Let $ A,B,C,$ and $ D$ denote four points in space and $ AB$ the distance between $ A$ and $ B$, and so on. Show that \[ AC^2+BD^2+AD^2+BC^2 \ge AB^2+CD^2.\]
Problem
Source: 1975 USAMO Problem 2
Tags: inequalities, geometry, parallelogram
Luis González
15.03.2010 06:59
If $M,N$ are the midpoints of $AB$ and $DC,$ according to a Casey's theorem we have: $AC^2 + BD^2 + BC^2 + AD^2 = AB^2 + CD^2 + 4MN^2$ The inequality obviously follows from $MN \ge 0.$
Wave-Particle
16.04.2017 05:46
$$(\overrightarrow{A}-\overrightarrow{C})^2+(\overrightarrow{B}-\overrightarrow{D})^2+(\overrightarrow{A}-\overrightarrow{D})^2+(\overrightarrow{B}-\overrightarrow{C})^2\ge (\overrightarrow{A}-\overrightarrow{B})^2 + (\overrightarrow{C}-\overrightarrow{D})^2$$$$\implies \overrightarrow{A}^2+ \overrightarrow{B}^2+ \overrightarrow{C}^2+ \overrightarrow{D}^2 - 2\overrightarrow{A}\overrightarrow{C}- 2\overrightarrow{B}\overrightarrow{D}- 2\overrightarrow{A}\overrightarrow{D}- 2\overrightarrow{B}\overrightarrow{C}+ 2\overrightarrow{A}\overrightarrow{B}- 2\overrightarrow{C}\overrightarrow{D}\ge 0$$$$\implies (\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}-\overrightarrow{D})^2 \ge 0.$$
Samusasuke
17.01.2018 05:07
Luis González wrote: If $M,N$ are the midpoints of $AB$ and $DC,$ according to a Casey's theorem we have: $AC^2 + BD^2 + BC^2 + AD^2 = AB^2 + CD^2 + 4MN^2$ The inequality obviously follows from $MN \ge 0.$ Casey's theorem only applies to planar quadrilaterals, the problem states four points in space
MATH1945
17.01.2018 17:08
Let $A=(x_1,y_1,z_1),B=(x_2,y_2,z_2),C=(x_3,y_3,z_3),D=(x_4,y_4.z_4)$
Then, by length formula, it's easy to see that
$AC^2+BD^2+AD^2+BC^2 - AB^2+CD^2 = x_1^2+x_2^2+x_3^2+x_4^2+2x_1x_2+2x_3x_4-2x_1x_3-2x_1x_4-2x_2x_3-2x_3x_4+y_1^2+y_2^2+y_3^2+y_4^2+2y_1y_2+2y_3y_4-2y_1y_3-2y_1y_4-2y_2y_3-2y_3y_4+z_1^2+z_2^2+z_3^2+z_4^2+2z_1z_2+2z_3z_4-2z_1z_3-2z_1z_4-2z_2z_3-2z_3z_4=(x_1+x_2-x_3-x_4)^2+(y_1+y_2-y_3-y_4)^2+(z_1+z_2-z_3-z_4)^2 \geq 0$
targo___
17.01.2018 18:23
Samusasuke wrote: Luis González wrote: If $M,N$ are the midpoints of $AB$ and $DC,$ according to a Casey's theorem we have: $AC^2 + BD^2 + BC^2 + AD^2 = AB^2 + CD^2 + 4MN^2$ The inequality obviously follows from $MN \ge 0.$ Casey's theorem only applies to planar quadrilaterals, the problem states four points in space Then consider the plane in which the all four point lies .As, three points are enough to make a plane ,so if the fourth point lie outside of that plane then is that figure quadrilateral ?
huashiliao2020
13.04.2023 23:11
Wave-Particle wrote: $$(\overrightarrow{A}-\overrightarrow{C})^2+(\overrightarrow{B}-\overrightarrow{D})^2+(\overrightarrow{A}-\overrightarrow{D})^2+(\overrightarrow{B}-\overrightarrow{C})^2\ge (\overrightarrow{A}-\overrightarrow{B})^2 + (\overrightarrow{C}-\overrightarrow{D})^2$$$$\implies \overrightarrow{A}^2+ \overrightarrow{B}^2+ \overrightarrow{C}^2+ \overrightarrow{D}^2 - 2\overrightarrow{A}\overrightarrow{C}- 2\overrightarrow{B}\overrightarrow{D}- 2\overrightarrow{A}\overrightarrow{D}- 2\overrightarrow{B}\overrightarrow{C}+ 2\overrightarrow{A}\overrightarrow{B}- 2\overrightarrow{C}\overrightarrow{D}\ge 0$$$$\implies (\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}-\overrightarrow{D})^2 \ge 0.$$ Nice solution! Yup, super basic vectors are the way to go, I did the same thing.