Area of triangle 2...

ABC is a right angled triangle with orthocentre at C. Given the circumradius as 'R' and the angle between the medians not passing through C be Ɵ. Find the area of triangle

Solution:
The Area of the triangle is

Area of Triangle...

ABC is a triangle P is a point on the plane of the triangle such that |AR(PAB)| = |AR(PAC)| = |AR(PBC)|. Find possible number of positions of P.

Solution:

The answer is Four

But how...???....think !!!...let me know if you want to know

Here follows a problem which troubles (possibly)..

FIND THE SHORTEST LINE SEGMENT WITH ITS ENDS ON TWO SIDES OF A GIVEN TRIANGLE WHICH BISECTS THE TRIANGLE

i have a solution to this problem i will post the same however i wish someone comes out with a solution

... it's been days and no reply has come...so, here is the solution

consider the figure below

Let PQ be the shortest line segment such that  Area of ΔCPQ is equal to Area of the Quadrilateral PQBA.

Let CP = x ,  CQ  = y , PQ =  u.

QM and BE are perpendiculars from Q and B on to CA, we can see that ΔCQM ||| ΔCBE, Then


Hence we will have  QM = ysin C and BE = a sin C



PQ bisects ΔABC implies



Using the Cosine formula for triangle CPQ





Let us assume this is equal to r. then



Upon simplification , we get


Observation : Minimum value of R.H.S is reached when the first etrm of L.H.S is 0. ie.,


This implies that r = ab(1-cosC) and consequently we get




Therefore length of PQ can be found as


Generalisation:

Find a+b-c, b+c-a, c+a-b, choose the least and next higher of them; for example if a+b-c and b+c-a are these two in that order, then P and Q must be chosen on BC and BA such that
 
and the shortest length PQ that bisects triangle abc will then be

S Ramanujan problem--for school kids

This is a problem posed by S Ramanujan .. try ...  

Here is the solution









so on.. we get