[/quote]
,
such that for every 
is satisfied? Explain the answer.
be a random point on the smaller arc 
,
be the intersection point of segments 
and 
.
are 
is the center of the square. Prove that points 
,
and 
such that
such that: 
be an integer, and let 
be positive real numbers such that 
.![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
colors, there exists a monochromatic regular hexagon.
is the smallest number such that if 
,
with two colors there exists a monochromatic arithmetic progression of length 
.
.
and 
.
points in plane. Prove that the number of isosceles triangles having their vertices among these 
.
.
and 
,
denotes the sum of all positive divisors of 
denotes the largest prime divisor of 
be a polynomial with real coefficients.
.
cannot have 2024 real roots.
and 
.
for all real numbers 
.
grid, one can find 
points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of 
!
denote the set of positive integers. Find all functions 
satisfying the condition
for all 
be a scalene triangle, 
its medial triangle, and 
)
).
be the circumcevian triangle of 
be the pedal triangle of 
be the reflection in 
of 
.
,
cyclically.
be the reflection in 
of 
,
cyclically.
be the reflection in 
of 
,
cyclically.
,
,
and the orthocentroidal circle of 
Y by babu2001, JustKeepRunning, Illuzion, Wizard_32, TheMathLife, anantmudgal09, Mr.Chagol, e_plus_pi, char2539, TheDarkPrince, MathInfinite, PCM_noob, FadingMoonlight, MathPassionForever, Euler1728, rmtf1111, Davrbek, Generic_Username, Fermat_Theorem, samuel, Carpemath, Mathman12334, wx_family2, Kayak, fjm30, buratinogigle, v_Enhance, bel.jad5, rkm0959, Satyanweshi_math1729, HolyMath, GeoMetrix, Drunken_Master, SHREYAS333, mathleticguyyy, Arhaan, amar_04, Supercali, EulersTurban, lilavati_2005, Polynom_Efendi, Aryan-23, Feridimo, NJOY, Pi-is-3, AmirKhusrau, kc5170, Purple_Planet, donfluffles1412, Atpar, Inconsistent, IAmTheHazard, psi241, srijonrick, Bassiskicking, OlympusHero, Kanep, TheMathCruncher_007, sotpidot, megarnie, centslordm, tiendung2006, rayfish, magicarrow, crazyeyemoody907, GeoKing, Jalil_Huseynov, mathmax12, EpicBird08, Lamboreghini, Adventure10, NO_SQUARES, CrimsonBlade273, Rounak_iitr, Eka01, Sedro, Funcshun840, iamnotgentle, wizixez, H_Taken
Let 
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and 
at 
and 
, respectively. The line through 
perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.
Prove that lines
and 
meet on the line through 
perpendicular to 
.
Proposed by Anant Mudgal, India
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and 
at 
and 
, respectively. The line through 
perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.Prove that lines
and 
meet on the line through 
perpendicular to 
.Proposed by Anant Mudgal, India
This post has been edited 2 times. Last edited by djmathman, Jul 17, 2019, 12:31 PM
so 
is cyclic. Now let 
.
are similar. Both quadrilaterals are cyclic, so it suffices to show the measures of the corresponding arcs on the two circles are equal. Clearly 
are both equal to 
,
are also equal. Also, 
and similarly 
,
with 
;
are collinear. Let 
be the antipode to 
be the midpoints of 
.
is obviously harmonic, so similarity implies 
is too. It's well-known that 
are collinear (say, by projecting harmonic bundle 
to 
and sends 
to the midpoint 
of 
.
be the foot of the projection from 
correspond in the inversion.
are concyclic. Let 
;
is cyclic with diameter 
,
.
are midpoints of 
this implies 
so 
.
so 
,
and we're done.
,
,
.
