20070306, 20:15  #1 
May 2003
10111110010_{2} Posts 
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.)

20070307, 12:30  #2 
"Phil"
Sep 2002
Tracktown, U.S.A.
2^{2}×3^{2}×31 Posts 
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]. 
20070307, 12:39  #3 
"Phil"
Sep 2002
Tracktown, U.S.A.
2^{2}×3^{2}×31 Posts 
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 
20070307, 15:29  #4 
May 2003
2·761 Posts 
Philmore,
Go ahead and solve for s. It should be a relatively simple looking formula. 
20070307, 16:27  #5 
"Lucan"
Dec 2006
England
14463_{8} Posts 

20070307, 18:06  #6 
"Phil"
Sep 2002
Tracktown, U.S.A.
2^{2}·3^{2}·31 Posts 
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".)

20070307, 19:50  #7  
"Phil"
Sep 2002
Tracktown, U.S.A.
1116_{10} Posts 
Quote:
s[sup]2[/sup] = (a[sup]2[/sup]+b[sup]2[/sup]+c[sup]2[/sup]+sqrt(3(a+b+c)(a+bc)(ab+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. 

20070308, 02:29  #8 
May 2003
2·761 Posts 
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.

20070308, 13:37  #9 
May 2003
2×761 Posts 
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, ZetaFlux 
20070308, 23:30  #10 
"Phil"
Sep 2002
Tracktown, U.S.A.
45C_{16} Posts 
I really am curious about the slick geometric solution, as a^{2} + b^{2} = c^{2} 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. 
20070308, 23:50  #11 
May 2003
2·761 Posts 
Here is a hint:
Try forming a right triangle. 
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