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Fixing It

A and B are two fixed points on a circle and RS is a variable diamater. What is the locus of the intersection P of AR and BS?

OK! Now Prove It

Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?

Summats Clear

Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1, 2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a - b) = ab.

Cyclic Triangles

Age 16 to 18
Challenge Level

Congratulations to Curt, Reigate College and to Andrei, Tudor Vianu National College, Bucharest, Romania for you solutions to this problem

When AB is a diameter angle $ACB$ is 90 degrees so we can use Pythagoras' Theorem. The area of the triangle is given by $\textstyle{1\over 2}ab = \textstyle {1\over 2}ch$ where $h$ is the length of the perpendicular from $C$ to $AB$. Then


Thus $(a+b)^2$ is a maximum when $h$ is a maximum and equal to the radius of the circle $\textstyle {1\over 2}c$ . So the maximum value of $a+b$ is $c\sqrt 2$.
triangle inscribed in circle with altitude h
When $AB$ is not a diameter we have (using the Cosine Rule):

$$\eqalign{ (a+b)^2 &= a^2 +b^2 + 2ab \cr &= c^2 + 2ab\cos \angle ACB +2ab \cr &= c^2 +2ab(1+\cos \angle ACB) \cr &=c^2 +4ab \cos^2 \textstyle {1\over 2}\angle ACB.}.$$

As the area of triangle $ACB$ is given by $\Delta = \textstyle {1\over 2} ab\sin \angle ACB$ we have

$$\eqalign { (a+b)^2 &= c^2 + {8\Delta \cos^2 \textstyle{1\over 2}\angle ACB \over \sin\angle ACB} \cr &= c^2 + 4\Delta \cot \textstyle {1\over 2}\angle ACB }.$$

If we keep $A$ and $B$ fixed and vary $C$ then, as $c$ and $\angle ACB$ are constant, and the area of the triangle $\Delta$ is a maximum when $h$ is a maximum, it follows that $a + b$ is a maximum when $h$ is a maximum, that is when $a=b$ and the altitude of the triangle drawn from $C$ to $AB$ is a line of symmetry of the triangle. In this case

$$a=b={c\over 2\sin \textstyle {1\over 2}\angle ACB}.$$

Conjecture: Let $Q$ be a variable cyclic quadrilateral in a circle of radius $r$. Then the area and the perimeter of $Q$ will be a maximum when $Q$ is a square; that is when each of the diagonals of the quadrilateral is a diameter and each diagonal bisects of the other diagonal at right angles.
cyclic quadrilaterals
Labelling $Q$ as $ABCD$, consider triangle $ABC$ with $AC$ fixed and $B$ varying. Let $B'$ be the position of $B$ when, by the previous result $AB + BC$ is a maximum, that is when $AB = BC$ Note that this also gives the maximum area of triangle $ABC$. Similarly by considering triangle $ADC$ with $AC$ fixed, we find $D'$ where $AD+DC$ is a maximum and $AD=DC$. Now $B'D'$ is a diameter, keep this fixed and consider triangles $B'CD'$ and $B'AD'$. The positions of $C$ and $A$ that maximise the perimeter and area are $C'$ and $A'$ where $A'C'$ is a diameter. Hence, for the perimeter and area of $Q$ to be a maximum all the sides of the quadrilateral must be equal making $Q$ a square. Hence the maximum perimeter is $4r\sqrt 2$ and the maximum area is $2r^2$.