## October 2020

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In the Math Corner of my October Diary I offered the following brainteaser.

: ABC is an equilateral triangle. Somewhere inside the triangle is a point distant exactly 3 inches from A, 4 inches from B, and 5 inches from C. What is the area of the triangle?Problem

Supplementary: Can such a point beoutsidethe triangle? If it can, what then is the triangle's area?

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**• My solution**

Straightforward trigonometry. Call the length of the triangle's side(s) *a*, and call angle ABP *θ* Applying
the cosine rule to triangle ABP:

3² =a² + 4² − 8acosθ

Now apply it to triangle BCP:

5² =a² + 4² − 8acos(60° −θ)

Two equations in two unknowns. We can eliminate *θ*. From the first equation there,
cos* θ* = (*a*² + 7) / 8*a*. Subtracting the square of that from 1 gives
sin²*θ* in terms of *a*.

Expanding the last term in the second equation, using
the formula for
cos (*α* − *β*), gives sin *θ* in terms of cos *θ* and *a*.
We already know cos *θ* in
terms of *a*, so now we have sin *θ* in terms of just *a*. Squaring that, we have *another* expression for
sin²*θ* in terms of *a*.

Putting the two expressions for sin²*θ* in terms of *a* on either side of an equals sign, we have an equation in just
*a*. The equation reduces to

a^{4}− 50a² + 193 = 0

Solving: *a*² = 25 ± 12√3.

Those two values, to four decimal places, are 45.7846 and 4.2154. The square roots are 6.7664 and 2.0531. Those are values of *a*,
the side length of the equilateral triangle. Plainly, by eye inspection, it's the first value that's appropriate here.

Since the area of an equilateral triangle is (√3 / 4) times the square of its side length, the answer is (25/4)√3 + 9, which works out to 19.8253 square inches.

For the supplementary: A minute or two's playing with drinking straws cut to length 3, 4, and 5 suggest that it *is* possible. As
before, call angle ABP *θ* and apply the same logic.

In spite of the fact that the cosine of angle CBP is now cos (60 + *θ*), not
cos (60 − *θ*), you end up at last with the same equation for *a*. Plainly the *other* value for
*a*² is the one that applies here:

a² = 25 − 12√3.

And now the area of the triangle is (25/4)√3 − 9, i.e. 1.8253 square inches.