Triangle inside an open disk

Question

Given an open disk $D=\{x \in \mathbb{R}^2 \mid |x-x_0|<r\}$ and fixed three points $x_1,x_2,x_3 \in D$, how can I show that the triangle (of vertices $x_1,x_2,x_3$) $\Delta=\{t_1x_1+t_2x_2+t_3x_3 \in \mathbb{R}^2 \mid t_1,t_2,t_3 \ge 0 \quad \land \quad t_1+t_2+t_3=1\}$ is all inside $D$, namely $\Delta \subseteq D$.

I know that I have to show that $|x-x_0|<r$ for all $x=t_1x_1+t_2x_2+t_3x_3 \in \Delta$ using the triangular inequality and the fact that $|x_1-x_0|<r,|x_2-x_0|<r,|x_3-x_0|<r$.

However after one hour I can't find any valid approach.

Any kind of hint would be very appreciated. Thank you!

Answer 1

Use the fact that\begin{align}|t_1x_1+t_2x_2+t_3x_3-x_0|&=|t_1x_1+t_2x_2+t_3x_3-(t_1+t_2+t_3)x_0|\\&\leqslant t_1|x_1-x_0|+t_2|x_2-x_0|+t_3|x_3-x_0|\\&<(t_1+t_2+t_3)r\\&=r.\end{align}

Answer 2

Hint, not a complete answer First, by subtracting $x_0$ from everything, you can reduce to the case where $x_0 = (0,0)$.

Second, try proving a lemma: if $P, Q \in D$, and $0 \le t \le 1$, then $(1-t)P + t Q \in D$.

The general statement then follows, because the lemma shows that every point on the edge from $x_1$ to $x_2$ is in $D$, and then every point of the triangle is a convex combination of some point on that edge with the point $x_3$. That's clear visually, but you need to turn that visual observation into a new way of writing the sum $t_1 x_1 + t_2 x_2 + t_3 x_3$.