You may also like

problem icon

Fitting In

The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

problem icon

Look Before You Leap

The diagonals of a square meet at O. The bisector of angle OAB meets BO and BC at N and P respectively. The length of NO is 24. How long is PC?

problem icon

Two Ladders

Two ladders are propped up against facing walls. The end of the first ladder is 10 metres above the foot of the first wall. The end of the second ladder is 5 metres above the foot of the second wall. At what height do the ladders cross?

Sitting Pretty

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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.