[/quote]
,
be a triangle with circumcircle 
and 
respectively denote the midpoints of the arcs 
that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
including 
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
be a circle with centre 
a convex quadrilateral such that each of the segments 
and 
is tangent to 
.
beyond 
,
beyond 
meets 
.
and 
beyond 
and 
,![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]](http://latex.artofproblemsolving.com/8/2/6/8261276053071cc6cc06dda824a883426bf1d4ac.png)
such that for every 
is satisfied? Explain the answer.
be a random point on the smaller arc 
be the intersection point of segments 
.
are 
, where 
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 
!
Y by adityaguharoy, EthanTAoPS, mathematicsy, Adventure10, Mango247, eggymath
Let 
be an isosceles trapezoid with 
. Let 
be the midpoint of 
. Denote by 
and 
the circumcircles of the triangles 
and 
, respectively. Let 
be the crossing point of the tangent to at 
with the tangent to at 
. Prove that 
is tangent to .
Jakob Jurij Snoj, Slovenia
be an isosceles trapezoid with 
. Let 
be the midpoint of 
. Denote by 
and 
the circumcircles of the triangles 
and 
, respectively. Let 
be the crossing point of the tangent to at 
with the tangent to at 
. Prove that 
is tangent to .Jakob Jurij Snoj, Slovenia
This post has been edited 2 times. Last edited by math90, Feb 23, 2019, 3:30 PM
Reason: Thanks #13
Reason: Thanks #13
be the midpoint of 
.
and 
be the tangents to 
and 
be the tangents to 
and 
at 
and 
respetively and let 
be on 
is tangent to 
and 
,
This clearly implies that the pencil 
is harmonic, thus 
and 
are concurrent. Analogously 
and 
lies on both 
be the intersection of tangents from 
.
is tangent to 
which would give from symmetry that 
is tangent to 
and we would be done.
be a triangle and 
such that 
is tangent to 
is a point on 
such that 
.
and 
is tangent to 
.
is tangent to 
and 
so ![\[\frac{AE}{EF}=\frac{CE}{EF} = \frac{DE}{DF}=\frac{DP'}{DF}.\]](http://latex.artofproblemsolving.com/6/9/2/6923fa6d068b7ddab61d407efe1b5a531bc1ce6a.png)
Also, ![\[\angle P'DF = \angle ECF + \angle CFE = \angle AEF.\]](http://latex.artofproblemsolving.com/d/1/2/d127371d7896a3c9de0c4a60b27cb620e21f01ba.png)
Therefore 
so 
and we are done. 
.
.
.
be the intersection of 
.
,
are tangent to 
and 
,
and 
are cyclic trapezoids. We use complex numbers with respect to 
.![[asy] pair C = dir(130); pair E = dir(210); pair D = dir(330); pair F = C*D/E; pair A = 2*E-C; pair B = 2*F-D; filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--D--cycle, lightred); draw(E--F, orange); draw(B--A, orange); draw(C--D, orange); draw(C--A, lightred); draw(B--D, lightred); pair P = 2*E*D/(E+D); draw(E--P--D, deepcyan); draw(E--D, deepcyan); draw(P--A, deepcyan); draw(circumcircle(E, F, A), dotted+deepgreen); pair O = origin; pair M = midpoint(E--D); draw(O--P, dotted+blue);
dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$D$", D, dir(D)); dot("$F$", F, dir(F)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$O$", O, dir(10)); dot(M);
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C = dir 130 E = dir 210 D = dir 330 F = C*D/E A = 2*E-C B = 2*F-D unitcircle 0.1 lightcyan / deepcyan A--B--C--D--cycle lightred E--F orange B--A orange C--D orange C--A lightred B--D lightred P = 2*E*D/(E+D) E--P--D deepcyan E--D deepcyan P--A deepcyan circumcircle E F A dotted deepgreen O = origin R10 M .= midpoint E--D R315 O--P dotted blue
*/ [/asy]](http://latex.artofproblemsolving.com/1/8/a/18a939e1b4d8fa056a652df77ec696d784e2c079.png)
,
,
,
.
as the intersection of the tangents to 
is tangent to ![\[ \frac{F-A}{P-A} = \frac{\frac{xz}{y} - (2y-x)}{\frac{2yz}{y+z} - (2y-x)} = \frac{y+z}{y} \frac{2xz-y(2y-x)}{2yz-(y+z)(2y-x)} = \frac{y+z}{y}. \]](http://latex.artofproblemsolving.com/7/c/a/7ca1114e69fd014d5e68d61d20ec21131333101a.png)
Hence 
equals 
,
.
