### 4 Dom

Use these four dominoes to make a square that has the same number of dots on each side.

### Just Rolling Round

P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?

### Areas and Ratios

Do you have enough information to work out the area of the shaded quadrilateral?

# Opening the Door

##### Age 14 to 18Challenge Level

Thank you very much to Pablo and Sergio from King's College Alicante in Spain, Amrit from Hymers College in the UK, Jacob and Michael from Cedar House School in South Africa, Alex from Bexley Grammar School in the UK and Niklas from Walton High School in the UK, who all sent in good work on this problem. The solution below draws on their observations and explanations.

Moving $z_1$ and $z_2$ about, there are lots of ways to make $z_3=z_1+z_2$ a real number, which lies on the real axis.

$z_1$ and $z_2$ need to be on either side of the real axis, and both the same vertical distance from the real axis. They can be directly above each other, but they don't have to be. Here are a couple of examples:

$(1+2i)+(1-2i)=2+0i$

$(3-3i)+(2+3i)=5+0i$

Notice that moving $z_1$ and $z_2$ horizontally gives other real sums:

$(-1-3i)+(4+3i)=3+0i$

$(-3-3i)+(1+3i)=-2+0i$

For these last three examples, $z_1=a_1-3i$ and $z_2=a_2+3i$ for some real numbers $a_1$ and $a_2$.
We can check this will always work:
$$z_1+z_2=(a_1-3i)+(a_2+3i)=a_1+a_2-3i+3i=(a_1+a_2)+0i$$
What if we had another pair in the form $a_1+bi$ and $a_2-bi$ (for some real number $b$)?

Then $$z_1+z_2=(a_1+bi)+(a_2-bi)=a_1+a_2+bi-bi=(a_1+a_2)+0i$$ So whenever $z_1=a_1+bi$ and $z_2=a_2-bi$, $z_1+z_2$ will be real.

How it works geometrically
When you add $z_2$ to $z_1$, you add the real parts, $a_1+a_2$, so $z_1$ is moved by $a_2$ along the real axis. The imaginary parts are also added, $b_1+b_2$, so $z_1$ is moved by $b_2$ along the imaginary axis.

This is the same as a translation by the vector from $0+0i$ to $z_2$:

$(1+2i)+(1-2i)=2+0i$

$(4+4i)+(-2-4i)=2+0i$

We can also prove algebraically that $z_1+z_2$ is only ever a real number if $z_1=a_1+bi$ and $z_2=a_2-bi$.
To do this, we should begin by letting $z_1$ and $z_2$ be any complex numbers, so $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$.
Then: $$z_1+z_2=(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$$
This is real only if the imaginary part is equal to $0i$,
so $(b_1+b_2)i=0i$, so $b_1+b_2=0$, which means $b_2=-b_1$.
So $z_1=a_1+bi$ and $z_2=a_2-bi$.

Adding two complex numbers to get an imaginary number is very similar.
Here are some examples:

$(2+i)+(-2+3i)=0+4i$

$(3+2i)+(-3+i)=0+3i$

$(3-4i)+(-3-i)=0-5i$

Similarly, notice that if $z_1=a+b_1i$ and $z_2=-a+b_2i$ then $z_1+z_2=0+(b_1+b_2)i$, which is imaginary (if $b_1+b_2=0$, then $z_1+z_2=0+0i$, which is real - but $0+0i$ is considered to be both real and imaginary).

Geometrically, now the horizontal distances to the imaginary axis are equal and opposite. This means that translation by $z_2$ 'cancels out' the real component of $z_1$:

We can prove that $z_1$ and $z_2$ must be of the form $a+b_1i$ and $-a+b_2i$ for their sum to be imaginary just like we proved the similar result for the sum to be real:
Let $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$.
Then: $$z_1+z_2=(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$$
This is imaginary only if the real part is equal to $0$, so $a_1+a_2=0$, so $a_2=-a_1$.
Therefore $z_1=a+b_1i$ and $z_2=-a+b_2i$.