AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 
. Show that if 
is a prime less than 
for some positive integer 
, then there exists 
such that 
.
problems in a meeting. For each 
, each leader will talk about the 
-th problem for -th minutes. The chair can assign one leader to talk about a problem of his choice, but he would have to wait for the leader to complete the entire talk of that problem before assigning the next leader and problem. The next leader can be the same leader. The next problem can be a different problem. Each leader’s longest idle time is the longest consecutive time that he is not talking.
so that the chair can ensure that the longest idle time for any leader does not exceed .
+2 w
be real numbers such that 
and 
Find the value of 
be real numbers such that 
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 
，and any rectangular strip of width 1 covers at most two points of S.
lie in a square of area 
. Prove that among the triangles there are 
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 . Prove that 
for every 
.
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 
.
and 
. Prove that
![$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$](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 and 
. Prove that
![$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$](http://latex.artofproblemsolving.com/6/9/5/695cc35986a391fdc15e4863d361ce373fbbf64c.png)
. Let 
be a point on side 
, and 
be a point on the extension of 
such that 
. Let 
be the circumcenter of 
, and 
be a point on the side extension of 
satisfying 
. Line BP intersects AC at point Q. Prove that 
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
over all pairs of positive integers 
.
be a triangle. Construct rectangles 
, 
, and 
outside such that 
. Let 
and 
intersect at 
and define 
similarly. Prove that line 
bisects 
.
be imaginary, and 
, 
, 
, 
, 
, 
. Since 
, triangle 
is homothetic to the triangle 
, 
, 
, so the median has slope 
. We will show that the intersection of the line through 
and 
intersects at a point that is the same after swapping 
and 
and multiplying by 
. Assume without loss of generality 
, so we need to find the intersection of the line through and 
and the line through 
and 
. The intersection is
and and multiplying by , which implies that this point is and lies on the line through 
parallel to the line through and the midpoint of , so bisects .
and 
and a point , let 
denote 
, where 
is the point on such that 
.
, 
, and 
be the centers of , , and , respectively. Notice that a homothety of scale factor at sends 
to 
, and symmetric statements apply to 
and 
. It suffices to show 
, as this shows 
is the -median of . Since 
and similar, it suffices to show![\[f(A,\overline{O_AO_C},\overline{O_BO_C})+f(B,\overline{O_AO_C},\overline{O_BO_C})=-f(A,\overline{O_AO_B},\overline{O_BO_C})-f(C,\overline{O_AO_B},\overline{O_BO_C})\]](http://latex.artofproblemsolving.com/b/1/a/b1a1b44efd8c6270c2f2f6abf0b3e5e7b9ce31e8.png)
Let 
and 
be the midpoints of 
and 
, respectively. Then, it suffices to show![\[f(M,\overline{O_AO_C},\overline{O_BO_C})=-f(N,\overline{O_AO_B},\overline{O_BO_C}).\]](http://latex.artofproblemsolving.com/c/4/c/c4c9045cd39924e35452bd32774463691157f7ba.png)
Let the lines parallel to through and intersect 
and 
at and 
, respectively, and let be the circumcenter of . Then, we have
as desired. 
is a regular polygon with 
diagonals. Randomly select a triangle whose three vertices are the three vertices of polygon 
,![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
be the centers of the three rectangles, so 
,
,
are similar isosceles triangles. Notice 
and similar, so the problem is equivalent to 
.
,
be the circumcenter, centroid, and orthocenter. Recall that 
,
,
concur at point 
,
passing through 
,
,
.
.
such that lines from 
,
,
,
,
are concurrent. (This is also the conic used to prove Sondat's theorem). 
,
since 
and 
.
gives 
.
bisects 
.
and 
be the reflections of 
are equally spaced and lines 
are equally spaced, we have that 
.
lies on 
is parallelogram)
,
are parallel and equally spaced, so we are done.
be the circumcircle of 
and 
similarly. Let 
be the radical center of 
.
.
,
.
is the radical axis of 
,
and 
.![[asy]
size(7cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
real t = 0.3;
real t = 0.3;
pair A = (1,4);
pair B = (0,0);
pair C = (4,0);
pair A_1 = B + t*dir(-90)*(C-B);
pair A_2 = C + t*dir(90)*(B-C);
pair B_1 = C + t*dir(-90)*(A-C);
pair B_2 = A + t*dir(90)*(C-A);
pair C_1 = A + t*dir(-90)*(B-A);
pair C_2 = B + t*dir(90)*(A-B);
pair A_prime = extension(A_1,B_2,A_2,C_1);
pair B_prime = extension(B_1,C_2,B_2,A_1);
pair C_prime = extension(C_1,A_2,C_2,B_1);
pair Y = extension(A,A+B_prime-C_prime,A_prime,B_prime);
pair Z = extension(A,Y,A_prime,C_prime);
pair R = extension(A,foot(A,B_1,C_2),B,foot(B,A_2,C_1));
fill(B--A_1--A_2--C--cycle, mediumgray);
fill(C--B_1--B_2--A--cycle, mediumgray);
fill(A--C_1--C_2--B--cycle, mediumgray);
draw(A--B--C--cycle, linewidth(1));
draw(A_1--B_2, black);
draw(B_1--C_2, black);
draw(C_1--A_2, black);
draw(Z--A, black, EndArrow(position=0.5));
draw(A--Y, black);
draw(B_prime--C_prime, black, EndArrow(position=0.5));
draw(B--A_1--A_2--C--cycle, black);
draw(C--B_1--B_2--A--cycle, black);
draw(A--C_1--C_2--B--cycle, black);
draw(A--R, gray);
draw(B--R, gray);
draw(C--R, gray);
dot("$A$", A, dir(101));
dot("$B$", B, dir(-142));
dot("$C$", C, dir(-27));
dot("$A_1$", A_1, dir(-136));
dot("$A_2$", A_2, dir(-45));
dot("$B_1$", B_1, dir(-9));
dot("$B_2$", B_2, dir(81));
dot("$C_1$", C_1, dir(121));
dot("$C_2$", C_2, dir(-149));
dot("$A'$", A_prime, dir(-172));
dot("$B'$", B_prime, dir(-53));
dot("$C'$", C_prime, dir(66));
dot("$Y$", Y, dir(-20));
dot("$Z$", Z, dir(-143));
dot("$R$", R, dir(45));
[/asy]](http://latex.artofproblemsolving.com/9/9/a/99aa3098517e1ecf955f864d71c1fe04883ff11e.png)
respectively. Then the problem is equivalent to showing that 
.
,
,
,
and 
.
.
.
,
