# VMO 1962 Problem 4

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 , sa = -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)