## UKMT Maths Challenge Q16

Above is a photo of question 16 from yesterday’s UKMT Senior Maths Challenge. And what I fantastic question it is!

It shows (incase you can’t make it out on the picture) a rectangle and a circle which have the same centre. The two shorter sides of the rectangle are tangents to the circle, and the rectangle is 6×12. The task is to find the area that is inside both.

After the challenge had been completed one if my Y13s came to ask about it, and it took me a little while to work it out. I played around with some similar triangles, thinking that I would need to calculate the bit of the rectangle outside the circle and subtract them from 72. Then I had the realisation that I could easily split it up into triangles and sectors that I could calculate the area of with great ease.

First I sketched:

Showing the rectangle split into areas and sectors. I then considered the triangles, if I split then into right angled triangles I had a hypotenuse of 6 and a height of 3, so the cosine of the top angle must be 0.5, which means the angle must be pi/3.

I then used the sine of this angle ((rt3)/2) to calculate the opposite side as 3rt3. Then thought “why didn’t I use Pythagoras’s Theorem?!” A quick check using Pythagoras’s Theorem shows us the side is rt27, which simplifies to 3rt3. The area of the triangle then must be 9rt3.

Knowing that the angles at the centre in the two triangles are 2pi/3, that angles round a point add up to 2pi and that the sectors are congruent means we can deduce the angle if each sector is pi/3. We also know the area of a sector is (r^2(theta))/2, which gives the area of each sector as 6pi.

Finally we sum the areas and get 18rt3 + 12pi.

The student had looked at the answers and estimated them using the knowledge that rt3 is around 1.7 and pick is a but more than three. This meant he could eliminate c,d and e, but he didn’t get to the final answer.

Reblogged this on ryanjhay79.