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we got this question in class and I am having a lot of trouble understanding how to go about it!

Question: Show that if Heron's formula is true for every triangle in which one of the sidelengths equals to 1, then it is true for every triangle.

My approach currently: So I basically said that given a triangle with one side length equal to 1, the other two sides can be of any other measurement provided that their sum is greater than 1. Such a set of triangles can generate all triangles since we can multiply any real number to all three sides, which gives a scalar of such triangles. Since Heron's formula works for the most basic set of triangles, then for scalars of such triangles, Heron's formula works as well.

user70871
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    Quite well explained. The "sum greater than $1$" part is not relevant. I think symbols would help. The point is that if the formula gives a certain answer $A$ for sides $x$, $y$, and $z$, then, substituting in the formula, we find that it gives answer $k^2A$ for side $kx$, $ky$, and $kz$. – André Nicolas Apr 04 '13 at 01:04
  • If I attempted the algebraic proof using symbols then it should give me what I've written down then right? – user70871 Apr 04 '13 at 01:08
  • @AndréNicolas, actually, it's quite necessary to restrict them so that the sum is greater than 1, because otherwise you'll get numbers that don't form a triangle (if the sum is equal to 1, then you get a straight line. Less than 1, and the other two sides cannot meet). In fact, there needs to be an additional restriction - that the difference between the two also can't be greater than 1 in magnitude. – Glen O Apr 04 '13 at 01:10
  • Yes. I think you have a correct understanding of the reason. But if this is an answer that is being graded, you want to make sure that you get full credit for understanding! – André Nicolas Apr 04 '13 at 01:11
  • Let the sides of the triangle be 1,a,b and assume (which you can, from your question) that Heron works. Then apply Heron on the values λ, λa, λb. You are able to factor out a power of λ from the squareroot, leaving the original Heron formula multiplied with a scalar in λ. That should do it in my view. – imranfat Apr 04 '13 at 01:13
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    It is certainly true that if one of the sides is $1$, there are restrictions on the other two sides. But that has nothing to do with the fact that if the Heron formula gives the right answer for triangles in which one side is $1$, it must give the correct answer for all triangles. And about your last comment, Heron will give a $\lambda^2$, for there will be "four" $\lambda$'s inside the square root. – André Nicolas Apr 04 '13 at 01:13

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[This is a community wiki answer to resolve the question]

From André Nicolas's comments, we see that if we take a triangle with sides $1, a, b$ and semiperimeter $S$ for which the Heron formula is known to work, and then enlarge it by a scale factor of $\lambda$, then for the enlarged triangle (with sides $\lambda, \lambda a, \lambda b$) we would have

$$\text{semiperimeter} = \frac{\lambda + \lambda a + \lambda b}{2} = \lambda S.$$

So that for the enlarged triangle, the Heron formula would give

$$\sqrt{\lambda S(\lambda S - \lambda)(\lambda S - \lambda a)(\lambda S - \lambda b)} = \sqrt{\lambda^4S(S-1)(S-a)(S-b)} = \lambda^2\sqrt{S(S-1)(S-a)(S- b)}$$

which is the correct expression for the area of the enlarged triangle. Thus, if Heron works for triangles with one side of length $1$, then it works for all triangles, since clearly every possible triangle is an enlargement of such a triangle (any triangle $p, q, r$ being an enlargement of $1, q/p, r/p$).

Old John
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