Pythagorean Games
In a Right triangle ABC with side lengths a,b and c.AM,BN and CQ are medians of the triangle.The points M,N and Q are connected to give a new triangle T1, medians MM1, NN1 and QQ1 are drawn again for the triangle T1 and the points M1,N1 and Q1 are also connected to give a new triangle T2. The process continues to infinity. if the sum of areas of all new triangles(T1,T2, T3 ,...) is H. and the area of the original tringle ABC is F. find 3H - F.
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
**We can assume that the right angle is A with coordinates (0,0) and the coordinates of B,C are ( a,0) , (0,b) such that c² = a² + b² , c is the hypotenuse. Now N,M and Q are mid points , their coordinats are (a/2,b/2) , (0,b/2)and (a/2,0) respectively are the vertices of T1 . T1 is also a right triangle (at N). The area of T1 = ab/8. Similarly
The area of T2 = ab/32 . The area of T3 = ab/128. T1+T2+T3 +… = ab/8 ( 1+1/4+1/16+… ) = ab/6 = H. F = ab/2. 3H-F = 0 .