Weighted sum of points within unit circle

complex numbersgeometry

Say you have two vectors within the unit circle, $r_1 = [x_1, y_1]$, $r_2 = [x_2, y_2]$. How can I prove that the sum $r_3 = \alpha_1*r_1 + \alpha_2*r_2$ is within the unit circle if $\alpha_1 + \alpha_2 = 1$ and $\alpha_{1,2} \geq 0$.

First of all, is it correct that the sum should be within the unit circle?

I tried using the sum of squares inequality, saying that the points are vectors originating in origo, but I did not get anywhere with that.

Best Answer

We have that

  • $|r_1|^2=x_1^2+y_1^2\le 1$
  • $|r_2|^2=x_2^2+y_2^2\le 1$
  • $r_1\cdot r_2=x_1x_2+y_1y_2\le 1$
  • $\alpha_1+\alpha_2=1\implies \alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2=1$

and

$$r_3 = \alpha_1\cdot r_1 + \alpha_2\cdot r_2=(\alpha_1\cdot x_1 + \alpha_2\cdot x_2,\alpha_1\cdot y_1 + \alpha_2\cdot y_2)$$

and therefore

$$|r_3|^2=r_{3,x}^2+r_{3,y}^2=(\alpha_1\cdot x_1 + \alpha_2\cdot x_2)^2+(\alpha_1\cdot y_1 + \alpha_2\cdot y_2)^2=$$

$$=\alpha_1^2\cdot (x_1^2+y_1^2) + \alpha_2^2\cdot (x_2^2+y_2^2)+2\alpha_1\alpha_2\cdot (x_1x_2+y_1y_2)\le$$

$$\le \alpha_1^2 + \alpha_2^2+2\alpha_1\alpha_2=1$$

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