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This problem is from A ProblemText in Advanced Calculus by J. M. Erdman (Chap. 16: The Heine-Borel Theorem, p. 91).

Use the Cantor intersection theorem to show that the medians of a triangle are concurrent.

This is the problem that surprised me the most in the full text. Any suggestions?


Here's the statement of the theorem we are supposed to use:

Theorem (Cantor intersection theorem). If $(A_n)$ is a nested sequence of nonempty closed bounded subsets of $\mathbb R^n$, then $\bigcap\limits_{n=1}^\infty A_n$ is nonempty. Furthermore, if $\operatorname{diam} A_n \to 0$, then $\bigcap\limits_{n=1}^\infty A_n$ is a single point.

My thoughts.

Of course, I know some standard proofs of this result (using plane geometry or vectors). But here, we are required to use the Cantor intersection theorem.

Fix an arbitrary triangle $\triangle$. I originally started out with $A_1 := \triangle$. My goal was to assume that the medians are not concurrent, and derive a contradiction. It is clear that if the medians are not concurrent, then they form a mini-triangle inside $A_1$. Should I define that triangle to be $A_2$ and proceed recursively to define $A_3, A_4, \ldots$? This idea is the only one I could think of. However, this does not seem good enough: I cannot see any contradiction arising from the conclusion that the intersection of all the $A_n$'s is nonempty (or a singleton set).

Can you suggest a better approach?

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    $A_iB_{i+1}A_{i+1}C_{i+1}$ is parallelogram, thus the diagonals half each other ;)2011-11-06

1 Answers 1

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Try apply the Cantor intersection theorem to the triangles made by the midpoints of the edges.

If $ABC$ is your triangle, then $A_1:=A, B_1:=B, C_1:=C$; and inductively $A_{i+1}$ is the midpoint of $B_iC_i$; $B_{i+1}$ is the midpoint of $A_iC_i$, $C_{i+1}$ is the midpoint of $A_iB_i$...

The fixed point of the Theorem should be $G$. It should follow immediately from the fact that $A_iA_{i+1}$ is median in both $A_iB_iC_i$ and $A_{i+1}B_{i+1}C_{i+1}$.

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    +1 Thanks, @N.S. Now, it's fully clear. Your first reply was very helpful to me; perhaps you can add it into the answer.2011-11-06