# Unusual proof.

• Feb 1st 2006, 01:12 PM
ThePerfectHacker
Unusual proof.
I was thinking of an unusual problem in geometry. Given triangle ABC. Define a "rank" as a line-segment drawn from any vertex to any point on the opposite side. To prove: That any two ranks intersect.

Why, do I need this? When math books on geometry prove for example that medians pass though a common point they assume that they intersect first. But if it were not always true then by contropositive their proof is destroyed. Thus, to prove that the medians pass through a common point we must first prove that the medians intersect somewhere.

I believe I have a proof to this from analytic geometry. Construct a triangle on \$\displaystyle (0,0),(a,b),(c,0)\$ where these points are not collinear. Then you could perhaps use elementary algebra to prove that such a point exists. But if anyone has a geometric proof I would greatly appreciate it.
• Feb 2nd 2006, 06:24 AM
Do you mean to specify ranks FROM DIFFERENT VERTICES?
• Feb 2nd 2006, 07:30 AM
CaptainBlack
Quote:

Originally Posted by ctelady
Do you mean to specify ranks FROM DIFFERENT VERTICES?

Ranks from the same vertex intersect trivially, don't they?

RonL
• Feb 2nd 2006, 07:58 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I was thinking of an unusual problem in geometry. Given triangle ABC. Define a "rank" as a line-segment drawn from any vertex to any point on the opposite side. To prove: That any two ranks intersect.

Why, do I need this? When math books on geometry prove for example that medians pass though a common point they assume that they intersect first. But if it were not always true then by contropositive their proof is destroyed. Thus, to prove that the medians pass through a common point we must first prove that the medians intersect somewhere.

I believe I have a proof to this from analytic geometry. Construct a triangle on \$\displaystyle (0,0),(a,b),(c,0)\$ where these points are not collinear. Then you could perhaps use elementary algebra to prove that such a point exists. But if anyone has a geometric proof I would greatly appreciate it.

Consider a triangle \$\displaystyle ABC\$, and a "rank" \$\displaystyle a_q\$ from vertex \$\displaystyle A\$ which meets \$\displaystyle BC\$ at \$\displaystyle Q\$
between \$\displaystyle B\$ and \$\displaystyle C\$, and another "rank" \$\displaystyle b_r\$ from vertex \$\displaystyle B\$ which meets \$\displaystyle AC\$ at \$\displaystyle R\$
between \$\displaystyle A\$ and \$\displaystyle C\$.

As \$\displaystyle \angle CAB = \angle CAQ + \angle QAB\$, \$\displaystyle \angle CAQ< \angle CAB\$.
Similarly \$\displaystyle \angle RBA< \angle CBA\$. So \$\displaystyle \angle CAQ + \angle RBA<180^{\circ}\$
and so the "ranks" are not parallel, and so meet.

RonL
• Feb 2nd 2006, 10:55 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Consider a triangle \$\displaystyle ABC\$, and a "rank" \$\displaystyle a_q\$ from vertex \$\displaystyle A\$ which meets \$\displaystyle BC\$ at \$\displaystyle Q\$
between \$\displaystyle B\$ and \$\displaystyle C\$, and another "rank" \$\displaystyle b_r\$ from vertex \$\displaystyle B\$ which meets \$\displaystyle AC\$ at \$\displaystyle R\$
between \$\displaystyle A\$ and \$\displaystyle C\$.

As \$\displaystyle \angle CAB = \angle CAQ + \angle QAB\$, \$\displaystyle \angle CAQ< \angle CAB\$.
Similarly \$\displaystyle \angle RBA< \angle CBA\$. So \$\displaystyle \angle CAQ + \angle RBA<180^{\circ}\$
and so the "ranks" are not parallel, and so meet.

RonL

CaptainBlack, that is not a bad demonstration however I made the same mistake! I said a line-segment. Thus, the fact that these "ranks" are not parallel does not prove anything, because they have finite length.
• Feb 2nd 2006, 08:10 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
CaptainBlack, that is not a bad demonstration however I made the same mistake! I said a line-segment. Thus, the fact that these "ranks" are not parallel does not prove anything, because they have finite length.

For most purposes you only need that the lines intersect, not that they
intersect inside the triangle. In fact the example you quote only requires
that there be a point of intersection not that it be inside the triangle.

However I'm sure that we can show that the point of intersection is inside
the triangle. (Once we decide what it means for a point to be inside a
triangle, and develop some theorems based on the definition we plump
for)

RonL
• Feb 3rd 2006, 10:06 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
For most purposes you only need that the lines intersect, not that they
intersect inside the triangle. In fact the example you quote only requires
that there be a point of intersection not that it be inside the triangle.

However I'm sure that we can show that the point of intersection is inside
the triangle. (Once we decide what it means for a point to be inside a
triangle, and develop some theorems based on the definition we plump
for)

RonL

I agree with you that showing that extended "ranks" interesect is the important fact, not the "ranks" themselves.

I already in the first post said we may show it algebraically buy constructing it in the x-y plane, but this is not fun. And messy
• Feb 3rd 2006, 11:02 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I agree with you that showing that extended "ranks" interesect is the important fact, not the "ranks" themselves.

I already in the first post said we may show it algebraically buy constructing it in the x-y plane, but this is not fun. And messy

It might be more tricky than you think. Consider what it means for a point
to be inside a triangle (or better still any simple closed curve).

You may well be making an assumption about how you know a point is
inside the triangle.

RonL
• Feb 4th 2006, 02:40 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
It might be more tricky than you think. Consider what it means for a point
to be inside a triangle (or better still any simple closed curve).

You may well be making an assumption about how you know a point is
inside the triangle.

RonL

I believe that to show that they intersect inside, you need to show that a unique solution exists on a certain interval then the problem of showing that they exist inside is resolved.
• Feb 10th 2006, 03:00 AM
CaptainBlack
Quote:

Originally Posted by Rich B.
Gentlemen:

Def(triangle interior): Let "I" denote "interior of triangle(ABC)".
then, I = {P,Q in Plane(ABC) : sgmnt(PQ) intrsct trngl(ABC) = {} for all P,Q in I.

This is OK for the triangle, but it would be better that the interior be
defined for more general simple closed curves than the triangle.

RonL
• Feb 10th 2006, 07:44 PM
Rich B.
Okay?
Def(interior of a triangle): Let point P lie in [Plane(ABC) - triangle(ABC)]. P lies in the Int(tri(ABC)) iff there exist points Q and R s.t. {Q,R} s.set tri(ABC) and P lies in sgmt(QR). [Restated: The interior of tri(ABC) is the set of all points, P, such that P lies between two points on the triangle (recall definition of betweenness). Of course, P lies on infinitely many segments, the endpoints of which lie on the triangle but, one is sufficient.]

Case 1: Two ranks "emanating" from the same triangle vertex intersect, trivially, "on" the triangle.

Case 2: By definition, the endpoints of a triangle's rank are elements of that triangle. Hence, according to the definition of "triangle interior", if sgmt(AR) is a rank of tri(ABC), then, sgmt(AR) - {AR} s.set Int(tri(ABC)). By PPT (previously proved theorem), the intersection of two distinct ranks is non-empty (Ron L.). Hence, provided those ranks do not share a common endpoint (case 1), it follows that such intersection lies within the triangle's interior, thus completing the proof.

Regards,

Rich B.