.
as 
.
are concyclic
.
.
is tangent to 
.
is tangent to 
.
,
tangent to 
.
as unit circle and let 
.
is tangent to 
.
and 
and as 
we have 
so 
So we have 
Now 
and 
Hence we are done.
,
meet 
be the midpoints of 
.
is tangent to 
are collinear. Since 
,
,
. We that 
by spiral similarity, Thus
, but since 
,
.
?
be the intersection of tangents to 
at points 
be the intersection of tangents to 
at points 
.
as the interesection of lines 
and 
are similar (since 
and 
)
,
is cyclic. From 
we obtain that 
.
implies similarity 
and 
.
Analagously, triangles 
and 
similar to 
,
Now, note that
But points 
,
.
.
at 
concur and the tangents to 
at 
concur. But 
and 
are tangent, so the problem follows.
are cyclic. If 
,
.
be the lines tangent to 
are concurrent. In fact, we'll even show that 
concur.
.
and 
is the 
H
.
is the midpoint of 
,
by the Midpoint Theorem. Inverting back, we see that 
is tangent to 
,
concur.
.
are tangent to 
such that 
and 
are isosceles and 
so 
is isosceles trapezoid 
(because 
)
as desired
.
hit 
at 
.
to 
are collinear.
is tangent to 
.
so 
.
is a parallelogram, so 
.
.
is cyclic![\[\measuredangle JFI=\measuredangle PFI=\measuredangle FBA=\measuredangle FDC =\measuredangle FDE+\measuredangle EDC=\measuredangle FEI+\measuredangle JEF=\measuredangle JEI\]](http://latex.artofproblemsolving.com/9/8/4/984d4fded19dda1110f5c780b205e19e4b1663b5.png)
.
and 
.![\[-1=(PJ;FG)\overset{\infty_{FH}}{=}(PI;HE)\overset{F}{=}(FA;HE)\]](http://latex.artofproblemsolving.com/b/d/c/bdc1f0ef7c9ae9f351d24a73363dbf5c6ca1cd95.png)
Hence 
is a symmedian and thus 
and 
and 
.
.
.
.![\[ f(C)-f(D) = (CK \cdot CA - CE \cdot CK) - (DF \cdot DK - 0) = CK \cdot EA - DF \cdot DK= CK \cdot EC - DF \cdot DK = 0, \]](http://latex.artofproblemsolving.com/6/d/e/6de386e1216d6feca8ce05fb24c828ed559e4d94.png)
as desired.
.
is an isosceles trapezoid, the center of 
lies on 
,![\[ \angle PKA = \angle PFA = \angle FEA, \]](http://latex.artofproblemsolving.com/3/2/e/32e8ba8da231a7eda9f2e454212ef386ad717541.png)
which implies the desired parallelism.
.
.
,
,
,
as the unit circle and redefine 
and 
are tangent to 
However clearly we have,
Also we have,
Then dividing we have,
Taking the conjugate this is simply,
and so we find 
proving the tangency.
and 
Then, 
are collinear as:![\[\measuredangle BFE = \measuredangle BAE = \measuredangle ACD = -\measuredangle EFD.\]](http://latex.artofproblemsolving.com/7/1/7/717d28d63b1554fedd4bdaa7ad5fd8872c334eac.png)
as![\[\measuredangle BFE = \measuredangle BAC = \measuredangle BDC = \measuredangle FDC.\]](http://latex.artofproblemsolving.com/1/8/7/1872672f518cbdf135517687d56b7a0c62a49241.png)
as the unit circle.
