# 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)

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

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

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.