be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
and 
.
.
lie on a circle 
and point 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
.
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 
are chosen uniformly and independently at random on sides 
and 
respectively, of equilateral triangle 
Which of the following intervals contains the probability that the area of 
is less than half the area of 
![$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$](http://latex.artofproblemsolving.com/5/8/7/5874157b23f9d385f707b696eee3d0780fd8a45b.png)
Y by MathRook7817, NaturalSelection, LostDreams, NO_SQUARES
Let 
and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
so let 
.
,
and deal with 
for 
has a coefficient equal to 
.
for all 
,
choices. By pigeonhole, there exists 
such that 
and 
have no 
term, for some 
.
and 
.
Then 
has no 
has no 
.
has no 
has no 
term.
distinct real roots.
,
t
distinct real roots for all 
.
trivial.
consecutive real roots there exists a local extremum, where 
is 
pairs of consecutive real roots so 
distinct real roots, completing the induction step.
has no 
or 
term, so 
are distinct real numbers. Thus 
will still satisfy the problem condition, hence we may WLOG assume 
.
-
as 
.
satisfy the following property: for each subset 
of size 
,
such that 
.
such subsets, by pigeonhole there are two distinct subsets 
and 
with 
.
and 
,
;![\[ p_d (S_1) = p_d (S_2) \implies (r_{k+1}-r_1)p_{d-1}(S_3) = 0 \implies p_{d-1}(S_3) =0 \]](http://latex.artofproblemsolving.com/c/b/b/cbbfb2d692b621810dffe3634cb1fea24c482b11.png)
since 
by assumption. Furthermore, we have![\[ 0 = p_d(S_1) = r_1 p_{d-1}(S_3) + p_d(S_3) \implies p_d(S_3) = 0.\]](http://latex.artofproblemsolving.com/d/6/4/d64110f93f2e662ec5487dffd1379a6a2e4b941c.png)
But it turns out to be impossible to have 
:
If 
we just need to consider a subset of these roots in the 
case to get our contradiction. Thus assume 
we define 
to be the degree 
for all 
Also let 
Then by Vieta's Formulas, we are given that for all 
Suppose WLOG that 
Some algebraic manipulation gives 
This means that the polynomial 
has two consecutive coefficients equal to 
has 
distinct real roots as well, along with two consecutive coefficients which are equal to 
contradiction as there can be no double roots.
roots of 
The set of degree 
Say an equation is \textbf{\emph{type} 
of set 
}![\[\sum_{\stackrel{S\subseteq \{r_1, \dots, r_m\}}{|S| = \ell}}\prod_{r\in S}r = 0\]](http://latex.artofproblemsolving.com/7/3/b/73bcca1583fcb59b43b2f5a551101da8ffaa0f74.png)
In other words, it is a cyclic sum of degree 
if 
is type 
of 
equations. However, there are only 
(
are missing from the first and second equations, respectively. Also assume the degree is 
This gives:
equation of 
must be 0(since otherwise, 
)
we get that the type 
is 0, or:
we are done. The base case is obvious.
;
,
until 
denote the 
varaibles 
,
,
.![\[ \sigma_i(x_1, x_2, \dots, x_n, y) = \sigma_i(x_1, x_2, \dots, x_n) + y\sigma_{i-1}(x_1, x_2, \dots, x_n). \]](http://latex.artofproblemsolving.com/6/9/4/6944d04a4acef711b4e9b50bcce7c8345195cb95.png)
,
with 
such that the corresponding coefficient of 
which is 
,![\[ \sigma_j(r_1, r_2, \dots, r_{n-2}, r_{n-1}) = \sigma_j(r_1, r_2, \dots, r_{n-2}, r_n) = 0. \]](http://latex.artofproblemsolving.com/a/7/2/a725b72faa1ce8fc620365a7f53b5a39ef4f2667.png)
Using the identity above, and assuming all roots are distinct, we obtain that 
,
.![\[ \prod_{i=1}^{n-2} (x-r_i), \]](http://latex.artofproblemsolving.com/4/7/e/47ecb038cb02052b84ce3b94605ebf2c292d4b07.png)
we are done by Descartes' rule of signs (a quick combinatorial argument shows that in fact we can only get up to 
real roots.)
such that 
has no nonreal roots. We can also assume 
case is trivial as the constant term must be nonzero, so fix 
.
