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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

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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?

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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?

Folding Squares

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

Talei from Poltair Community School and Sports College sent in this excellent solution to the problem, showing the diagonals are split in the ratio 1:2 (or equivalently that the smaller part is a third of the whole)
Annotated diagram of square

FD = 2EF and FC = 2AF because . . .
We can prove that triangle FCD is an enlargement, by a scale factor of 2, of the triangle AEF.
The triangles' angles are the same - at point F the crossing lines give each triangle an identical angle because they are opposite each other. The line AC joins parallel lines to give an identical angle to each triangle because they are alternate, and the line ED works in the same way.
Therefore, because AE must be exactly half the length of CD, each line in triangle AEF will be half of its enlarged equivalent in triangle FCD.
FD = 2EF because FD is the enlarged equivalent line of EF in triangle AEF.
FC = 2AF because FC is the enlarged equivalent line of AF in triangle AEF.

Talei's method using similar triangles can also be applied to the general case for a pallelogram and arbitary intersection point:


Annotated diagram of parallelogram

$$\mathrm{\angle afe = \angle cfd \textrm{ (opposite angles) and } \angle eaf = \angle dcf \textrm{ (alternate / 'Z-' angles)}}$$
$$\mathrm{\therefore \triangle aef \textrm{ is similar to }\triangle cdf \Rightarrow \frac{cf}{af} = \frac{cd}{ae} \Rightarrow cf = \frac{cd}{ae}af}$$
$$\mathrm{ac = af + cf \Rightarrow ac = af \left( 1 + \frac{cd}{ae} \right)}$$
$$\mathrm{\therefore \ af = \frac{1}{1 + \frac{cd}{ae}} \textrm{ multiplying through by ae and using cd = ab} \Rightarrow af = \frac{ae}{ae + ab} ac}$$