Def'n: A set U of points in the plane is a convex set if whenever A,B are distinct points in U, then the segment $\overline{AB}$ is entirely contained in U.
Question: Show that the inside of a triangle is a convex set.
I think what this wants me to do is take a triangle $DEF$ and then make an arguement using the "crossbar theorem"
Crossbar theorem: In the triangle ABC Let $\angle BAC $ be an angle, and let D be a point in the interior of the angle. Then the ray $\overrightarrow{AD}$ must meet the segment $\overline{BC}$
If i apply this theorem to all three angles then it shows it should imply that my set of points U is sitting inside the triangle? how do i show that is the case and show that the result follows?
$\endgroup$ 32 Answers
$\begingroup$Here is a more geometric proof:
$1).$ Half-planes are convex: Let $H$ be a half-plane, bounded by the line $l.$ Consider three distinct points $P,R,Q\in H$ such that $R$ is between $P$ and $Q$. If $R\notin H$ then $R$ lies on the other side of $l$ by the same sides lemma, so that $\overline {PR}$ intersects $l$ in a point $T$ such that $P,T,R$ are colinear, and $T$ is between $P$ and $R$. I will denote this situation by writing $PTR$. Now, if $PTR$ and $PRQ$ then by the Betweeness Principle, $PTQ$ which implies, on appealing to the same sides lemma once more, that $P$ and $Q$ are on opposite sides of $l,$ which is a contradiction.
$2).$ The intersection of a finite number of convex sets is convex: consider $A\cap B$, where $A$ and $B$ are any two convex sets. Let $P,Q$ be points in $A\cap B.$ Then, the segment $\overline {PQ}\subseteq A$ and $\overline {PQ}\subseteq B\Rightarrow \overline {PQ}\subseteq A\cap B.$ Now use induction to prove the general case.
$3).$ By definition, the interior of an angle is the intersection of two half-planes, and the interior of a triangle $\Delta ABC$ is the intersection of the interiors of $\angle A,\angle B$ and $\angle C.$ Now, by $2).$, we know that interiors of angles are convex. We will use this fact to prove that triangles are convex: consider the triangle $\Delta ABC$ and let $P,Q$ be distinct points in the interior of $\Delta ABC$ such that $PRQ$ for any point $R$. We need to show that $R$ is in the interior of $\Delta ABC$. But this is trivial, because if $P$ and $Q$ are interior to $\Delta ABC$ then they are by definition, interior to each of $\angle A,\angle B$ and $\angle C$.
$\endgroup$ 1 $\begingroup$If you don't want to take the result as a definition, maybe you can argue this way:
$1).$ It's easy to see that any open half-plane in $\mathbb R^2$ is convex:
By definition, $H$ is a half-plane in $\mathbb R^2\Leftrightarrow H=\left \{ (x,y):Ax+By+C<D \right \}$ for some fixed constants $A,B,C,D$. It is convenient to write this as $H=\left \{ \vec x:f(\vec x)<D \right \}$ where $\vec x=(x,y)$ and $f(\vec x)=Ax+By+C<D.$ Now using the definition of convexity, check by direct substitution, that $f(t\vec x+(1-t)\vec y)\le tf(\vec x)+(1-t)f(\vec y)$ whenever $\vec x,\ \vec y\in H$ and $t\in [0,1]$.
$2).$ The interior of a triangle is the intersection of three open half-planes.
$3).$ The intersection of a finite number of convex sets is convex:
To prove this, it suffices to consider $A\cap B$, where $A$ and $B$ are any two convex sets; the general result follows by induction. Let $P,Q$ be points in $A\cap B.$ Then, the segment $\overline {PQ}\subseteq A$ and $\overline {PQ}\subseteq B\Rightarrow \overline {PQ}\subseteq A\cap B.$
$\endgroup$ 11
