**Problem 4 of the Vietnam Mathematical Olympiad 1962**

Let be given a tetrahedron ABCD such that triangle BCD equilateral and AB = AC = AD. The height is h and the angle between two planes ABC and BCD is α. The point X is taken on AB such that the plane XCD is perpendicular to AB. Find the volume of the tetrahedron XBCD.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Let’s find the area of triangle XDC denoted (XDC) and the length segment XB.

Let a be the side length of triangle BDC and b = AB = AC = AD.

Draw the altitude AM to BC and apply Pythagorean’s theorem, we have AM² = b² - a² \over 4

or AM = 1 \over 2 \sqrt{4b²-a²} and BC² = a² = CX² + BX²

Since M is the midpoint of BC and ABC is an isosceles triangle with AB = AC and triangle BCD is equilateral, α = ∠AMD.

We also have tan∠ABC = AM \over BM = CX \over BX = 1 \over a \sqrt{4b²- a²}

Now solve the two equations, we have BX = a² \over 2b , CX = a \over 2b \sqrt{4b²- a²}

Now apply Heron’s formula for (XDC), taking into account that CX = DX, we have (XDC) = \sqrt{s (s-a)(s-CX)²}

where s = a \over 2 + CX = a \over 2 + a \over 2b \sqrt{4b²- a²} , s – CX = a \over 2 , s – a = -a \over 2 + a \over 2b \sqrt{4b²- a²}

After some computations, (XDC) = a² \over 4b \sqrt{3b²- a²}

The volume of the tetrahedron XBCD, by definition, is

V = 1 \over 3(XDC) × BX = a^4 \over 24b^{2} \sqrt{3b²- a²}

Furthermore, h² = AM² - HM² = AM² - ( DM \over 3)² = 3b²- a² \over 3 or h = \sqrt{3b²- a² \over 3} .

The volume is now V = a^4h \sqrt{3} \over 24b² (*)

But tanα = h \over HM = 6h \over a \sqrt{3} or a = 6h \over tanα \sqrt{3}

And b² = AM² + BM² = h² + HM² + a² \over 4 = h² + a² \over 3

Or b² = h² + 4h² \over tan²α

Substitute the values of a and b² into (*), we have

The volume of the tetrahedron XBCD = 6h³ \sqrt{3} \over tan²α (tan²α + 4)