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

$U = \set {p + s t \mathbf v + \paren {1 - s} t \mathbf w : s \in \openint 0 1, t \in \R_{>0} }$

is a convex set.


Proof

Let $p_1, p_2 \in U$.

Then for $i \in \set {1, 2}$, $p_i = p + s_i t_i \mathbf v + \paren {1 - s_i} t_i \mathbf w$ for some $s_i \in \openint 0 1, t_i \in \R_{>0}$.


InteriorOfConvexAngle.png

Without loss of generality, 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:

\(\ds q\) \(=\) \(\ds p + s_1 t_1 \mathbf v + \paren {1 - s_1} t_1 \mathbf w + s \paren {p + s_2 t_2 \mathbf v + \paren {1 - s_2} t_2 \mathbf w - p - s_1 t_1 \mathbf v - \paren {1 - s_1} t_1 \mathbf w}\) for some $s \in \openint 0 1$
\(\ds \) \(=\) \(\ds p + \paren {\paren {1 - s} s_1 t_1 + s s_2 t_2} \mathbf v + \paren {\paren {1 - s} \paren {1 - s_1} t_1 + s \paren {1 - s_2} t_2} \mathbf w\)
\(\ds \) \(=\) \(\ds p + \dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r r \mathbf v + \dfrac {t_1 + s t_2 - s t_1 - \paren {1 - s} s_1 t_1 - s s_2 t_2} r r \mathbf w\) where $r = t_1 + s \paren {t_2 - t_1}$
\(\ds \) \(=\) \(\ds p + \dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r r \mathbf v + \paren {1 - \dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r} r \mathbf w\)

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


We have:

$\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r > 0$

and so:

\(\ds 1 - \dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r\) \(=\) \(\ds \dfrac {\paren {1 - s} \paren {1 - s_1} t_1 + s \paren {1 - s_2} t_2} r\)
\(\ds \) \(>\) \(\ds 0\)

It follows that:

$\dfrac {\paren {1 - s} s_1 t_1 + s s_2 t_2} r \in \openint 0 1$

Then $q \in U$.

It follows by definition that $U$ is convex.

$\blacksquare$