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Topic: Triangle Area (Read 1611 times) |
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Barukh
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Given an arbitrary triangle ABC, three points are drawn at its sides so that they cut off 1/3 of the corresponding side. Find the area of the triangle XYZ in terms of the area of ABC.
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Grimbal
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Re: Triangle Area
« Reply #1 on: Oct 19th, 2005, 6:00am » |
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1/7?
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towr
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Re: Triangle Area
« Reply #2 on: Oct 19th, 2005, 6:11am » |
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1/9th?
(let's all take a guess  )
[e]Maybe I shouldn't have used a bad, sketchy drawing and an estimate,  [/e]
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| « Last Edit: Oct 19th, 2005, 6:17am by towr » |
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Barukh
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Re: Triangle Area
« Reply #3 on: Oct 19th, 2005, 6:16am » |
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One of you is right. Got a proof?
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towr
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Re: Triangle Area
« Reply #4 on: Oct 19th, 2005, 6:19am » |
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on Oct 19th, 2005, 6:16am, Barukh wrote:| One of you is right. Got a proof? |
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Nope, just a very bad drawing on a scrap of paper, and the audacity to use it to make a guess.
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towr
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Re: Triangle Area
« Reply #5 on: Oct 19th, 2005, 6:26am » |
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A good drawing and invariance under skewing/resizing proves Grimbal's ratio though.
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Eigenray
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Balance the triangle at Z
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| « Last Edit: Oct 19th, 2005, 4:23pm by Eigenray » |
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Barukh
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Re: Triangle Area
« Reply #7 on: Oct 21st, 2005, 1:00am » |
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Eigenray, not exactly clear what you had in mind.
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towr
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If you look at the attached drawing, you can see the pinkish colored area is the same size as the blue triangles (you can rearrange the parts cut by a blue line, to form it into a triangle)
And the red traingle, which is what we're interested in, is 1/7th of the pinkish area, and thus 1/7th of a blue traingle.
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Eigenray
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Re: Triangle Area
« Reply #9 on: Oct 21st, 2005, 7:34pm » |
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on Oct 21st, 2005, 1:00am, Barukh wrote:| Eigenray, not exactly clear what you had in mind. |
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Put weights 1,2,4 at vertices B,A,C respectively. Since the trriangle balances on both lines BZ and CZ, it balances at Z.
Combine the weights at A and C to a weight 6 as shown, say at E. Since it's still balanced at Z, BZ:ZE = 6:1.
Then area BZC = 6/7 (area BEC) = 6/7 * 1/3 = 2/7 of the area of ABC. Similarly we can get two more triangles each with 2/7 of the area, leaving just triangle XYZ with 1/7 the remaining area.
We can also find the length ratios along each line, as well as the relative areas of each region. Combining the weights at A and B to weight 3 at F, say, shows CZ:ZF = 3:4. Then it follows CZ:ZX:XF = 3:3:1. Then BZF = 4/7 BCF = 4/7*2/3 = 8/21. Since ABY=2/7 (symmetry), and AFX = 1/7 AFC = 1/7*1/3 = 1/21, we can conclude BYXF = 5/21, and therefore again XYZ=1/7.
I guess the diagram wasn't very clear.
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| « Last Edit: Oct 21st, 2005, 7:35pm by Eigenray » |
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Neelesh
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Re: Triangle Area
« Reply #10 on: Oct 23rd, 2005, 1:57am » |
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I am still not clear how you have arrived at the solution.
What is meant by "balancing a triangle"?
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Barukh
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Re: Triangle Area
« Reply #11 on: Oct 23rd, 2005, 6:33am » |
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Nice solutions from both Eigenray and towr! The solution I was aware of is identical to towr's.
on Oct 23rd, 2005, 1:57am, Neelesh wrote:I am still not clear how you have arrived at the solution.
What is meant by "balancing a triangle"? |
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I think Eigenray means the following: let certain masses be put on the vertices of the triangle. Where is the center of gravity of such a configuration? Eigenray showed that if you put masses of 1, 2, 4 units onto vertices B, A, C respectively, the center of gravity of such 7-unit mass configuration will be at point Z.
Very elegant, don't you think?
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| « Last Edit: Oct 23rd, 2005, 6:34am by Barukh » |
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Eigenray
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Re: Triangle Area
« Reply #12 on: Oct 23rd, 2005, 12:39pm » |
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Barycentric coordinates were discovered by Mobius in 1827. I learned about them in high school as "mass points", mostly as a trick for solving the type of geometry problems that frequently arise in mathletes competitions.
towr's solution is quite elegant. (Although the "pinkish colored area" is almost completely invisible on my LCD monitor unless I look at it funny; I only noticed it was there today!) But what if, instead of trisecting the side lengths, we [pi]-sected them? 
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SMQ
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Re: Triangle Area
« Reply #13 on: Oct 23rd, 2005, 8:14pm » |
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Well, following your earlier methodology and generalizing to AC/EC = n where n >= 2:
Place weights of 1, n-1, and (n-1)2 at B, A and C respectively. Combine the weights at A and C to a weight of n(n-1) at E. The area of BZC is then n(n-1)/[n(n-1)+1] * 1/n = (n-1)/(n2-n+1). This leaves the area of xyz as 1 - 3(n-1)/(n2-n+1) = (n-2)2/(n2-n+1).
For n = [pi] xyz ~ 1/5.9298736928
--SMQ
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--SMQ
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Eigenray
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Re: Triangle Area
« Reply #14 on: Oct 23rd, 2005, 11:08pm » |
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Indeed. Generalizing further, one obtains Routh's Theorem.
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