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.
The diagram shows a regular dodecagon. What is the size of the marked angle?
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.
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?