);
dot((9.924343193909214,-0.8781989574561639),dotstyle);
dot((4.367476888693955,2.066465238469869),linewidth(4.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/6/e/6/6e65a394b6410c2e129558418c00bf0fad30ccfa.png)
.
,
exists. So coupled with this, we get 
(call this 
)
, hence 
,
is a parallelogram. Let 
and 
.
and 
,
.
is the radius of the circumcircle of 
,
.
is isosceles, 
so 
.
.
is a right triangle, 
and 
so 
as 
is cyclic. Therefore, 
,
Define 
to be the circles 
Let 
Clearly 
by the conditions. Also, 
implies that 
Then 
is a parallelogram. Further, since 
and 
hence 
Since 
which is independent of the position of 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -1086.5975564873966, xmax = 1190.1243804115663, ymin = -879.117740078539, ymax = 608.3405920287947; /* image dimensions */
draw(arc((-586.1713229414271,-304.6162131768956),108.41533032852203,92.83654263451801,108.26645574818497)--(-586.1713229414271,-304.6162131768956)--cycle, linewidth(0.2) + red);
draw(arc((-603.0349055618302,35.73566225078713),108.41533032852203,-35.66215225285588,-20.232239139188955)--(-603.0349055618302,35.73566225078713)--cycle, linewidth(0.2) + red);
/* draw figures */
draw(circle((-353.0915854617429,10.809060298607557), 392.19799485807476), linewidth(0.4));
draw((-586.1713229414271,-304.6162131768956)--(-120.01184798205874,326.2343337741107), linewidth(0.4));
draw((-120.01184798205874,326.2343337741107)--(337.5007033357593,-310.9150475089905), linewidth(0.2));
draw((337.5007033357593,-310.9150475089905)--(-586.1713229414271,-304.6162131768956), linewidth(0.4));
draw((-120.01184798205874,326.2343337741107)--(-124.3353098028339,-307.76563034294304), linewidth(0.4));
draw((-728.1424189633996,125.50927590610821)--(21.959248039913632,-103.89115530889305), linewidth(0.2));
draw((-586.1713229414271,-304.6162131768956)--(-728.1424189633996,125.50927590610821), linewidth(0.4));
draw((-728.1424189633996,125.50927590610821)--(-124.3353098028339,-307.76563034294304), linewidth(0.4));
draw(circle((21.959248039913632,-103.89115530889305), 640.4009214688908), linewidth(0.2) + linetype("2 2"));
draw((-603.0349055618302,35.73566225078713)--(354.3642859561624,-651.2669229366732), linewidth(0.4));
draw((337.5007033357593,-310.9150475089905)--(354.3642859561624,-651.2669229366732), linewidth(0.4));
draw((-603.0349055618302,35.73566225078713)--(337.5007033357593,-310.9150475089905), linewidth(0.4));
draw((-586.1713229414271,-304.6162131768956)--(354.3642859561624,-651.2669229366732), linewidth(0.4));
draw((-603.0349055618302,35.73566225078713)--(-586.1713229414271,-304.6162131768956), linewidth(0.4));
draw((-124.3353098028339,-307.76563034294304)--(337.5007033357593,-310.9150475089905), linewidth(0.2));
draw((-603.0349055618302,35.73566225078713)--(-728.1424189633996,125.50927590610821), linewidth(0.2));
/* dots and labels */
dot((-353.0915854617429,10.809060298607557),dotstyle);
label("$N$", (-377.5612961388625,42.4125677139054), NE * labelscalefactor);
dot((-120.01184798205874,326.2343337741107),dotstyle);
label("$A$", (-110.85958353069825,348.1437992403399), NE * labelscalefactor);
dot((-120.0118479820587,326.2343337741107),linewidth(4pt) + dotstyle);
dot((-586.1713229414271,-304.6162131768956),linewidth(4pt) + dotstyle);
label("$B$", (-622.5799426813222,-360.89246110819965), NE * labelscalefactor);
dot((-124.3353098028339,-307.76563034294304),dotstyle);
label("$M$", (-141.2158760226844,-365.2290743213406), NE * labelscalefactor);
dot((337.5007033357593,-310.9150475089905),linewidth(4pt) + dotstyle);
label("$C$", (346.6531104556648,-293.6749563045155), NE * labelscalefactor);
dot((21.959248039913632,-103.89115530889305),dotstyle);
label("$X$", (56.1000251752257,-98.52736171317432), NE * labelscalefactor);
dot((21.95924803991367,-103.89115530889306),linewidth(4pt) + dotstyle);
dot((-728.1424189633996,125.50927590610821),linewidth(4pt) + dotstyle);
label("$Y$", (-783.0346315675349,129.14483197672368), NE * labelscalefactor);
dot((-603.0349055618302,35.73566225078713),linewidth(4pt) + dotstyle);
label("$T$", (-652.9362351733084,12.056275221919), NE * labelscalefactor);
dot((354.3642859561624,-651.2669229366732),linewidth(4pt) + dotstyle);
label("$L$", (368.33617652136917,-696.9799851266205), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/e/7/3/e73dea5e48a9336cf6d41ab4dd3a9a29fa3a5b4a.png)
be reflection of 
.
