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### Just Rolling Round

P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?

### Tri-split

A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?

### Polycircles

Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?

# Tracking Points

##### Stage: 4 Challenge Level:

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.

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