Archive for March, 2018

Quarter circle problem

March 2, 2018 2 comments

Here is a problem I have had on my classroom wall for a long time. I have a large display of problems on there that sometimes the students present me solutions to. This is one no one yet had done and I had not attempted.

Last week I was discussing the problem wall with a colleague and this jumped out at me, so I thought I would give it a try. It took rather longer than I’d care to admit, to be honest. I set off on a few false starts and came up with some incorrect solutions due to an incorrect solution I’d made. After a while I gave up and left it a few days before tackling it afresh. When I retackled it the process was much shorter and gave me a lovely concise solution. I will explain some of my incorrect thought processes first, and then I will explain how I got my solution. Before reading on, why not have a go yourself – I’d love to hear how you approached it either in the comments or via social media.

The problem:

I’m not one hundred percent sure where I sourced this question, I think it’s from the great solvemymaths website – but if it’s not please let me know.

The first thing I did was a sketch; this led me to see that what we were dealing with was a quarter circle inside a square with a triangle. It had to be a square due to tangents from a point being equal and radii being equal. I also spotted that said triangle was an 8, 15, and 17 Pythagorean triple. These were observations that would be key when I eventually got round to solving the problem.

Then I made my mistake that caused a lot of issues. I marked the point that the triangle was tangent to the circle as the midpoint of the hypotenuse. Looking back now this is such a daft thing to do. I was pretty tired and must have briefly confused tangents and chords I guess. Either way, a silly and costly mistake. Using this I tried a coordinate geometry approach and got numbers that didn’t make sense. I knew the radius would have to equal 15 + y, but I was getting values less than 15 and y could not be negative as it was just a scalar length. I tried this approach a few times from different angles but each came up the same. I was convinced my algebra was correct, so the mistake must have been somewhere else. I left it for another day. Here are some of my incorrect workings:


he correct way:

When I came back to the problem, I had a clearer head and as soon as I sketched it I could see the way to answer it.

The point where the hypotenuse was tangent to the circle was not the midpoint, but it could be defined in other ways. Using tangents from a point from the points where the tangent intersects the sides of the square we can see that the hypotenuse must be the sum of the distances from said intersection points to the corners that are also tangent points to the circle. I.e. the lengths I marked x and z. Thus we know that x + z is 17, as that is the length of the hypotenuse of an 8, 15, 17 triangle. We also know, as it’s a square, that x + 8 is equal to the radius (labelled y in my diagram) and that z + 15 is also equal to the radius (y). Thus we have three equations in three unknowns so an easily solvable system that gives us the answer of y = 20. And hence area = 400pi.

A nice concise solution in the end to a lovely problem that caused me far too much headache. I’m off to kick myself some more…..

Categories: Maths
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