What is graph of this combination?

Question

If $x=(0,0),y=(1,0),z=(0,1)$ then what is the graph of the combination $$t_1^{1/2}x+t_2^{1/2}y+t_3^{1/2}z,$$ where $t_1+t_2+t_3=1$ and $0\leq t_1,t_2,t_3\leq 1$. Thank you.

Answer 1

Let $S$ be the set of points $(a,b) \in \mathbb{R}^2$ which can be expressed as $$(a,b) = x\sqrt{t_1} + y\sqrt{t_2} + z\sqrt{t_3}$$ where $$x = (0,0),\;y=(1,0),\;z=(0,1)$$ $$t_1 + t_2 + t_3=1$$ $$t_1,t_2,t_3 \ge 0$$ Let $D$ be the set of points of the standard unit disk which are in the first quadrant, including the boundary points.

Claim $S=D$.

First we show $S \subseteq D$.

Let $(a,b) \in S$. Then $$(a,b) = x\sqrt{t_1} + y\sqrt{t_2} + z\sqrt{t_3}$$ $$x = (0,0),\;y=(1,0),\;z=(0,1)$$ $$t_1 + t_2 + t_3=1$$ $$t_1,t_2,t_3 \ge 0$$ hence $$(a,b) = \left(\sqrt{t_2}, \sqrt{t_3}\right)$$ $$t_2 + t_3\le1$$ $$t_2,t_3 \ge 0$$ which implies $$a^2 + b^2 \le 1$$ $$a,b \ge 0$$ so $(a,b) \in D$, and thus, $S \subseteq D$, as claimed.

Next we show $D \subseteq S$.

Let $(a,b) \in D$. Then $$a^2 + b^2 \le 1$$ $$a,b \ge 0$$ Letting$\;\;t_2=a^2,\;\;t_3=b^2,\;\;t_1=1-t_2-t_3,\;\;$we have $$t_1 + t_2 + t_3 = 1$$ $$t_1,t_2,t_3 \ge 0$$ and \begin{align*} (a,b) &= \left(\sqrt{t_2},\sqrt{t_3}\right)\\[4pt] &= x\sqrt{t_1} + y\sqrt{t_2} + z\sqrt{t_3}\\[4pt] &\in S\\[4pt] \end{align*} hence $D \subseteq S$, as claimed.

Therefore $S=D$.

Answer 2

Note $f(t_1, t_2, t_3) = t_1^{1/2}x + t_2^{1/2}y + t_3^{1/2}z$, you can notice that $f(t_1, t_2, t_3) = (\sqrt{t_2}, \sqrt{t_3})$.

Now, $t_1$ doesn't really matter anymore and you can consider $$T = \left\{ (t_2, t_3) ~\middle|~ t_1 + t_2 + t_3 = 1 \wedge 0\leq t_1, t_2, t_3 \leq 1\right\} = \left\{ (t_2, t_3) ~\middle|~ t_2 + t_3 \leq 1 \wedge 0\leq t_2, t_3 \leq 1\right\}$$ which is a triangle with corners $x$, $y$ and $z$.

Now, what you are interested in is $f(T)$. First consider the image of the sides of your triangle. You easily see that for the sides $A = [x,y]$ and $B = [x,z]$, the image is the same as the original. But for the side $C = [y,z]$, you get something different. For $C$, just parametrize it as $$C = \left\{ (t_2,1-t_2) ~\middle|~ 0\leq t_2 \leq 1\right\}.$$ You then see that $$f(C) = \left\{ (\sqrt{t_2},\sqrt{1-t_2}) ~\middle|~ 0\leq t_2 \leq 1\right\}$$ which is the quarter perimeter of the circle of radius 1 and centre $x$, from $y$ to $z$.

So in the end, you get $f(T)$ is the area delimited by $f(A)$, $f(B)$ and $f(C)$ which is simply the area of the quarter circle described previously.