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Fitting In

The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

Pinned Squares

What is the total number of squares that can be made on a 5 by 5 geoboard?

Zig Zag

Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?

Just Opposite

Age 14 to 16
Challenge Level

Just using coordinates and the clue in the original diagram there is a short and sweet solution sent in by Jo from Leeds.

By concentrating on the geometry, she made the algebra very simple. With the clue that $p-q = a-c$ and $p+q=b-d$ she was able to find the area of the square $ABCD$:

If the vertices were $(6,7)$ and $(3,2)$, the surrounding square would have side $(7-2)$.

The inner square would have a side of $6-3 =3$.
The area of the tilted square is therefore $5 \times 5 - \frac{(5 \times 5)-(3 \times 3)}{2} =17 $ sq units and this can be generalised.

In the algebraic case.

Area $= (b-d)^2- \frac{((b-d)^2- (a-c)^2)}{2} $
example square

Michael Gray from Madras College solved the first part using vectors.

The midpoint of the square is given by $M=((a+c)/2, (b+d)/2)$. Hence
\begin{eqnarray} \mathbf{CM} &=& \vec{m} - \vec{c} \\ &=& \frac{1}{2} {a-c \choose b-d} \end{eqnarray}
The vector $\mathbf{MD}$ is the vector $\mathbf{CM}$ rotated by $90^o$ anticlockwise and so:
\begin{eqnarray} \mathbf{MD} &=& \frac{1}{2} {-(b-d) \choose a-c} \\ \mathbf{OD} &=& \mathbf{OM} + \mathbf{MD} \\ &=& \frac{1}{2} {a+c \choose b+d} + \frac{1}{2} {-(b-d) \choose a-c} \\ &=& \frac{1}{2} {a-b+c+d \choose a+b-c+d} \end{eqnarray}
and this gives the coordinates of the point $D$. The vector $\mathbf{OB}$, giving the coordinates of $B$, is found in a similar way $$\mathbf{OB} = \frac{1}{2} {a+b+c-d \choose -a+b+c+d}.$$