,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
to the side 
and 
,
,
be the incenter, 
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
Y by Amir Hossein, Snakes, AlastorMoody, Inconsistent, Greenleaf5002, microsoft_office_word, samrocksnature, HWenslawski, mathematicsy, megarnie, Adventure10, Mango247, Mogmog8, lian_the_noob12, deplasmanyollari, Exponent11, ItsBesi
Let 
be the circumcenter of an acute triangle 
. Line 
intersects the altitudes of through 
and 
at 
and 
, respectively. The altitudes meet at 
. Prove that the circumcenter of triangle 
lies on a median of triangle .
be the circumcenter of an acute triangle 
. Line 
intersects the altitudes of through 
and 
at 
and 
, respectively. The altitudes meet at 
. Prove that the circumcenter of triangle 
lies on a median of triangle .This post has been edited 2 times. Last edited by fastlikearabbit, Jul 10, 2018, 11:03 AM
and 
at 
and 
respectively. Then 
is perpendicular to 
.
,
is perpendicular to 
,
and 
are concyclic, implying 
,
.
be the midpoint of 
is a trapezoid with 
perpendicular to the bases, it is easy to verify that 
,
goes through the midpoint of 
be feet of altitudes from 
of 
respectively. Let 
and let 
be the circumcircle of 
which is centered at 
.
be other tangent to 
but 
,
are colinear. By Brokard's theorem on 
,
.
.
hence we are done.
But 
so circumcircle of 
.
be intersection point of 
to 
repectively, where 
is point 
and 
.
.
.
are concyclic.
and 
.
collinear. But 
and 
,
and 
.
,
and 
are isogonal. Thus 
correspond to 
,![\[ BH : HX : XP : PE = CQ : QY : YH : HF. \]](http://latex.artofproblemsolving.com/d/5/6/d5613efea62f2298d030b682881651ff4ce4b7f5.png)
In other words ![\[ \frac{BX}{XE} = \frac{CY}{YF}. \]](http://latex.artofproblemsolving.com/e/9/7/e977e20b85532b42356d281f6600c6a25b314387.png)
![[asy]size(200); pair A = dir(115); pair B = dir(220); pair C = dir(320); pair H = orthocenter(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); pair D = foot(A, B, C); draw(A--B--C--cycle, red);
pair P = extension(A, origin, B, H); pair Q = extension(A, origin, C, H);
pair S = circumcenter(H, P, Q); pair M = midpoint(B--C);
pair X = midpoint(H--P); pair Y = midpoint(H--Q); draw(B--H--F, red); draw(P--E, red); draw(Q--C, red);
draw(P--H--Q--cycle, heavycyan); draw(X--S--Y, heavycyan); draw(A--M, dashed+heavygreen); draw(D--A--P, dotted+red); pair O = origin;
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$D$", D, dir(D)); dot("$P$", P, dir(340)); dot("$Q$", Q, dir(Q)); dot("$S$", S, dir(10)); dot("$M$", M, dir(M)); dot("$X$", X, dir(110)); dot("$Y$", Y, dir(Y)); dot("$O$", O, dir(50));
/* TSQ Source:
A = dir 115 B = dir 220 C = dir 320 H = orthocenter A B C R270 E = foot B A C F = foot C A B D = foot A B C A--B--C--cycle 0.1 lightred / red
P = extension A origin B H R340 Q = extension A origin C H
S = circumcenter H P Q R10 M = midpoint B--C
X = midpoint H--P R110 Y = midpoint H--Q B--H--F red P--E red Q--C red
P--H--Q--cycle 0.1 lightcyan / heavycyan X--S--Y heavycyan A--M dashed heavygreen D--A--P dotted red O = origin R50
*/ [/asy]](http://latex.artofproblemsolving.com/e/7/7/e7728d1d301cd7761af44a19414259a1836ec3c6.png)
with that directed ratio, the intersection 
and 
,
,
.
.
.
