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Difference between revisions of "1966 IMO Problems/Problem 3"

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<cmath>\textrm{Volume}(MNOP) = [MNO] \cdot x_1 \cdot \frac{1}{3} + [MNP] \cdot x_2 \cdot \frac{1}{3} + [MOP] \cdot x_3 \cdot \frac{1}{3} + [NOP] \cdot x_4 \cdot \frac{1}{3}</cmath>
 
<cmath>\textrm{Volume}(MNOP) = [MNO] \cdot x_1 \cdot \frac{1}{3} + [MNP] \cdot x_2 \cdot \frac{1}{3} + [MOP] \cdot x_3 \cdot \frac{1}{3} + [NOP] \cdot x_4 \cdot \frac{1}{3}</cmath>
 +
<cmath> = [MNO] \cdot \frac{(x_1 + x_2 + x_3 + x_4)}{3}</cmath>
 
<cmath>\therefore\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.</cmath>
 
<cmath>\therefore\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.</cmath>
  

Revision as of 19:27, 9 December 2022

Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

Solution

We will need the following lemma to solve this problem:

$\emph{Lemma:}$ Suppose there is a point in a regular tetrahedron $MNOP$ such that the distances from this point to the faces $MNO$, $MNP$, $MOP$, and $NOP$ are, respectively, $x_1$, $x_2$, $x_3$, and $x_4$. Then, the value $x_1 + x_2 + x_3 + x_4$ is constant.

$\emph{Proof:}$

We will compute the volume of $MNOP$ in terms of the areas of the faces and the distances from the point to the faces:

\[\textrm{Volume}(MNOP) = [MNO] \cdot x_1 \cdot \frac{1}{3} + [MNP] \cdot x_2 \cdot \frac{1}{3} + [MOP] \cdot x_3 \cdot \frac{1}{3} + [NOP] \cdot x_4 \cdot \frac{1}{3}\] \[= [MNO] \cdot \frac{(x_1 + x_2 + x_3 + x_4)}{3}\] \[\therefore\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.\]

This value is constant, so the proof of the lemma is complete.

$\emph{Proof of problem statement:}$

Let our tetrahedron be $ABCD$, and the center of its circumscribed sphere be $O$. Construct a new regular tetrahedron, $WXYZ$, such that the centers of the faces of this tetrahedron are at $A$, $B$, $C$, and $D$.

It is trivial that if we have a point that has the least sum of the distances to $A$, $B$, $C$, and $D$, then it is inside $ABCD$.

Suppose that we have a point inside of $ABCD$. Let this point be $P$. Then,

\[OA + OB + OC + OD = \sum \textrm{Distances from }O\textrm{ to faces of }WXYZ\] \[= \sum \textrm{Distances from }P\textrm{ to faces of }WXYZ \leq PA + PB + PC + PD,\]

with equality only occurring when $AP$, $BP$, $CP$, and $DP$ are perpendicular to the faces of $WXYZ$, meaning that $P = O$. This completes the proof. $\square$

~mathboy100

See Also

1966 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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