Segments of a polygon

[isaaclimdc]

Hi all,

I have this math problem where an equilateral triangle has each of its sides divided into ratio 2:1, and a smaller equilateral drawn within it from the intersecting lines. There is also a similar problem, but relating to a square. My objective is to figure out a relationship between the ratio of the areas of the two triangles to "n" (the number of divisions of each side)

In the document attached, the example given is when divided into 1:2 parts. The ratio of the areas for the triangle there is 7.0.

Has anyone heard of this problem before and knows how to do it?

I would greatly appreciate any help given.

Thanks,
Isaac.

 

[HallsofIvy]

For the square, it is fairly easy, isn't it, to use the Pythagorean theorem to calculate the length of the sides of the inner square, relative to the sides of the outer square.

 

[dodo]

And the Law of Cosines can do a similar job for the Triangle, since you have two sides and the angle between them (60 degrees).

 

[isaaclimdc]

so far for the triangle I've gotten the rule ((n-1)^2+n)/((n-2)^2) and it works. However, I'm asked to somehow prove this. Anybody have any idea how to?

 

[dodo]

Your attachment is now available, and now I understand why you get such a high ratio (7) for the triangle: you are doing a different thing than for the square. (Note that the square is not joining point B-E with a segment, for example.)

I would have expected the triangle to be formed by joining the newly formed points at 2:1 (as you did for the square), rather than throwing lines from the original vertices. For example, with a ratio of 1:1, joining the midpoints with lines splits the triangle into 4 identical ones (and thus the area ratio is 4), while using your construction, midpoints at 1:1 will produce no inner triangle, but only a single point where the three lines intersect.

Are you sure that the triangle is not meant to be split as you did for the square, instead?

 

[isaaclimdc]

Yeah thanks for that... i got that wrong.. the triangle is right, but the square is wrong. The lines join the segments to the vertices of the polygon. Anyone has any idea how to prove the rule i found?

 

[dodo]

Well, the ideas have already been mentioned. Your segments will produce a lot of inner triangles; work your way with them to calculate the sides of the inner triangle or square, and thus its area.

In your case, you'd typically use either Pitagoras' rule (for the square) of the Law of Cosines (for the triangle) to calculate new sides, and then the Law of Sines to calculate new angles, so that you can continue calculating new sides on other inner triangles.

 

[shankar2298]

Hi Issac

Hi Issac,

thanks for posting this. Are u doing I.B maths. I got the same thing for my maths portfolio.

 

[shankar2298]

Can anybody please explain me in detail about how to use the Law of cosines and what is 'n' in
the rule ((n-1)^2+n)/((n-2)^2)

Thank you

 

[shankar2298]

Can anybody please explain me in detail about how to use the Law of cosines and what is 'n' in
the rule ((n-1)^2+n)/((n-2)^2) and explain me this rule.

Thank you

 

[slayer845]

omg i just go this portfolio as well...how do you prove that rule? and i don't really see a relationship between the ratios of the sides with the ratios of the triangles. I've looked at all the measurements but can't see any kind of pattern...please help.

Thnx

 

[ravi5555]

so far for the triangle I've gotten the rule ((n-1)^2+n)/((n-2)^2) and it works. However, I'm asked to somehow prove this. Anybody have any idea how to?

can you help me with this portfolio?

 

[ravi5555]

For the square, it is fairly easy, isn't it, to use the Pythagorean theorem to calculate the length of the sides of the inner square, relative to the sides of the outer square.

Hi all,

I have this math problem where an equilateral triangle has each of its sides divided into ratio 2:1, and a smaller equilateral drawn within it from the intersecting lines. There is also a similar problem, but relating to a square. My objective is to figure out a relationship between the ratio of the areas of the two triangles to "n" (the number of divisions of each side)

In the document attached, the example given is when divided into 1:2 parts. The ratio of the areas for the triangle there is 7.0.

Has anyone heard of this problem before and knows how to do it?

I would greatly appreciate any help given.

Thanks,
Isaac.

can you help me with this portfolio?

 

[kevinchan7]

so far for the triangle I've gotten the rule ((n-1)^2+n)/((n-2)^2) and it works. However, I'm asked to somehow prove this. Anybody have any idea how to?

im also doing this portfolio!

hmmm anyone know how to get to the above formula?
im kind of stuck after drawing the triangles for 1:2, 1:3, 1:4, 1:5 as i can't seem to find any relation between the side ratios and the area ratios...

a similar example or a detailed guide on homework to reach the formula above would be nice :D
hehe ^_^
thx in advance :D

 

[mixtape]

im also doing this portfolio!

hmmm anyone know how to get to the above formula?
im kind of stuck after drawing the triangles for 1:2, 1:3, 1:4, 1:5 as i can't seem to find any relation between the side ratios and the area ratios...

a similar example or a detailed guide on homework to reach the formula above would be nice :D
hehe ^_^
thx in advance :D

I am also doing this portfolio right now, and I don't know how to proceed, so what if we work together? All I have found is that the ratio of the lengths squared (the ratio of the "real" lengths if your draw it) equals the ratio of the areas. But that is a very simple thing, there's got to be more to it, there's got to be a connection between that and the ratio the lines are cut. As you probably understand i could really use some help. :P

 

[minima]

polyhedra dissection

--------------------------------------------------------------------------------

Hi,

I have made a cube dissection, please check my short videos

www.youtube.com/elusivecube

I use 176 magnets that never repel so there is a absolute freedom in connections,
By far the best attributes are that I can build a whole concentric hierarchy of solids
starting with a irregular tetrahedron and ending up with the rhombic dodeca hedron.

I can build the 7 basic crystall latice structures, well actually all 14 are possible.
Have this set for about a decade and still discovering new structuresw all space fillers, all at the lovest common denominator. "Do more with less"

All the build solids are build by the least number of volume identical blocks, in a very near future I will demonstrate Hilbert's third problem.

Smiths College published a article by Marjorie Senechal 'Which tetrahedrons fill space" according to the article this problem is stil not completely solved, but I believe I do have the solution, so please watch my next realese.

Thank you
frank

 

[barom]

the equation that i got is (n^2+n+1)/(n-1)^2
the length of the inner triangle is (n^2-1)/(n^2+n+1)^(1/2)
and the area for the inner triangle is ((3)^(1/2)/4) x ((n^2-1)^2/n^2+n+1))
from this you could get the above equation.
the area for the big triangle is (3)^(1/2)/4 x (n+1)^2

 

[jz123]

Thanks barom for all of your help.

I would like to request one more thing. Could you please explain briefly, if possible, on how you got the equations and especially on how you got (n^2+n+1) for the equations.

Thanks,
Regards,
jz123.

 

[fellbell]

HL maths coursework is so depressing :(

 

[IB math]

Hey,
I'm doing the math portfolio as well and I've gotton all the ratios but I'm stuck on how to get the equation. How did you derive the equation (n^2+n+1)/(n-1)^2?

 

[desy.kris]

hello I am doing this portfolio too, for a pre-IA task. not formal, only for a practice. and i don't know how to get the conjecture, could any of you help me please? I'm really stuck and don't know what to do

 

[desy.kris]

I got the conjecture and proved it already, but my teacher said I need to find the formula for the length of the side of the inner triangle to prove it once more time because at first I used approximation for the values of the ratios. can anybody help me to find the formula for the length of the side of the inner triangle? I have found a formula using sine but that aint accurate enough :( how can I arrive to Barom's formula?