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# 1 cm thick rings are hanging from a peg. top rings outside diameter is 20cm. outside diameter of each of outer?

rings is 1cm less then the one above it. bottom rings outside diameter is 3 cm. So how would I find the distance from the top of the top ring to the bottom of the bottom ring?

### 1 Answer

- jeffreyLv 79 years agoFavorite Answer
Answer ... 170 cm ...

First, determine the total number of rings …

Let … i = number of the ring … i = 1 is the bottom ring R (1)

… i = 2 is the ring R (2) above the bottom ring … etc …

Let … ODR (i) = outer diameter of ring R (i) … then

… ODR (i + 1) = ODR (i) + 1 in cm … i = 1 , 2 , 3 , ⋯

For i = 1 … ODR (1 + 1) = ODR (2) = ODR (1) + 1 …

and since ODR (1) = 3 … ODR (2) = 3 + 1 = 4 …

For i = 2 … ODR (2 + 1) = ODR (3) = ODR (2) + 1 …

Substituting … ODR (2) = ODR (1) + 1 … we get …

… ODR (3) = [ ODR (1) + 1 ] + 1 = ODR (1) + 2 = 5 …

For i = 3 … ODR (3 + 1) = ODR (4) = ODR (3) + 1 …

Substituting … ODR (3) = ODR (1) + 2 … we get …

… ODR (4) = [ ODR (1) + 2 ] + 1 = ODR (1) + 3 = 6 …

That is … for i = 1 , 2 , 3 , we have …

… ODR (2) = ODR (1) + 1 = 4 …

… ODR (3) = ODR (1) + 2 = 5 …

… ODR (4) = ODR (1) + 3 = 6

Generalizing … ODR (n) = ODR (1) + ( n – 1 ) = n + 2 …

Then … ODR (n) = 20 for the topmost ring corresponds to

… ODR (n) = n + 2 = 20 … --> … n = 20 – 2 = 18 …

That means there are 18 rings in all …

Consider next the outside diameter of the ring (ODR) expressed

in terms of the inner diameter of the ring (IDR) … since each ring is

1 cm thick … ODR (1) = IDR (1) + 2 = 3 … --> … IDR (1) = 3 – 2 = 1 ...

… ODR (2) = IDR (2) + 2 = 4 … --> … IDR (2) = 4 – 2 = 2 …

… ODR (3) = IDR (3) + 2 = 5 … --> … IDR (3) = 5 – 2 = 3 …

Generalizing … IDR (n) = n in cm …

Now … the rings hang in such a way that any ring R(i) rests on the ring

R(i+1) above it, so that 1 cm thickness of R(i+1) extends below the

upper inner edge of R(i) , while the ring R(i -1) below ring R(i) hangs also

on R(i), so that 1 cm thickness of R(i -1) also extends above the lower inner

edge of R(i) …that means … IDR(i) = 1 + Δ(i) + 1 = 2 + Δ(i) … Δ(i) = separation

distance between the outer diameters of rings R(i+1) and R(i -1) inside the inner

diameter of ring R(i) …i = 2 , 3 , 4 , etc...For i = 2 … Δ(2) = IDR(2) – 2 = 2 – 2 = 0 …

so that inside ring R(2) , R(3) and R(1) touch each other …for the distance of the top

of ring R(i) from the top of ring R(i -1) inside ring R(i) , we find that …

… DTR(i) = 1 + 1 + Δ(i) = 2 + Δ(i) … i = 2 , 3 , 4 , ⋯ , 18 …for i = 2 , we have …

… DTR(2) = 2 + Δ(2) = 2 … since … Δ(2) = 0 … for i = 3 , we have …

… DTR(3) = 2 + Δ(3) … Δ(3) = IDR(3) – 2 = 3 -2 = 1 …

… DTR(3) = 2 + 1 = 3 …for i = 4 , we have … DTR(4) = 2 + Δ(4) …

… Δ(4) = IDR(4) – 2 = 4 -2 = 2 … DTR(4) = 2 + 2 = 4 … Notice that DTR(n) = n …

Finally … distance of top ring from bottom ring is … 18 + 17 + 16 + ⋯ + 3 + 2 = 170