Evaluate the aspect ratio of a triangle

Jelly, 7 bytes

SH_÷@HP

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Explanation

enter image description here

Let’s read this chain:

  • The implicit argument is a list [a, b, c].

  • First we read S. This takes the sum: a + b + c.

  • Then, we read H. This halves it: (a + b + c)/2. (This is s.)

  • Then, we read a dyad _ (subtract), followed by another dyad. This is a hook: it lacks a right argument, so it receives the argument to this chain, [a, b, c], giving us [s-a, s-b, s-c]. (This is the fifth chain pattern in the table here.)

  • Then, we read the dyad-monad pair ÷@H. This is a fork: ÷@ is division with the arguments flipped, and H is halve, so our working value gets Half the argument to this chain ÷’d by it. This vectorizes; we’re left with [(a/2)/(s-a), (b/2)/(s-b), (c/2)/(s-c)]. (This is the second chain pattern in the table here.)

  • Finally, we take the product with P, getting us abc/(8(s-a)(s-b)(s-c)).

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Jelly, 6 bytes

This answer is based on Emigna's 05AB1E answer. Many thanks to Dennis and Lynn for their help in figuring this answer out. Golfing suggestions welcome! Try it online!

S_Ḥ⁸÷P

Ungolfing

           Implicit argument [a, b, c].
S          Take the sum, a+b+c or 2*s
  Ḥ        Take the double, [2*a, 2*b, 2*c].
 _         Vectorized subtract, giving us [2*(s-a), 2*(s-b), 2*(s-c)].
   ⁸÷      Vectorized divide the initial left argument, the input [a, b, c],
             by [2*(s-a), 2*(s-b), 2*(s-c)].
     P     Take the product giving us the aspect ratio, abc/8(s-a)(s-b)(s-c).

Jelly, 6 bytes

S÷_2Pİ

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How it works

S÷_2Pİ  Main link. Argument: [a, b, c]

S       Sum; compute 2s := a + b + c.
 ÷      Divide; yield [2s ÷ a, 2s ÷ b, 2s ÷ c].
  _2    Subtract 2; yield [2s ÷ a - 2, 2s ÷ b - 2, 2s ÷ c - 2].
    P   Product; yield (2s ÷ a - 2)(2s ÷ b - 2)(2s ÷ c - 2).
     İ  Invert; yield 1 ÷ (2s ÷ a - 2)(2s ÷ b - 2)(2s ÷ c - 2).