,
.![[asy] size(300); pair A = dir(135); pair B = dir(200); pair C = dir(340); pair H = orthocenter(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); pair D = foot(A, B, C); draw(A--B--C--cycle, red);
pair P = extension(A, origin, B, H); pair Q = extension(A, origin, C, H);
pair S = circumcenter(H, P, Q); pair M = midpoint(B--C);
draw(B--H--F, red); draw(P--E, red); draw(Q--C, red);
draw(P--H--Q--cycle, heavycyan); draw(A--M, dashed+heavygreen); draw(D--A--P, heavycyan); pair O = origin; pair T = extension(B, C, A, A+dir(90)*A); draw(unitcircle, red); draw(circumcircle(H, P, Q), heavycyan); draw(A--T--B, red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$D$", D, dir(D)); dot("$P$", P, dir(340)); dot("$Q$", Q, dir(Q)); dot("$S$", S, dir(10)); dot("$M$", M, dir(M)); dot("$O$", O, dir(50)); dot("$T$", T, dir(T));
/* TSQ Source:
A = dir 135 B = dir 200 C = dir 340 H = orthocenter A B C R270 E = foot B A C F = foot C A B D = foot A B C A--B--C--cycle 0.1 lightred / red
P = extension A origin B H R340 Q = extension A origin C H
S = circumcenter H P Q R10 M = midpoint B--C
B--H--F red P--E red Q--C red
P--H--Q--cycle 0.1 lightcyan / heavycyan A--M dashed heavygreen D--A--P heavycyan O = origin R50 T = extension B C A A+dir(90)*A unitcircle red circumcircle H P Q heavycyan A--T--B red
*/ [/asy]](http://latex.artofproblemsolving.com/2/d/1/2d1ba8da0cc8bce6cee493251cfe1321ad716ed1.png)
to 
,
,
)
,
.![\[ \measuredangle MAO = \measuredangle SAP = -\measuredangle OTB = \measuredangle MTO. \]](http://latex.artofproblemsolving.com/c/6/a/c6a35a92e25b9c7b946473b7ea01a1deccefab58.png)
Consequently, 
is cyclic. This implies 
as desired.
in the unit circle, so 
.
of 
equals 
and the antipode 
of 
,
is 
Of course 
.
,
,
is 
(unit circle).
,
is 
.
,
is 
.
of 
,
,
is 
.
equals 
and twice that for the 
.
is real, which is clear since it is equal to its conjugate.
and 
are tangent to 
and 
are also tangent to 
.
lie on the radical axis of these two circles and they are collinear.
be the 
and that 
via angle chasing, so 
.
be the circumcenter of 
.
are isogonal conjugates, a bit of angle chasing shows that 
and 
,
,
.
.
be the foot of the perpendicular from 
is cyclic, so 
.
,
.
,
are collinear as desired.
be the harmonic conjugate of 
and 
be the reflection of 
is tangent to 
implying the conclusion.
are tangent to 
Thus 
is a harmonic quadrilateral, and 
so 
are collinear and 
implies 
collinear, as desired.
![[asy]
size(7cm);
pair A = dir(130);
pair B = dir(200);
pair C = dir(-20);
pair F = foot(C,A,B);
pair E = foot(B,A,C);
pair H = extension(C,F,B,E);
pair O = (0,0);
pair P = extension(A,O,B,E);
pair Q = extension(A,O,C,F);
pair M = (B+C)/2;
pair O_1 = circumcenter(H,P,Q);
pair H_1 = orthocenter(H,P,Q);
draw(A--B--C--cycle, red);
draw(arc(circumcenter(A,P,B), abs(circumcenter(A,P,B)-A), -110, 80), lightblue);
draw(arc(circumcenter(A,Q,C), abs(circumcenter(A,Q,C)-A), 190, 280), lightblue);
pair D = foot(A,B,C);
draw(A--Q, orange);
draw(B--E^^C--F^^A--D, orange);
draw(circumcircle(A,B,C), blue);
draw(O_1--P^^O_1--Q, orange);
draw(A--M, mediumred+dotted);
draw(circumcircle(H,P,Q), heavygreen);
draw(P--H_1, orange);
dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$D$", D, S);
dot("$O$", O, NE);
dot("$H$", H, S);
dot("$P$", P, W);
dot("$Q$", Q, S);
dot("$E$", E, NE);
dot("$F$", F, W);
dot("$O'$", O_1, SW);
dot("$H'$", H_1, S);
dot("$O'$", O_1, SW);
dot("$M$", M, S);
[/asy]](http://latex.artofproblemsolving.com/c/d/c/cdcf5e2bf0f15394d41c1214cbeeb601f0f7c45d.png)
and 
be the foots of the altitudes from 
and 
so 
.