,
,
We compute
![\[\frac{\frac{1}{e^2} - \frac{1}{cd}}{(\frac{1}{d} + \frac{1}{e})(\frac{1}{c}-\frac{1}{e})} = \frac{cd-e^2}{(e+d)(e-d)} = \frac{a-p}{a-f} \div \frac{e-a}{e-f}.\]](http://latex.artofproblemsolving.com/2/e/1/2e144edf7dc25e65b04f3606de6928498413c573.png)
giving us 
as needed.
is parallel to the real axis. Then, 
.![\[a=2e-c\]](http://latex.artofproblemsolving.com/0/f/8/0f89e6df3032e86bc540119077be8993babb3d74.png)
and,![\[b=-\overline{a}=-\overline{2e-c}=\frac{e-2c}{ce}\]](http://latex.artofproblemsolving.com/b/8/0/b80da8d0030226d1baacd6e82d47f3977fc3b3c3.png)
Let 
)
Now, we wish to show 
.![\[\arg \left(\frac{e-a}{p-a}\right)=\arg\left(\frac{e-b}{a-b}\right)\]](http://latex.artofproblemsolving.com/4/5/b/45b5282e706b028874de9dd08cb80bbbae1c94b3.png)
Thus, we must have 
.
Now, we also have that,
Thus, 
and indeed, 
which implies that 
Redefine 
Since 
we have 
so 
Let 
cyclic so 
but this implies 
are tangent to 
Similarly we get 
are tangent to 
so we are done.
![[asy]
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,
cyclic.
congruent to 
as the unit circle and let 
Now redefine 
and due to reflection 
,
Therefore 
thus 
and let 
.
but notice that 
so 
we similarly get 
,
and the conclusion follows from the quadrilateral 
and noting that 
.
with point 
and 
and 
so LOS in 
and 
gives the desired conclusion.
and 
so 
is cyclic. This gives 
so 
,
.![\[\frac{de(d-f)}{(d+e)(2de-ef-2df+d^2)}\in\mathbb{R}.\]](http://latex.artofproblemsolving.com/3/0/b/30b58adf8f03e2f385864c3349853759e7e214e3.png)
This is not hard to do.
.
,
.
.![\[p'=\frac{2de}{d+e}=\frac{2e}{1+ec}.\]](http://latex.artofproblemsolving.com/4/1/7/417ae11a5454f876e23ac868dc5ddbc2d5412645.png)
Then,![\[P=P'\iff P'\in AA \iff \angle P'AE=\angle ABE,\]](http://latex.artofproblemsolving.com/b/5/8/b586a1cf9dd5d4b6f5de97c2a1d668d8072a9f68.png)
so![\[\frac{\left(\frac{p'-a}{e-a}\right)}{\left(\frac{a-b}{e-b}\right)}=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]](http://latex.artofproblemsolving.com/a/e/2/ae2144bdb801dc46cef362dae3d3d7337d3c9c8a.png)
Conjugating,![\[\left(\overline{\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}}\right)=\frac{\left(\frac{1}{c}\right)\left(\frac{e^2c+e-2c}{e^2c}\right)\left(\frac{-2e^2+ec+1}{e^2c}\right)}{\left(\frac{1+ec}{ec}\right)\left(\frac{e-c}{ec}\right)\left(\frac{2c-e-2e^2c+ec^2}{e^2c^2}\right)},\]](http://latex.artofproblemsolving.com/9/a/b/9ab4459c48b173ef5767c636fe71efa40884090c.png)
![\[=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]](http://latex.artofproblemsolving.com/1/6/d/16dda72c8c369411b68e74350377f435462bf002.png)
They are equal, so 
.
to 
.
and 
of 
and 
and showing with lengths that 
.
as well.
to 
.
,
and 
are similiar, implying 
,
be the center of 
and 
First, I claim that the midpoint of 
is 
Since the perpendicular bisector of 
we know that 
We also have that the distance from 
and 
are the same, and hence 
be 
such that 
We can compute 
Let 
for some 
Then, we have that 
Thus, 
and hence 
is 
Now, define 
are both tangent to 
We know that 
.
We will do this by proving that 
We have that:
Thus, we have that they are perpendicular, and hence we are done.
Then, let 
It follows that the complex numbers representing 
and 
are 
and 
respectively.
is tangent to 
because this will imply that 
First, note that the complex number representing 
Also, since 
Then, in order to show that 
is tangent to 
We have
and we find that 
so we are done. 
