**Problem 4 of the Vietnam Mathematical Olympiad 1986**

Let ABCD be a square of side 2a. An equilateral triangle AMB is constructed in the plane through AB perpendicular to the plane of the square. Point S moves on AB such that SB = x. Let P be the projection of M on SC and E, O be the midpoints of AB and CM, respectively.

(a) Find the locus of P as S moves on AB.

(b) Find the maximum and minimum lengths of SO.

**Solution by Steve Dinh, aka Vo Duc Dien**

Let θ =∠MCP, η =∠ECP and I be the midpoint of CE. The lengths of the other segments are calculated to be

CA = CM = 2 \sqrt{2} a, CE = a \sqrt{5} , IC = \frac {1} {5} CE = \frac {1} {2} a \sqrt{5} , ME = a \sqrt{3} ,

SE = x – a, MS = \sqrt{(x - a)² + 3a²} , SC = \sqrt{x² + 4a²}

(a) Applying the law of cosines, we have

MS² = CM² + SC² - 2CM · SC cosθ or

(x - a)² + 3a² = 8a² + x² + 4a² - 4a \sqrt{2(x² + 4a²)} cosθ

or cosθ = \frac{x + 4a}{2\sqrt{2(x² + 4a²)}}

but cosθ = \frac{CP}{CM}; therefore, CP = \frac{a(x + 4a)}{ \sqrt{x² + 4a²}}

Again, the law of cosines gives us

SE² = CE² + SC² - 2 CE · SC cosη or

(x – a)² = 5a² + x² + 4a² - 2a \sqrt{5(x² + 4a²)} cosη

or cosη = \frac {x + 4a}{ \sqrt{5(x² + 4a²)}}

IP² = IC² + CP² - 2 IC · CP cosη or

IP² = \frac{5a²}{4} + \frac{a²(x + 4a)²}{x² + 4a²} - a \sqrt{5} \frac{a(x + 4a)}{ \sqrt{x² + 4a²}} \frac{x + 4a}{\sqrt{5(x² + 4a²)}} = \frac{5a²}{4}

or IP = \frac {1}{2} a \sqrt{5}

which is constant, and the locus of P is part of the circle with center at I and the radius of \frac {1}{2} a \sqrt{5}

that passes through point E and is from B to Q where Q is the intersection of the circle and CA.

(Hubert Shutrick has observed that this conclusion follows from the fact that, since ∠MPC is always right, P lies on the sphere with CM as a diameter, i.e., the sphere with center O and radius OC (= OM = OP).)

(b) Since I and O are the midpoints of CE and CM, respectively, IO || ME, and the plane containing the three points

M, C and E is perpendicular with the plane of the square, IO is then perpendicular with CE and SO² = IO² + SI².

But IO = \frac {1}{2} ME = \frac {1}{2} a \sqrt{5} is fixed; the extreme values of SO depend on SI.

As S moves on AB, SI is minimum when S is at the midpoint of EB (SI = a) and is maximum when S is at A when

SI² = AI² = AF² + FI² where F is the midpoint of AD. Let G be the midpoint of AC.

SI² = AF² + (FG + GI)² = a² + (a + \frac {a}{2} )² = \frac {13a²}{4} and

SO²_{max} = \frac {5a²}{4} + \frac {13a²}{4} = \frac {9a²}{2} or SO_{max} = \frac{3a}{\sqrt{2}} and

SO²_{min} = \frac {5a²}{4} + a² = \frac {9a²}{4} or SO_{min} = \frac {3a}{2}