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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

Triangle Incircle Iteration

Keep constructing triangles in the incircle of the previous triangle. What happens?


M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.


Age 14 to 16 Challenge Level:

This solution was sent in by Liwei Deng, aged 17, of The Latymer School in London.

First of all, join up the centres of the three circles. This will form a triangle with side lengths $x+9$, $x+4$ and $13$.
Then draw vertical and horizontal lines as shown to create three right-angled triangles (shaded on diagram).
Taking the unknown radius to be $x$, we can mark on the lengths shown in green.

We can now use Pythagoras' Theorem on the two yellow right-angled triangles to find the missing sides. The upper one is $6 \sqrt{x}$,

and the lower one is $ 4 \sqrt{x}$ .

Now we move to the pale blue shaded triangle.
The horizontal side can be found by subtracting:

$$ 6\sqrt{x} - 4\sqrt{x} = 2\sqrt{x}.$$

The vertical side can also be found by subtracting: as we know that the height of the rectangle is $2x$, this side must be $2x-9-4 = 2x-13$.
Using Pythagoras' Theorem on this triangle, we find that

$$ (2\sqrt{x})^2 + (2x-13)^2 = 13^2 $$ $$ 4x + 4x^2 - 52x + 169 = 169 $$ $$ 4x^2 - 48x = 0 $$ $$ 4x(x - 12) = 0 $$ $$ x = 12 $$.