»  Solutions to puzzles in my VDARE.com monthly Diary

  October 2020


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

Problem:  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?

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


• 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 − 8a cos θ

Now apply it to triangle BCP:

5² = a² + 4² − 8a cos(60° − θ)

Two equations in two unknowns. We can eliminate θ. From the first equation there, cos θ = (a² + 7) / 8a. Subtracting the square of that from 1 gives sin²θ 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 sin²θ in terms of a.

Putting the two expressions for sin²θ in terms of a on either side of an equals sign, we have an equation in just a. The equation reduces to

a4 − 50a² + 193 = 0

Solving:  a² = 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 a² 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.