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
$$(a+b)^2=a^2+b^2+2ab=c^2+2ch.$$
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 $c\sqrt 2$.
When $AB$ is not a diameter we have (using the Cosine
Rule):
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
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.
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 $2r^2$.