See complete Problem 70
Squares Inscribed in a Triangle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 70
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See complete Problem 70
Squares Inscribed in a Triangle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
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