If n = 1 then s↓1= bh/(b+h), as showed in the problem proposed 69 (with proof posted in September 9, 2009). If n = 2 we have s↓2= s↓1h↓2/(s↓1+h↓2), where h↓2 is the height of the triangle with vertex B and base s↓1. From h↓2= h - s↓1 we have h↓2= h - bh/(b+h) = h↑2./(b+h). So,
With the help of Problem 69, or just simply by similat triangle, it is easily found that s1=bh/(b+h)
By similar triangle again, (h-s1-s2)/h=s2/b hs2=bh-bs1-bs2 (h+b)s2=bh-bs1 By putting back s1=bh/(b+h) and with simplification, the desired s2 is achieved The rest can be deducted by MI easily
Proof.
ReplyDeleteIf n = 1 then s↓1= bh/(b+h), as showed in the problem proposed 69 (with proof posted in September 9, 2009). If n = 2 we have s↓2= s↓1h↓2/(s↓1+h↓2), where h↓2 is the height of the triangle with vertex B and base s↓1.
From h↓2= h - s↓1 we have h↓2= h - bh/(b+h) = h↑2./(b+h). So,
s↓2= s↓1h↓2/(s↓1+h↓2) = (bh/(b+h))(h↑2./(b+h))/((bh/(b+h)+(h↑2./(b+h)) = bh↑2./(b+h)↑2.
Therefore, by induction we have s↓n= bh↑n./(b+h)↑n.
QED, Ianuarius.
In addition to the above proof by Unknown on September 11, 2009, find the finalization of the proof by induction here.
ReplyDeleteWith the help of Problem 69, or just simply by similat triangle, it is easily found that s1=bh/(b+h)
ReplyDeleteBy similar triangle again, (h-s1-s2)/h=s2/b
hs2=bh-bs1-bs2
(h+b)s2=bh-bs1
By putting back s1=bh/(b+h) and with simplification, the desired s2 is achieved
The rest can be deducted by MI easily