b u m p...

b u m p...

Let areas $S_0 = [P_1OP_2]$, $S_1 = [P_0OP_2]$, and $S_2 = [P_0OP_1]$, and let vectors $p_0 = (a_0, b_0)$, $p_1 = (a_1, b_1)$, and $p_2 = (a_2, b_2)$. We claim that $S_0p_0 + S_1p_1 + S_2p_2 = \vec{0}$. Since $O$ is in the interior of $\triangle P_0P_1P_2$, $P_1$ and $P_2$ must be on different sides of line $OP_0$. So the components of $p_1$ and $p_2$ which are perpendicular to $p_0$ must be in opposite directions. Furthermore, the ratio of their magnitudes is $\frac{OP_1 \cdot sin\angle P_0OP_1}{OP_2 \cdot sin\angle P_0OP_2} = \frac{1/2 \cdot OP_0 \cdot OP_1 \cdot sin\angle P_0OP_1}{1/2 \cdot OP_0 \cdot OP_2 \cdot sin\angle P_0OP_2} = \frac{S_2}{S_1}$. Thus, $S_1p_1 + S_2p_2$ is parallel to $p_0$, which implies $S_0p_0 + S_1p_1 + S_2p_2$ is parallel to $p_0$. Similarly, $S_0p_0 + S_1p_1 + S_2p_2$ is parallel to $p_1$ and $p_2$. Therefore, $S_0p_0 + S_1p_1 + S_2p_2$ must be the zero vector $\vec{0}$. If $S_2, S_1, S_0$ are in a geometric sequence, then $S_1 = S_2r$ and $S_0 = S_2r^2$ for some positive real $r$. So $S_0p_0 + S_1p_1 + S_2p_2 = \vec{0} \iff r^2p_0 + rp_1 + p_2 = \vec{0} \iff r^2a_0 + ra_1 + a_2 = r^2b_0 + rb_1 + b_2 = 0$, which means there exists a real number $x$ such that $a_0x^2 + a_1x + a_2 = b_0x^2 + b_1x + b_2 = 0$. If there exists a real number $x$ such that $a_0x^2 + a_1x + a_2 = b_0x^2 + b_1x + b_2 = 0$, then $x^2p_0 + xp_1 + p_2 = \vec{0}$. Then $$\vec{0} = S_2\vec{0} - \vec{0} = S_2(x^2p_0 + xp_1 + p_2) - (S_0p_0 + S_1p_1 + S_2p_2) = (S_2x^2 - S_0)p_0 + (S_2x - S_1)p_1$$but $p_0$ and $p_1$ cannot be of the same or opposite directions, so $S_2x^2 - S_0 = 0 \iff S_0 = S_2x^2$ and $S_2x - S_1 = 0 \iff S_1 = S_2x$, which implies $S_2, S_1, S_0$ are in a geometric sequence.

I think power of a point may help here