Aliens vs Barca

I usually enjoy the weekly puzzle on the Weekly Newsletter from Chris Smith (find him on Twitter @aap03102 – he will happily add you to his list of subscribers).

The puzzle in the most recent edition was a corker:

I loved this because there are so many ways of doing it.  I won’t quibble over whether two ways really are ‘different’ or not and will list as many as I and various students have come up with.  (Yr 11 thoroughly enjoyed it too!)

In all diagrams CN is labelled as a.  All explanations are sketched rather than presented as full proofs.

1) Similar Triangles

Triangles SNM and CNS are similar (look at the angles).  This means that a/180 = 180/135

Rearranging gives a = 240 and the diameter is 375.

2) Pythagoras 1

In triangle SNM the hypotenuse is 225. 

Angle MSC is a right angle (angle in a semi-circle), so we have two further right-angled triangles: CNS and MSC.  We can use Pythagoras in both of them:

(a+135)² = b² + 225²

b² = a² + 180²

Eliminate b² and solve to get a=240.

A brief excursus – 3,4,5 triangles.

The numbers that Chris chose had some meaning (this puzzle appeared in newsletter number 375 and the answer is … 375).  It is also nice that they are numbers I recognise (usually in the form of angles).  135 = 3x45 and 180 = 4x45, which means we have a 3,4,5 triangle, so the hypotenuse is 5x45 = 225.  In my handwritten notes for these solutions I made the arithmetic easier by using e = 45 and calling the sides 3e, 4e and 5e.

3) Pythag 2

Draw in a radius (labelled r):

Do Pythagoras on triangle ONS and work out r.  Then double it.

4) Pythag 3

This is really the same as version 3 but with different labels.

Use Pythagoras on the triangle ONS

5) Intersecting Chord Theorem

This is related to version 1.

The intersecting chord theorem states that CN x NM = SN x NU

This gives 135a = 180 x 180

6) Area of a triangle

Looking at the big triangle, the area is ½ (a+135) x 180

If we treat CS as the base then it is also ½ b x 225.

Equate these and then use Pythagoras on CNS (a^2 + 180^2 = b^2) as a second equation and solve for a.

7) More similar triangles – using cos.

OT is the perpendicular bisector of SM.

We know SM = 225, so TM is half of that, 112.5.  Cos(M) = 135/225 (from triangle SNM).  But in triangle OTM we have cos(M) = 112.5/r

8) Trig 1

Work out the size of angle α (inverse tan of 180/135).

Angle OSM = α (isosceles triangle)

Angle NSM = 90 – α
Angle OSN = α – (90 – α)

So angle OSN = 2α – 90.

Use this angle, trigonometry and the side SN to work out either ON or OS.

9) Trig 2

We know what angle α is (using trig in triangle SNM) and we know that SM is 225 (Pythag).  We also know that angle CSM is a right angle (angle in a semi-circle).  Use trig in that triangle to get the diameter.

10) Trig 3

We could also find the length of CS using trig (as in version 9) and can then use Pythagoras to find CM.

11) Similar Triangles again

Triangle SNM is similar to triangle CSM. 

225/135 = diameter / 225.

How many different bits of maths did we use?

Many of these methods overlap, but it is rather nice that there are so many different aspects of maths that are involved.  It was also good (with Yr 11) to be able to focus on methods and explanation, rather than on finding the answer.

I think this can involve:

• Circle theorems and other angle rules
• Pythagoras
• Trigonometry
• Area of a triangle
• Similar triangles

Many thanks to Chris for a great problem!