Interesting Property of a Trapezium

Prove that the straight line that passes through the point of intersection of the diagonals of a trapezium and through the point of intersection of its non-parallel sides, bisects each of the parallel sides of the trapezium.

Area of a Triangle

Solution:

Geometry: Problem involving orthocentre

ABC  is a triangle with AB = 13; BC = 14 and CA = 15. AD and BE  are the altitudes from A and B  to BC and AC respectively. H is the point of intersection of AD and BE. Find the ratio HD/HB. 

Solution: 

Cone on a Sphere : Geometry

A hollow right circular cone rests on a sphere as shown in the figure. The height of the cone is 4 metres and the radius of the base is 1 metre. The volume of the sphere is same as that of the cone. What is the distance between the centre of the sphere and the vertex of the cone?

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A Triangle With a 45 Degrees Angle in Square : A very hard problem from Geometry

Let P and Q be the points where AN and AM intersect the diagonal BD, respectively. It is noted that ∠QAN=∠QDN=45∘.Thus, AQND is a cyclic quadrilateral, therefore implying that ∠AQN=90∘.

This in turn implies that ΔAQN is a right-isoceles triangle with AN as the hypotenuse. Thus, AQ=AN2–√.

The same argument can be employed to show that AP=AM2–√.

Now, since triangles APQ and AMN share a common vertex angle A, we have, Area(ΔAPQ)Area(ΔAMN)=AQ⋅APAM⋅AN. However, from the conclusions in the previous two paragraphs, we have AQAN=APAM=12–√.

This therefore implies that Area(ΔAPQ)Area(ΔAMN)=12, or equivalently, Area(ΔAPQ)Area(MNPQ)=1, as claimed by the problem.

The problem, due to V. Proizvolov, appeared in Kvant - a popular Russian magazine - (#1, 2004, M1895), with a solution in a later issue (#4, 2004).