### Triangle Incircle Iteration

Start with any triangle T1 and its inscribed circle. Draw the triangle T2 which has its vertices at the points of contact between the triangle T1 and its incircle. Now keep repeating this process starting with T2 to form a sequence of nested triangles and circles. What happens to the triangles? You may like to investigate this interactively on the computer or by drawing with ruler and compasses. If the angles in the first triangle are a, b and c prove that the angles in the second triangle are given (in degrees) by f(x) = (90 - x/2) where x takes the values a, b and c. Choose some triangles, investigate this iteration numerically and try to give reasons for what happens. Investigate what happens if you reverse this process (triangle to circumcircle to triangle...)

### 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?

# Fitting In

##### Stage: 4 Challenge Level:

Well done Freddie Manners (Packwood Haugh School) for this solution of the first part of the problem. Perhaps someone else can solve the second part.

If the radius of the circle is $r$, the length of each side of the square $r\sqrt{2}$ (Pythagoras' theorem).

Take a triangle from centre of circle to mid-point $HG$, mid-point $HG$ to $G$, and $G$ to centre. This is a right-angled triangle.

If the length of the sides of the small square is $x$, the sides of this triangle are:

$x/2$, $x + r\sqrt{2}/2$ and $r$.

By pythagoras, ${({1\over 2}x)^2} + {( {r\sqrt2\over2} + {x})^2} = {r^2}$ ${x^2\over4} + {2r^2\over4} + {xr\sqrt2} + {x^2} = {r^2}$ ${x^2} + {2r^2} + {4xr\sqrt2} + {4x^2} = {4r^2}$ ${5x^2} + {4xr\sqrt2} = {2r^2}$ Substitute $y = r\sqrt{2}$ (length of the side of the large square)

${5x^2} + {4xy} = {y^2}$ ${5x^2} + {4xy} - {y^2} ={ 0}$ ${x} = {{-4y \pm\sqrt{{(4y)^2} + {20y^2}}}\over 10}$ ${x} = {{-4y \pm\sqrt{36y^2}} \over 10}$ ${x} = {{-4y \pm 6y}\over 10}$ ${10x} = {2y or -10y}$

Ratio must be positive, therefore $10x = 2y$, therefore $y = 5x$
QED

Solution to part B.

Submitted by Samantha Gooneratne, Colombo International School, Sri Lanka. Well done Samantha! Her teacher also solved the problem using similar triangles, which I have include below Samantha's solution.

Let $X$ be the mid point of $PQ$, $C$ the center of the circle and $r$ the radius.

 $RX + XC = r$ Hence $PR \sin 60^{\circ}+ CM \cos 60^{\circ}=r$ ${{\sqrt{3}\over 2}+ {{1 \over2}r} = {r}}$ Therefore ${ {\sqrt{3}.PQ = {r}}}$ Now ${{LM} ={2(r \sin 60)}}$ so ${{LM} ={\sqrt{3}.r}}$ ${{LM} = {\sqrt{3}.(\sqrt{3}.PQ)}}$ ${{LM} = {3PQ}}$

Solution using similar triangles:

$NR$ is a diameter so $NLR = 90^{\circ}$
but $NLM = 60^{\circ}$ so $RLP = 30^{\circ}$
Now $LPR = 120^{\circ}$ so $LRP = 30^{\circ}$
Hence $LP = PR$ but $PR = PQ$
so $LP = PQ = QM$
$LM = 3 PQ$