Characterization of Interior of Triangle

Theorem
Let $\triangle$ be a triangle embedded in $\R^2$.

Denote the vertices of $\triangle$ as $A_1, A_2, A_3$.

For $i \in \left\{ {1, 2, 3}\right\}$, put $j = i \bmod 3 + 1$, $k = \left({i+1}\right) \bmod 3 + 1$, and:


 * $U_i = \left\{ {A_i + st \left({A_j - A_i}\right) + \left({1-s}\right) t \left({A_k - A_i}\right) : s \in \left({0\,.\,.\,1}\right), t \in \R_{>0} }\right\}$

Then:


 * $\displaystyle \operatorname{Int} \left({\triangle}\right) = \bigcap_{i \mathop = 1}^3 U_i$

where $\operatorname{Int} \left({\triangle}\right)$ denotes the interior of the boundary of $\triangle$.

Proof
From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\operatorname{Int} \left({\triangle}\right)$ is well-defined.

Interior is Subset
Let $q \in \operatorname{Int} \left({\triangle}\right)$.

Let $i \in \left\{ {1, 2, 3}\right\}$, and put $j = i \bmod 3 + 1$, $k = \left({i+1}\right) \bmod 3 + 1$.

Let $S_i$ be the side of $\triangle$ that is adjacent to $A_i$ and $A_j$, let $S_j$ be the side adjacent to $A_j$ and $A_k$, and let $S_k$ be the side adjacent to $A_k$ and $A_i$.

Define $\mathbf v = A_j - A_i \in \R^2$, $\mathbf w = A_k - A_i \in \R^2$, and define $\mathbf u = \mathbf v - \mathbf w$.

Define two rays $\mathcal L = \left\{ {q + s \mathbf u: s \in \R_{\ge 0} }\right\}$, and $\mathcal L' = \left\{ {q + s' \left({- \mathbf u}\right) : s' \in \R_{\ge 0} }\right\}$.

As both rays are parallel to $S_j$, $\mathcal L$ or $\mathcal L'$ can only cross $S_j$ if $S_i$ and $S_k$ lie on opposite sides of $S_j$.

This would imply that there is a non-convex angle in $\triangle$.

As Sum of Angles of Triangle Equals Two Right Angles shows, this is impossible, as a non-convex angle is larger than two right angles.

Then, neither $\mathcal L$ nor $\mathcal L'$ crosses $S_j$.

As $\mathcal L \cup \mathcal L'$ is a straight line, and $\mathcal L \cap \mathcal L' = \left\{ {q}\right\}$, it follows that $\mathcal L$ and $\mathcal L'$ cannot both intersect the same side.

Then Jordan Polygon Interior and Exterior Criterion shows that $\mathcal L$ and $\mathcal L'$ each crosses one of the sides $S_i$ and $S_k$.

When we denote the intersection points $p_1, p_2 \in \R^2$, we have:


 * $p_1 = A_i + r_1 \mathbf v = q + r \left({\mathbf v - \mathbf w}\right)$
 * $p_2 = A_i + r_2 \mathbf w = q + r' \left({\mathbf v - \mathbf w}\right)$

for some $r_1, r_2 \in \left({0\,.\,.\,1}\right)$, where either $r, -r' \in \R_{>0}$ or $-r, r' \in \R_{>0}$.

Subtracting the two equations gives:


 * $r_1 \mathbf v - r_2 \mathbf w = \left({r - r'}\right) \left({\mathbf v - \mathbf w}\right)$

which can be rearranged as:


 * $\left({r_1 + r' - r}\right) \mathbf v - \left({r_2 + r' - r}\right) \mathbf w$

As $\mathbf v$ and $\mathbf w$ are direction vectors for the adjacent sides $S_i$ and $S_k$, they cannot be parallel, so $v$ and $w$ are linearly independent.

It follows that $0 = r_1 + r' - r = r_2 + r' - r$, so $r_1 = r_2$.

Adding the two equations gives:


 * $2 A_i + r_1 \mathbf v + r_2 \mathbf w = 2q + \left({r + r'}\right) \left({\mathbf v - \mathbf w}\right)$

which can be rearranged to give an expression for $q$:

This shows that $q \in U_i$.

