You have four bars, which have different integer lengths.You construct a triangle using 3 of them.
When you make a new triangle by substituting one of the bars with the fourth one, you realize that these two triangles are similar triangles.
Question: What is the minimum possible total length of these four bars?
[Post your answers with explanation in the comments]
5 comments:
4 each wid length 1
2 triangles r eqilateral wid side 1
Let length of bars be a,b,c,d
Now, a/b=b/c=c/d = z
and, a # b # c # d
also, a+b > c;
and, b+c > d
(sum of 2 sides of triangle more than the 3rd side)
c = dz
b = cz = dz2
a = bz = dz3
Min(a+b+c+d)
= dz3 + dz2 + dz + d
= d(1 + z + z2 + z3)
= d(1-z4)/(1-z)
= d(1-z)(1+z)(1+z2)/(1-z)
= d(1+z)(1+z2)
a # b # c # d
=> z # 1
and a+b > c
=> dz2(1+z) > dz
=> z(1+z) > 1
Minimum 15 units i.e. 1,2,4 and 8
Taking 1st triangle :
Bar lengths - 1,2,4 (ratio=1:2:4)
Taking 2nd triangle :
(Bar of 1 unit lenth is replaced by bar of 8 unit length)
Bar lengths - 2,4,8 (ratio=1,2,4)
As, side ratio is same in both cases, these triangles are similar !
HOPE U HAV GOT ANS.
65
lengths 8,12,18,27 units
Minimum 15 units i.e. 1,2,4 and 8
is not possible as
124 and 248 can not make triangle.
check at
http://www.orkut.co.in/Main#CommMsgs?cmm=50230760&tid=5381924481234269138&na=1&nst=1
Ya prerna is right.
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