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) + 3a} , SC = \sqrt{x + 4a}

(a) Applying the law of cosines, we have

MS = CM + SC - 2CM · SC cosθ or

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

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

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

Again, the law of cosines gives us

SE = CE + SC - 2 CE · SC cosη or

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

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

IP = IC + CP - 2 IC · CP cosη or

IP = \frac{5a}{4} + \frac{a(x + 4a)}{x + 4a} - a \sqrt{5} \frac{a(x + 4a)}{ \sqrt{x + 4a}} \frac{x + 4a}{\sqrt{5(x + 4a)}} = \frac{5a}{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 SO = IO + SI.

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

SI = AI = AF + FI where F is the midpoint of AD. Let G be the midpoint of AC.

SI = AF + (FG + GI) = a + (a + \frac {a}{2} ) = \frac {13a}{4} and

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

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