### Circumspection

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.

### Lawnmower

A kite shaped lawn consists of an equilateral triangle ABC of side 130 feet and an isosceles triangle BCD in which BD and CD are of length 169 feet. A gardener has a motor mower which cuts strips of grass exactly one foot wide and wishes to cut the entire lawn in parallel strips. What is the minimum number of strips the gardener must mow?

### Long Short

A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot be greater than sqrt2. Find a quadrilateral of this type for which s2 is approximately sqrt3and show that s2 is always less than sqrt3. Find a quadrilateral of this type for which s3 is approximately 2 and show that s2 is always less than 2. Find a quadrilateral of this type for which s4=2 and show that s4 cannot be greater than 2.

# Triangle Incircle Iteration

##### Stage: 4 Challenge Level:

Meg offers this solution:

The tangent of a circle is at right-angles to the radius of the circle. That is, if you join the centre point of the circle to a point where the circle meets the outer triangle, it makes an angle of $90 ^{\circ}$ with the side of the triangle.

The bisector of the angles of triangle $ABC$ will all pass through the centre of the circle.

From this we know that
$OAZ = \frac {a}{2}$ and $OZA = 90 ^{\circ}$
Hence $AOZ = 90- \frac{a}{2}$
Now consider triangle XOZ. This triangle is isosceles, so $OXZ = XZO = \frac{180-(180-a)}{2} = \frac{a}{2}$

By similar arguments
$OXY = OYZ = \frac{b}{2}$ and $OYZ = OZY = \frac{c}{2}$ Hence the new angles of the triangle are

$ZXY= \frac{a}{2}+\frac{b}{2}$

$XYZ = \frac{b}{2} + \frac{c}{2}$

$YZX =\frac{c}{2} + \frac{a}{2}$

$a+b+c=180$

Hence $\frac{a+b}{2}= 90 - \frac{c}{2}$

It follows that $ZXY =90 - \frac {c}{2}$.

A similar argument can be followed for $XYZ$ and $YZX$. If you continue drawing triangles within circles, the angles will decrease as shown here:

Triangle 1: a
Triangle 2: $90 - \frac{a}{2}$
Triangle 3: $90- \frac{90-\frac{a}{2}}{2}$ =$90 - \frac{90}{2} + \frac{a}{4}$
Triangle 4: $90- \frac{90-\frac{90}{2}+\frac{a}{4}}{2}= \frac{3}{4}.90 - \frac{a}{8}$

When you continue this iteration, you can demonstrate that the $a$ term becomes less and less significant, and the sum of the rest of the terms tends to 60 degrees. Hence the triangle tends to an equilateral triangle.