Interior of Convex Angle is Convex Set

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Theorem

Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$.


Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle.


Then the set

$U = \left\{ {p + st \mathbf v + \left({1-s}\right) t \mathbf w : s \in \left({0\,.\,.\,1}\right) , t \in \R_{>0} }\right\}$

is a convex set.


Proof

Let $p_1 ,p_2 \in U$.

Then for $i \in \left\{ {1, 2}\right\}$, $p_i = p + s_i t_i \mathbf v + \left({1 - s_i}\right) t_i \mathbf w$ for some $s_i \in \left({0\,.\,.\,1}\right) , t_i \in \R_{>0}$.

WLOG assume that $t_1 \le t_2$.

Suppose that $q \in \R^2$ lies on the line segment joining $p_1$ and $p_2$, so:

\(\displaystyle q\) \(=\) \(\displaystyle p + s_1 t_1 \mathbf v + \left({1 - s_1}\right) t_1 \mathbf w + s \left({ p + s_2 t_2 \mathbf v + \left({1 - s_2}\right) t_2 \mathbf w - p - s_1 t_1 \mathbf v - \left({1 - s_1}\right) t_1 \mathbf w }\right)\) for some $s \in \left({0\,.\,.\,1}\right)$
\(\displaystyle \) \(=\) \(\displaystyle p + \left({ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}\right) \mathbf v + \left({ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}\right) \mathbf w\)
\(\displaystyle \) \(=\) \(\displaystyle p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \dfrac{t_1 + st_2 - st_1 - \left({1 - s}\right) s_1 t_1 - s s_2 t_2}{r} r \mathbf w\) where $r = t_1 + s \left({t_2 - t_1}\right)$
\(\displaystyle \) \(=\) \(\displaystyle p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \left({ 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} }\right) r \mathbf w\)

As $t_1 \le t_2$, it follows that $r \in \R_{>0}$.

We have $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}> 0$, and:

\(\displaystyle 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}\) \(=\) \(\displaystyle \dfrac{ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}{r}\)
\(\displaystyle \) \(>\) \(\displaystyle 0\)

It follows that $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} \in \left({0\,.\,.\,1}\right)$.

Then $q \in U$.

By definition of convex set, it follows that $U$ is convex.

$\blacksquare$