Home > Maths > Interesting radius problem

Last week a friend of mine sent me this interesting puzzle from Ed Southall (@solvemymaths). I was out and about when I received it so was unable to make a start on it right away but did have some initial thoughts. Initially I assumed that the chord on the vertical side of the square was half the side and this led me to assume that it would be a problem I could solve using either similar triangles or chord bisectors. However, when I later sat down to have a go at it I realised that this assumption was incorrect, there was nothing to indicate that was what was happening on that side of the square and if this info were needed to answer the question Ed would have included it.

My next thought was that I could place the construction onto a set of axis and use a bot of coordinate geometry to solve it. At this point I decided this was impossible as there would be too many variables (3) to solve with 2 points. *

Today, still having no luck, I had a bit of time so I thought I would try to construct it and see what happened. First I constructed a square and marked the mid point and found a point that was equidistant to draw the circle. Then it hit me – any point on the bisector of the line would give me a circle through these 2 points. How could I construct it? So I looked back at the initial puzzle and the answer was there, shouting at me. I had missed earlier (*) that there is another point on the edge of the circle. This meant we were dealing with a circle through three given points, not only would I be able to contstruct it now, I wouldn’t even need to as I could go back to the coordinate geometry approach I had chalked off before. First I set the bottom left corner of the square as the origin, this gave me the three points on the circumference as (0,0) , (0.5,0) and (1,1).

(x-a)^2 +(y-b)^2 = r^2

at (0,0)

(-a)^2 + (-b)^2 = r^2

a^2 + b^2 = r^2 (1)

at (0.5,0)

(0.5-a)^2 + (-b^2) = r^2

0.25 – a + a^2 + b^2 = r^2

Sub in (1)

So 0.25 = a (2)

at (1,1)

(1-a)^2 + (1-b)^2 = r^2

a^2 – 2a + 1 + b^2 – 2b + 1 = r^2

Sub in (1)

2 – 2a – 2b = 0

1 – a = b

Sub in (2)

b = 0.75

At this point I realised fractions made more sense, so using (1):

(1/4)^2 + (3/4)^2 = r^2

r = rt10 / 4

There you have it. A nice little puzzle that took me far longer than it should have. Often when I solve these puzzles others then pop up with nicer more concise solutions. If you did it a different was I would love to know.

After I had finished I thought I would try doing the construction anyway, I’m not amazing at construction and I’ve been working to improve that so I had a crack. Pretty close correct to 2dp which is as close as I could measure:  Categories: Maths