Area of a Triangle as a Function of the Lengths of its 3 Sides

[a1call]

Area of a Triangle as a Function of the Lengths of its 3 Sides.

If you already know the answer please keep quiet.
If you don't, try to come up with the function before googling it.

[Batalov]

All you need is a piece of paper, draw a triangle as below,
then write:
b^2 = x^2 + y^2
c^2 = ... (fill the blank)
Area = a*y / 2

Now: solve for y and get rid of x, and you will have area as a function of a, b, c.
It is, you know, seventh grade level effort. You can be done in under 5 minutes.
Is that even a puzzle?

[a1call]

That is not a general function. For different types of triangles the solutions will be different. For example for a right angle triangle the height can be equal to a side, so the algo will have to be adjusted.
A hint would be that the solution is independent of a base and a height( so you don't need to draw the triangle to specify the base and the height).
The other hint is that the solution is 2k years old (but you shouldn't underestimate the difficulty.).

[LaurV]

We have read your "dotted line mystery" posts in the other forum (after Ernst gave us the link in another thread on this forum), and we know about your struggle to rediscover the Heron formula, which we knew for about 40 years by now... but what Serge said is correct, and it sets you on the right track. Just make all the calculus to the end, and THINK. Stop bullshitting around.

[science_man_88]

Originally Posted by a1call

That is not a general function. For different types of triangles the solutions will be different. For example for a right angle triangle the height can be equal to a side, so the algo will have to be adjusted.
A hint would be that the solution is independent of a base and a height( so you don't need to draw the triangle to specify the base and the height).
The other hint is that the solution is 2k years old (but you shouldn't underestimate the difficulty.).

you can break basically any triangle into right triangles that it does work ... for even a right triangle can have this done in theory all Serge did was split the hypotenuse ( a in this case is the longest side) and make two right triangles for which 1/2* base * height works the height is y in both cases the base of one right triangle is x the other is (a-x) so we have a total sum of 1/2*(a-x)*y+1/2*x*y=1/2*((a-x)+x)*y = 1/2*a*y = (a*y)/2

[a1call]

The 2k years old general solution:
Heron's Formula
https://www.mathsisfun.com/geometry/herons-formula.html

[CRGreathouse]

Originally Posted by a1call

The 2k years old general solution

I think we're being asked to derive that from first principles, but only if we didn't know about it already.

[Dr Sardonicus]

Originally Posted by a1call

Area of a Triangle as a Function of the Lengths of its 3 Sides.

If you already know the answer please keep quiet.
If you don't, try to come up with the function before googling it.

I learned this in high school geometry class.

Some things can be discerned without knowing the answer. Assume there is an algebraic formula F(a, b, c). Then

1) F(a, b, c) = F(a, c, b) = F(b,c,a) = F(b, a, c) = F(c, a, b) = F(c, b, a)

2) If k is a positive constant, then F(k*a, k*b, k*c) = k^2 * F(a, b, c)

3) Assuming 0 < a <= b <= c, and c = a + b, then F(a, b, c) = 0.

4) If 0 < a <= b <= c, and c < a + b, then F(a, b, c) is not a positive real number.

[a1call]

Interesting elaboration Dr Sardonicus. I particularly like number 2 and 4.
I have boasted about having an exceptionally good (long term) memory. Yet I do not recall ever being taught this formula in school (only base times 1/2 the height). So I thought perhaps I was not the only one and some others might be as impressed by the ancients' ingenuity as I was.

[science_man_88]

Originally Posted by a1call

Interesting elaboration Dr Sardonicus. I particularly like number 2 and 4.
I have boasted about having an exceptionally good (long term) memory. Yet I do not recall ever being taught this formula in school (only base times 1/2 the height). So I thought perhaps I was not the only one and some others might be as impressed by the ancients' ingenuity as I was.

I probably heard it at one point on this forum or reading but that's about it that I know of ( Admittedly, I missed a lot of school growing up ( probably spent as many hours driving in or visiting the hospital as I did school)).

[Batalov]

Two trivial notes:

Originally Posted by a1call

That is not a general function. For different types of triangles the solutions will be different.

Just in case this is not obvious to you: the area of a triangle with sides 5,6,9 is exactly the same as the area with sides 9,5,6. Triangles don't come labeled with "a", "b", "c" glued on to them. You can call "a" any side. Hint: always use "a" for the longest side for the purpose of the initial drawing. Now go back to the post #2 and try to prove to us that if "a" is the longest side, then the top vertex can "hang over" to the side (or even approach vertical lines perpendicular to (0,0) or (a,0) ). It can't! (or else b or c will become > a)

When you do what is spelled out for you in post #2, you will find that the obtained solution has a, b and c entered symmetrically.

Originally Posted by a1call

The other hint is that the solution is 2k years old (but you shouldn't underestimate the difficulty.).

... so you think people 2000 years ago couldn't do what is shown in post #2?? Wow! What amazing depth of ignorance! The opposite is true: people 2000 years ago were exactly the same as today. If anything, (average) people now only became stupider! (They don't have to work as hard as "ancients" - they have calculators, smart phones, microwaves, airplanes etc etc. so they really don't have to think anymore. They just press buttons.)