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 2007-03-06, 20:15 #1 Zeta-Flux     May 2003 7×13×17 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.)
 2007-03-07, 12:30 #2 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts The area will be given by $$\frac{(a^2+b^2+c^2)\sqrt{3}+3\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}}{8}$$. Since this equals $$\frac{s^2\sqrt{3}}{4}$$, we can easily solve for $$s$$.
 2007-03-07, 12:39 #3 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3×373 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
 2007-03-07, 15:29 #4 Zeta-Flux     May 2003 7·13·17 Posts Philmore, Go ahead and solve for s. It should be a relatively simple looking formula.
2007-03-07, 16:27   #5
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by philmoore 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!

2007-03-07, 18:06   #6
philmoore

"Phil"
Sep 2002
Tracktown, U.S.A.

3·373 Posts

Quote:
 Originally Posted by davieddy 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".)

2007-03-07, 19:50   #7
philmoore

"Phil"
Sep 2002
Tracktown, U.S.A.

111910 Posts

Quote:
 Originally Posted by Zeta-Flux 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.

 2007-03-08, 02:29 #8 Zeta-Flux     May 2003 154710 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.
 2007-03-08, 13:37 #9 Zeta-Flux     May 2003 7·13·17 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, Zeta-Flux
 2007-03-08, 23:30 #10 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 100010111112 Posts 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.
 2007-03-08, 23:50 #11 Zeta-Flux     May 2003 110000010112 Posts Here is a hint: Try forming a right triangle.

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