,
and so ![\[ P = \frac{A-R}{1-R\overline{A}} = \frac{\frac{2yz}{y+z} + \frac{yz}{x}} {1 + \frac{yz}{x} \cdot \frac{2}{y+z}} = \frac{yz(2x+y+z)}{2yz+x(y+z)}. \]](http://latex.artofproblemsolving.com/3/3/4/334af39b32b7748631b7bcd7cdcbb12f74e30fd5.png)
We now compute 
Similarly ![\[ O_C = \frac{x-y}{x+y} \cdot \frac{yz(2x+y+z)}{(z-y)(x+z)}. \]](http://latex.artofproblemsolving.com/2/1/f/21f7cb46c9a09d01deddba9c98c46019e0c94ef5.png)
Therefore, subtraction gives ![\[ O_B-O_C = \frac{yz(2x+y+z)}{(x+y)(x+z)(y-z)} \left[ (x-z)+(x-y) \right] = \frac{yz(2x+y+z)(2x-y-z)}{(x+y)(x+z)(z-y)}. \]](http://latex.artofproblemsolving.com/5/1/7/517f3a25ea01049617e918f3c1aa9d31c6538180.png)
It remains to compute 
.
we have 
so 
.
Thus 
This gives 
as needed.
.
is cyclic. It allows us to make next
are any points on sides 
.
,
.
,
.
.
pass through mixtilinear touchpoint 
.
be the reflection of 
is 
are concyclic.
again at 
.
so 
.
so 
.
,
.
which means 
.
be the center of 
.
.
, it suffices to show that
We compute the first term first, note that
For the second term, we use Power of Point twice.
And the last term note that 
so,
Thus multiplying all together, the resulting terms
as 
,
intersect 
,
intersect 
,
,
,
,
and 
is isoceles, and we can prove 
,
,
,
is a rhombus
is and isoceles trapezoid, and that 
and 
are congruent, and from that we can prove that 
the unit circle, where 
represent 
.
and cyclic permutations. Let 
for some real constant 
must be pure imaginary,
So 
.
It follows that the circumcenter 
of 
is located at
Intermediate steps are omitted so the reader does not need to suffer the factorization process like I did. Then we have 
It will suffice to show that 
is pure imaginary. In particular, 
So it suffices that the ratio 
which is clear as the expression equals the negative of its own conjugate.
.
and 
are similar. Since 
symmedian of 
,
symmedian of 
,
.
.
.
to 
Queue point of 
,
,
![[asy]
import geometry;
size(10cm);
defaultpen(fontsize(9pt));
pen pri; pri=RGB(31, 191, 184);
pen sec; sec=RGB(5, 113, 108);
pen tri; tri=RGB(25, 120, 165);
pen qua; qua=RGB(3, 17, 99);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pen qfil=invisible;
pair O, A, B, C, I, D, E, F, R, P, Q, K, D_prime;
O=(0, 0);
A=dir(125);
B=dir(220);
C=dir(320);
I=incenter(A, B, C);
D=foot(I,B,C);
E=foot(I,A,C);
F=foot(I,A,B);
R=2*foot(I,D,foot(D,E,F))-D;
P=2*foot(I,A,R)-R;
Q=intersectionpoints(circle(B,F,P), circle(C,E,P))[1];
K=extension(D,I,P,Q);
D_prime=2I-D;
fill(D--E--F--cycle,sfil);
filldraw((path)(A--B--C--cycle), white+0.1*pri, pri);
filldraw(circumcircle(B, P, F), tfil, tri);
filldraw(circumcircle(C, E, P), tfil, tri);
filldraw(circumcircle(B, I, C), sfil, sec);
filldraw(circumcircle(A, B, C), fil,pri);
filldraw(incircle(A, B, C), fil, pri);
draw(R--D,tri+dashed);
draw(P--K,tri+dashed);
draw(D--K,tri+dashed);
draw(A--K,tri+dashed);
dot("$A$",A,dir(110));
dot("$B$",B,dir(200));
dot("$C$",C,dir(-15));
dot("$K$",K,dir(-5));
dot("$I$",I,dir(-30));
dot("$D$",D,S);
dot("$E$",E,dir(75));
dot("$F$",F,dir(130));
dot("$R$",R,NW);
dot("$P$",P,dir(120));
dot("$Q$",Q,dir(-20));
dot("$D'$",D_prime,dir(45));
[/asy]](http://latex.artofproblemsolving.com/a/1/4/a1476ee173037fd0cbdfb02c95b44d0784b4bc5e.png)
which proves the claim. 
and 
are isogonal.
and 
where 
lies on 
.
