# Perimecenter

• Jun 15th 2006, 05:03 PM
ThePerfectHacker
Perimecenter
I believe I have discovered a new triangle center.

Triangle ABC
From A draw AP to BC such as the perimeters of two triangles are equal
From B draw BP to AC such as the perimeters of two triangles are equal
From C draw CP to AB such as the perimeters of two triangles are equal
The three lines intersect at a common point.

I will not post a proof before you have the fun of solving it.

Has anyone ever seen this? Or is this new? Cuz, I never did I searched online for triangular centers.
• Jun 15th 2006, 05:17 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
I believe I have discovered a new triangle center.

Triangle ABC
From A draw AP to BC such as the perimeters of two triangles are equal
From B draw BP to AC such as the perimeters of two triangles are equal
From C draw CP to AB such as the perimeters of two triangles are equal
The three lines intersect at a common point.

I will not post a proof before you have the fun of solving it.

Has anyone ever seen this? Or is this new? Cuz, I never did I searched online for triangular centers.

Although I don't know a proof of it, I did find this idea by just drawing (I found that if you went to the center of the opposite side of the triangle to the angle you started, then all the angle lines would intersect at one point). I used it too make my original equation located here . The equation in its unsimplified form was...

$\frac{\tan\left(\frac{180-\frac{360}{s}}{2}\right)\cdot\frac{d}{2}\cdot\frac {d}{2}}{2}\cdot2s$

The center applies for all triangles, even things like scalene, and all regular polygons.

But I always just assumed it had been discovered before.

I'm a senior member now :D
• Jun 15th 2006, 09:07 PM
malaygoel
Quote:

Originally Posted by Quick
Although I don't know a proof of it, I did find this idea by just drawing (I found that if you went to the center of the opposite side of the triangle to the angle you started, then all the angle lines would intersect at one point).

If you proceed in the quoted manner, the perimeters may not be equal.
The one-point you found is centroid.(Is it???)
• Jun 16th 2006, 02:41 AM
Quick
Quote:

Originally Posted by malaygoel
If you proceed in the quoted manner, the perimeters may not be equal.
The one-point you found is centroid.(Is it???)

I suppose your right. It is only equal when all sides are the same.
• Jun 16th 2006, 07:44 AM
ThePerfectHacker
Quote:

Originally Posted by malaygoel
If you proceed in the quoted manner, the perimeters may not be equal.
The one-point you found is centroid.(Is it???)

Negative, it is not the centroid.
If you draw a line such that the AREA's would be equal then it would be a centroid.
But overhere you are concered about the perimeter.
• Jun 16th 2006, 09:27 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
Negative, it is not the centroid.
If you draw a line such that the AREA's would be equal then it would be a centroid.
But overhere you are concered about the perimeter.

I was referring to Quick's response(post 2 of the thread) in which the centre was found by joining a vertex to the midpoint of other side.
• Jul 17th 2006, 04:18 PM
Quick
• Jul 17th 2006, 04:24 PM
ThePerfectHacker
The solution is extremely simple, once you do the following trick. Find, the length of all the same sides.

Meaning, after you have drawn the six perimeter line you divided the each of the triangles side into two parts. Find the length of those parts given that the length is a,b,c of its sides.
• Jul 17th 2006, 07:22 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
The solution is extremely simple, once you do the following trick. Find, the length of all the same sides.

Meaning, after you have drawn the six perimeter line you divided the each of the triangles side into two parts. Find the length of those parts given that the length is a,b,c of its sides.

May I remind you Hacker, that I have not taken geometry (this is unfortunate, as I'm incredibly curious and want to get farther in math), the only thing I know is the little "hints" and "ideas" I found in my algebra book. I can find the angle and length of everything, but I have no idea how to prove that the lines intersect.
• Jul 17th 2006, 07:27 PM
ThePerfectHacker
Quote:

Originally Posted by Quick
May I remind you Hacker, that I have not taken geometry (this is unfortunate, as I'm incredibly curious and want to get farther in math), the only thing I know is the little "hints" and "ideas" I found in my algebra book. I can find the angle and length of everything, but I have no idea how to prove that the lines intersect.

No, it is basic geometry.
You have ABC
From A draw a line D to BC to create equal perimeter.
From B draw a line E to AC to create equal perimeter.
From C draw a line F to BC to create equal perimeter.

Given AB=c, AC=b, BC=a
Find the lengths of,
BC,BD,EA,CE,FB,FC
• Jul 17th 2006, 07:41 PM
Quick
Is this the diagram you're talking about?
• Jul 17th 2006, 07:48 PM
ThePerfectHacker
Good
Now try to find the length of the segments through a,b,c.

If you can do that, I will show you a killer theorem (Seriousously this one is one of my favorite in geometry it is so awesome) which will easily solve it.
• Jul 17th 2006, 08:16 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
Good
Now try to find the length of the segments through a,b,c.

If you can do that, I will show you a killer theorem (Seriousously this one is one of my favorite in geometry it is so awesome) which will easily solve it.

BC=a (obvious)
BD= $\frac{a-(c-b)}{2}$
EA= $\frac{b-(c-a)}{2}$
CE= $\frac{b+(c-a)}{2}$
FB= $\frac{c-(a-b)}{2}$
FC=I'm having a horrible time trying to link this segment to a simple formula using only a,b,c any hints?

P.S. I realize that FC is part of two triangles, and I know there sides, I just can't figure out how to use that information to find FC (I wish sin,cosine, and tangent, worked for non-right triangles)

P.S2 I realize I can use the angles to figure out the length, but that wouldn't be in terms of a,b,c