T is a triangle. Show that there is a triangle T' whose sides are equal to the medians of T, and that T'' is similar to T.
Solution
Let L, M be the midpoints of BC, AC respectively. Take B' as the reflection of B in M, so that AB'CB is a parallelogram. Let N be the midpoint of B'C. Then MN is parallel to AB' and half its length. Hence it is also equal and parallel to BL. So MNLB is a parallelogram and hence LN = BM. Finally, AN has the same length as the median from C. So ALN has sides equal to the medians of ABC. Thus the triangle T certainly exists.
Note that MNCL is a parallelogram, so X is the midpoint of LN. Hence AX is a median of T. But X is the midpoint of MC, so AX = (3/4) AC. Now LN is the median from B (in length), and we have shown that the median to it is (3/4) the length of the side opposite to B. Similarly for the other sides. So T'' has sides equal to those of T times (3/4). In particular, it is similar to T.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03