we observe 
,![[asy]
import geometry;
size(7cm);
defaultpen(fontsize(9pt));
pen pri; pri=RGB(31, 191, 184);
pen sec; sec=RGB(5, 113, 108);
pen tri; tri=RGB(25, 120, 165);
pen qua; qua=RGB(3, 17, 99);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pen qfil=invisible;
pair I, D, E, F, A, B, C, X, Y, P, Q, D_prime, K;
I=(0, 0);
D=dir(110);
E=dir(220);
F=dir(320);
A=((E+F)/2);
B=((D+F)/2);
C=((D+E)/2);
X=((D+A)/2);
Y=intersectionpoints(circle(D,E,F),line(D, A))[0];
P=intersectionpoints(circle(X,abs(X-D)), circle(D, E, F))[1];
Q=intersectionpoints(circle(P, E, C), circle(B, P, F))[0];
D_prime=2I-D;
K=foot(A,D,D_prime);
filldraw((path)(D--E--F--cycle), white+0.1*pri, pri);
filldraw(circumcircle(D, E, F), sfil, pri);
filldraw(circumcircle(P,A,D), sfil, tri);
filldraw(circumcircle(P,Q,Y), sfil, tri);
draw(circumcircle(P,Q,C), dashed+sec);
clipdraw(circumcircle(P,Q,B), dashed+sec);
draw(D--Y,tri);
draw(Q--B,tri);
draw(D--D_prime,tri);
draw(A--K,tri);
draw(D_prime--P,tri+dashed);
dot("$A^*$",A,dir(-120));
dot("$B^*$",B,dir(40));
dot("$C^*$",C,dir(-110));
dot("$I$",I,dir(20));
dot("$D$",D,N);
dot("$E$",E,dir(-100));
dot("$F$",F,dir(-40));
dot("$P$",P,dir(130));
dot("$Q^*$",Q,dir(W));
dot("$X$",X,dir(240));
dot("$Y$",Y,dir(-40));
dot("$D'$",D_prime,dir(-50));
dot("$K$",K,dir(-15));
[/asy]](http://latex.artofproblemsolving.com/d/2/5/d25b23e501c5932b5b5849bfcd2d5dc7a923646b.png)
.
is cyclic. This follows as,
which proves the claim. 
and 
,
and 
.
by homothety.
.
as,
where we have 
as both 
because 
because,
so we have proven the claim. 
from the picture.
lies on 
.
and thus the claim is proven. 
implies 
are concurrent.
,
,
.
and so on. Define 
,
is power of point 
is linear function, we have 
.
.
.
.
.
and 
.
and similar equality with segment 
.
,
,
is harmonic quadrilateral, so 
and it ie enough to prove that 
or 
,
)
and 
is orthogonal 
are concyclic
are concyclic, 
are concyclic
are concyclic
are concyclic
and 
are concyclic (this is obvious)
are concyclic
be 
,
as 
.
(so the antipode of 
,
,
.
and 
.
collinear.
we get 
.
and 
are symmetric about line 
is a median so 
is a symmedian. This implies that 
by projecting 
onto 
is tangent to 
is the bisector, and 
from 
,
.
and 
denote the inverses of 
,
and the foot from 
be the midpoints of 
,
,
and 
are cyclic. Then, ![\[\measuredangle PQB'=\measuredangle PFB'=\measuredangle PFD=\measuredangle PED=\measuredangle PEC'=\measuredangle PQC'\]](http://latex.artofproblemsolving.com/7/d/5/7d53afe464cf8f392bc13c4b1d1241c20291df72.png)
so 
collinear. Also, 
so 
collinear. A homothety at 
to 
as both are perpendicular to 
as desired.
. We now invert with respect to the incircle and restate in terms of 
with center 
.
such that 
,
is cyclic.
is a cyclic quadrilateral with diameter 
are cyclic and finally even more angle chasing shows that 
is cyclic.
,
are collinear.
is median on 
and 
is 
s![\[\measuredangle KPE=\measuredangle KDE=\frac{\measuredangle C}{2}=\measuredangle FPA=\measuredangle MPE\]](http://latex.artofproblemsolving.com/0/5/b/05b00df055c33cbfb4b882400cc257b0ddcde346.png)
Hence 
.
respectively. 
is the foot of the altitude from 
are the midpoints of 
respectively and 
intersects 
intersects 
and 
.
then 
are concyclic.
.![\[\measuredangle PQF=\measuredangle PBF=\measuredangle PBA=\measuredangle PCA=\measuredangle PCE=\measuredangle PQE\]](http://latex.artofproblemsolving.com/3/a/b/3ab100d48e1465aefe067c10faa4a634e00f7093.png)
Thus, 
are collinear
are collinear.
