Geometry shortlist problem in IMO 2014

Geometry

Problem 1

The points P and Q are chosen on the side BC of an acute-angled triangle ABC so that \angle PAB=\angle ACB and \angle QAC=\angle CBA. The points M and N are taken on the rays AP and AQ, respectively, so that AP = PM and AQ = QN. Prove that the lines BM and CN intersect on the circumcircle of the triangle ABC. (See : Solution Problem 1)

Also Read:

Problem 2:

Let ABC be a triangle. The points K, L, and M lie on the segments BC, CA, and AB, respectively, such that the lines AK, BL, and CM intersect in a common point. Prove that it is possible to choose two of the triangles ALM, BMK, and CKL whose inradii sum up to at least the inradius of the triangle ABC. ( Read : Solution Problem 2)

Problem 3:

Let \Omega and O be the circumcircle and the circumcentre of an acute-angled triangle ABC with AB a BC. The angle bisector of \angle ABC intersects \Omega at M\neq B. Let \Gamma be the circle with diameter BM. The angle bisectors of \angle AOB and \angle BOC intersect \Gamma at points P and Q, respectively. The point R is chosen on the line PQ so that BR = MR. Prove that BR || AC. (Here we always assume that an angle bisector is a ray). (Read : The Solution )

Problem 4:

Consider a fixed circle \Gamma with three fixed points A, B, and C on it. Also, let us fix a real number \lambda \in (0,1). For a variable point P\notin {A,B,C} on \Gamma, let M be the point on the segment CP such that CM=\lambda .CP. Let Q be the second point of intersection of the circumcircles of the triangles AMP and BMC. Prove that as P varies, the point Q lies on a fixed circle. (Read : The Solution)

Problem 5:

Let ABCD be a convex quadrilateral with \angle B=\angle D=90^0. Point H is the foot of the perpendicular from A to BD. The points S and T are chosen on the sides AB and AD, respectively, in such a way that H lies inside triangle SCT and

\angle SHC=\angle BSC=90^0, \angle THC=\angle DTC=90^0

Prove that the circumcircle of triangle SHT is tangent to the line BD. (Read : Solution)

Problem 6:

Let ABC be a fixed acute-angled triangle. Consider some points E and F lying on the sides AC and AB, respectively, and let M be the midpoint of EF. Let the perpendicular bisector of EF intersect the line BC at K, and let the perpendicular bisector of MK intersect the lines AC and AB at S and T, respectively. We call the pair (E, F) interesting, if the quadrilateral KSAT is cyclic. Suppose that the pairs (E1, F1) and (E2, F2 ) are interesting. Prove that

\frac{E_1.E_2}{AB}=\frac{F_1.F_2}{AC}

Read : Solution Problem 6

Problem 7:

Let ABC be a triangle with circumcircle \Omega and incentre I. Let the line passing through I and perpendicular to CI intersect the segment BC and the arc BC (not containing A) of \Omega at points U and V , respectively. Let the line passing through U and parallel to AI intersect AV at X, and let the line passing through V and parallel to AI intersect AB at Y . Let W and Z be the midpoints of AX and BC, respectively. Prove that if the points I, X, and Y are collinear, then the points I, W, and Z are also collinear.

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