[/quote]
,
be a positive integer, and let 
be nonnegative integers such that 
Prove that![\[
\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.
\]](http://latex.artofproblemsolving.com/c/4/9/c49d0ccfef69402d05dc44d721172498c605cf7e.png)
Note: 
for all nonnegative integers 
.
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
and write the numbers 
down on a board. A move consists in erasing any two numbers 
and 
,
and 
on the board, and then removing repetitions (e.g., if the board contained the numbers 
,
and 
,
)
with real coefficients of degree 
)![\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]](http://latex.artofproblemsolving.com/6/5/e/65e5e4eefcadd5d0019b4929b3472cdf203c13aa.png)
has no real roots.
and 
,
,
the pair 
if for any elements 
we have 
.
be the greatest possible number of elements of a set that escapes the pair 
.
with incircle 
be the incenter and 
be the point where 
.
with 
and 
be the orthocenter of 
,
on 
also satisfies 
.
,
satisfying 
the following inequality holds:
of the triangle 
the orthogonal projection of 
on 
.
and 
are such that 
is a paralellogram, find the locus of point 
when 
.
be the circle tangent to 
passing through 
be the circle tangent to 
be the circle tangent to 
passing through 
.
its circumcenter and 
its circumcircle. Let 
and 
be points on 
,
.
,
.
.
and the tangent to 
are concurrent and that 
.
Y by Davi-8191, PiMath12345, phoenixfire, super.shamik, sotpidot, HamstPan38825, son7, icematrix2, HWenslawski, samrocksnature, Jc426, Instance, math31415926535, mathematicsy, megarnie, centslordm, Ruy, Lamboreghini, nguyenducmanh2705, Stuffybear, Adventure10, Mango247, A21, Rounak_iitr
Quadrilateral 
is inscribed in circle 
with 
and 
. Let 
be a variable point on segment 
. Line 
meets again at 
(other than 
). Point 
lies on arc 
of such that 
is perpendicular to 
. Let 
denote the midpoint of chord 
. As varies on segment , show that moves along a circle.
is inscribed in circle 
with 
and 
. Let 
be a variable point on segment 
. Line 
meets again at 
(other than 
). Point 
lies on arc 
of such that 
is perpendicular to 
. Let 
denote the midpoint of chord 
. As varies on segment , show that moves along a circle.This post has been edited 2 times. Last edited by djmathman, Apr 29, 2015, 12:36 AM
made something interesting, but I am not too sure.
,
and 
are concyclic?
,
,
.
and 
be on the unit circle. We claim 
is the center. It follows from standard formulas that ![\[ x = \frac 12 \left( s + t - 1 + s/t \right) \]](http://latex.artofproblemsolving.com/a/a/0/aa0ca8b1326d61191d60ef17474baa3d7af51d15.png)
thus ![\[ 4\operatorname{Re} x + 2 = s + t + \frac 1s + \frac 1t + \frac st + \frac ts \]](http://latex.artofproblemsolving.com/7/9/d/79d090a92c6969e4472777dcb102c7c3fec4a0f4.png)
which depends only on 
,![\[ 4\left\lvert z-\frac{s+t}{2} \right\rvert^2 = \left\lvert s+t+1 \right\rvert^2 = 3 + (4\operatorname{Re}x+2) \]](http://latex.artofproblemsolving.com/4/c/d/4cdfc896ae4d92513bbdbe8a8abf359daf0d73f5.png)
does not depend on 
is the entire difficulty of the problem -- it's totally unexpected that the center doesn't even depend on 
is fixed. So if 
is the midpoint of 
then 
is fixed, i.e. the power of 
of 
is fixed. Let 
be the radius of 
.
.
.
is the centroid of 
with 
.
.
centered at 
thing, so I didn't write it down.
where 
but didn't explicity show that 
where 
.
depends only on the real part of 
'
.
is cyclic by shooting lemma and has center 
.
is fixed, and hence 
,
is independent of 
and 
is fixed, 
is the nine-point center of 
,
as the midpoint of ![\[AX\cdot AS'=AN_9^2-\frac{R^2}{4},\]](http://latex.artofproblemsolving.com/3/8/5/3855e2026fb039cb2a3c60e60fe30d02d2bb33e4.png)
![\[\frac{AX\cdot AS}{2}=AN_9^2-\frac{R^2}{4},\]](http://latex.artofproblemsolving.com/e/0/6/e065d302ae6605caed7773ea4d1464e6938b5136.png)
![\[AN_9^2=\frac{AY^2-YO^2}{2}+\frac{3R^2}{4}.\]](http://latex.artofproblemsolving.com/4/4/b/44bca5cf5146c8072f0f114bbea248b032315c31.png)
Thus, 
does not depend on the position of 
.
.