.![\[\measuredangle T'QF=\measuredangle PQF=\measuredangle PBA=\measuredangle PT'A\]](http://latex.artofproblemsolving.com/3/4/4/344fe4afd13bd5161925118bf501e06784fc8762.png)
This yields 
.
is the angle bisector of 
.
and 
.![\[-1\overset{?}{=}(Q(XY)_{\infty},\overline{QFE};\overline{QPT},QH)\overset{XY}{=}(XY_{\infty},Y;X,H)\]](http://latex.artofproblemsolving.com/8/3/c/83ca7efc560668cd246233c272a5deeee7b3796c.png)
.
with respect to 
.
and 
are collinear.
we get that 
are concyclic. Also 
and 
are isosceles trapezoids. By 
we have![\[\measuredangle PTA=\measuredangle PSA=\measuredangle DSH=\measuredangle DT'H=\measuredangle H'TD'=\measuredangle H'TA\]](http://latex.artofproblemsolving.com/b/6/2/b623e0b22f4d9986b77af2edd01161b53b7ddfab.png)
This yields 
are collinear. Hence 
.![\[\measuredangle PQH=2\measuredangle PQF=2\measuredangle PBA=\measuredangle POA=180-\measuredangle HOP\]](http://latex.artofproblemsolving.com/f/4/c/f4c85ad767bfceb98998cd575ecaeea57255cb33.png)
As desired
is cyclic. Points 
,![\[\angle PQC = \angle PEC = \angle PFB = \angle PQB.\]](http://latex.artofproblemsolving.com/c/c/d/ccddb8c57355e35d227a450ab2f6f3bb697f1186.png)
Now, 
finishes.
meet 
and we will also prove 
.
are isogonal so it suffices for ![\[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle BAT)}{\sin(\angle CAT)}=\frac{\sin(\angle CAS)}{\sin(\angle BAS)}\]](http://latex.artofproblemsolving.com/4/b/4/4b4a27edf7a58841aa19db32fd02dbe628b14798.png)
By ratio lemma ![\[\frac{\sin(\angle CAS)}{\sin(\angle BAS)}=\frac{CS}{BS}\cdot\frac{AB}{AC}=\frac{p-b}{p-c}\cdot\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/0/3/a/03adbb3514a34778a73cad428ba35bb8e588c2c2.png)
By trig ceva ![\[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle AFR)}{\sin(\angle RFE)}\cdot\frac{\sin(\angle FER)}{\sin(\angle REA)}=\left(\frac{\sin(\angle FDR)}{\sin(\angle EDR)}\right)^2=\left(\frac{\sin C/2}{\sin B/2}\right)^2\]](http://latex.artofproblemsolving.com/f/8/8/f88c5c46cc6cad9322b8c9169f2513c3a20264f6.png)
One can easily check that these are equal so we get ![\[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle EDF= \measuredangle BIC\]](http://latex.artofproblemsolving.com/9/a/1/9a151055dc9fcb16dbd1396072cd9b4768353466.png)
So 
and 
are the midpoints of 
,
is the midpoint of 
and 
are the intersections of 
with the circle with diameter 
and 
,
and 
is cyclic. Now look at the problem from the perspective of 
is the nine-point circle. The problem translates to 
being cyclic.
.
.
(which is centered at 
and since 
is a parallelogram, we get that 
we have 
so 
is an isosceles trapezoid, which suffices.
so 
is actually the perpendicular bisector of 
.![\[\measuredangle DIM'=\measuredangle M'IP= \measuredangle K'IN'\]](http://latex.artofproblemsolving.com/1/0/b/10bca1fa7c33aafa0df0ba52aa5feecccf6a4750.png)
And since 
we get that 
is cyclic (well known lemma). 
.
and ![\[\measuredangle PDK'=\measuredangle PDE+\measuredangle EDK'=\measuredangle PDE+90^\circ - \measuredangle DFE=90^\circ-\measuredangle PED\]](http://latex.artofproblemsolving.com/6/7/c/67ca8eb9d62cb4626cf8e1ffa7eb42fa3cd4d7e2.png)
Therefore, using that 
we get ![\[\measuredangle M'K'P=90^\circ-\measuredangle PDK'=\measuredangle PED=\measuredangle C'Q'P. \ \ \ \blacksquare\]](http://latex.artofproblemsolving.com/7/9/6/796dcf4d72f6b93d2114c454169357d475de2b19.png)
be the reflection of 
be the feet from 
be the intersections of 
with 
be the second intersections of 
with 
and 
.
are collinear. Next, 
by Iran lemma, and by symmetry 
.
