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 the circumcircle of a scalene triangle 
. Let 
be a circle internally tangent to in 
. Tangents from 
touch in 
and 
, such that lies in the interior of 
. Similarly, tangents from 
touch in 
and 
, such that lies in the interior of .
and 
concur on the bisector of 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
, 
and 
are the midpoints of sides 
and 
respectively. Point lies inside the triangle. Let 
. Prove that 
.
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
, let 
be the circumcentre, 
be the orthocentre, and 
be the centroid.
satisfy the following condition:
is constant.
. Find 
.
such that numbers 
are both perfect squares.
such that ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
be an acute triangle inscribed in a circle of center . If the altitudes 
intersect at and the circumcenter of 
is 
, prove that 
is a parallelogram.
is a perfect square number.
is good if there exists a permutation 
of 
such that 
is perfect square for all 
. Determine all the good numbers in the set 
.
of positive integers such that 
and 
are both powers of 
.
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
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).
.
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 
.