What's the source ?
Very tough. I got the solution, but took lot of time !

consider the right angled triangle formed in the co-ordinate plane :


*
**
***
****
*****
......... and so on 51 times

The number of integral points within the triangle =
Total no. of points - the points on edges of triangle = (1+2+3+....+51) - all the points on the permiter of triangle
= 1326 - 150 = 1176.

Equation of the hypotenuse : y=-x+50
[You can get this by substituting (50,0) and (0,50) in y= mx+c]
This means that every point on the hypotenuse which has a y-co ordinate as integer has the x-coordinate as integer.
Lets get the reqd number for smaller triangles
Its a right angled isoseles traingle
Common side : Points within triangle formed
3 : 1
4 : 3(1+2)
5 : 6(1+2+3)

The pattern found for a triangle with common side=n is sum of positive integers upto (n-2)
Thus for 50, required no.of points = sum of numbers upto 48
= 48*49/2
= 1176

The triangle formed on x-y plane is right angled isoceles

I tried to apply this to triangle with co-ordinates (0,0),(4,0) and (0,4)
refer to the image below.



it can be noticed that there are 3 integer points which are inside the triangle (marked blue)
the other one's are either outside the triangle or on the hypotenuse and the equal sides
The total number of integer points formed inside the square (Two right angled triangles combined)
are (intersections at straight lines in the square) = (4-1)*(4-1) = 9
and integer points that are inside the triangle are = 9 - 6 = 3
This can be written as [9-(4-1)]/2 = 3

This can be checked for co-ordinates (0,0),(3,0) and (0,3)
For this number of integer co-ordinates inside the triangle are = [(3-1)(3-1) - (3-1)]/2 = 1

Generalizing-
[(n-1)*(n-1) - (n-1)]/2 = (n-1)(n-2)/2
for right angled triangle with length 50, n - 1 = 49 and n - 2 = 48
(49*48)/2 = 1176

beat_gmat_09 wrote:What's the source ?
Very tough. I got the solution, but took lot of time !

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