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Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Solutions were received from Mary from Birchwood High School; Sana, Jenny, Chris and Rosion of Madras College, St. Andrews; as well as Andrei from School 205 Bucharest.

Some hard work went into these solutions but a little more reflection at the start of the problem could have made it much easier and reduced the consequent errors that appeared.

Triangle ABC with AB = 25, AC = 40 and BC=39. Dropa perpendicular from A to BC meeting BC at P - this divides BC into two parts of lengths BP=7 and PC=32. AP is perpendicular to BC and AP = 24 (Pythagoras theorem)

The first thing to look for and notice is Pythagorean triples "hidden" in the sides.

Triangle ABC has AB = 25, AC = 40 and BC=39.

Draw a line from A to BC meeting BC at P so that it divides BC into two parts of lengths BP=7 and PC=32.

AP is perpendicular to BC and AP = 24 (Pythagoras' theorem).

N.B. In triangle APB, $AP^2 = AB^2 - PB^2$

In triangle APC, $AP^2 = AC^2 - PC^2$

From the centre O of the circle draw linesOA and OC and the perpendicular from O to AC to meet AC at Q.

From the centre O of the circle draw linesOA and OC and the perpendicular from O to AC to meet AC at Q.

Let angle AOQ = x °.

Therefore angle COQ is x (by symmetry)

Angle ABC = 1/2 Angle AOC = x (angle at centre and circumference)

The radius of the circle is OA.

In triangle AOQ
AO = 20/sin x

In triangle ABP
sin x = 24/25

Therefore OA = (20 x 25)/24 = 125/6

Therefore the diameter is 125/3 units (exactly)