IMO 1995 SL G7

Origin: LAT | Category: Geometry

Problem

O is a point inside a convex quadrilateral ABCD of area S. K, L, M, and N are interior points of the sides AB, BC, CD, and DA respectively. If OKBL and OMDN are parallelograms, prove that \sqrt S \geq\sqrtS1 + \sqrtS2, where S1 and S2 are the areas of ONAK and OLCM respectively.

Solution

If O lies on AC, then ABCD, AKON, and OLCM are similar; hence AC = AO+OC implies \sqrt S = \sqrtS1 +\sqrtS2. Assume that O does not lie on AC and that w.l.o.g. it lies inside triangle ADC. Let us denote by T1, T2 the areas of parallelograms KBLO, NOMD respectively. Consider a line through O that intersects AD, DC, CB, BA respectively at X, Y, Z, W so that OW/OX = OZ/OY (such a line exists by a continuity argument: the left side is smaller when W = X = A, but greater when Y = Z = C). The desired inequality is equivalent to T1 + T2 \geq2\sqrtS1S2. Since trian- gles WKO, OLZ, WBZ are similar and WO + OZ = WZ, we have \sqrtSWKO + \sqrtSOLZ = \sqrtSWBZ = \sqrtSWKO + SOLZ + T1, which im- plies T1 = 2\sqrtSWKOSOLZ. Simi- larly, T2 = 2\sqrtSXNOSOMY . Since OW/OZ = OX/OY , we have SWKO/SXNO = SOLZ/SOMY . A B C D O K L M N S1 S2 T1 T2 X Y Z W Therefore we obtain T1 + T2 = 2

SWKOSOLZ + 2

SXNOSOMY = 2

(SWKO + SXNO)(SOLZ + SOMY ) \geq2

S1S2. Second solution. By an affine transformation of the plane one can trans- form any nondegenerate quadrilateral into a cyclic one, thereby preserving parallelness and ratios of areas. Thus we may assume w.l.o.g. that ABCD is cyclic. By a well-known formula, the area of a cyclic quadrilateral with sides a, b, c, d and semiperimeter p is given by S =

(p −a)(p −b)(p −c)(p −d) . Let us set AK = a1, KB = b1, BL = a2, LC = b2, CM = a3, MD = b3, DN = a4, NA = b4. Then the sides of quadrilateral AKON are ai, the sides of CLOM are bi, and the sides of ABCD are ai + bi (i = 1, 2, 3, 4). If p and q are the semiperimeters of AKON and CLOM, and xi = p−ai, yi = q −bi, then we have S1 = \sqrtx1x2x3x4, S2 = \sqrty1y2y3y4, and S =

(x1 + y1)(x2 + y2)(x3 + y3)(x4 + y4) . Thus we need to show that

4\sqrtx1x2x3x4 + 4\sqrty1y2y3y4 \leq (x1 + y1)(x2 + y2)(x3 + y3)(x4 + y4) . By setting yi = tixi we reduce this inequality to 1 + 4\sqrtt1t2t3t4 \leq (1 + t1)(1 + t2)(1 + t3)(1 + t4) . One way to prove the last inequality is to apply the simple inequality 1 + \sqrtuv \leq

(1 + u)(1 + v) to \sqrtt1t2, \sqrtt3t4 and then to t1, t2 and t3, t4.