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Consider any triangle ABC. Connect the midpoints of each of the three sides. The inscribed triangle is equal to the other three triangles and they are all congruent. It turns out that the medians of the larger triangle are also the medians of the smaller, centrally inscribed, triangle.

(This is where it gets dicey)

I'd like to say that at this point, you could repeat this process on the smaller centrally inscribed triangle and then continue to do so infinitely. I'd then argue that at infinity, the points of the smallest triangle would be the same point. Tada, the medians are concurrent. Is this valid?

This was done for fun, I'd appreciate not being buried in a blizzard of Algebra or Calculus. Intuitively, does this work?

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Yes, it can be done, with a small blizzard of analysis. The intuition is good. The only problem is that to carry out details, we need to go well beyond elementary geometry. One would avoid the definitely dicey "at infinity" and replace it with the notion of limit. –  André Nicolas Dec 5 '12 at 5:51
    
Cool. I haven't learned any analysis yet, but have a reasonable grasp of what passes as "calculus" in a university curriculum. If you think that I won't be out of my depth with a small blizzard of analysis, I'd love to hear the explanation. Otherwise, if you want to reformat your comment as an answer, I'll accept it. –  Josh Infiesto Dec 5 '12 at 5:54
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2 Answers 2

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The intuition is good, it is a nice argument, and it can be made tight.

The only problem is that to carry out details, we need to go well beyond elementary geometry.

One needs to bypass the definitely dicey "at infinity" and replace it with the notion of limit. Let $T_0$ be the original triangle, $T_1$ the "central triangle" of $T_0$, $T_2$ the central triangle of $T_1$, and so on. We have what is technically called a nested sequence of triangles. If $d_n$ is the diameter of $T_n$, however one cares to define diameter, then $\lim_{n\to\infty}d_n=0$.

The three medians of $T_0$ pass through $T_n$, as you observed. It is a standard result that the intersection of the nested sequence $(T_n)$ consists of a single point, and the medians must all pass through this point.

You could tweak the proof to obtain the result that the meeting point of the medians divides each median in $2$ to $1$ ratio.

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Thanks, I thought it was pretty. The limits make sense. I had thought that I might be able to get the 2:1 result. I'll have to look at that more. –  Josh Infiesto Dec 5 '12 at 7:04
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The technical issue is that the fact that the medians are concurrent can be proved using a far weaker set of axioms than the ones we implicitly used. We used the Continuity Principle, essentially equivalent to the fact our geometry is over $\mathbb{R}^2$. Bu that is the ordinary geometry of the intuition, so it's OK. –  André Nicolas Dec 5 '12 at 7:15
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Your solution is clever, however...

In the case of "classic" plane geometry (straightedge-and-compass geometry) constructions which involve infinite iterations aren't usually allowed. If they were, I could describe a construction, involving an infinite number of steps / iterations, to trisect an angle.

Are you are restricting yourself to this limited sense of geometry?

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I was not. I was mostly hoping to satisfy myself in proving that they meet in a point. I wanted to make sure that my argument wasn't fallacious. –  Josh Infiesto Dec 5 '12 at 6:28
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