Someone tweeted me this puzzler a couple of days ago (I forget who it was, apologies):

“If the square has side 2cm, what is the radius of the circle?”

I had a stab at it today and thought I’d share my process, which may well be awful, but as Bob Hoskins famously didn’t quite say, “it’s good to share”

My first thought was “it’s a circle, I need a radius, where can I get one”. So I drew in a few lines:

I know that where the dotted diagonal lines cross is the centre of the circle, and I drew in the bold black line because I know I can calculate it, which might come in handy.

I thought I better have a visual reminder of what lengths are equal, in the hope it might prompt a new thought:

Then I just stared at it for a while.

I couldn’t visualise how calculating the black diagonal line would help me, even though I could do it. So I drew in another line as I thought about it:

Which helped me spot the blue arrow thing which kicked in a thought about circle theorems. At this point, note I didn’t have a direction I was going in, I was just trying stuff out. Testing the waters to see where it’s shallow. I thought about the circle theorem thing, and eventually decided against it, and ditched the idea. However, looking at the diagram with the two angles on it made me realise the direction I wanted to take:

I want to work with this triangle, because it has the radius (twice, bonus) as its side length, and I can use the length that I know I can calculate (the longest side of the green triangle, which was mentioned in the first diagram). So now all I need is the angles inside the green triangle and off I go. And to get those, I can use the big red bottom left triangle:

So now I have some ugly trig to work through:

[/insert elevator music whilst ugly trig is being worked through]

and out pops r = 1.25 cm.

The nice answer makes me suspect there’s a far nicer approach to this. I’ll ponder it some more.

*Update: A few people have sent me their approaches, which are predictably simpler than mine. Interestingly, some are asking why I ‘chose to solve it in this complicated way’. I didn’t choose to! This is where my first thoughts took me. Anyway, here are some alternative approaches:

And from the comments section:

“If you let the height of the square be r + h, where r is the radius of the circle and the width of the square to be 2w (w = 1), you can form a triangle with sides w and h and hypotenuse r. From here do a little Pythagoras and the answer of 1.25 drops out quite quickly. “

Saw something like this with my Year 10s last year.

If you let the height of the square be r + h, where r is the radius of the circle and the width of the square to be 2w (w = 1), you can form a triangle with sides w and h and hypotenuse r. From here do a little Pythagoras and the answer of 1.25 drops out quite quickly.

I can send a snapshot of my scribbles, but I doubt you’ll need it

Jeremy

Having got the circle theorem idea in my head just by looking at the diagram, I pushed all the way through on that, and ended up using the sine rule to get the answer.

There is a circle theorem that says that if two chords intersect then the product of their segments are equal. The theorem can be proved quite easily using similar triangles. So if the diameter of the circle is drawn in from the midpoint of the bottom of the square it will cut the top of the square in two segments each of length one and is itself cut into two unequal segments of 2 and x. By the theorem x must equal 1/2 since 2 times x = 1 times 1. So the diameter of the circke is 2.5 so its radius is 1.25.

Pingback: Quizzino della domenica: quasi Vitruvio | Notiziole di .mau.

Hey, I find it very interesting to know about this problem and its solution..

Let A be the point at which the square makes a tangent to the circle. Draw the diameter from this point. Let this cut through the top of the square at C and meet the circle again at D. Let B be the top left corner of the square. By Pythagoras AB is sqrt(5). Ratio of AB : AC is sqrt(5) : 2. Since angle ABD is a right angle (angle in a semi circle) and angle at A is in common triangles ABC and ABD are similar. So ratio of AD : AB is also sqrt(5):2. Since AB is sqrt(5) then AD (the diameter) is sqrt(5)*sqrt(5)/2 = 5/2. Hence radius is 5/4.

(Ed could have continued his original proof in a similar way since the little green triangle containing the radius is also similar to these)