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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?

Faust
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  • Everything would be easiert when defining triangles as convex hulls of three points in general position. :D – M. Winter Oct 25 '17 at 13:30
  • Taking the required result as an axiom? – Faust Oct 25 '17 at 13:43
  • It's always one possible way, right? But I guess your triangle is just defined by naming the three points? What is your definition of being inside a triangle? – M. Winter Oct 25 '17 at 14:08

2 Answers2

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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 $\overline {PQ}\subseteq B\Rightarrow \overline {PQ}\subseteq A\cap B.$

Matematleta
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  • This is certainly a possible proof but I can absolutely not understand how OP can accept this as the desired answer. His question was clearly asking for a purely geometric proof ala Euclid's Elements and not one using analysis. In some sense you also just choose a convenient enough definition, but also just not convenient enough for OP to call it "using the result as a definition". – M. Winter Oct 25 '17 at 14:39
  • I suppose the OP should specify which definition he's using. (OT: Way too much downvoting on this site. I know a lot of people who won't post here because of the downvoting. The idea here is to exchange ideas, not to engage in oneupsmanship). – Matematleta Oct 25 '17 at 14:57
  • Yes he should, or he did by accepting your answer. Or he himself had no idea what definitions he is using, which is the worst kind of OP because every answer fits him, even though most of them are no answers for his initial problem. Especially when the problem is from a book and a specific set of tools was expected to be used. I would say new users are quite fast in accepting answers without double checking, but Faust is not very new. To the downvotes: Once I read "most users do not vote on the quality of a post but according to the votes the answer should have." – M. Winter Oct 25 '17 at 15:04
  • My class does not have a very clear layout and there are no pre reqs for it at my university but i have used advanced methods in other problems. My approach would be the same as my book only defines what a convex set is in this problem as stated above. When i asked a grad student the response i received was the exact same as the posted answer. Would i prefer a geometric type of approach of course! That would be in the spirit of the problem and i would certainly prefer it but i understand the given answer and don't want to reject it as i didn't state it had to be a geometric argument. @M.Winter – Faust Oct 25 '17 at 19:30
  • @Faust I understand your point. I still think the problem is interesting and I am going to look for a geometric solution. Sometimes it is just sad that taking a long time to give a clean answer will not be appreciated, but what a mathematician would I be when I am not sometimes doing math for the math's sake alone? ;) – M. Winter Oct 25 '17 at 19:35
  • I honestly believe it can be solved using the definition of an angle combined with the definition of a triangle and applying the crossbar theorem to all three of those angles im just not sure how to formulate it with such a loose Def'N of what a convex set is. There is actually an unanswered question involving a similar request about quadrilaterals here where OP requests a geometric argument and has been unanswered for quite awhile. https://math.stackexchange.com/questions/245434/geometry-proof-convex-quadrilateral?rq=1 – Faust Oct 25 '17 at 19:44
  • @Faust I would add that you additionally need definitions for "being inside a triangle" and "being inside an angle". If you can provide all these definitions I will give it another look and you are probably right that we can solve it using the Crossbar theorem. – M. Winter Oct 26 '17 at 05:14
  • @Faust: if you'll accept as a definition that the interior of an angle is the intersection of two half-planes, and that half-planes are convex (which I think can be proved in a purely geometric way), then I think I have a proof. If you'd like to see it, I will post it as an answer as it is too long for the comments section. – Matematleta Oct 26 '17 at 14:17
  • Imagine it isn’t convex. Then there exist 2 points A, B in the interior such that the segment AB is not contained in the interior. That means it meets a side. But two lines meet in at most one point. Now use plane separation to get a contradiction? think that will work? @M.Winter – Faust Oct 26 '17 at 15:01
  • @Faust Your reasoning still misses the very meaning of "being inside". I think this definition is necessary to prove that the segment $\overline{\mathrm{AB}}$ intersects a side of $\Delta \mathrm{DEF}$ (or why do you know this?). I cannot follow after the plane separation part. But maybe it is a way. You can always try to answer your own question if you think you have a solution. – M. Winter Oct 26 '17 at 15:35
  • I think the pertinent definitions all follow from the one definition of half-plane, which, in turn, only uses an easy application ot the betweeness axiom – Matematleta Oct 26 '17 at 16:24
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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 $\angle C$.

Matematleta
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