I am dealing with the test of the OBM (Brasilian Math Olimpyad), University level, 2017, phase 2.
As I've said at others topics (questions 1 and 2, this last yet open, here), I hope someone can help me to discuss this test.
The question 3 says:
Let be $X=\{(x,y)\in\mathbb{R}^2|y\geq 0, x^2+y^2=1\}\cup \{(x,0),-1\leq x \leq 1\}$ the border of a semi-disc closed with radius $1$.
a) Let be $n>1$ an integer and $P_1,P_2,…,P_n\in X$. Prove that exists a permutation $\sigma:\{1,2,…,n\}\rightarrow\{1,2,…,n\}$ such that
$\sum^n_{j=1}|P_{\sigma(j+1)}-P_{\sigma(j)}|^2\leq 8$where we define $\sigma(n+1)=\sigma(1)$.
b) Determine the sets $\{P_1,P_2,…,P_n\}\subset X$ such that for all permutation $\sigma:\{1,2,…,n\}\rightarrow\{1,2,…,n\}$ ,
$\sum^n_{j=1}|P_{\sigma(j+1)}-P_{\sigma(j)}|^2\geq 8$
where we define $\sigma(n+1)=\sigma(1)$.
Well. I draft the solution as following:
We'll show that the permutation such that $P_{\sigma(1)}P_{\sigma(2)}…P_{\sigma(n)}$ is a convex polygon
respect the inequality.
We'll call $\sigma_n$ one of these permutations to $\{P_1,P_2,…,P_n\}$ and define $S_n=\sum^n_{j=1}|P_{\sigma(j+1)}-P_{\sigma(j)}|^2$.
These notations will help us in our proof by induction.
So:
1) The case $n=2$ (trivial)
2) The case $n=3$ is my problem
3) To indution, I used the following result:
All of the convex polygon with more than $3$ sides have at least one internal angle $\geq 90^o$ (the inequality is strict to $n>4$)
I've proved this result and I've combined it with the fact that the on a triangle with sides $a,b,c$ such that the angle between $a$ and $b$ is $\geq 90^o$, we have $a^2+b^2\leq c^2$.
I've wrote a long proof trying combine these results and it's a little dificulty to me write it here today, but if someone want, I can try.
Well, as I've said, my problem is with $n=3$, particularly, acutangles triangles enrolled on $X$.
Maybe this is simples, but I'm trying and couldn't solve… I hope someone could help me. Or, maybe, give an other ideia to the solution.
The item b), I did as following:
From a), we have to find the sets $\{P_1,P_2,…,P_n\}$ such that $S_n\boxed{=}8$.
$\{(\pm1,0)\}$ is trivial and the sets of type $\{P_1,(\pm1,0)\}$ with $P_1$ on the semicircle above too, because we have a rectangle triangle and can use Pytagoras.
I've proved that I cannot have a point between $(-1,0)$ and $(1,0)$. Also, the polygon with more an angle $>90^o$ don't respect, by the argument of item a). So, we must only analyze rectangles. I did this analyse and didn't find any set.
Conclusion: $\{(\pm1,0)\}$ and the sets of type $\{P_1,(\pm1,0)\}$ with $P_1$ on the semicircle above.
What do you think? Thanks very much.
Best Answer
Let us prove that for arbitrary 3 points placed on a semicircle of unit radius, the sum $S$ of squares of their distances is less then or equal to 8.
Case 1: all three points on the diameter
It's easy to show that 3 arbitrary points shown on the left have smaller $S$ compared to the special case shown on the right ($AB<AB'$, $AC<AC'$, $BC<B'C'$
For the three points on the right:
$$S=x^2+(2-x)^2+2^2=x^2+4-4x+x^2+4=8-2x(2-x)$$
Obviously $x\le2$ so $S\le8$.
Case 2: Two points on the diameter, one point above on the circle.
Arbitrary case is shown on the left. For every such case it is possible to find a similar case, with one point on the diameter moved to the end of it, that has bigger $S$. For example, if ve move point $A$ to the left end of the diameter $BA'>BA$, $CA'>CA$. Now look at the picture on the right and triangles $A'BC$ and $A'BC'$. We want to prove that $S(A'BC)<S(A'BC'):$
$$S(A'BC)=c^2+a'^2+(2-x)^2=c^2+(a^2+x^2-2ax\cos\alpha)+4-4x+x^2=$$
$$S(A'BC)=c^2+a^2+4+2x^2-2ax\cos\alpha-4x=S(A'BC')-2x(2-x)-2ax\cos\alpha\le S(A'BC')$$
Note that $S(A'BC')=8$.
Case 3: Two points on circumference, one point on the diameter
For the triangle shown on the left, it is always possible to move one point to the end of the diameter and create a triangle that has a bigger $S$. For example, if you move point $A$ of triangle $ABC$ to point $A'$: $BA'>BA$, $CA'>CA$. So $S(ABC)\lt S(A'BC)$ and according to case (2), $S(A'BC)\le8$
Case 4: All three points on circumference
This case is trivial. Such triangle has smaller $S$ compared with triangle $A'BC'$ and according to case (2) $S(A'BC')=8$.