# APMO 2003 Problem 2

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) Its 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