Using Pythagoras’ theorem on each of the right-angled triangles VAB, VBC and VCA, we obtain the following values for AB, BC and CA.

AB = √(a² + b²), BC = √(b² + c²), CA = √(c² + a²).

To find the area of triangle ABC we need to know the sine of one of the angles; let’s choose the angle at A.

Using the cosine rule we have

cos A = (√(c² + a²)²+ √(a² + b²)²– √(b² + c²)²)/(2√(c² + a²) √(a² + b²))

=(2a² + b² + c² – (b²+ c²))/( 2√(c² + a²) √(a² + b²))

= a²/(√(c² + a²) √(a² + b²))

Now we can find the value of sin A using 1 – cos² A.

sin² A = 1 – [a²/(√(c² + a²) √(a² + b²))]²

= 1 - a⁴/((c² + a²) (a² + b²))

= ((c² + a²) (a² + b²) – a⁴)/((c² + a²) (a² + b²))

= (a²b² + b²c²+ c²a²) /((c² + a²) (a² + b²))

So, sin A = √(a²b² + b²c²+ c²a²)/ (√(c² + a²) √(a² + b²))

Therefore, the area of triangle ABC is

½ x AB x AC x sin A

= ½ √(a² + b²)) x √(c² + a²)x√(a²b² + b²c²+ c²a²)/ (√(c² + a²) √(a² + b²))

= ½ √(a²b² + b²c²+ c²a²)

We have found that the set of values a = 2, b = 4 and c = 6 give an integer area (equal to 14).

We’d be interested to know if you find any more!

Thanks to Aarnav, Y11, for suggesting the following way to generate suitable sets of values.

We want ½ √(a²b² + b²c²+ c²a²) to be an integer, say x.

Note that in order for the square root to be divisible by 2, at least two of a, b and c must be even.

Let us suppose that a and b are any two even numbers we choose.

Then a²b² + b²c²+ c²a²= 4x²

c²(b²+ a²)= 4x² - a²b²

c²(b²+ a²)= (2x + ab)(2x – ab)

Let us choose again, that c² = 2x – ab and b²+ a² = 2x + ab.

Eliminating x (e.g. by subtracting these two equations) we get

c² – (a² + b²) = – 2ab

c² = a² + b² – 2ab = (a – b)²

Therefore c = |a – b|, so the set of integers a, b, and |a – b| will produce an integer value for x.

Similarly, if we choose that c² = 2x + ab and b²+ a² = 2x – ab, and work through to find c in the same way, we get

c = a + b

so a, b, a + b will also produce an integer value for x.

Challenge 310: A Particular Pyramid