MathDB

1

Part of 2012 India IMO Training Camp

Problems(4)

A geometry problem

Source:

6/9/2012
Let ABCABC be an isosceles triangle with AB=ACAB=AC. Let DD be a point on the segment BCBC such that BD=2DCBD=2DC. Let PP be a point on the segment ADAD such that ∠BAC=∠BPD\angle BAC=\angle BPD. Prove that ∠BAC=2∠DPC\angle BAC=2\angle DPC.
geometrytrigonometrygeometry unsolved
Problem with a trapezium-proving two lines are parallel

Source:

7/17/2012
Let ABCDABCD be a trapezium with AB∥CDAB\parallel CD. Let PP be a point on ACAC such that CC is between AA and PP; and let X,YX, Y be the midpoints of AB,CDAB, CD respectively. Let PXPX intersect BCBC in NN and PYPY intersect ADAD in MM. Prove that MN∥ABMN\parallel AB.
geometrytrapezoid
Angle equality in cyclic quadrilateral

Source:

7/22/2012
The cirumcentre of the cyclic quadrilateral ABCDABCD is OO. The second intersection point of the circles ABOABO and CDOCDO, other than OO, is PP, which lies in the interior of the triangle DAODAO. Choose a point QQ on the extension of OPOP beyond PP, and a point RR on the extension of OPOP beyond OO. Prove that ∠QAP=∠OBR\angle QAP=\angle OBR if and only if ∠PDQ=∠RCO\angle PDQ=\angle RCO.
invariantgeometrycyclic quadrilateralradical axispower of a pointgeometry proposed
quadrilateral ABCD without parallel sides is circumscribed

Source: ARO 2005 - problem 11.7

4/30/2005
A quadrilateral ABCDABCD without parallel sides is circumscribed around a circle with centre OO. Prove that OO is a point of intersection of middle lines of quadrilateral ABCDABCD (i.e. barycentre of points A, B, C, DA,\,B,\,C,\,D) iff OA⋅OC=OB⋅ODOA\cdot OC=OB\cdot OD.
geometryincenterparallelogramtrigonometrycomplex numbersRussia