### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

### Intersecting Circles

Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?

### Square Pegs

Which is a better fit, a square peg in a round hole or a round peg in a square hole?

# An Unusual Shape

##### Stage: 3 Challenge Level:

There were lots of solutions submitted and most of you got the right answer, that the goat was tied up on a long side of the shed, 5 feet from the bottom left corner of the diagram in the question. There were a couple of different ways of working this out:

Stuart has fixed the rope 5 feet from the bottom left hand corner - This is because there is 20 feet to the top right corner which is where the goat got to and how long the rope is.
(thanks, Thomas from Wilson's School)

Stuart fixed the rope directly above the lowest point of the lower semi-circle as it was the only place that would let the goat eat all of the grass that it did on the diagram.

(Hussein, also from Wilson's School)

The goat is tethered 5 metres in on the side where the majority of grass has been eaten. Therefore the goat can eat the full half on its walls side (South side). The rope can then bend round east for the 10 metres along the wall then north and another 10 metres along the side. On the west side the rope goes 5 metres west then 10 metres along the west wall and then the remaining 5 metres of the rope is used eating as far as it can along the north wall. The goat then eats the grass that it can reach in those regions.

(a nice explanation from James, also from Wilson's School)

Danielle from Darrington C of E Primary School used a picture to help explain her answer.

Well done to Tim, Michael, Muntej, Jamie, Charlie and Kartik, who all got this right.

Kartik and Tim went on to work out the area of grass eaten by the goat. Both of them did this by splitting the area up into several shapes, all of which were halves or quarters of circles, and adding the areas together.

I think the area available to the goat was 904 foot squared. This is because if you look at the shape, you can split it into 3 quarter circles and one semi-circle. You then work out the area of the different parts and add them all up.

A few of you considered the problem of whether the goat would have a larger area to graze if the location of where it was tied changed.

Kartik provided a beautifully worked set of examples to show that it did vary and went on to suggest that the grazing area is greatest when the goat is tethered at a corner.

Elliott and Oliver also showed that tethering the goat to a corner increased its grazing area.

If you put the goat on the bottom right corner of the shed then the area =

$(20\times 20 \times \pi) \times 0.75 = 300 \pi$
$+ {{10 \times 10 \times \pi }\over {4}} = {25 \pi}$
$+ {{5 \times 5 \times \pi }\over {4}} = {6.25 \pi}$
$300 + 25 + 6.25 = 331.25 \pi$
$= 1040.65 ft^2$

Oliver from Colchester Royal Grammar School considered a different approach to this.

The area eaten is 287.5$\pi$ $ft^2$. If there was no shed in the way the total possible area for the goat to eat is:

$A= \pi \times 20 \times 20=400\pi$

So the shed is stopping the goat eat an area of 112.5$\pi$ feet squared of grass. This uneatable bit could be reduced by tying the goat at the corner of the shed because the goat is further away from the centre of the shed. The area of available grass is now:

$A= (3 \times {{\pi x 20 x 20}\over{4}}) + ({{\pi x 10 x 10}\over{4}}) +({{\pi x 5 x 5}\over{4}})$
$A= 300\pi + 25\pi + 6.25\pi$
$A = 331.25\pi$
which is roughly 1040.65 feet squared.

Alastair, Ed and Llewellyn from St Peter's College also came to the same conclusion.

Aswaath's (Garden International School, Malaysia) solutions include a formula to show that you get the greatest grazing area when the goat is tethered at the corner.

But can we be sure that tethering the goat to the corner provides it with the maximum possible grazing area for any length of rope?

Maybe you could try to prove this.