,
is cyclic and 
.
.
lie on the circle with diameter 
.
by Iran lemma, so by symmetry 
and thus ![\[\measuredangle PZP_C=\measuredangle PZE=\measuredangle PFE=\measuredangle PX_CP_C\]](http://latex.artofproblemsolving.com/0/7/4/0744b511398448a78df97c554ff20a074fd74f65.png)
as desired.![\[FR_C\cdot FE=FP\cdot FX_C=FP_B\cdot FZ=EP_B\cdot ED=EQ_C\cdot EF,\]](http://latex.artofproblemsolving.com/5/d/b/5db577bfc3dbb01c65f64fb06fd577a38c236bae.png)
since 
by Iran lemma. This implies 
,
.
.
and line 
and 
we get the involution is reflection over 
must also pass through 
be the circular points at infinity. Construct the cubic 
through 
.![\[((BDIF\mathcal I\mathcal J)\cup\overline{AEC})\cap((CDIE\mathcal I\mathcal J)\cup\overline{AFB})\]](http://latex.artofproblemsolving.com/c/2/5/c25d0bd6a0f20204cc7284f2a0ecabbe865968e0.png)
we see 
.![\[B+F+P+\mathcal I+\mathcal J=P+\mathcal I+\mathcal J-A=C+E+P+\mathcal I+\mathcal J\]](http://latex.artofproblemsolving.com/5/b/f/5bfb8e8551ad9a3c0c3f8a1cc540106739e376f7.png)
we get 
.![\[B+F+\mathcal I+\mathcal J+P+Q=0=B+F+\mathcal I+\mathcal J+D+I,\]](http://latex.artofproblemsolving.com/b/0/4/b0481a4c3a0a99f977680c245612e1e090013cbb.png)
we get 
.![\[B+D+I+F+\mathcal I+\mathcal J=0=B+F+A+D+I+X,\]](http://latex.artofproblemsolving.com/4/2/d/42ddd1fdd440494a7626196f96ef07f23842c51e.png)
so 
.![\[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle FEP+\measuredangle PFE=\measuredangle FPE=\measuredangle FDE=\measuredangle BIC,\]](http://latex.artofproblemsolving.com/1/e/9/1e904e132621d75b7722cbf2060bedf811edf03c.png)
so 
.
are coconic, 
,![\[E+F=\mathcal I+\mathcal J=A+X.\]](http://latex.artofproblemsolving.com/8/5/a/85a571df813ac39063b1512e6d2e3cb903f0956d.png)
Thus 
intersect on the line at infinity, as desired.
Let 
Let 
and let 
We start with a preliminary claim:
and 
then 
Let 
be the line through 
and let 
It is clear that 
Projecting this harmonic bundle through 
so 
passes through 
are collinear.
and 
say 
gets sent to the second intersection of 
say 
Similarly define 
intersects 
then 
intersects 
at 
then 
are collinear.
and 
are cyclic and 
be the 
are concyclic, and let 
Redefine 
with 
We will show that this 
is cyclic.
Clearly 
Finally, say by the incircle diameter lemma, we see that 
is tangent to 
This proves our claim.
is cyclic.
is tangent to 
The circle 
This if 
then 
Therefore, our inverted claim is as follows:
such that 
Let 
Let 
be the midpoints of 
respectively, Prove that 
is cyclic.
the claim thus follows by Reim's Theorem.
is cyclic. This equates our two definitions of 
are collinear.
so symmedian properties give 
But 
and 
Therefore, 
and we are done!
.
and the antipode of 
.
.
where 
is the reflection of 
and 
then by angle chasing we have 
So 
is cyclic. Also by PoP we have 
Which gives 
is cyclic and so 
.
is cyclic. Let 
and 
.
and 
.
which would imply the result.
and 
which gives 
is a parallelogram and so 
.
.
.
And 
so we must prove that 
and this is true since
And we are done.
