AoPS Academy
lines lines are used as roads by the townsfolk. In the past, the roads in Linetown used to be two-way, but this often led to residents accidentally cycling back to where they started. roads one-way such that it is impossible for any resident to start at a house, follow the roads in the correct directions, and end up back at the original house. In how many ways can Turbo achieve this?
such that for all 
, 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
, let 
, 
, and 
denote the feet of the altitudes from 
, 
, and 
, respectively, and let 
denote the circumcenter of 
. Points 
and 
denote the projections of and , respectively, onto 
, and 
. There exists a point 
such that 
and 
. If 
and 
lies on 
such that 
, prove that line 
bisects 
.
+1 w
such that the inequality ![\[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| \leq M(a^{2}+b^{2}+c^{2})^{2}\]](http://latex.artofproblemsolving.com/0/3/6/036776e42f5f5bcc79638c64a5fe2a0488e342f2.png)
holds for all real numbers 
,
and 
.
with equality case 
and its permutation . So who can prove this ?![\[(a-b)(b-c)(c-a)(a+b+c)\]](http://latex.artofproblemsolving.com/f/e/d/fed5089d2af351f9f08fb09e739d798505400e83.png)
So just let 
then the inequality becomes ![\[|xyzs|\leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\]](http://latex.artofproblemsolving.com/b/2/9/b295ae363f08481a39f845c49023b2db8f5f789a.png)
With the property that 
have the same sign. Let 
be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller)
.
and the inequality becomes: ![\[|2x^{3}s|\leq \frac{M}{9}(s^{2}+6x^{2})^{2}\]](http://latex.artofproblemsolving.com/1/8/e/18e44c3115015f8b12438b5c7192c06ec5403558.png)
Now using the AM-GM inequality for 
we get ![\[(s^{2}+6x^{2})^{2}\geq 16\sqrt{8s^{2}x^{6}}=32\sqrt 2sx^{3}\]](http://latex.artofproblemsolving.com/9/a/7/9a788445acf7864595ba035bb6ca5129fb9e57be.png)
So 
works.
we have 
.
?
as before and suppose 
with 
.
,
.
so
and 
with 
and ![\[\mid xyzs \mid \leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\]](http://latex.artofproblemsolving.com/e/1/0/e103289ca9fb73b12a08f35d057ce781522014bd.png)
after our substitution, if 
,
and that 
and 
,
on the unit sphere, and using Lagrange multipliers. After a great deal of algebra, I found the critical points; most led to zero values, leaving only one possibility (and permutations) for the maximum.
Method 
.
hence we hav
with simple AM-GM we will only get 
which is by far to big.
hence we hav
to use that and also you forgot the square after 
you did applied. But this is most beautiful solution I ever seen to this problem it's just awesome congratulation.
and is possible for
.
,
.
and note the following bounds.
However, after a full expansion,
Hence,
as desired.
,![\[M = \max_{\substack{(a,b,c)\in\mathbb{R}^3\\(a,b,c)\ne(a,a,a)}}\frac{|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})|}{(a^2+b^2+c^2)^2}.\]](http://latex.artofproblemsolving.com/4/e/7/4e74dccd29feecfa031ff08d1e7ecaf1ecc2838d.png)
Note that![\[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| = |(a-b)(b-c)(c-a)(a+b+c)|.\]](http://latex.artofproblemsolving.com/1/1/1/111897f73ab3e06b05c699ee168bf155ec11dc73.png)
This motivates reparameterizing with 
,
,
,
,![\[(a^2+b^2+c^2)^2 = \frac{1}{81}(3t^2+\sum(x-y)^2)^2,\]](http://latex.artofproblemsolving.com/4/5/c/45c014306a395d6173ebbbb77554122c5fec5d5f.png)
so we have![\[M = 81\cdot\max_{\substack{(x,y,z,t)\in\mathbb{R}^4\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyzt|}{[3t^2+\sum(x-y)^2]^2}.\]](http://latex.artofproblemsolving.com/3/5/6/356b11ff902dea9b3a6b3851b7b7dbf7169b718e.png)
Fixing 
,
since the thing we are maximizing stays the same if we negate 
.
