### 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

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

### At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

# Kite in a Square

##### Stage: 4 Challenge Level:

Olivia from St Paul's Girls' School sent us this solution and Niharika from Leicester sent us this solution.  Well done to both of you.

Below are the three "muddled up methods" with all the steps in the correct order.

Similar triangles

Let ABCD be a unit square.

Line $AD$ is parallel to line $MF$, so $\angle EDA$ and $\angle EMF$ are equal, and $\angle EAD$ and $\angle EFM$ are equal (alternate angles).

As line $AC$ intersects line $MD$ at point $E$, the two opposite angles $\angle MEF$ and $\angle AED$ are equal.

Therefore, $\triangle AED$ and $\triangle FEM$ are similar.

The line $MF$ is half the length of $AD$.

Therefore, the line $EH$ is half the length of $PE$.

$PH$ has length $\frac{1}{2}$ units, so $PE$ has length $\frac{1}{3}$ units and $EH$ has length $\frac{1}{6}$ units.

$\triangle MEF$ has area $\frac{1}{2}(\frac{1}{2}\times\frac{1}{6}) = \frac{1}{24}$ sq units.

Therefore, the shaded area $MEFG = \frac{1}{24} \times 2 = \frac{1}{12}$ sq units.

Pythagoras

Assume that the sides of the square are each $2$ units long.
Thus, $DJ$ and $FJ$ are each $1$ unit long.

By Pythagoras, $DF$ has length $\sqrt 2$.

Area of $\triangle DFE = \frac{DF \times EF}{2}$
$= \frac{\sqrt 2 \times EF}{2} = \frac{EF}{\sqrt 2}$

$(EH)^2+(HF)^2=(EF)^2$
$EH = HF$
$(EH)^2 = \frac{1}{2}(EF)^2$
$EH = \frac{EF}{\sqrt 2}$

Area of $\triangle MEF = \frac{1}{2}(1 \times EH) = \frac{1}{2}(\frac{EF}{\sqrt 2})$

So the shaded area $MEFG$ is equal to the area of $\triangle DFE$.

The area of $\triangle DMC = 2$ sq units.
The area of $\triangle DFC = 1$ sq unit.
Thus the combined area of $\triangle DFE$, $\triangle CFG$ and
shaded area $MEFG$ is $1$ sq unit.

Areas of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ are equal
so each must have an area of $\frac{1}{3}$ sq units.

The total area of the square is $4$ sq units, so the shaded area is
$\frac{1}{12}$ the area of the whole square.

Coordinates

Consider a unit square drawn on a coordinate grid.

The line joining $(0,0)$ to $(1,1)$ has equation $y=x$.
The line joining $(0,0)$ to $(\frac{1}{2},1)$ has equation $y=2x$
The line joining $(0,1)$ to $(1,0)$ has equation $y=1-x$.

The point $(a,b)$ is at the intersection of the lines
$y=2x$ and $y=1-x$.

So $a = \frac{1}{3}, b=\frac{2}{3}$.

one of which has vertices $(\frac{1}{3},\frac{2}{3}), (\frac{1}{2},\frac{1}{2}), (\frac{1}{2},1)$.
The perpendicular height of the triangle is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.
Area of the triangle $= \frac{1}{2}(\frac{1}{2} \times \frac{1}{6}) = \frac{1}{24}$
Therefore the shaded area is $2 \times \frac{1}{24} = \frac{1}{12}$