if $P_n$ is the midpoint of the line segment joining $P_{n−2}$ and $P_{n−3}$, then prove area converges to 0

Question

Consider a sequence $P_1,P_2,…$ of points in the plane such that $P_1,P_2,P_3$ are non-collinear and for every n≥4,$P_n$ is the midpoint of the line segment joining $P_{n−2}$ and $P_{n−3}$.

Let L denote the line segment joining P1 and P5. Prove that the area of the triangle formed by the points $P_n,P_{n−1},P_{n−2}$ converges to zero as n goes to infinity.

I thought of proving the question using coordinate system taking an arbitrary triangle but after $P_4$ and $P_5$ calculating $P_6$ disrupts the symmetry. Please suggest a way to navigate the problem.

Answer 1

Let $O$ be an arbitrary point of the plane. Consider vectors $v_n:=\overrightarrow{OP_n}$. Then, for all $n\ge 3$ we have $$ v_{n}=\frac{1}{2}(v_{n-2}+v_{n-3}) $$ or $$ v_{n}-v_{n-1}=-(v_{n-1}-v_{n-2})-\frac{1}{2}(v_{n-2}-v_{n-3}). $$ The area of the triangle $P_{n-2}P_{n-1}P_{n}$ equals the half of length of the cross product of $v_{n}-v_{n-1}$ and $v_{n-1}-v_{n-2}$. Put $u_n=v_{n}-v_{n-1}$. Then, for all $n\ge 3$ we have $$ u_{n}=-u_{n-1}-\frac{1}{2}u_{n-2}. $$ Now it suffices to prove that $[u_{n-1},u_{n}]\to 0$ when $n\to\infty$.

Can you continue now?

Answer 2

For $n = 4, 5, 6, \ldots$, since $P_n$ is the midpoint of the segment joining the points $P_{n-3}$ and $P_{n-2}$, we can conclude that $$ \vec{OP_n} = \frac{1}{2} \left( \vec{OP_{n-3}} + \vec{OP_{n-2}} \right). \tag{1} $$

So \begin{align} & \ \ \ \mbox{area } \triangle P_{n-2} P_{n-1} P_n \\ &= \frac{1}{2} \left\lvert \vec{ P_{n-2} P_n } \times \vec{ P_{n-2} P_{n-1} } \right\rvert \\ &= \frac{1}{2} \left\lvert \left( \vec{OP_n} - \vec{OP_{n-2} } \right) \times \left( \vec{OP_{n-1}} - \vec{OP_{n-2} } \right) \right\rvert \\ &= \frac{1}{2} \left\lvert \left[ \frac{1}{2} \left( \vec{OP_{n-3}} + \vec{OP_{n-2}} \right) - \vec{OP_{n-2} } \right] \times \left( \vec{OP_{n-1}} - \vec{OP_{n-2} } \right) \right\rvert \\ &= \frac{1}{2} \left\lvert \frac{1}{2} \left( \vec{OP_{n-3}} - \vec{OP_{n-2}} \right) \times \left( \vec{OP_{n-1}} - \vec{OP_{n-2}} \right) \right\rvert \\ &= \frac{1}{4} \left\lvert \vec{ OP_{n-1} } \times \vec{OP_{n-2}} + \vec{ OP_{n-2} } \times \vec{OP_{n-3}} + \vec{ OP_{n-3} } \times \vec{OP_{n-1}} \right\rvert \\ & = \ldots \ldots \end{align}

To be honest, I'm not sure how to proceed from here. The above calculation might show you some way.