.
polynomials 
for integers 
.
,
to have some coefficient equal to ![\[r_{k+1} \in \left\{\alpha_{k-2}, \frac{\alpha_{k-3}}{\alpha_{k-2}}, \dots, \frac{\alpha_0}{\alpha_1}\right\}.\]](http://latex.artofproblemsolving.com/c/4/3/c4391a633b7bd99c9efcdcf06784184cf76623fa.png)
values in this set. Fix the value of 
.
polynomials 
.
and 
,![\[\frac{B_i}{B_{i+1}} = r_{k+1}\]](http://latex.artofproblemsolving.com/d/4/4/d445dd8d7bdd2d71bacb18007e8f3f795ec178cf.png)
is a linear equation in 
.
has at most one solution for 
and 
must have different 
such that 
and 
and rearranging gives 
,
,
forms a geometric series of ratio 
and so 
.
.
distinct roots that have the same coefficient equal to 
t
t
subsets of 
denote the degree 
.
be the set of roots. Then for every subset 
for some 
.
is a family of 
.
in 
sets outside of 
not in 
sets in 
to be the family of 
vanishes. Then there are 
.
such that![\[e_i(r_1,r_2,\dotsc,r_k)=e_i(r_1,r_2,\dotsc,r_{k-1},r_k')=0.\]](http://latex.artofproblemsolving.com/6/5/4/65493474ee6be306e68267a4525f9eb71a6fde6a.png)
Now we expand these by writing![\[e_i(r_1,r_2,\dotsc,r_{k-1})+r_ke_{i-1}(r_1,r_2,\dotsc,r_{k-1})=e_i(r_1,r_2,\dotsc,r_{k-1})+r_k'e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0,\]](http://latex.artofproblemsolving.com/7/7/1/771984e5e44c4450813a08f2be0f9286a3fe4048.png)
and since 
,![\[e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0.\]](http://latex.artofproblemsolving.com/f/4/e/f4e4ba4a4715fe022fdcae81c899b256d7652b1c.png)
is a set of real numbers such that 
for some ![\[e_i(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-1}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=0,\]](http://latex.artofproblemsolving.com/9/f/9/9f9551c3271e058203e79e08edc1d748776bb41f.png)
from which we may deduce that![\[e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2.\]](http://latex.artofproblemsolving.com/c/7/9/c79d3854db78998729783daadb389328df034cb6.png)
,
can vanish, for then two consecutive elementary symmetric polynomials must vanish. By "On the nonnegativity of generalized discriminants of quadratics",![\[e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2\ge \frac{i(k-i)}{(i-1)(k-i-1)}e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2}).\]](http://latex.artofproblemsolving.com/4/a/0/4a06b883523756f1888d66bf9610c302726d6780.png)
Since 
is strictly positive, we have a contradiction.
then it happens that all of 
happen to have one coefficient equal to 
such that 
are missing the same coefficient 
for 
(this is because none of the roots are equal and none can be zero), let these be called 
respectively and let 
be 
(it doesn't matter anyway lmao), then we have that 
and 
which the means that 
is missing the 
is missing the 
then we do in fact have that 
is missing both the 
coefficient. (Now we can wlog say 
then it has 
but also it has two consecutive missing coefficients so write it as 
(essy to check that if 
this is trivial and if 
then you have cuadratics of the form 
which shows pairwise sum of roots is zero and thus all are zero, a contradiction!), so for 
notice that 
and that you can set it so that least degree term on 
is greater than 
.
-
but also 
and these polynomials must have all of their roots real by chaining the fact that 
has a non-zero leading coefficient and thus if it were the zero polynomial we are done by the same argument as if it weren't which we will see now.
,
,
and thus by taking derivative w.r.t. 
for all 
for 
,
we have LHS is zero while RHS isn't, contradiction!.
.
.
is a polynomial which must have only real distinct roots, but it is divisible by 
,
.
.
.
.
,
.
for some distinct nonzero roots 
,
,
.
be 
,
,
,
and 
)
coefficient.
be 
both of which have zero as an 
coefficients. We now prove the following claim, which finishes the problem:
be a polynomial with distinct nonzero real roots. Then 
be a polynomial with 
has 
distinct roots because between each two roots of 
distinct roots. But since there are two consecutive zero coefficients, eventually some multiple derivative of 
