sequence of triangles by connecting midpoints, centroid interior wanted

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sequence of triangles by connecting midpoints, centroid interior wanted

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Source: Chile 2007 L1 p2

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#1 h May 26, 2019, 8:39 AM • 2 Y

Y by Adventure10, Mango247

From a triangle $T = \triangle ABC$ , we build the triangle $T_1 = \triangle A_1B_1C_1$ whose vertices they are the midpoints of the sides of $T$ . The triangle $T_2 = \triangle A_2B_2C_2$ is constructed from $T_1$ in a way analogue. We build the triangles $T_3, T_4,..., T_{2007}$ . Prove that the center of gravity $G$ of the triangle $T$ is inside the triangle $T_{2007}$ .

This post has been edited 2 times. Last edited by parmenides51, May 26, 2019, 8:43 AM
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#2 h May 26, 2019, 9:53 AM • 1 Y

Y by Adventure10

The center of gravity of all the $T_i's$ are their centroids.
All the triangles will have same centroid

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#3 h May 27, 2019, 6:50 AM • 1 Y

Y by Adventure10

Here's an inductive proof (Strong induction on $T_i$ )
Claim: The centroid of all the triangle $T_i$ are same and hence the center of mass
Remark: 2007 is not special
Proof: We induct on $i$ and leave the base cases as an exercise to the reader. Now assume that the centroid of $T,T_1,T_2,\ldots,T_{n-1}$ is same for some positive integer $n\ge 1$ . Let the vertices $A_{n-1},B_{n-1},C_{n-1}$ of triangle $T_{n-1}$ have complex numbers $a,b,c$ respectively. Therefore, the COM of all the triangles $T,T_1,T_2,\ldots,T_{n-1}$ will be $\frac{a+b+c}{3}$ by our inductive argument. Obviously, the vertices of $T_n$ will contain the complex numbers $\frac{a+b}{2},\frac{b+c}{2},\frac{c+a}{2}$ and hence COM(centroid) of $T_n$ will be $\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{3}=\frac{a+b+c}{3}$ which is same as the COM of triangles $T,T_1,T_2,\ldots,T_{n-1}$ completing our induction hypothesis.

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