*Given a triangle of side lengths 13,14, and 15. What is the radius 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 = \(\sqrt{s(s-a)(s-b)(s-c)}\) $$\sqrt{21\cdot6\cdot7\cdot9} = 84$$
- 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.

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 but a good extension that for a right triangle's in circle radius:

$$r = \frac{ab}{a+b+c} = s - c $$

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 but a good extension that for a right triangle's in circle radius:

$$r = \frac{ab}{a+b+c} = s - c $$

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