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        riddles >> medium >> Triangle Area
        
(Message started by: Barukh on Oct 19^th, 2005, 1:05am)

Title: Triangle Area
Post by Barukh on Oct 19^th, 2005, 1:05am

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.

Title: Re: Triangle Area
Post by Grimbal on Oct 19^th, 2005, 6:00am

[hide]1/7[/hide]?

Title: Re: Triangle Area
Post by towr on Oct 19^th, 2005, 6:11am

[hide]1/9th[/hide]?
(let's all take a guess ;D)

[e]Maybe I shouldn't have used a bad, sketchy drawing and an estimate, ::)[/e]

Title: Re: Triangle Area
Post by Barukh on Oct 19^th, 2005, 6:16am

One of you is right. Got a proof?

Title: Re: Triangle Area
Post by towr on Oct 19^th, 2005, 6:19am

on 10/19/05 at 06:16:20, Barukh wrote:

One of you is right. Got a proof?

Nope, just a very bad drawing on a scrap of paper, and the audacity to use it to make a guess.

Title: Re: Triangle Area
Post by towr on Oct 19^th, 2005, 6:26am

A good drawing and invariance under skewing/resizing proves Grimbal's ratio though.

Title: Re: Triangle Area
Post by Eigenray on Oct 19^th, 2005, 4:11pm

[hide]Balance the triangle at Z[/hide]

Title: Re: Triangle Area
Post by Barukh on Oct 21^st, 2005, 1:00am

Eigenray, not exactly clear what you had in mind.

Title: Re: Triangle Area
Post by towr on Oct 21^st, 2005, 3:01am

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.

Title: Re: Triangle Area
Post by Eigenray on Oct 21^st, 2005, 7:34pm

on 10/21/05 at 01:00:42, Barukh wrote:

Eigenray, not exactly clear what you had in mind.

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.

Title: Re: Triangle Area
Post by Neelesh on Oct 23^rd, 2005, 1:57am

I am still not clear how you have arrived at the solution.
What is meant by "balancing a triangle"?
???

Title: Re: Triangle Area
Post by Barukh on Oct 23^rd, 2005, 6:33am

Nice solutions from both Eigenray and towr! The solution I was aware of is identical to towr's.

on 10/23/05 at 01:57:52, Neelesh wrote:

I am still not clear how you have arrived at the solution.
What is meant by "balancing a triangle"? ???

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?

Title: Re: Triangle Area
Post by Eigenray on Oct 23^rd, 2005, 12:39pm

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? :)

Title: Re: Triangle Area
Post by SMQ on Oct 23^rd, 2005, 8:14pm

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

Title: Re: Triangle Area
Post by Eigenray on Oct 23^rd, 2005, 11:08pm

Indeed. Generalizing further, one obtains Routh's Theorem (http://mathworld.wolfram.com/RouthsTheorem.html).