,![\[f(t) = \frac{t}{(3t^2+a)^2}.\]](http://latex.artofproblemsolving.com/e/6/7/e67e0a04fedbc43b6a6c14766a43a180e282f414.png)
Note that 
and 
as 
,
is at a critical point of 
somewhere in 
.![\[f'(t)=0 \iff (3t^2+a)^2 = t\cdot(2(3t^2+a))\cdot(6t)\iff 3t^2+a = 12t^2\iff t=\sqrt{a}/3.\]](http://latex.artofproblemsolving.com/5/f/3/5f3b4f2395698ae73d6a93783a5e7de06b590942.png)
Thus, the maximum of ![\[f(\sqrt{a}/3) = \frac{\sqrt{a}/3}{16a^2/9} = \frac{3}{16}a^{-3/2},\]](http://latex.artofproblemsolving.com/c/9/2/c923ceb5c991b262806dfc5f88fa3bb69cbf0867.png)
so the answer to the problem is![\[M = \frac{243}{16}\cdot\max_{\substack{(x,y,z)\in\mathbb{R}^3\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyz|}{\left[\sum(x-y)^2\right]^{3/2}}.\]](http://latex.artofproblemsolving.com/c/9/1/c910168e15de0756e4ccabe06a9ecb8dd71f4b9e.png)
Note that replacinig 
does not change the expression within the maximum, so we can WLOG assume 
.![\[M = \frac{243}{16}\cdot\max_{x,y\ge 0}\frac{xy(x+y)}{[6x^2+6xy+6y^2]^{3/2}}.\]](http://latex.artofproblemsolving.com/b/8/b/b8b48404793773001da41997d13c2380184de73b.png)
By de-homogenizing, we see that![\[M = \frac{243}{16}\cdot\max_{x\ge 0}\frac{x+1/x}{[6(x^2+1/x^2-1)]^{3/2}} = \frac{243}{16}\cdot\max_{\alpha\ge 2}\frac{\alpha}{[6(\alpha^2-1)]^{3/2}},\]](http://latex.artofproblemsolving.com/a/0/8/a085345976d54ee62a7031721242130c74e08d0c.png)
and it is easy to check that![\[g(\alpha)=\frac{\alpha}{(\alpha^2-1)^{3/2}}\]](http://latex.artofproblemsolving.com/9/9/f/99f92c51065454460bea85c3857c370bfae2c05e.png)
is maximized at 
if we restrict to 
.![\[M = \frac{243}{16}\cdot\frac{2}{6^{3/2}\cdot 3^{3/2}} = \frac{243}{16}\cdot\frac{1}{27\sqrt{2}} = \boxed{\frac{9}{16\sqrt{2}}}.\]](http://latex.artofproblemsolving.com/6/0/4/6045ae6d6b1a334914c3c016c8b99851de87991c.png)
WLOG let 
.
,
Notice that 
,
,
.
which by AM-GM on the RHS gives ![$$\frac{M}{9}(2r^2+2r^2+2r^2+s^2)^2 \ge \frac{4M}{9}\left(\sqrt[4]{8r^6s^2}\right)^2 = \frac{8M\sqrt{2}}{9}r^3s \ge 2r^3s$$](http://latex.artofproblemsolving.com/4/3/8/4380be5f6eeaa40a8e34164df65918f5a03f2123.png)
with equality when 
.
.
.
gives 
.
.
.![\[|wxyz|\le \frac{M}{9}(w^2+x^2+4y^2+z^2)^2.\]](http://latex.artofproblemsolving.com/f/3/3/f333068891631b8dc9d53025b2ced466d7379a4e.png)
are 
,
(if this is not true, then changing 
keeps the inequality exactly the same).
