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A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?

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Areas and Ratios

What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.

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Six Discs

Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?

Symmetric Angles

Stage: 4 Short Challenge Level: Challenge Level:2 Challenge Level:2
As the figure has rotational symmetry of order $4$, $ABEF$ is a square.

Area $ABEF=4\times$area$\triangle BDA=4\times \frac{1}{2}BD \times DA=2(DB)^2=24$cm$^2$, so $BD=\sqrt{12}$cm$=2\sqrt{3}$cm.

As $ABEF$ is a square, $\angle ABD=45^{\circ}$ so $\angle CBD=45^{\circ} -15^{\circ} =30^{\circ}$.

Since $\tan {30^{\circ}}= \frac{CD}{BD}=\frac{CD}{2\sqrt{3}}$, we have $CD=2\sqrt{3}\tan{30^{\circ}}$cm.

Now consider the following equilateral triangle with side lengths $2$:

The vertical line is perpendicular to the base-line and so bisects both the angle at the top vertex and the base-line.

Consider the left right-angled triangle. Pythagoras' theorem gives $a=\sqrt{3}$ and then we have $\tan{30^{\circ}}=\frac{1}{a}=\frac{1}{\sqrt{3}}$

Therefore $CD=2\sqrt{3}\tan{30^{\circ}}$cm$=2$cm.

This problem is taken from the UKMT Mathematical Challenges.

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