1975 USAMO Problems/Problem 2

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Problem

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

Solution

If we project points $A,B,C,D$ onto the plane parallel to $AB$ and $CD$ , $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ are coplanar:

Let $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$ . We wish to prove that $a^2+b^2+c^2+d^2\ge m^2+n^2$ . Let us fix $\triangle BCD$ and the length $AB$ and let $A$ vary on the circle centered at $B$ with radius $m$ . If we find the minimum value of $a^2+b^2$ , which is the only variable quantity, and prove that it is larger than $m^2+n^2-c^2-d^2$ , we will be done.

First, we express $a^2+b^2$ in terms of $c,d,m,\theta,\phi$ , using the Law of Cosines:

$a^2+b^2=c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta)$

$\implies (a^2+b^2-c^2-d^2-2m^2)^2$ $=4m^2(c^2\cos^2(\theta)+d^2\cos^2(\phi-\theta)+2cd\cos(\theta)\cos(\phi-\theta))$ .

$a^2+b^2$ is a function of $\theta$ , so we take the derivative with respect to $\theta$ and obtain that $a^2+b^2$ takes a minimum when $c\sin(\theta)-d\sin(\phi-\theta)=0$

$\implies c^2\sin^2(\theta)+d^2\sin^2(\phi-\theta)-2cd\sin(\theta)\sin(\phi-\theta)=0$ . $\begin{eqnarray*}\implies(a^2+b^2-c^2-d^2-2m^2)^2&=&4m^2(c^2+d^2+2cd(\cos(\theta)\cos(\phi-\theta)-\sin(\theta)\sin(\...$