O.I.M. polygon inequality

Question

I am trying to prove an inequality which was used to prepare the Romanian O.I.M. team. I seem to lack ideas on how to tackle this problem.

We take a convex polygon $P_1\ldots P_{n+2}$ and consider two points Inside, $M$ and $M'$. We wish to show the following:

$$\sum_{i=1}^{n+2} (MP_i -M'P_i) < n\cdot MM'$$

I would gladly accept any indications or any help whatsover. Thank you.

Answer 1

First, look at the perpendicular bisector $\ell$ of $MM'$. Let $H$ be the closed half plane containing $M'$ with respect to $\ell$. If there are $k$ points of the polygon not in $H$, then we can easily see that $$\sum_{i=1}^{n+2}\,\left(MP_i-M'P_i\right) <(n+2-k)\cdot MM'$$ holds (this is because $MP-M'P<0$ if $P$ is not in $H$ and $MP-M'P\leq MM'$ for any point $P$ on the plane). If $k\geq 2$, then we are done. Hence, we only have to consider the case where exactly one point of the polygon is not in $H$. Suppose that this point is $P_1$. Now, if the straight line $MM'$ goes through $P_1$, then simply pick its neighboring vertex $P_2$. If not, then the line $MM'$ must intersect the interior of either $P_1P_2$ or $P_1P_{n+2}$. Assume that the straight line $MM'$ intersects the interior of $P_1P_2$.

Suppose that $m:=MM'$, $P_1M=a$, $P_2M=b$, $\theta:=\pi-\angle P_1MM'$, and $\phi:=\angle\,P_2MM'$. We have that $0\leq \theta\leq\phi<\frac{\pi}{2}$. That is, $\cos(\theta)\geq \cos(\phi)$, so that $$M'P_2=\sqrt{b^2+m^2-2bm\cos(\phi)}\geq\sqrt{b^2+m^2-2bm\cos(\theta)}=M'P_2'\,,$$ where $P_2'$ is the point on the ray $P_1M$ such that $P_2'M=P_2M$ and that $P_1$, $M$, $P_2'$ lie in this order. Now, on the triangle $M'P_1P_2'$, we have that $$P_1P_2'=MP_1+MP'_2=MP_1+MP_2$$ and that $$M'P_1+M'P_2' >P_1P_2'=MP_1+MP_2\,,$$ by the Triangle Inequality. As $P_2'M'\leq P_2M'$, we conclude that $$M'P_1+M'P_2 \geq M'P_1+M'P_2' > MP_1+MP_2\,.$$ Now, for all $i=3,4,\ldots,n+2$, $MP_i-M'P_i\leq MM'$. Hence, $$\sum_{i=1}^{n+2}\,\left(MP_i-M'P_i\right) \leq n\cdot MM'+\left(MP_1+MP_2-M'P_1-M'P_2\right)<n\cdot MM'\,.$$

Answer 2

Thank you Batominovski for your clear answear. I just wanted to add that I think that $M'P_1+M'P_2 > MP_1+MP_2$ can be proven just with the triangular inequality.

Clearly in the last situation the point $M$ is inside the triangle $P_1P_2M'$ so we have $A=(P_2M)\cap (P_1M')$ that lies on the segment $[P_1M']$. We thus have

$$MP_1 < P_1A + AM\\ AP_2<AM'+M'P_2$$

So we get

$$MP_1+MP_2<P_1A+AM+MP_2=P_1A+AP_2<P_1A+AM'+M'P_2=M'P_1+M'P_2$$

which is what we were looking for.