Geometry

Geometry Problems of RMO’17

Q.1. Let AOB be a given angle less than 180^{\circ} and let P be an interior point of the angular region determined by \angle AOB. Show, with proof, how to construct, using only ruler and compass, a line segment CD passing through P such that C lies on the way OA and D lies on the ray OB, and CP:PD=1:2.

IMG-20171008-WA0035

Analysis: Draw a line parallel to OB through C. Let it intersect OP (extended) at K. \triangle CPK and \triangle DPO are similar which implies OP : PK = DP : PC = 2:1.

Construction: Choose K on extended OP such that OP : PK=2:1. Draw a line parallel to OB through K. It will intersect OA at C (required). Then CP will cut OB at D (required).

Q.5. Let \Omega be a circle with a chord AB which is not a diameter. \Gamma_{1} be a circle on one side of AB such that it is tangent to AB at C and internally tangent to \Omega at D. Likewise, let \Gamma_{2} be a circle on the other side of AB such that it is tangent to AB at E and internally tangent to \Omega at F. Suppose the line DC intersects \Omega at X \neq D and the line FE intersects \Omega at Y \neq F. Prove that XY is a diameter of \Omega.

 

IMG-20171008-WA0036

Label the centres as P and Q, as shown in the figure. Let O be the centre of the main circle.

\angle PCD=\angle PDC=\angle OXD gives XO\parallel CP. Now, CP\bot AB, so we get XO\bot AB. Thus XO is nothing but the perpendicular bisector of AB. Similarly YO is also the perpendicular bisector of AB. Hence X, O, Y are collinear. Hence done.

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