IMO 1988 SL 13

Origin: GRE

Problem

In a right-angled triangle ABC, let AD be the altitude drawn to the hypotenuse and let the straight line joining the incenters of the triangles ABD, ACD intersect the sides AB, AC at the points K, L respectively. If E and E1 denote the areas of the triangles ABC and AKL respectively, show that E E1 \geq2.

Solution

Let AB = c, AC = b, \angleCBA = \beta, BC = a, and AD = h. Let r1 and r2 be the inradii of ABD and ADC respectively and O1 and O2 the centers of the respective in- circles. We obviously have r1/r2 = c/b. We also have DO1 = \sqrt 2r1, DO2

\sqrt 2r2, and \angleO1DA

\angleO2DA = 45◦. Hence \angleO1DO2 = 90◦and DO1/DO2

c/b from which it follows that \triangleO1DO2 ∼ \triangleBAC. A B C D K L O1 O2 P We now define P as the intersection of the circumcircle of \triangleO1DO2 with DA. From the above similarity we have \angleDPO2 = \angleDO1O2 = \beta = \angleDAC. It follows that PO2 \parallelAC and from \angleO1PO2 = 90◦it also fol- lows that PO1 \parallelAB. We also have \anglePO1O2 = \anglePO2O1 = 45◦; hence \angleLKA = \angleKLA = 45◦, and thus AK = AL. From \angleO1KA = \angleO1DA = 45◦, O1A = O1A, and \angleO1KA = \angleO1DA we have \triangleO1KA ∼= \triangleO1DA and hence AL = AK = AD = h. Thus E E1 = ah/2 h2/2 = a h = a2 ah = b2 + c2 bc \geq2 . Remark. It holds that for an arbitrary triangle ABC, AK = AL if and only if AB = AC or ∡BAC = 90◦.