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 $\frac{1}{2} ab = \frac{1}{2} ch$ where $h$ is the length of the perpendicular from $C$ to $AB$ . Then

$(a + b)^{2} = a^{2} + b^{2} + 2 ab = c^{2} + 2 ch .$

Thus $(a + b)^{2}$ is a maximum when $h$ is a maximum and equal to the radius of the circle $\frac{1}{2} c$ . 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):

\begin{matrix} (a + b)^{2} & = a^{2} + b^{2} + 2 ab \\ = c^{2} + 2 ab \cos ∠ ACB + 2 ab \\ = c^{2} + 2 ab (1 + \cos ∠ ACB) \\ = c^{2} + 4 ab \cos^{2} \frac{1}{2} ∠ ACB . \end{matrix} .

As the area of triangle ACB is given by

Δ = \frac{1}{2} ab \sin ∠ ACB

we have

\begin{matrix} (a + b)^{2} & = c^{2} + \frac{8 Δ \cos^{2} \frac{1}{2} ∠ ACB}{\sin ∠ ACB} \\ = c^{2} + 4 Δ \cot \frac{1}{2} ∠ ACB \end{matrix} .

If we keep

A

and

B

fixed and vary

C

then, as

c

and

∠ ACB

are constant, and the area of the triangle

Δ

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

AB

is a line of symmetry of the triangle. In this case

a = b = \frac{c}{2 \sin \frac{1}{2} ∠ 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.

Labelling

Q

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

4 r \sqrt{2}

and the maximum area is

2 r^{2}