[/quote]
,
be the set of real numbers. Determine all functions 
such that ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
and 
.
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
be real numbers such that 
Prove that 
Where 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of 
Prove that 
,
,
,
can be expressed by a sum of positive integers 
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
and 
.
equally spaced points on a circular track 
of circumference 
in some order, each label used once. Initially, Bunbun the Bunny begins at 
.
,
,
,
to 
,
of 
is a diameter of 
with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and 
board with a number written in each cell. Every two neighbour rows sum up to at least 
,
.
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
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 midpoint
is the median as well as altitude of the triangle
is cyclic with 
being the diameter
be the midpoint of 
.
is the midpoint of 
,
.
.
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 
wrt 
and 
and 
the reflexion of 
.
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 
in 
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 
be the root that lies within angle 
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 
then 
also 
is cyclic.
is the midline of 
As desired.
is (showed in black) a cute shape, which surprisingly is independent of 
:![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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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 
be the reflection of 
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.
.
(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.