and 
.
and 
,
and 
are both tangent to 
and 
.
and 
are tangent to ![\[\angle O'PA = 180^\circ - \angle OPQ = 180^\circ - \angle HPH' = 90^\circ + \angle BAC \]](http://latex.artofproblemsolving.com/4/d/d/4dd57d39e791eec39d2624b8b9b4c89393ee31ec.png)
but 
showing that 
so by Lemma, 
refers to directed angles.
.
Similarly, 
.![\[\measuredangle AHQ = \measuredangle AHB + \measuredangle PHQ = -\measuredangle ACB - \measuredangle BAC = -\measuredangle ACB = \measuredangle HPQ.\]](http://latex.artofproblemsolving.com/3/9/8/39808eb41ac3f730f771683d5cea4373e2d2cae0.png)
such that 
is tangent to 
.
.
be the spiral similarity that maps 
are concyclic.![\[\measuredangle OAM = \measuredangle PAK = - \measuredangle BTO = \measuredangle OTM.\]](http://latex.artofproblemsolving.com/3/8/3/3830cf51ddc29a40c66d79732c75110b78c4abb3.png)
.
.
; remark 
and 
thus 
through 
we have 
but if we denotes the 
we deduce 
hence 
are colinear .
is the harmonic conjugate of 
with respect to 
be the orthic triangle, and let 
and are oppositely oriented.
Then,
In a similar way, 
.
is tangent to 
and the Tangent-Chord theorem finishes.
,
.
with 
.
,
be the circumcenter of triangle 
and 
respectively, it suffices to show that 
.
,
,
implies 
concyclic. Similarly, if 
,
are concyclic. But now 
implies 
and similarly 
,
is a parallelogram, 
is the 
.
and 
as desired.
and 
are tangent to 
)
and 
gives us that 
.
(inversely oriented), so 
is a symmedian in 
,
passes through the circumcenter of 
.
is cyclic, hence 
so ![[asy]
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pen zzttqq = rgb(0.6,0.2,0.);
draw((9.043957767876451,1.5088386351661338)--(7.,-5.)--(15.,-5.)--cycle, linewidth(0.8) + zzttqq);
/* draw figures */
draw((9.043957767876451,1.5088386351661338)--(7.,-5.), linewidth(0.8) + zzttqq);
draw((7.,-5.)--(15.,-5.), linewidth(0.8) + zzttqq);
draw((15.,-5.)--(9.043957767876451,1.5088386351661338), linewidth(0.8) + zzttqq);
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draw((7.718094898849119,-2.713277693501062)--(15.,-5.), linewidth(0.8));
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draw((9.043957767876451,1.5088386351661338)--(11.581626409158005,-3.926534878779204), linewidth(0.8));
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draw((-4.897172507611289,-5.)--(7.,-5.), linewidth(0.8));
draw((9.043957767876451,1.5088386351661338)--(-4.897172507611289,-5.), linewidth(0.8));
draw((-4.897172507611289,-5.)--(11.,-2.680762783903952), linewidth(0.8));
draw((9.043957767876451,1.5088386351661338)--(11.,-5.), linewidth(0.8));
draw((9.043957767876451,1.5088386351661338)--(9.043957767876451,-5.), linewidth(0.8));
draw((11.,-2.680762783903952)--(11.,-5.), linewidth(0.8));
/* dots and labels */
dot((9.043957767876451,1.5088386351661338),dotstyle);
label("$A$", (8.90118908016921,1.7943760105806144), NE * labelscalefactor);
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dot((15.,-5.),dotstyle);
label("$C$", (15.223802392918426,-5.28287179433544), NE * labelscalefactor);
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/* end of picture */[/asy]](http://latex.artofproblemsolving.com/5/a/d/5adf80708b2411c9440e8b771b1ac84d323fb395.png)
and let 
.
