### Just Opposite

A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

### Beelines

Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses?

# Pentagon

##### Stage: 4 Challenge Level:

We received good solutions from several people. Let's start by looking at the solution sent in by Tom of Wolgarston High School.

Firstly, let's consider the triangular problem. If instead of coordinates we use vectors to describe the triangle, then we can write ${\bf p}_1$, ${\bf p}_2$ and ${\bf p}_3$ to describe the corners of the triangle (which we don't know yet) and ${\bf m}_1$, ${\bf m}_2$ and ${\bf m}_3$ to describe the midpoints of the lines.

If we can find a way of expressing each vector ${\bf p}_1$, ${\bf p}_2$ and ${\bf p}_3$ using just the vectors of the midpoints, then we can locate the corners.

By the standard vector laws, the midpoint between any two points descbribed by vectors is the average of those vectors. So we get the following equations:
1. ${\bf m}_1 = \frac{1}{2}({\bf p}_1 + {\bf p}_2)$
2. ${\bf m}_2 = \frac{1}{2}({\bf p}_2 + {\bf p}_3)$
3. ${\bf m}_3 = \frac{1}{2}({\bf p}_3 + {\bf p}_1)$
Now we solve the simultaneous equations to find expressions for the corners. For ${\bf p}_1$ add equations 1 and 2 and then subtract equation 3. This gives the following:
${\bf p}_1 = {\bf m}_1 + {\bf m}_2 - {\bf m}_3$

and similarly for ${\bf p}_2$ and ${\bf p}_3$ we get:
${\bf p}_2 = {\bf m}_2 + {\bf m}_3 - {\bf m}_1$
${\bf p}_3 = {\bf m}_3 + {\bf m}_1 - {\bf m}_2$

Thus we have expressed all the triangle's vertices as expressions of their midpoints.

See if you can continue from here using the same method for pentagons.

After looking at pentagons Tom then went on to look at the quadrilaterals and he found that he could no longer solve the simultaneous equations in the same way. Ben Kenny noticed that if an arrangement of midpoints produceded a quadrilateral then this quadrilateral was not necessarily unique. For example:

Both quadrilaterals have the same midpoints but different vertices.

Do you notice anything special about when we can find a quadrilateral? Try connecting the midpoints. Can you say anything interesting about the inner quadrilateral? Try using Tom's method and see what properties you might be able to gain by looking at the simultaneous equations.