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Source: Iran TST 2008

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#1 h May 25, 2008, 2:47 PM • 2 Y

Y by Stranger8, Adventure10

Let $P_1,P_2,P_3,P_4$ be points on the unit sphere. Prove that $\sum_{i\neq j}\frac1{|P_i-P_j|}$ takes its minimum value if and only if these four points are vertices of a regular pyramid.

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#2 h Jul 25, 2008, 9:34 AM • 2 Y

Y by Adventure10, Mango247

Official Solution:
From the QM-HM inequality for the six numbers $|P_i-P_j|$ we have $\sum_{i\neq j}\frac{1}{|P_i-P_j|}\geq C\left(\sum_{i\neq j}|P_i-P_j|^2\right)^{-\frac 12}$ where $C$ is a constant. The equality holds iff $|P_i-P_j|$ 's are all equal, which is satisfied when $P_i$ 's are the vertices of a regular tetrahedron.

We now try to find an upper bound for $\sum_{i\neq j}|P_i-P_j|^2$ . Considering $P_i$ 's as vectors in 3D space, then this sum becomes $3\sum_i(P_i.P_i)-2\sum_{i\neq j}(P_i.P_j)$ . Adding and subtracting $\sum_i(P_i.P_i)$ we get $4\sum_i(P_i.P_i)-(\sum_iP_i).(\sum_iP_i)$ . But $P_i.P_i$ is a constant. Therefor the expression is maximized when $\sum_iP_i=0$ for which it is equal to $4$ . This condition is again satisfied when $P_i$ 's are the vertices of a regular tetrahedron.

Therefore, the condition of minimality of $\sum_{i\neq j}\frac 1{|P_i-P_j|}$ becomes that, $|P_i-P_j|$ 's should be equal for different $i,j$ 's and that $\sum_iP_i=0$ . But if $|P_i-P_j|$ has a constant value, then every face of the tetrahedron is an equilateral triangle, which itself shows that the tetrahedron is regular.

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#3 h Oct 24, 2008, 5:18 AM • 2 Y

Y by Adventure10, Mango247

Omid Hatami wrote:

Official Solution:
...
We now try to find an upper bound for $\sum_{i\neq j}|P_i - P_j|^2$ . Considering $P_i$ 's as vectors in 3D space, then this sum becomes $3\sum_i(P_i.P_i) - 2\sum_{i\neq j}(P_i.P_j)$ ...

I don't see this; isn't it $2\left(\sum_i|P_i|^2-\sum P_i\cdot P_j\right)$ ?

I guess it doesn't really matter in your proof.
Mine was similar, except I used Cauchy engel form:
$\sum (P_i-P_j)^{-1} \leq \frac{C}{\sum |P_i-P_j|}$
so we need upper bound for $\left(\sum |P_i-P_j|\right)^2\leq C' \sum |P_i-P_j|^2$ then end same as above

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#4 h Apr 17, 2013, 2:12 PM • 2 Y

Y by Adventure10, Mango247

Suppose $P_i=(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})$ for all $i\in\{1,2,3,4\}$ . Now call $f(i,j)=|P_i-P_j|$ .

Certainly we need to minimize $\sum \frac {1}{f(i,j)}$ .

$\sum f^2(i,j)=\sum(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$ .

And now so $\sum f^2(i,j)=4-\sum (\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$ .

Now, $(\sum f^2(i,j))(6)\geq (\sum f(i,j))^2\geq (\frac {36}{\sum \frac {1}{f(i,j)}})^2$ -

(**)

Now so $\sum \frac {1}{f(i,j)}$ is minimized when $\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=0$ .

Now so clearly we've two cases for when $\sum \frac {1}{f(i,j)}$ can be minimized, one if all of $P_i$ lies on a circle and vertices of a square, and if not then vertices of a regular tetrahedron. If first case would br true then, equality wouldn't hold for the inequality of (**). Thus we must have those are vertices of regular tetrahedron since $f(i,j)$ is constant for any $i,j$ .

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