Anyone?

[akatookey]

Does anyone have any ideas on how to prove Hero's formula or Area=(1/2)ab*sin(c)? This is a very interesting question and I've started toying around with it...I'll work on it more over teh weekend but anyone have any ideas??? Sigh...I was never very good with trig proofs

[TripleM]

I vaguely remember seeing the proof of Hero's formula before but I'm not entirely sure how. On the other hand, the proof of A=1/2 ab sin C is relatively easy: just draw in one altitude of the triangle and use A=1/2 bh.

I'll have another think about Hero's formula.

[Eve]

[rrusczyk]

Extra challenge: can you get to Hero(n) without all that nasty trig?

[i/3]

From what i know (and check the outstanding planetmath web site at http://planetmath.org/?op=getobj&from=objects&name=HeronsFormula) this is not the Heron formula.

[ComplexZeta]

Heron's original proof relied on constructing the incenter and then showing that the area of a triangle is the semiperimeter times the inradius. The proof is quite involved and difficult to come up with. It is also extremely clever. It can be found in William Dunham's Journey Through Genius.

[Xbizkitl]

Hero's formula can be found my manipulating the area sine formula, if I remember correctly.

Let me try it and get back to you in the morning.

[andy17null]

x biz kil l, your sig is inaccurate. In Hoc Signo Vinces means "by this sign you shall conquer, not prosper. I love nitpicking



-Isaac

[Kamior]

I did this one in 6th grade, before I knew about trig or Hero(n)'s formula (I didn't convert it to s(s-a)..., I did that later).

Start with triangle ABC. Label the sides a, b, and c. Call the altitude dropped to c d, and call the line segment from the point of intersection of d and c to corner B x. Then we have two right triangles, with equations:

x^2 + d^2 = a^2
d^2 + (c-x)^2 = b^2

Subtracting the second from the first, we get -2cx + c^2 = b^2 - a^2. Solving for x, we get x = (c^2+a^2-b^2)/(2c). Plugging that into the first equation, we get ((c^2+a^2-b^2)/(2c))^2 + d^2 = a^2. Moving everything to the right and expanding, we get
d^2 = a^2 - (c^4 + b^4 + a^4 +2 (ac)^2 - 2(ab)^2 - 2(bc)^2)/(4c^2)
d^2 = (-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(4c^2)
Instead of playing around with this, go to area = cd/2. Then area^2 = ((cd)^2)/4, and we can plug in d:

Area^2 = (-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(16)
Area = sqrt(-c^4 - b^4 - a^4 + 2(ac)^2 + 2(ab)^2 + 2(bc)^2)/(16)

This was as far as I got in 6th grade. It can be factored into
sqrt(((a+b+c)/2)((a+b-c)/2)((a-b+c)/2)((-a+b+c)/2))
Then defining the semiperimeter gives Hero(n)'s formula.

I've also read about the incircle method, but I prefer to use my own proof.