Thiessen weights for semicircular catchment

Setup: Semicircle of radius $r$ in the upper half-plane, area $A = \pi r^2/2$. Gauges: $O(0,0)$, $P_1(r/2, r/2)$, $P_2(-r/2, r/2)$.

        y
        |   P2 .  | . P1
        |   (-r/2,r/2) (r/2,r/2)
   -----+----O-----+---- x
       -r          r

Step 1: Perpendicular bisectors

• Between $P_1$ and $P_2$: they are at the same height, so the bisector is $x = 0$.
• Between $O$ and $P_1$: points equidistant from both satisfy $x^2+y^2 = (x-r/2)^2 + (y-r/2)^2$, which gives $x + y = r/2$.
• Between $O$ and $P_2$: similarly $-x + y = r/2$.

Step 2: Polygon of the gauge at O

The area closer to $O$ than to $P_1$ or $P_2$ is bounded by the x-axis, $y = r/2 - x$ and $y = r/2 + x$. This is a triangle with vertices $(-r/2, 0)$, $(r/2, 0)$ and $(0, r/2)$ (all inside the semicircle):

$$A_O = \tfrac{1}{2} \times r \times \frac{r}{2} = \frac{r^2}{4}$$ $$W_O = \frac{A_O}{A} = \frac{r^2/4}{\pi r^2/2} = \frac{0.5}{\pi}$$

Step 3: Remaining two gauges

The rest of the semicircle is divided by $x = 0$ into two equal halves (by symmetry):

$$A_{P_1} = A_{P_2} = \frac{1}{2}\left(\frac{\pi r^2}{2} - \frac{r^2}{4}\right)$$ $$W_{P_1} = W_{P_2} = \frac{1}{2}\left(1 - \frac{0.5}{\pi}\right) = 0.5 - \frac{0.25}{\pi}$$

Numerically: $W_O = 0.159$, $W_{P_1} = W_{P_2} = 0.4204$; the sum $= 1.000$. Hence proved.