Friday, October 20, 2017

Heron's Formula

The MathCounts guide for the year arrived today and I was looking over the problems. The following one caught my eye.

Given a triangle of side lengths 13,14, and 15. What is the radius of the largest circle inside it?

This is a slightly convoluted way of saying what is the radius of  the incircle. I then thought about it for a moment and came up with the following approach.

  • Find the area with Heron's Formula  = \(\sqrt{s(s-a)(s-b)(s-c)}\)  $$\sqrt{21\cdot6\cdot7\cdot8} = 84$$
  • The area is also \(\frac{1}{2} r \cdot 2s\)  So in this case that means r = 4.

My second thought was I don't think any of the kids in Math Club know Heron's formula. How would they approach this?  And then I realized they could drop an altitude and compute it via the Pythagorean theorem.  That would get them to being able to find the area from the standard formula \(\frac{1}{2}\)base x height.

Note: if you were lucky you might pick the base to be 14 and then find its made up of 2 Pythagorean triples the 5-12-13 and 9-12-15 but I'm going to ignore that possibility in favor of some more investigation.

From here   \(14^2 - x^2 = 15^2  - (13-x)^2\)   All the x^2 terms cancel out and you're left with a simple linear equation that reduces to \(x = \frac{70}{13}\) .

From there you reapply the Pythagorean theorem to find:
$$h=\sqrt{14^2 -\frac{70^2}{13^2}} = 7 \sqrt{\frac{4\cdot 13^2 - 100}{13^2}}$$
$$  =  7 \frac{\sqrt{576}}{13} = \frac{168}{13} $$

Now you can compute the area =  \(\frac{1}{2} \cdot 13 \cdot  \frac{168}{13} = 84 \) again.

And then I had an epiphany (which I'm sure has occurred in many textbooks.)  This would be a great intro to actually derive Heron's formula.  After finishing the concrete problem redo it with side lengths of a, b, and c.    

Repeating our steps. 

  \(b^2 - x^2 = c^2  - (a-x)^2\)     The x^2 terms cancel out again leaving:

$$ x = \frac{a^2 + b^2 - c^2}{2a} $$

You again apply the Pythagorean theorem to find the altitude:

$$ h = \frac{\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}}{2a}$$

This looks complex at first but if stare at it long enough there are a lot of differences of square here that we can take advantage of.  This would be a good time to review 

$$a^2 - b^2 = (a-b)(a+b)$$

So first  \(4a^2b^2 - (a^2 + b^2 - c^2)^2 = (2ab + (a^2 + b^2  - c^2))(2ab  - (a^2 + b^2 -c^2))\)
You then can combine the a and b terms to get:

$$ ((a+b)^2  - c^2)(c^2 - (a-b)^2)$$

That's nifty because we can reapply the difference of squares formula again:

$$((a+b)+c)((a+b)-c)(c + (a -b))(c - (a - b))$$

Putting this back into the height formula and with a little rearranging we get

$$ Area  = \frac{1}{2} a \cdot \frac{ \sqrt{(a+b+c)(a+b -c)(a+c-b)(b+c-a)}}{2a} $$

We're now ready to substitute the semiperimeter s  = \(\frac{(a+b+c)}{2}\) in which results in:

$$ Area  = \frac{1}{2} a \cdot \frac{ \sqrt{(2s)(2s -2c)(2s-2b)(2s - 2a)}}{2a} $$
$$           =  \frac{1}{2} a \cdot \frac{ \sqrt{ 16 \cdot  s(s -c)(s-b)(s - a)}}{2a} $$
$$           =  \sqrt{s(s -c)(s-b)(s - a)} $$

I'm guessing this would take almost the full hour especially if I let the kids experiment on their own first.

As a followup: something nifty happens when you investigate the Pythagorean triple triangles. The numbers in the formula are always the side lengths of the 2 triangles and the square/incircle radius that make them up.

This is not a coincidence and its easy to prove. Another good extension for a right triangle's in circle radius:

$$r = \frac{ab}{a+b+c} = s - c $$  

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