### Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

### The Root Cause

Prove that if a is a natural number and the square root of a is rational, then it is a square number (an integer n^2 for some integer n.)

### Good Approximations

Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.

# Spirostars

##### Stage: 5 Challenge Level:

Chris, who did not reveal his school, sent the following:

Working in degrees, if $\theta$ divides 360 then the shape will be a regular polygon. Else if $\theta$ is rational, the shape will be a star. If $\theta$ is irrational then a 'donut' shape will eventually be drawn, i.e. a star with infinitely many points.

Consider theta being irrational. No matter how many turns are made, the pen will never be pointing in the same direction so cannot repeat its path.

To complete the solution to this problem, consider the diagram:

 First think: 'why do the points lie on a circle?' We should not take for granted that they do so we must first prove it. Draw the perpendicular bisectors of the line segments $AB$ and $BC$, then the intersection of the perpendicular bisectors is equidistant from $A$, $B$ and $C$ and thus it is the centre of the circle with $AB$ and $BC$ as chords. All the perpendicular bisectors of the line segments in the path will meet in a single point $O$ equidistant from the endpoints of the line segments. Therefore all the line segments are chords of a single circle with centre $O$. The angle of turn between the equal chords $AB$ and $BC$ in the path is $\angle PBC = \theta$. Triangles $AOB$ and $BOC$ are isosceles and $\angle OAB = \angle OBA = \angle OBC = \alpha$. Then (using angles on a straight line and angles in a triangle) $\theta = 180 - 2\alpha = \angle AOB$.

Consider radii drawn from the centre to the ends of $n$ consecutive chords in the path....

Now can you prove that the path is closed if and only if $\theta$ is a rational multiple of 360 degrees? Consider radii drawn from the centre to the ends of $n$ consecutive chords in the path. The sum of angles between the radii is equal to the total angle turn $n\theta$ so the path is closed if and only if the total angle turn is an integer multiple of 360 degrees, that is if and only if $n\theta = 360m$ or, equivalently $\theta$ is a rational multiple of 360 degrees.

So consider $\theta = 360 {p\over q}$ where $p$ and $q$ are positive integers and $p< q$. If the angle of turn is more than 360 degrees, complete revolutions can be discounted, so effectively $\theta = 360 {p*\over q }$ produces the same path as $\theta = 360 {p\over q}$ where $p* \equiv p$ mod $q$.

If $p$ is a factor of q, (say $q = kp$) then the path is a regular $k$ sided polygon because the total angle turned when $k$ chords are drawn is $k \times {360\over k}= 360$ degrees.

If $p$ and $q$ are not coprime then $sp=tq$ for some integers $s$ and $t$ so that the total angle turn with s chords is $s \times 360{p\over q} = 360t$ which is an integer multiple of 360 degrees so the path is a star with $s$ points. If $p$ and $q$ are coprime then the path is a $q$ pointed star.