,
and 
.![\[w(2y-w)\le y^2\]](http://latex.artofproblemsolving.com/5/6/d/56d29ffc0d232caa7678d794bb225d7737aed60e.png)
so the LHS increases.
,![\[2y^3z\le \frac{M}{9}(2y^2+2y^2+2y^2+z^2)^2\]](http://latex.artofproblemsolving.com/c/e/0/ce063aaa214823e0b4428574293a8e609c1cbef2.png)
![\begin{align*}
2y^2+2y^2+2y^2+z^2 \\
\ge 4\sqrt[4]{8y^6z^2} \\
\implies (y^2+y^2+4y^2+z^2)\ge 16\sqrt{8y^6z^2} \\
=32\sqrt{2}y^3z
\end{align*}](http://latex.artofproblemsolving.com/1/d/6/1d6815e80b1337aeacaa308a898559da86a72d3c.png)
gives 
,
and 
So the inequality is: 
Assume 
By AM-GM: 
equality when 
by AM-GM: 
Combination of 
and 
imply, 
i.e. the least value, when 
The opposite case, that is 
can be shown similar. 
.
.
.
= 
.
,
.
at the same rate, the only terms that change are the 
term and the 
term. Now take the derivative with respect to 
(more precisely, take the limit as 
approaches 
.
. In the equality case, this should be equal to zero (We know there has to be an equality case because the inequality is homogenous and the equality will not be reached by a limit, because we can simply rescale downwards. Sorry if this is a little handwavy). Thus, we substitute A=0 into our inequality and eliminate 
.
subject to the constraint 
.
.
subject to the constraint. We now apply Lagrange multipliers on 
. Cyclically summing this and rearranging gives that 
.
back in and subtracting it from both sides, we get that 
, which rearranges to 
. If it is not the case that 
and cyclically sum to get a contradiction. If they are in arithmetic progression, WLOG 
,
,
,
be 
,
,
.![\[p^2+q^2+r^2+s^2=3(a^2+b^2+c^2)\]](http://latex.artofproblemsolving.com/b/4/a/b4a4d78feb6c6e6bfc127a529a6bebbd65a2f233.png)
we have the right-hand side as 
.
.![\[f(s)=\frac{(p^2+q^2+r^2+s^2)^2}{9s|pqr|}=\frac{(x+s^2)^2}{ys}\]](http://latex.artofproblemsolving.com/6/7/0/670bbcbd7f377c5d29bb9f3f7419b6ab1a5bb920.png)
Then, we have![\[f(s)=\frac{s^4+2s^2y+x^2}{ys}=\frac1y(s^3+2sx+x^2s^-1)\]](http://latex.artofproblemsolving.com/9/7/4/974ea3587429e16e8eca5bae860b85dd6c94aaac.png)
![\[f'(s)=\frac1y(3s^2+2x-x^2s^-2)=\frac1{s^2y}(3s^2-x)(s^2+x)\]](http://latex.artofproblemsolving.com/2/a/3/2a32f2fb805c95867485b1f417f42cadd11a3956.png)
Clearly, 
we when 
.
and 
,
.
be a monic cubic with roots 
,
and 
are 
'
.
,
so![\[y_{1,2}=\pm \frac{s^3}{\sqrt2}-pqr\]](http://latex.artofproblemsolving.com/e/4/7/e474c4cffdc3a2951a36fe19de2c35baae3c4e98.png)
Therefore, 
,
as desired. Thus, 
.
.![\[(a,b,c)=(\tfrac{\sqrt{2}}{3}+1,\frac{\sqrt{2}}{3}),\frac{\sqrt{2}}{3})+1\]](http://latex.artofproblemsolving.com/4/8/5/485d5e1f2c3326c5695ccc440318861086f849a3.png)
Yields 
which yields the desired equality.