Let A be one of the two points of intersection of two circles with centers X and Y. The tangents at A to these two circles meet the circles again at B,C. Let the point P be located so that PXAY is a parallelogram. Show that P is the circumcentre of triangle ABC.

Solution: Since PXAY is a parallelogram, we have PX∥AY. As AY⊥AB, it follows that PX⊥AB. Since AB is a chord of the circle with center X, we conclude that PX is in fact the perpendicular bisector of AB. Similarly, PY is perpendicular bisector of AC. Thus the perpendicular bisectors of two sides, AB and AC, … Continue reading Let A be one of the two points of intersection of two circles with centers X and Y. The tangents at A to these two circles meet the circles again at B,C. Let the point P be located so that PXAY is a parallelogram. Show that P is the circumcentre of triangle ABC.

Let D, E, F be points on the sides BC, CA, AB respectively of a triangle ABC such that BD = CE = AF and ∠BDF = ∠CED = ∠AFE. Prove that ABC is equilateral.

INMO 2011 Solution 1: Solution 2: Here is a pure geometric solution. Consider the triangle BDF, CED and AFE with BD, CE and AF as bases. The sides DF, ED and FE make equal angles θ with the bases of respective triangles. If B ≥ C ≥ A, then it is easy to see that … Continue reading Let D, E, F be points on the sides BC, CA, AB respectively of a triangle ABC such that BD = CE = AF and ∠BDF = ∠CED = ∠AFE. Prove that ABC is equilateral.

Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all sides of the quadrilateral, then find its radius.

RMO 2006 Solution: Let P, Q, R, S be the points of contact of in-circle with the sides AB, BC, CA, DA respectively. Since AD is perpendicular to AB and AB || DC we see that, AP=AS=SD=DR=r, radius of the in-circle. Let BP=BQ=y and CQ=CR=x . Using AB=3CD, we get (r+y)=3(r+x)