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Let $C, D, E$ be three non-degenerate triangles in $\mathbb R^2$. Let $c, a, b$ be the vertices of $C$, let $d, a, b$ be the vertices of $D$, and let $e, a, b$ be the vertices of $E$. I want to show that there is one point contained in the interior of at least two of the given triangles.

Here are my thoughts: If two of the triangles are the same then we're done, so suppose without loss of generality that $C$ and $D$ are different. Think of the triangles as simplices. By definition, this means that the sets $\{c - a, b - a\}$, $\{d - a, b - a\}$, and $\{e - a, b - a\}$ are linearly independent. Since $C$ and $D$ are different, this must mean that $\{c - a, d - a\}$ is also linearly independent, hence it is a basis of $\mathbb R^2$. This means I can write $e - a = \gamma(c-a) + \delta(d-a)$ for some appropriate $\gamma,\delta$. Now any point on the triangle $E$ can be written as $\alpha a + \beta b + \epsilon e$ where $\alpha, \beta, \epsilon$ are nonnegative and sum to 1. Using the fact that $e - a = \gamma(c-a) + \delta(d-a)$, we can write any point on the triangle $E$ as $(\alpha + \epsilon(1-\gamma-\delta))a + \beta b + \gamma\epsilon c + \delta\epsilon d$. I was hoping to make this latter sum into a convex sum with just $a, b, c$ or $a, b, d$ by picking $\epsilon$ appropriately and therefore showing that $E$ overlaps with $C$ or $D$, but alas that doesn't work.

So, is this the right way of going about this? Or is there a better approach?

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I don't see a question mark here. Is your question "How do I prove this?", or "Can someone give me a hint in the right direction?", or something more specific like "Can this line of reasoning be salvaged?" For what it's worth, I don't think picking $\{c-a, d-a\}$ as a basis is going to help, because it "forgets about" the relative position of $b$. You probably want to formalize in some way the notion that the line through $a$ and $b$ divides the plane into two halves, so two of the three points $c$, $d$ and $e$ must be in the same half. – Rahul Mar 19 '11 at 21:08
Point taken. I have added an explicit question. I like the idea of dividing the plane as you suggested and will think about it some more. – echoone Mar 20 '11 at 3:05
up vote 2 down vote accepted

Draw a small circle centered at the midpoint of $[a,b]$. Each of the triangles $C$, $D$ and $E$ contains one half of this circle. Now apply the pigeon hole principle.

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That is very clever. Thanks. – echoone Mar 21 '11 at 16:40

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