This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Triangles Between Squares

Let \(T_a\) represent the \(a\)

^{th}triangle number. This means that \(T_a=\frac{1}{2}a(a+1)\).Suppose that for some integer \(n\), \(n^2 \leq T_a <(n+1)^2\). This means that:

$$n^2 \leq T_a$$
$$n^2 \leq \frac{1}{2}a(a+1)$$
$$2n^2 \leq a^2+a$$
But for every positive integer \(a \leq a^2\), so:

$$2n^2 \leq 2a^2$$
$$n^2 \leq a^2$$
\(n\) and \(a\) are both positive integers, so:

$$n \leq a$$
Now consider \(T_{a+2}\):

$$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$
$$=\frac{1}{2}(a^2+5a+6)$$
$$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$
$$=\frac{1}{2}a(a+1)+2a+3$$
$$=T_a+2a+3$$
We know that \(a \geq n\) and \(T_a \geq n^2\), so:

$$T_a+2a+3 \geq n^2+2n+3$$
$$>n^2+2n+1 = (n+1)^2$$
And so \(T_{a+2}\) is not between \(n^2\) and \((n+1)^2\). So if a triangle number \(T_a\) is between \(n^2\) and \((n+1)^2\) then the next but one triangle number \(T_{a+2}\) cannot also be between \(n^2\) and \((n+1)^2\). So there cannot be more than two triangle numbers between \(n^2\) and \((n+1)^2\).

##### Extension

Given an integer \(n\), how many triangle numbers are there between \(n^2\) and \((n+1)^2\)?

#### Sine

Cosine can be drawn the same way as sine but starting B at the top of the circle.

Tangent can be drawn by giving the following instructions:

A. Stand on the spot.

B. Walk around A in a circle, holding this string to keep you the same distance away.

C. Make a straight line with A and B, staying on the line tangent to the circle through B's starting point.

D. Walk in a straight line perpendicular to C's line.

E. Stay in line with C and D.

##### Extension

Could people be used to draw graphs of secant, cosecant and cotangent?

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