# Medians of a triangle are concurrent: prove using the Cantor intersection theorem

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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I’d let $A_{n+1}$ be the triangle whose vertices are the midpoints of the sides of $A_n$. It’s clear that you’ll get a single point of intersection, and similar triangles should let you show pretty easily that the vertices of the $A_n$ form three collinear sets. (That has to be the oddest use of the Cantor intersection theorem that I’ve seen, though!) –  Brian M. Scott Nov 6 '11 at 6:24
@Brian I'm clear till the single point of intersection bit (call this point $G$). Can you elaborate on the remaining part? Do I use $G$ to show that the vertices of $A_n$ "form collinear sets"? –  Srivatsan Nov 6 '11 at 6:27
@Brian I agree with your comment about the oddest use of Cantor intersection. It's as if that the author really starts off with a reasonable standard geometry (similar triangles) proof, and then injects some analysis into it... –  Srivatsan Nov 6 '11 at 6:28
@Srivatsan: I posted a solution some time ago, the last line tells you how to prove that the vertices are colinear ;) –  N. S. Nov 6 '11 at 6:32
$A_{n+2}$ is similar to $A_n$ with sides parallel to those of $A_n$. $A_{n+1}$ also has sides parallel to these, but it’s inverted. I’ve not thought it through carefully, but it appears that you can use similar triangles to show that corresponding vertices of the triangles are collinear; the three lines are of course the original medians. –  Brian M. Scott Nov 6 '11 at 6:37

## 1 Answer

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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Thanks! This looks helpful, but I need some more explanation in the last line. Is $G$ the single point of intersection of all the triangles? And what exactly follows from that fact that $A_i A_{i+1}$ is median of both $A_iB_iC_i$ and $A_{i+1}B_{i+1}C_{i+1}$? –  Srivatsan Nov 6 '11 at 6:36
Since $A_iB_{i}$ is half of $A_{i-1}B_{i-1}$ it is easy to argue that the intersection is a single point. Lets call this point $G$. Now, the fact that $A_iA_{i+1}$ is median in both triangles, it follows by induction that the line supporting all medians $A_iA_{i+1}$ is the same... You can now argue that $G$ must be on this line.... Or you could simply apply the cantor intersection Theorem directly to the intervals $A_iA_{i+1}$ and then again to $B_iB_{i+1}$ and $C_iC_{i+1}$. Actually this last idea might be better :) –  N. S. Nov 6 '11 at 6:41
@Srivatsan The fact that $A_iA_{i+1}$ is median in both triangles follows immediately from $A_iB_{i+1}A_{i+1}C_{i+1}$ parallelogram. –  N. S. Nov 6 '11 at 6:43
+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. –  Srivatsan Nov 6 '11 at 6:46