Bleaug adds new meaning to "laziness". Hmmm, do your lazy tricks lazy scientists to solve this one, huh? Although not confirmed, the similarity of the terms with Fox 3 is striking. Enjoy...

Oh, one more thing: if one of the corners "must" always be on the right angle, then there should be further limitations on a,b,c right triangle. Even if that condition was removed, the formula for x will remain the same, won't it?

In left white triangle:

ReplyDeletex/sin(b)=a/sin(120)

In bottom white triangle:

x/sin(a)=b/sin(120)

then, multipliying:

x²=4*(a*b*sin(a)*sin(b))/3

Apply sin(a)=a/c and sin(b)=b/c (c=hypothenuse).

César Lozada

Venezuela

César's use of sin rule and pythagoras reveals the answer. There should be other approaches as well. Let's list the lazy observations for this case:

ReplyDelete1. x=f(a,b) is formula-symmetric for a and b, due to rotation.

2. the unit of x is "length", in other words, units of a, b, and x are identical, as it should be.

3. a=0 or b=0 => x=0

4. x <= min{a,b}

Oh, one more thing: if one of the corners "must" always be on the right angle, then there should be further limitations on a,b,c right triangle. Even if that condition was removed, the formula for x will remain the same, won't it?

ReplyDeleteFrom the area of right triangle

ReplyDelete(sqrt(a^2+b^2)*x*sqrt(3)/2)/2=a*b/2

Newzad yours looks like the easiest:

ReplyDeleteA(right triangle) / A(equilateral = hypotenuses / x

(Because their heights are equal)

=> [ab/2] / [x^2 sqrt(3)/4] = sqrt(a^2 + b^2)/x

gives the answer.

Thank you.