be the side lengths of 
.
by Pythagoras.
,
by Pythagoras.
by Ptolemy over 
(
.
is known, hence both the angles is known.
.
! That means it is independent of 
(Geo is a lie without bash)
![[asy]
size(9cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.4));
pair A = dir(55), B = dir(235), M = dir(305), C = 2M-B, X = dir(355), D = dir(175), T = intersectionpoint(D--M,arc(X,abs(B-X),90,225)), E = 2M-T;
dot("$A$", A, dir(55));
dot("$B$", B, dir(235));
dot("$M$", M, dir(270));
dot("$C$", C, dir(20));
dot("$X$", X, dir(5));
dot("$D$", D, dir(175));
dot("$T$", T, dir(210));
dot("$E$", E, dir(280));
draw(A--B--C--A--M--X^^unitcircle, purple);
draw(T--X^^E--X^^B--X, blue);
draw(T--B--E--C--T, orange);
draw(X--A--D--2B-D^^D--E, heavygreen);
draw(arc(circumcenter(T,B,E),abs(circumcenter(T,B,E)-T),165,315), magenta);
[/asy]](http://latex.artofproblemsolving.com/c/8/3/c833602748eb6c4b95a9ad00ae03f83023e9aec9.png)
meet 
.
by SAS, so 
.
so we are done. 
,
is tangent to 
which does not depend on 
whose diagonals meet at 
.![[asy] pair A = dir(180); pair B = dir(0); pair M = dir(35); pair C = 2*M-B; pair X = dir(100);
filldraw(A--B--C--cycle, invisible, red); draw(A--M, red); filldraw(unitcircle, invisible, blue);
pair Y = -X; draw(CP(X, B), mediumgreen); pair T = IP(M--Y, CP(X, B)); pair S = 2*M-T; draw(S--T, mediumgreen); draw(B--X--M, mediumgreen);
pair N = midpoint(B--T); draw(B--T, deepgreen); draw(T--X--S, mediumgreen); draw(X--N, deepgreen);
filldraw(circumcircle(X, M, T), invisible, dotted+deepgreen);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$M$", M, dir(M)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$T$", T, dir(T)); dot("$S$", S, dir(S)); dot("$N$", N, dir(N));
/* TSQ Source:
A = dir 180 B = dir 0 M = dir 35 C = 2*M-B X = dir 100
A--B--C--cycle 0.1 lightred / red A--M red unitcircle 0.1 cyan / blue
Y := -X CP X B mediumgreen T = IP M--Y CP X B S = 2*M-T S--T mediumgreen B--X--M mediumgreen
N = midpoint B--T B--T deepgreen T--X--S mediumgreen X--N deepgreen
circumcircle X M T 0.1 yellow / dotted deepgreen
*/ [/asy]](http://latex.artofproblemsolving.com/7/b/4/7b4965b2b27f900ee2b6c423c9d3fa36131d5a66.png)
![\[ \angle CTM = \angle MSB = \angle TSB = \frac{1}{2} \angle TXB = \angle NXB. \]](http://latex.artofproblemsolving.com/3/5/9/359d1f3d40edad997f74d2b2d5e54053dcad5d2e.png)
Also, noting that 
are cyclic (as 
)![\[ \angle MTB = \angle MTN = \angle NXM. \]](http://latex.artofproblemsolving.com/c/f/2/cf25f21e3b1b51cc004d8eb6351185cca5e5e0c1.png)
Thus 
which is fixed.
