SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Question

Let BC be a chord of a circle (O) such that BC is not a diameter. Let AE be a diameter perpendicular to BC such that A belongs to the larger arc BC of (O). Let D be a point on the larger arc BC of (O) which is different from A. Suppose that AD intersects BC at S, ED intersects BC at T. Let F be the midpoint of ST and I be the second intersection of the circle (ODF) with BC. 1. Let the line passing I and parallel to OD intersect AD and ED at M and N respectively. Find the maximum value of the area of triangle MDN when D moves on the larger arc BC of (O) (such that D A). 2. Prove that the perpendicular from D to ST passes through the midpoint of MN. FIGURE_0

Accepted Answer

1) First, note that ADE = 90^ then DO, DF are two medians of the triangles ADE, SDT. Then ODF = ODT + FDT = OET + FTD = 90^. Hence, OIF = 180^ - ODF = 90^ which implies that I is the midpoint of BC. Since ODE is isosceles triangle then INE, IMA are also isosceles which implies that IE = IN, IM = IA. Hence MN = IM - IN = IA - IE = const. FIGURE_0 Two triangles DMN and DAE are similar with the constant ratio, then to maximize the area of DMN, we have to maximize the area of triangle ADE. We have [ADE] = 12 DA DE DA^2 + DE^22 = AE^22. The equality occurs when DA = DE or D lies on circle such that ADE is isosceles right triangle. 2) Denote P as the midpoint of MN then PDN = PND = INE = IEN, thus DP AE or the perpendicular line from D to ST passes through the midpoint of MN.