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

Revision as of 15:49, 25 September 2024

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:

$\emph{Lemma:}$ Let $MNOP$ be a regular tetrahedron, and $T$ a point inside it. Let $x_1, x_2, x_3, x_4$ be the distances from $T$ to the faces $MNO, MNP, MOP$ , and $NOP$ . Then, $x_1 + x_2 + x_3 + x_4$ is constant, independent of $T$ .

$\emph{Proof:}$

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

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

because the areas of the four triangles are equal. ( $[ABC]$ stands for the area of $\triangle ABC$ .) Then

$\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.$

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 $D$ .

For any point $P$ in $ABCD$ ,

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

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

Remarks (added by pf02, September 2024)

1. The text of the Lemma needed a little improvement, which I did.

2. The Solution above is not complete. It considered only points $P$ inside the tetrahedron, but the problem specifically said "any other point in space".

3. I will give another solution below, in which I will address the issue I mentioned in the preceding paragraph.

Solution 2

TO BE CONTINUED. SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.

(Solution by pf02, September 2024)

@@ Line 12: / Line 12: @@
 <math>\emph{Proof:}</math>
-We will compute the volume of <math>MNOP</math> in terms of the areas of the faces and the distances from the point to the faces:
+We will compute the volume of <math>MNOP</math> in terms of the areas of the faces and the
+distances from the point <math>T</math> to the faces:
 <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>
-because the areas of the four triangles are equal.  Then
+because the areas of the four triangles are equal.  (<math>[ABC]</math> stands for
+the area of <math>\triangle ABC</math>.)  Then
 <cmath>\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.</cmath>

1966 IMO (Problems) • Resources
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Difference between revisions of "1966 IMO Problems/Problem 3"

Revision as of 15:49, 25 September 2024

Contents

Solution

Remarks (added by pf02, September 2024)

Solution 2

See Also