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Old 2007-03-06, 20:15   #1
Zeta-Flux
 
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Default Triangle puzzle

You are given an equilateral triangle. There is some fixed point in the interior, and straight lines of length a, b, and c are drawn from the corners of the triangle to this point. Find the area of the triangle in terms of a,b,c. (Alternatively, find the length of one of the sides.)
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Old 2007-03-07, 12:30   #2
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The area will be given by
[tex]\frac{(a^2+b^2+c^2)\sqrt{3}+3\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}}{8}[/tex].
Since this equals [tex]\frac{s^2\sqrt{3}}{4}[/tex], we can easily solve for [tex]s[/tex].
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Old 2007-03-07, 12:39   #3
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Once again, I am finding the simultaneous use of spoiler tags with tex markup creating problems. Suggestions, anyone?

The area is ((a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup])sqrt(3)+3*sqrt(2(a[sup]2[/sup]b[sup]2[/sup]+a[sup]2[/sup]c[sup]2[/sup]+b[sup]2[/sup]c[sup]2[/sup])-(a[sup]4[/sup]+b[sup]4[/sup]+c[sup]4[/sup])))/8
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Old 2007-03-07, 15:29   #4
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Philmore,

Go ahead and solve for s. It should be a relatively simple looking formula.
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Old 2007-03-07, 16:27   #5
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Quote:
Originally Posted by philmoore View Post
Once again, I am finding the simultaneous use of spoiler tags with tex markup creating problems. Suggestions, anyone?
No, but I couldn't manage to uncover it!
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Old 2007-03-07, 18:06   #6
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Quote:
Originally Posted by davieddy View Post
No, but I couldn't manage to uncover it!
To see what should be there, go back to my post #2 above, hit QUOTE, then in the Reply to Thread page, remove the spoiler tags, add a minimum of one character of your own, and then hit the "Preview Post" button. (But please don't "Submit Reply".)
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Old 2007-03-07, 19:50   #7
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Quote:
Originally Posted by Zeta-Flux View Post
Go ahead and solve for s. It should be a relatively simple looking formula.
I'm not quite sure how simple is relatively simple. How about:

s[sup]2[/sup] = (a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup]+sqrt(3(a+b+c)(a+b-c)(a-b+c)(-a+b+c)))/2
or:
s[sup]2[/sup] = (a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup]+sqrt(6(a[sup]2[/sup]b[sup]2[/sup]+a[sup]2[/sup]c[sup]2[/sup]+b[sup]2[/sup]c[sup]2[/sup])-3(a[sup]4[/sup]+b[sup]4[/sup]+c[sup]4[/sup])))/2
or:
s[sup]2[/sup] = (a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup]+sqrt(3(a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup])[sup]2[/sup]-6(a[sup]4[/sup]+b[sup]4[/sup]+c[sup]4[/sup])))/2
I like the first, reminiscent of Heron's formula, but I don't see how, if I take the square root to solve for s, I get any significant simplification in any of the three versions.
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Old 2007-03-08, 02:29   #8
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Hmmm... maybe I was thinking of the special case when b=c. I'll have to go ask the person who presented this puzzle to me.
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Old 2007-03-08, 13:37   #9
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Now I remember what I was missing. One is supposed to assume that (a,b,c) is a pythagorean triple (like 3,4,5). That way, you don't need to use Bramagupta's formula. There is a slick geometric solution.

Cheers, and sorry for the harded puzzle,
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Old 2007-03-08, 23:30   #10
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I really am curious about the slick geometric solution, as a2 + b2 = c2 now leads to the algebraic solution:

s[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup] + a*b*sqrt(3)
or: s[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup] - 2*a*b*cos(150[sup]0[/sup])

By the way, my solution was strictly analytic, but I am hoping that this hint given by the law of cosines will help in finding a more geometric solution.


Nice problem, at any rate, both the original stated version and this new variation.
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Old 2007-03-08, 23:50   #11
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Here is a hint:

Try forming a right triangle.
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