[/quote]
,
and 
.
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](http://latex.artofproblemsolving.com/2/2/e/22e6ec2f98777db80e39333542664e903663bf81.png)
![$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$](http://latex.artofproblemsolving.com/f/c/a/fca343ee701eeff4c309b31beb3426ec96b49b55.png)
Let 
.
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
is a regular polygon with 
diagonals. Randomly select a triangle whose three vertices are the three vertices of polygon 
be a sequence of positive integers with 
,
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
.
is a prime less than 
for some positive integer 
,
such that 
.
problems in a meeting. For each 
,
-
so that the chair can ensure that the longest idle time for any leader does not exceed 
be real numbers such that 
and 
Find the value of 
and 
Find the value of 
?
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
lie in a square of area 
.
that intersect in a region of area not less than 
.
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
for every 
.
,![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
be a finite set of real numbers, 
be a real number, and 
be 2025 non-zero real numbers. Define![\[A =
\left\{
(x_1, x_2, \cdots, x_{2025}) : x_1, x_2, \cdots, x_{2025} \in X \text{ and } \sum_{i=1}^{2025} \lambda_i x_i = d
\right\},\]](http://latex.artofproblemsolving.com/2/e/1/2e18b6d25bfaa3e3d52f7efe5e4cba7a9cb87db9.png)
![\[B =
\left\{
(x_1, x_2, \cdots, x_{2024}) : x_1, x_2, \cdots, x_{2024} \in X \text{ and } \sum_{i=1}^{2024} (-1)^i x_i = 0
\right\},\]](http://latex.artofproblemsolving.com/4/1/f/41f6b126cc3cf24757b37147906c767d49a0fdd9.png)
![\[C =
\left\{
(x_1, x_2, \cdots, x_{2026}) : x_1, x_2, \cdots, x_{2026} \in X \text{ and } \sum_{i=1}^{2026} (-1)^i x_i = 0
\right\}.\]](http://latex.artofproblemsolving.com/d/e/6/de6586b88004e3f5aa7774525b00ee975c072ff9.png)
Show that 
.
.
be a point on side 
,
such that 
.
,
be a point on the side extension of 
satisfying 
.
.
,
,
,
jump is equal to 
.
over all pairs of positive integers 
.
Y by
Let 
be a given triangle with circumcenter 
and orthocenter 
. Let 
and 
be the feet of the perpendiculars from 
and 
to the opposite sides, respectively. Let 
be the reflection of 
with respect to 
. Prove that 
is a cyclic quadrilateral.
Proposed by Imad Uddin Ahmad Hasin
be a given triangle with circumcenter 
and orthocenter 
. Let 
and 
be the feet of the perpendiculars from 
and 
to the opposite sides, respectively. Let 
be the reflection of 
with respect to 
. Prove that 
is a cyclic quadrilateral.Proposed by Imad Uddin Ahmad Hasin
intersect ![\[
\angle OAC = 90^\circ - \angle B = \angle PAF.
\]](http://latex.artofproblemsolving.com/7/c/a/7caca0e0958cda8742a8458a91901ab0411c369f.png)
![\[
\angle AFP = \angle AHE = 90^\circ - \angle BAD = \angle B.
\]](http://latex.artofproblemsolving.com/4/e/4/4e4eaf75c9765fb39dfb0df2ad220cc27ab77fae.png)
This implies:![\[
AO \perp EF,
\]](http://latex.artofproblemsolving.com/0/0/0/000333016d25fba3389566af9fa7a7dd464c2709.png)
and thus 
be the midpoint of 
.
is cyclic.![\[
AO \cdot AP = AQ \cdot AF.
\]](http://latex.artofproblemsolving.com/5/1/b/51b9473a79b3fdbee7e691f6e158b3ca48cecec3.png)
Multiplying both sides by ![\[
AO \cdot (2AP) = (2AQ) \cdot AF.
\]](http://latex.artofproblemsolving.com/8/d/a/8dac4f97a8555f5a9e112fe4e764b99d73ae82f4.png)
Since 
and 
,![\[
AO \cdot AA' = AC \cdot AF = AD \cdot AH \quad (\text{As } DHFC \text{ is cyclic}).
\]](http://latex.artofproblemsolving.com/2/f/8/2f8b5b2a92ba92d374f90dabf69467b91b12b733.png)
This directly implies that 
is cyclic.
and 
.
and 
swap and an easy angle chase also shows that 
and the result follows by PoP.
be the midpoint of 
and 
.
as 
.![\[\angle EAA'=\angle BAD=\angle CAO\implies A, O, A' \text{ are collinear.}\]](http://latex.artofproblemsolving.com/6/0/6/606fcfde1dccea0185b74f40e3d6685f16e2878a.png)
We know that, 
as 
and 
is the 
.
.![\[\triangle DAO\sim\triangle XAO'\implies \frac{AD}{AO}=\frac{AX}{AO'}=\frac{2AX}{2AO'}=\frac{AA'}{AH}\implies AD\cdot AH=AA'\cdot AO\implies H,O,A',D\text{ are concyclic by \textbf{Power of Point}}\]](http://latex.artofproblemsolving.com/5/7/b/57b407e85e381a9a25aa4b355bfa8113271b7118.png)
.
implies 
is cyclic. Thus, we should show 
.
implying 
