Tor-tiled Triangles

Geometry Level 4

Equilateral triangle $A_1B_1C_1$ has side length $1$ . A median is drawn from point $A_1$ to hit $B_1C_1$ at $A_2$ . $B_1$ is renamed $C_2$ and $A_1$ is renamed $B_2$ . Now, the same maneuver is done to triangle $A_2B_2C_2$ to produce triangle $A_3B_3C_3$ , and so on, until infinity. Points $A_n,B_n,C_n$ ultimately coincide to point $P$ as $n$ approaches infinity. $AP^2+BP^2+CP^2$ can be expressed as $\dfrac{p}{q}$ for positive coprime integers $p,q$ . Find $p+q$ .

The answer is 51.

1 solution

Jack D'Aurizio
Apr 15, 2014

It is easy to see that the triangle $A_5 B_5 C_5$ is equilateral, too, so the limit point $P$ is just the perspector of $ABC$ and $A_5 B_5 C_5$ . Let $P_A,P_B,P_C$ be the intersections of $AP, BP, CP$ with $BC,AC,BC$ : it is not difficult to see that $\frac{CP}{PA}=\frac{CP}{PB}=2$ while $\frac{AP}{PB}=1$ . Let $m_c$ be the length of $CP$ : by Van Obel's theorem we have $\frac{CP}{AP_C}=4$ , and Pythagoras' theorem gives: $PA^2+PB^2+PC^2 = 2 AP^2+2 PP_C^2+CP^2 = \frac{1}{2}+\frac{2}{25}m_C^2+\frac{16}{25}m_C^2 = \frac{1}{2}+\frac{18}{25}\cdot\frac{3}{4}=\frac{26}{25}.$

If my understanding is correct, then this is what the diagram should look like:

https://i.imgur.com/C1eMFvv.png

If this is not right, can someone post the actual diagram?

Jon Haussmann - 7 years, 1 month ago

Yes, except for we are drawing medians, not altitudes. The altitude problem also seems interesting to explore, though.

Daniel Liu - 7 years, 1 month ago

Right, medians. I had altitudes stuck in my head. Thanks for clarifying.

Jon Haussmann - 7 years, 1 month ago

Tor-tiled Triangles

The answer is 51.

1 solution

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