As $i \in \left\{ {1, 2, 3}\right\}$ was arbitrary, it follows that $\displaystyle q \in \bigcap_{i \mathop = 1}^3 U_i$.

Interior is Superset
Let $\displaystyle q \in \bigcap_{i \mathop = 1}^3 U_i$.

For $i \in \left\{ {1, 2, 3}\right\}$, define $j, k \in \left\{ {1, 2, 3}\right\}$, the sides $S_i, S_j, S_k$ of $\triangle$, and their direction vectors $\mathbf v, \mathbf u, \mathbf w$ as in the section above.

As $q \in U_i$, it follows that $q = A_i + st \mathbf v + t \left({1-s}\right) t \mathbf w$ for some $s \in \left({0\,.\,.\,1}\right)$ and $t \in \R_{>0}$.

Let $\mathcal L = \left\{ {q + r \left({- \mathbf v}\right) : r \in \R_{\ge 0} }\right\}$ be a ray with start point $q$.

If $\mathcal L$ crosses the side $S_i$ that $\mathcal L$ is parallel to, then the intersection point is:


 * $A_i + r_1 \mathbf v = q - r \mathbf v$

for some $r_1 \in \left[{0\,.\,.\,1}\right], r \in \R_{\ge 0}$, which can be rearranged as:


 * $q = A_i + \left({r_1 - r}\right) \mathbf v + 0 \mathbf w$

As $\mathbf v$ and $\mathbf w$ are linearly independent, this implies $\left({1-s}\right) t = 0$.

Then either $t = 0$ or $s = 1$, which is impossible, so $\mathcal L$ does not intersect $S_i$.

However, we will show that $\mathcal L$ crosses $S_k$, as the intersection point is:

for some $r_2 \in \left[{0\,.\,.\,1}\right]$, $r \in \R_{>0}$, which implies:


 * $\mathbf 0 = \left({st - r}\right) \mathbf v + \left({t \left({1-s}\right) - r_2}\right) \mathbf w$

As $v$ and $w$ are linearly independent, we have $0 = st - r$ and $0 = t \left({1-s}\right) - r_2$.

Then $r = st \in R_{>0}$, and $r_2 = t \left({1-s}\right) \in \R_{>0}$.

We must show that $r_2 < 1$.

As $q \in U_k$, we have $q = A_k + s't' \left({- \mathbf w}\right) + t' \left({1-s'}\right) \mathbf u$ for some $s' \in \left({0\,.\,.\,1}\right)$, $t' \in \R_{>0}$.

As $A_k = A_i + \mathbf w$, we have:


 * $q - r \mathbf v = A_i + \mathbf w - s't' \mathbf w + t' \left({1-s'}\right) \mathbf v - t' \left({1-s'}\right) \mathbf w - r \mathbf v$

As $q - r \mathbf v = A_i + r_2 \mathbf w$, we rearrange the equality to obtain:


 * $\mathbf 0 = \left({t' \left({1-s'}\right) - r}\right) \mathbf v + \left({1 - r_2 - s't' - t' \left({1-s'}\right) }\right) \mathbf w$

As $\mathbf v$ and $\mathbf w$ are linearly independent, this gives two equations:


 * $0 = t' \left({1-s'}\right) - r$
 * $0 = 1 - r_2 - s't' - t' \left({1-s'}\right) $

Adding these equations gives:


 * $s't' + r = 1 - r_2$

As $s't' > 0$ and $r > 0$, it follows that $r_2 < 1$.

Then, we have shown that $\mathcal L$ crosses $S_k$.

Let $\mathcal L' = \left\{ {q + r \mathbf v : r \in \R_{\ge 0} }\right\}$ be a ray with start point $q$.

An argument similar to the one above shows that $\mathcal L'$ crosses the side $S_j$.

As $\mathcal L \cup \mathcal L'$ is a straight line, $\mathcal L$ and $\mathcal L'$ cannot both intersect the same side.

It follows that $\mathcal L$ has one crossing of the boundary of $\triangle$, so the parity of $q$ is $\operatorname{par} \left({q}\right) = 1$.

From Jordan Polygon Interior and Exterior Criterion, it follows that $q \in \operatorname{Int} \left({\triangle}\right)$.

The result now follows from Equality of Sets.