and 
,
.
we have that 
and 
.
.
,
we have that 
.
so 
is cyclic.
and 
which proves our desired collinearity.
.
is the radical axis of 
and 
are tangent to 
,
,
and call it 
.
)
,
.
,
,
,
,
,
so we are done.
,
be the midpoints of 
.
and 
,
,
lie on 
and 
,
and 
to meet sides 
Hence by AA similarity our claim holds.
This statement holds as 
and 
which gives us AA similarity.
and 
are tangent to 
we get 
so 
is the radical axis of the two circles.
I claim that 
are tangent to 
is tangent to 
is tangent to 
;
.
is parallel to 
we are done. From LoS to 
and 
we will got 
.
at 
.
gives 
.
and 
.
be the centers of 
respectively. Define 
Notice that 
and 
Therefore 
so 
It follows that 
and 
is cyclic. Thus 
Hence 
are collinear and we are done. QED
denote the 
H
and 
are cyclic.
.![\[\measuredangle PAH_a = \measuredangle PAB + \measuredangle BAH_a = \measuredangle OAB + \measuredangle BAH_a = \measuredangle HBC + \measuredangle CBH_a = \measuredangle PBC + \measuredangle BCH_a = \measuredangle PBH_a\]](http://latex.artofproblemsolving.com/1/3/e/13ea963dcd79deb149a0df2058d4f287d6acaaf0.png)
which implies that points 
and line 
.![\[\measuredangle PHQ = \measuredangle BHC = \measuredangle CAB\]](http://latex.artofproblemsolving.com/b/f/e/bfecf745a9cce33b9cda64f8369248c77c8a322b.png)
and![\[\measuredangle HQP = 90 + \measuredangle BAQ = 90 + \measuredangle BAO = \measuredangle BCA\]](http://latex.artofproblemsolving.com/2/6/c/26ccf78ede5ad9d04006be8b3e653100acc7a9f2.png)
which implies that ![\[\measuredangle O_1PB = \measuredangle O_1PH = \measuredangle ABO = \measuredangle OAB = \measuredangle PAB\]](http://latex.artofproblemsolving.com/6/9/5/6955b8221c7e5ba265f981d28e698e1e7e00ab8c.png)
which implies that 
is indeed tangent to ![\[\text{Pow}_{(ABP)}(O_1) = O_1P^2 = O_1Q^2 = \text{Pow}_{(ACQ)}(O_1)\]](http://latex.artofproblemsolving.com/5/5/8/558ea8e276cd1e96a8364e567754961f76563550.png)
which implies that 
or the 
and that 
,
where 
with letters in that order. In this map, we get that 
,
.![\[ \measuredangle MAO = \measuredangle XAP = -\measuredangle OTB = \measuredangle MTO. \]](http://latex.artofproblemsolving.com/a/1/f/a1fa78a9d82118a8dfb922c98d63c9da4f3519bc.png)
Which implies that 
holds true to tangency and thus 
,
and i proved with Humpty - Point.Let 
be the circumcenter of 
and similarly , 
.
On the other hand 
and this implies 
and it's easy to see that information implies 
is cyclic , and similarly , 
implies 
is cyclic.
this means 
is tangent to 
.
this means 
is tangent to 
.
by PoP the powers of 
i.e, 
