Two triangles $ ABC$and $ A'B'C'$ are positioned in the space such that the length of every side of $ \triangle ABC$ is not less than $ a$, and the length of every side of $ \triangle A'B'C'$ is not less than $ a'$. Prove that one can select a vertex of $ \triangle ABC$ and a vertex of $ \triangle A'B'C'$ so that the distance between the two selected vertices is not less than $ \sqrt {\frac {a^2 + a'^2}{3}}$.
Problem
Source: 17-th Iranian Mathematical Olympiad 1999/2000
Tags: vector, geometry proposed, geometry
BaBaK Ghalebi
06.01.2009 21:48
let $ O$ be an arbitrary point in the space,and for any point $ X$ let $ x$ denote the vector from $ O$ to $ X$.let $ S = \{A,B,C\}$,$ \sigma_1 = a + b + c$ and $ \sigma_2 = a\cdot b + b\cdot c + c\cdot a$.define $ S',\sigma_1',\sigma_2'$ analogously in terms of $ A',B'$ and $ C'$. given $ (P,P')\in S\times S'$.note that $ PP'^2 = |p|^2 - 2p\cdot p' + |p'|^2$.summing over all 9 possible pairs yields the total $ t$ with: \begin{eqnarray*}t & = & \sum_{P\in S}3|p|^2+\sum_{P'\in S'}3|p'|^2-2\sigma_1\cdot\sigma_1' \\ & = & \sum_{P\in S}3|p|^2 + \sum_{P'\in S'}3|p'|^2 + \left(|\sigma_1 - \sigma_1'|^2 - |\sigma_1|^2 - |\sigma_1'|^2\right) \\ & \geq & \sum_{P\in S}3|p|^2 + \sum_{P'\in S'}3|p'|^2 - |\sigma_1|^2 - |\sigma_1'|^2 \\ & = & \left(\sum_{P\in S}2|p|^2 - 2\sigma_2\right) + \left(\sum_{P'\in S'}2|p'|^2 - 2\sigma_2'\right) \\ & = & |a - b|^2 + |b-c|^2 + |c - a|^2 + |a' - b'|^2 + |b' - c'|^2 + |c' - a'|^2 \\ & = & AB^2 + BC^2 + CA^2 + A'B'^2 + B'C'^2 + C'A'^2 \\ & \geq & 3(a^2 + a'^2)\end{eqnarray*} thus one of the nine distances is greater or equal to: $ \sqrt {\frac t9}\geq\sqrt {\frac {a^2 + a'^2}3}$ as desired.the equality case is never reached.because $ A,B$ and $ C$ do not all lie on a line,the nine values $ PP'^2$ we sum to find $ t$ cannot all be equal to each other.