We received many good solutions of this question. They either use the properties of the area of triangles or similar triangles to solve the problem.

Emile from CNDL and John from South Island School (Hong Kong) used the area of triangles to solve the problem. Emile writes:

$xy/ 2$ is the area of the triangle ABC. You can also divide the triangle in to two triangles AB and CBO with area $rx/2$ and $ry/2$ respectively. Thus, we have

$$xy/2=rx/2+ry/2\;.$$ Thus,

$$

\begin{align}

xy&=rx+ry\\

xy&=r(x+y)\\

1/r&=(x+y)/xy\\

1/r&=1/x+1/y\;.\\

\end{align}

$$

Rachel, Emily, Samantha and Kri from Millais School, Ciaran from St Patrick's School, Wilson's School Maths Club and Nathanael used similar triangles to solve this problem. This is the solution from Ciaran of St Patrick's school.

AB and BC are tangents to the circle, thus ON and OM are perpendicular to BC and AB. Thus triangle ONC and triangle ABC are similar since they share the same angle C and both have a right angle. $ON = r$ and $NC = y-r.$ Thus,

$$

\begin{align}

r/(y-r) &= x/y\\

ry &= xy -rx\\

ry + rx &= xy\\

xy&=r(x+y)\\

1/r&=(x+y)/xy\\

1/r&=1/x+1/y

\end{align}

$$

and the rest of the solution is the same as Emile's.

Congratulations to all of you who did it correctly.