,
.
Now, note that 
which does not depend on 
![[asy]
unitsize(3cm);
pair A, B, C, M, T, T1, X;
A = dir(90);
B = dir(210);
C = dir(330);
M = (B + C) / 2;
X = rotate(20, (A + B) / 2) * M;
T = IP(CP(X, B), Line(M, rotate(90, M) * X, 10), 0);
T1 = 2 * M - T;
draw(A--B--C--A^^unitcircle, heavyred);
draw(circumcircle(A, B, M), heavygreen);
draw(CP(X, T), blue);
draw(X--T^^X--B^^X--M^^X--T1^^T--T1, orange);
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$M$", M, S);
dot("$X$", X, E);
dot("$T$", T, NW);
dot("$T'$", T1, SE);
[/asy]](http://latex.artofproblemsolving.com/b/c/4/bc409ff859e00734663e0f0790a2ef03ca53e446.png)
is a parallelogram. Additionally, as 
,
is isosceles, so we have 
.
.
which is fixed, as desired. 
and midpoint of 
so that 
and 
is 
Since 
we should have 
on the unit circle, and upon substituting our expression for 
Let 
)
or 
.
,
which does not depend on 
,
which is constant done yay kill me now
![[asy]
//meh asy, as always...
size(210);
pair A=(0,6.5),B=(-3,0),C=(3,0),M=(0,0),T=(-3.246,3.0711);
pair X=(1.828,1.9321), TT=(3.246,-3.0711);
draw(A--B--C--cycle, orange);
draw(circumcircle(A,B,M), orange);
draw(T--B--TT--C--cycle, magenta);
draw(X--T, red);
draw(X--B, red);
draw(X--TT, red);
draw(X--M, orange);
draw(circumcircle(B,T,TT), red);
draw(T--TT, magenta);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$M$",M,S);
dot("$T'$",TT,S);
dot("$X$",X,NE);
dot("$T$",T,NW);
[/asy]](http://latex.artofproblemsolving.com/7/1/7/717054514fc4032603ddeac8305c902ec827fc58.png)
.
which does not depend on 
so 
thus 
.
.
,
which doesn't depend on 
Clearly 
hence 
is isosceles so 
so 
and 
is fixed and independent from 
is cyclic and 
Hence, 
and 
so 
.
,
is a rectangle. Since 
,
,
This implies that 
,
Let 
so 
is cyclic. Then, 
because 
Now, 
which is fixed so we're done.
be the reflection of 
meet 
,
,
is cyclic. Now, observe 
which implies 
.
is tangent to 
is the 
.
,
.
is clearly a parallelogram, 
which finishes. 
meet 
.
,
at 
is cyclic and 
.
.
because 
is a parallelogram. However, setting 
as the arc midpoint of 
,
is fixed regardless of 
(since 
-
)![\[\angle MTC=\angle NMT=\angle TXN=\angle NXB\]](http://latex.artofproblemsolving.com/c/d/f/cdf2019dff316133813e5699515df035ef486dbc.png)
and 
,
.
(the 
,
,
is a parallelogram. On the other hand, since 
,
is tangent to 
,
which is fixed. 
is a parallelogram. Then![\[2(\angle MTB-\angle CTM)=\angle BXS-\angle BTX=(\angle BXM+\angle MXS)-(\angle MXT-\angle BXM)=2\angle BXM=2\angle BAM\]](http://latex.artofproblemsolving.com/0/6/5/0652c8576d33cd6ec3f0fae45ce8bdfa56139168.png)
done.
