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.
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
+1 w
, such that for any positive integer 
and non-negative reals 
, the following inequality always holds:
Here 
and 
denote the median and arithmetic mean of 
, respectively.
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
be an isosceles triangle with 
, and 
an arbitrary point on side 
. The internal angle bisector of 
meets the circumcircle of 
again at 
, and the internal angle bisector of 
meets it again at 
. Show that lines 
, 
and 
are concurrent.
be non-negative real numbers. Prove that
be a triangle such that 
. Points 
lie on segments 
, respectively, such that lines 
and 
are the angle bisectors of triangle 
. Find the measure of angle 
.
satisfy 
prove in equality.
has circumcircle 
and circumcenter 
. A circle 
with center 
intersects the segment at points 
and 
, such that 
, , , and 
are all different and lie on line in this order. Let 
and 
be the points of intersection of and , such that , , , , and lie on in this order. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
. Prove that lies on the line 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and , 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
and sides of lengths , and respectively, the following inequality holds: 
.
such that 
holds for all triangles is 
. 
.
and sides of lengths , and respectively, the following inequality holds: .
, 
, 
and assuming that 
gives 
. Thus, our inequality can be written as: 
and using the substitutions 
and 
, where 
, it further reduces to:![\[x^3 + 2a^2x + 2b^2x + a^2b + ab^2 > ax^2 + bx^2 + 4ab x.\]](http://latex.artofproblemsolving.com/7/2/c/72cd6fcc810750ef6e409a2572fae5fff0603b35.png)
![\[\begin{array}{c}\displaystyle \left[\left(\frac {x^3}4 + a^2x\right) + \left(\frac {x^3}4 + b^2x\right) + (a^2x + b^2x)\right] + \left(\frac {x^3}2 + a^2b + ab^2\right) \\\\
> \left[ax^2 + bx^2 + 2abx\right] + 2abx,\end{array} \]](http://latex.artofproblemsolving.com/0/9/6/096f737fc5a22dc1d5977ca14c1f9aefffe145cd.png)
, which we can also prove by AM-GM. Indeed,![\[\frac {x^3}2 + a^2b + ab^2 \ge 3\sqrt[3]{\frac {a^3b^3x^3}2} = \frac {3\cdot abx}{\sqrt[3]{2}},\]](http://latex.artofproblemsolving.com/9/6/0/96075fdb988a97e7acab237473eb2592c470156b.png)
![$\frac 3{\sqrt[3]{2}} > 2$](http://latex.artofproblemsolving.com/0/1/5/015ceb0e6f605be87671b782e86e99b689b3e8e5.png)
gives the conclusion. More interestingly, we can also derive the optimal inequality by introducing parameters to split the terms and making use of AM-GM as previously shown. However, the computations are quite horrendous and I spent a great deal of time until I managed to find this optimal constant (because I was unable to see the elegant idea of DerJan to reduce it to a very simple calculus problem).
satisfying
Prove
.
and 
.
By CS and AM-GM
The derivative of the last line is
Thus, we get that the minimum is at 
,
so we're done.
and looking at the inequality: 
,![\[\begin{array}{c}x^3(1 - k) + 2a^2x + 2b^2 x + a^2b + ab^2 \ge ax^2 (1 + k) + bx^2 (1 + k) + abx (4 + k).\end{array}\]](http://latex.artofproblemsolving.com/d/a/b/dab6990125f7e313e255ff48100348f113620e53.png)
such that 
and 
and rewrite the latter inequality as:![\[\begin{array}{c}\displaystyle x^3(1 - k)\left(\frac m2 + \frac m2 + n\right) + (a^2x + b^2 x)(p + q) + a^2b + ab^2 \\\\
\ge ax^2 (1 + k) + bx^2 (1 + k) + abx (4 + k).\end{array}\]](http://latex.artofproblemsolving.com/2/d/2/2d2c038659bab888f31d2ab80164b94e6dd96cb9.png)
![\[\begin{array}{c}
\displaystyle x^3(1 - k)\cdot \frac m2 + a^2x\cdot p \ge 2ax^2 \sqrt{\frac {(1-k)mp}2} \equiv ax^2 (1 + k) \\\\
\displaystyle x^3(1 - k)\cdot \frac m2 + b^2x\cdot p \ge 2bx^2 \sqrt{\frac {(1-k)mp}2} \equiv bx^2 (1 + k) \\\\
\displaystyle x^3(1 - k)\cdot n + a^2 b + a b^2 \ge 3abx\sqrt[3]{(1 - k)n} \equiv abx(4 + k - 2 q),
\end{array}\]](http://latex.artofproblemsolving.com/9/9/1/991ecbd5f27f1cd41f3e66f03e0197878452113c.png)
.![\[\begin{array}{c}
\displaystyle mp = \frac {(1 + k)^2}{2(1 - k)} \\\\
\displaystyle 27(1 - k)n = (4 + k - 2q)^3.
\end{array}\]](http://latex.artofproblemsolving.com/5/a/e/5ae7a46f4e42adecf34578b0b9972bffc23417b9.png)
and plugging into the very last equation we finally obtain:![\[27(1 - k)(1 - m) = \left[k + \frac {(1 + k)^2}{(1 - k) m}\right]^3,\]](http://latex.artofproblemsolving.com/a/2/6/a260f7f486d98029d286594b40728fecf2397fcc.png)
for the optimal value 
by binary-searching up to 12 decimals until the real solutions for 
started to converge to the same number 
. It was quite a big hack, but it did work well in the end and I was really thrilled to derive the result in this rather complicated, yet funny way 
.