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Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

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Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

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Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?

Rotating Triangle

Stage: 3 and 4 Challenge Level: Challenge Level:1

Edward from Graveney School, Tooting, London sent this solution.
Let us make $a$ the radius of the largest circle centre $A$ etc. Then the lengths of the sides of the triangle are: $AB = a - b$, $AC = a - c$ and $BC = b + c$.
The perimeter of the triangle is: $$AB + BC + CA = (a - b) + (a - c) + (b + c)= 2a.$$ So the perimeter of the triangle is twice the radius of the large circle whatever the sizes of the small circles.