In the plane, consider a point $ X$ and a polygon $ \mathcal{F}$ (which is not necessarily convex). Let $ p$ denote the perimeter of $ \mathcal{F}$, let $ d$ be the sum of the distances from the point $ X$ to the vertices of $ \mathcal{F}$, and let $ h$ be the sum of the distances from the point $ X$ to the sidelines of $ \mathcal{F}$. Prove that $ d^2 - h^2\geq\frac {p^2}{4}.$
Problem
Source: IMO Shortlist 1996, G9
Tags: geometry, perimeter, inequalities, Pythagorean Theorem, geometric inequality, IMO Shortlist
pianoforte
11.08.2008 01:40
Let the segments from X to the vertices be $ a_1,a_2,a_3,\cdots,a_n$, numbered in increasing order clockwise. Let the segments from X to the sides be $ b_1,b_2,b_3,\cdots,b_n$, also numbered in increasing order clockwise, and such that $ b_1$ is to the side immediately clockwise to $ a_1$. For each side, the foot of $ b_i$ partitions it into two segments, that we shall label $ m_i$ between $ a_i$ and $ b_i$, and $ n_i$ between $ b_i$ and $ a_{i + 1}$ (where $ a_{n + 1}$ = $ a_1$).
Thus, $ \sum_{i = 1}^n a_i = d$, $ \sum_{i = 1}^n b_i = h$, $ \sum_{i = 1}^n (m_i + n_i) = p$. Then, we rewrite $ d^2 - h^2 = (\sum_{i = 1}^n (a_i - b_i))(\sum_{i = 1}^n (a_i + b_i))$. (I)
From the Pythagorean Theorem, $ \sqrt {a_i^2 - b_i^2} = m_i$ and $ \sqrt {a_{i + 1}^2 - b_i^2} = n_i$.
Thus, $ \sum_{i = 1}^n m_i = \sum_{i = 1}^n \sqrt {a_i^2 - b_i^2}$ and $ \sum_{i = 1}^n n_i = \sum_{i = 1}^n \sqrt {a_{i + 1}^2 - b_i^2}$.
Add them, $ \implies \sum_{i = 1}^n (m_i + n_i) = \sum_{i = 1}^n (\sqrt {a_i^2 - b_i^2} + \sqrt {a_{i + 1}^2 - b_i^2}) = p$, or
\[ (\sum_{i = 1}^n (\sqrt {(a_i + b_i)(a_i - b_i)} + \sqrt {(a_{i + 1} + b_i)(a_{i + 1} - b_i)}) )^2 = p^2
\]
.
By Cauchy-Schwarz's Inequality,
\[ (\sum_{i = 1}^n (a_i + b_i) + \sum_{i = 1}^n (a_{i + 1} + b_i))(\sum_{i = 1}^n (a_i - b_i) + \sum_{i = 1}^n (a_{i + 1} - b_i)) \ge (\sum_{i = 1}^n (\sqrt {(a_i + b_i)(a_i - b_i)} + \sqrt {(a_{i + 1} + b_i)(a_{i + 1} - b_i)}) )^2 = p^2
\]
The LHS is equivalent to $ (2(\sum_{i = 1}^n (a_i + b_i))(2(\sum_{i = 1}^n (a_i - b_i)) \ge p^2$. By (I), $ d^2 - h^2 \ge p^2/4$, as desired.
math154
01.06.2011 06:54
Hmm... I feel like the above solution might have some issues with directed segments, but I guess it probably works if we write upper bounds rather than equalities for $p$... Anyway, this seems really weak. Set $a_i=XA_i$ and $\theta_i=\measuredangle{A_iXA_{i+1}}$ directed counterclockwise so that $\sum\theta_i=2\pi$ (all angle, area, and length calculations for $h$ will be directed). Then we can easily calculate \begin{align*} d &= \sum{XA_i} = \sum{a_i} = \frac{1}{2}\sum(a_i+a_{i+1}) \\ \pm h &= \sum\frac{2[A_iXA_{i+1}]}{A_iA_{i+1}} = \sum\frac{a_ia_{i+1}\sin\theta_i}{\sqrt{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}} \\ p &= \sum_{A_iA_{i+1}} = \sum\sqrt{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}. \end{align*}By Cauchy-Schwarz, \begin{align*} 4(d^2-h^2) &= \sum\left(a_i+a_{i+1}-\frac{2a_ia_{i+1}\sin\theta_i}{\sqrt{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}}\right)\sum\left(a_i+a_{i+1}+\frac{2a_ia_{i+1}\sin\theta_i}{\sqrt{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}}\right) \\ &\ge \left(\sum\sqrt{\frac{(a_i+a_{i+1})^2(a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i)-4a_i^2a_{i+1}^2(1-\cos^2\theta_i)}{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}}\right)^2 \\ &\ge \left(\sum\sqrt{a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i}\right)^2 = p^2, \end{align*}where we have used the fact that \begin{align*} &(a_i+a_{i+1})^2(a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i)-4a_i^2a_{i+1}^2(1-\cos^2\theta_i)-(a_i^2+a_{i+1}^2-2a_ia_{i+1}\cos\theta_i)^2 \\ &= 2a_ia_{i+1}(a_i-a_{i+1})^2(1+\cos\theta_i) \\ & \ge 0. \end{align*}Equality holds for regular polygons. Edit: Actually, it seems that the auxiliary inequality is just the case in which $\mathcal{F}$ is a line segment.
pianoforte
01.06.2011 07:51
Aww, I'm hating when people bump threads where I've posted solutions in them from like 3 years ago, and I have either no idea how I came up with them or what I was talking about...