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Andre (age 15),
from Romania, tackled this problem. Here's his solution, which uses
vectors:
In the Cartesian system of axes, the
vector $\overrightarrow{AB}$, with the origin at (0,0,0) and end
point at (1,1,1) could be written as:
$\overrightarrow{AB}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ where
$\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are the unit vectors of
the axes Ox, Oy and Oz respectively.
Now, linking end to end 10 vectors
$\overrightarrow{AB}$ comes to multiplying the vector
$\overrightarrow{AB}$ by 10, so that one obtains a vector with the
origin at (0,0,0) and end point at (10,10,10).
For the vector $\overrightarrow{CD}$:
$\overrightarrow{CD}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}$. Adding up
2 such vectors, one easily sees that the resulting vector has the
origin at (0,0,0) and the end point at (2,6,4). Adding up 10 such
vectors, the origin remains the same and the end point is (10, 30,
20). For $n$ vectors, the end point is $(n, 3n, 2n)$.
If the starting point is (1,0,-1) and the
vector to be drawn is
$\mathbf{v}=n\mathbf{i}+n\mathbf{j}+n\mathbf{k}$, then the end
point will be at $(1+n, 0+n, -1+n)$, i.e.\ at $(n+1, n,
n-1)$.
If the origin is at $(a,b,c)$, the end
point is at $(n+a, n+b, n+c)$. If the vector is
$\mathbf{u}=n\mathbf{i}+3n\mathbf{j}+2n\mathbf{k}$, and its origin
is at (-3,0,2), its end point will be at $(n-3, 3n, 2n+2)$. If for
the same vector the origin is at $(a,b,c)$, its end point will be
at $(n+a, 3n+b, 2n+c)$.
If the vector of interest is now
$\mathbf{u}+\mathbf{v}=2n\mathbf{i}+ 4n\mathbf{j}+3n\mathbf{k}$, in
the general case of the origin at $(a,b,c)$, the end point will be
at $(2n+a,4n+b, 3n+c)$.
Stephen (from Framwellgate School,
Durham) solved the problem and went on to think about what happens
if you alternate segments. Here is what he sent us :
With AB followed by CD (if we count AB as segment 1 and CD as
segment 2) we can see that 2 segments makes (1,1,1)+(1,3,2)=(2,4,3)
therefore the nth segment when n is even is (n,2n,3n/2) because we
need to divide it by 2, as (2,4,3) is 2 segments. For n segments
when n is odd, n-1 is even so take the n-1th segment from last
coordinates to give (n-1,2(n-1),3(n-1)/2) and add (1,1,1), because
this is the first segment and therefore will be the last segment on
odd number of segments (n) - to get: (n,2n-1,(3n-1)/2) if starting
from (a,b,c) then we just need to add this coordinate to the two
solutions.