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.
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
satisfy that: 
and 
. Find the maximum: 
be a convex quadrilateral. Suppose that the circles with diameters and 
intersect at points 
and . Let 
and 
. Prove that the points , , and are concyclic. and are not the diagnols)
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be positive integers, where 
. It is given that there exist a positive integer 
such that
. Prove that the above equation is true for all positive integers 
. (Here 
is the sum of digits of 
taken in base 
).
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).
such that 
where 
.
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)
![\[27(1 - k)(1 - m) = \left[k + \frac {(1 + k)^2}{(1 - k) m}\right]^3,\]](http://latex.artofproblemsolving.com/a/2/6/a260f7f486d98029d286594b40728fecf2397fcc.png)
for the optimal value 
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 
.