Center Path

Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of the point X and prove your assertion.

This gives a short summary of the properties and theorems of cyclic quadrilaterals and links to some practical examples to be found elsewhere on the site.

Dodecagon Angles

Weekly Problem 50 - 2012
The diagram shows a regular dodecagon. What is the size of the marked angle?

Tied Up

Stage: 3 Challenge Level:

A one hectare field is in the shape of a right angled triangle. At the midpoint of each side is a post. Horace the goat is tethered at the midpoint of the hypotenuse, and two sheep, Sid and Sadie, at the midpoints of the other two sides. Each rope is just exactly long enough to allow the animal to reach the two corners at the ends of the side where they are tied up.

R. Davis of City of Norwich School even suggested that Sid and Sadie might produce a little lamb who would not be tied up and so could reach every point of the field.

Seriously though, it has to be the goat, but why? Michael Swarbrick-Jones, Y7, Comberton Village College and Peggy Brett, Fiona Conroy, Bella Heesom, Jennifer Simcock and Joanne Barker, Year 9 The Mount School York, all recognised that the goat was tethered at the centre of the circum-circle for the right angled triangle passing through all three vertices. This uses the theorem that if the angle subtended by a chord of a circle at the circumference is 90 degrees then the chord is a diameter of the circle.

Diagram by Joanne Barker

The ropes only allow the goats to get to points within the circles shown in Joanne's diagram. Only Horace's circle includes the whole field.

To prove that the sheep cannot get to all parts of the field Ian Green of Coopers' Company & Coburn School argues as follows: "The only way that it's possible for an animal to be able to reach all points on the field is if it can reach the opposite corner. Drawing a line from the animal to the opposite corner gives us three right-angled triangles:

Sally and Sid's ropes are equal in length to one of the shorter sides in their respective triangles so they cannot reach to the opposite vertex at the other end of the hypotenuse. As you can see, the only way for Sally or Sid to get to all points of the field is if the other two sides equal 0, making the triangle into a straight line. Therefore only Horace can reach all points on the field.

Many solutions sent in depended on special cases like the 3, 4, 5 triangle but another method which applies in general was used by James Page and Jack Adcock of Hethersett High School, Norwich and by T. Vogwill's of the City of Norwich School whose solution is given below:

"If you make the triangle into a rectangle by doubling it exactly, you get a rectangle with the halfway point of the hypotenuse as the centre. This means it is an equal distance to each corner so Horace the Goat can reach all points in the rectangle, and when you half it again he can reach every point in the triangle. The sheep can't."

This neatly uses the property that the diagonals of a rectangle are equal in length. It also uses the property that the corners of a rectangle are further from the centre than any other points in the rectangle so that Horace can but reach all points. Can you prove that the sheep cannot do so?