Four Points in Space

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Source: 1975 USAMO Problem 2

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#1 h Mar 15, 2010, 2:37 AM • 2 Y

Y by Adventure10, Mango247

Let $A,B,C,$ and $D$ denote four points in space and $AB$ the distance between $A$ and $B$ , and so on. Show that $AC^2+BD^2+AD^2+BC^2 \ge AB^2+CD^2.$

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#2 h Mar 15, 2010, 4:59 AM • 2 Y

Y by Adventure10, Mango247

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 $MN \ge 0.$

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#4 h Apr 16, 2017, 2:46 AM • 4 Y

Y by Wizard_32, targo___, Adventure10, Mango247

$(\overrightarrow{A}-\overrightarrow{C})^2+(\overrightarrow{B}-\overrightarrow{D})^2+(\overrightarrow{A}-\overrightarrow{D})^2+(\overrightarrow{B}-\overrightarrow{C})^2\ge (\overrightarrow{A}-\overrightarrow{B})^2 + (\overrightarrow{C}-\overrightarrow{D})^2$ $\implies \overrightarrow{A}^2+ \overrightarrow{B}^2+ \overrightarrow{C}^2+ \overrightarrow{D}^2 - 2\overrightarrow{A}\overrightarrow{C}- 2\overrightarrow{B}\overrightarrow{D}- 2\overrightarrow{A}\overrightarrow{D}- 2\overrightarrow{B}\overrightarrow{C}+ 2\overrightarrow{A}\overrightarrow{B}- 2\overrightarrow{C}\overrightarrow{D}\ge 0$ $\implies (\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}-\overrightarrow{D})^2 \ge 0.$

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#5 h Jan 17, 2018, 2:07 AM • 1 Y

Y by Adventure10

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

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#6 h Jan 17, 2018, 2:08 PM • 2 Y

Y by Adventure10, Mango247

solution

This post has been edited 1 time. Last edited by MATH1945, Jan 17, 2018, 2:08 PM
Reason: f

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#7 h Jan 17, 2018, 3:23 PM • 2 Y

Y by Adventure10, Mango247

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 ?

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#8 h Apr 13, 2023, 8:11 PM

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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.

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