.
respectively, and let 
,
,
by 
,
,![\[\text{Pow}_\Omega(M)-\text{Pow}_\omega(M)=\frac{\text{Pow}_\Omega(S)+\text{Pow}_\Omega(T)}{2},\]](http://latex.artofproblemsolving.com/d/a/2/da24b444331bda321778590f3d7348c1572cd7d2.png)
because 
.
.
at 
intersect 
and 
.
is constant. The rest is direct by Pythagoras:
since 
,
.
by Power of a Point. But 
by Pythagoras, and adding 
leaves 
,
.
,
is the nine-point circle of 
.
which is a fixed quantity. 
,
is the midpoint of 
is a parallelogram. We have 
as 
,
due to a homothety at 
.
is a fixed quantity, meaning 
![[asy]
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label("$T$", T[1], dir(270));
label("$Y$", Y, dir(90));
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label("$N$", N, SE);
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label("$K$", K, dir(90));
dot(O);
dot(A);
dot(P);
dot(B);
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[/asy]](http://latex.artofproblemsolving.com/e/9/6/e96fe26fe3add0170fad4ba7ca0b13d346e09e91.png)
.
we have that 
is orthogonal to 
.
,
is constant i.e. 
.
the locus of all points s.t. the powers w.r.t. the 
be points so that 
.
not lying on 
be 
at 
,
.
are on the same side of 
so that 
is a harmonic quadrilateral. Show that as ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/3/c/e/3ceca8ce94034c245bbe0cfc8ee348db99e40b69.png)
is a fixed line independent of 
at 
,
are collinear. Let 
intersect 
at 
.
,
.
are fixed lines, so it is enough to show that 
is a fixed line. Inspecting the diagram it appears as if 
,
,
is an isosceles trapezium)
so that 
.
intersect at 
.
.
,
.
,
.
thus 
.
,
And we are through. 
and 
.
and 
be the intersections of the line perpendicular to 
.
be an acute scalene triangle with 
and 
being the midpoints of 
the intersection of 
with 
,
and 
be the midpoints of sides 
and 
respectively. Show that the points, ![\[\measuredangle DAX = \measuredangle H_AAX = \measuredangle H_AA'A = \measuredangle QDH_A = \measuredangle XDA\]](http://latex.artofproblemsolving.com/f/2/0/f2022cfe56288aab0d58fdcc4a131c3efabfee0c.png)
so 
.
.
are tangent at ![\[XP \cdot XQ = XA^2 = XM_B \cdot XM_C\]](http://latex.artofproblemsolving.com/3/0/e/30e5b6f8eaa98a2aa79625c14a46933b9e3c09e0.png)
so quadrilateral 
is indeed cyclic.
.
since this is the 
m
and 
be the reflection of 
so it is also the midpoint of 
.
as it is the 
, 
and 
concur , at ![\[4NM_P \cdot NM_Q = DP \cdot DQ = DB \cdot DC = D'B \cdot D'C = 4NM_C \cdot NM_B\]](http://latex.artofproblemsolving.com/e/d/1/ed15e8144dca887146ecc3390afb86842be16aa6.png)
so quadrilateral 
is cyclic.
being the unit circle. and 
,
.
.
must have a common real part, in other words 
for some constant 
,
. Putting in our value of 
Now we wish to find the value of 
is fixed, so thus 
is fixed. Expanding gives 
Now, for this value to not depend on 
ergo 
.
.
be the feet from 
to the opposite sides. Note by a simple angle chase 
be 
.
is cyclic, as in the problem taking 
give two points from the locus.
is cyclic, as if 
,
by power of a point on nine point circle, semicircle on 
cyclic. We do this by forgotten coaxiality lemma.
and 
,
which is negative. The ratio is 
.
etc, so we're done as then 
,
,
and 
.
is constant, then 
is constant.![\[\frac{1}{2} (s + t + 1 - s \overline{t}) = x;\]](http://latex.artofproblemsolving.com/e/3/c/e3c9eb469a01d68de66451f2f9046c31e33d9f06.png)
therefore
where 
.
which is constant, as desired.
point circle of 
and 
where 
.
.![\[Pow_{N_9}(A) = AN_9^2 - \frac{R^2}{4} = AX\cdot AM_2 = \frac{AX \cdot AS}{2} = \frac{AQ^2}{2}\]](http://latex.artofproblemsolving.com/b/e/7/be7f4066dcbddb4bad80ad17b5cc66b771e50c4f.png)
Therefore, 
and 
passing through 
,
and 
and 
.
is cyclic.
,
and 
.
,
.
always bisects 
.
,
bisects 
and 
,
is an isosceles trapezoid. Therefore, the perpendicular bisector of 
be altitudes of 
.
is cyclic.
as 
is cyclic since its the nine point circle of 
is cyclic with diameter 
and by power of point 
.
and 
,
is cyclic where 
,
in which 
.
