Problem 2 of Asian Pacific Mathematical Olympiad 2003

Suppose ABCD is a square piece of cardboard with side length a. On a plane are two parallel lines l1 and l2, which are also a units apart. The square ABCD is placed on the plane so that sides AB and AD intersect l1 at E and F respectively. Also, sides CB and CD intersect l2 at G and H respectively. Let the perimeters of triangle AEF and triangle CGH be m1 and m2 respectively. Prove that no matter how the square was placed, m1 + m2 remains constant.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to the lovely Katrina Ngo)

Its easily seen that the two triangles AEF and CHG are similar. We have

HC / AE = GC / AF

Picks points M and N on BC and DC, respectively such that NH = AE and MG = AF.

We then have HC / NH = GC / MG or GH || MN

From H draw a line parallel to AD and intercept MN at L. Triangles AEF and HNL are congruent. Therefore

AE = NH, AF = LH = MG, EF = LN, GH = ML

m1 = AE + AF + EF

m2 = HC + GC + GH

m1 + m2 = AE + AF + EF + HC + GC + GH = NH + HC + GC + MG + ML + LN = NC + MC + MN

or m1 + m2 is the perimeter of triangle MCN.

From F draw a line perpendicular to and intercept GH at J, we have FJ = a as given by the problem. Similarly, from A draw a line perpendicular to and intercept MN at I, we have

FJ = AI = a

This proves that line MN is tangential to the circle with radius a and center A. Therefore the perimeter of triangle MCN equals BC + DC = 2a, or m1 + m2 is a constant.

Extension of the problem: Let K be the foot of incenter of incircle of triangle MCN to MN. Prove that

IK = MN 2 × KN