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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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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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No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.


Stage: 4 Short Challenge Level: Challenge Level:1

$\angle BAC = 90^{\circ}$

$A$, $B$ and $C$ are all equidistant from $D$ and therefore lie on a circle whose centre is $D$. $BC$ is a diameter of the circle and $\angle BAC$ is therefore the angle subtended by a diameter at a point on the circumference (the angle in the semicircle).

(Alternatively: let $\angle ACD = x$ and show that $\angle DAC = x,\, \angle ADB = 2x$ and $\angle DAB = 1/2 (180^{\circ}) - 2x = 90^{\circ} - x$. Hence $\angle BAC = x + 90^{\circ} - x = 90^{\circ}$.)

This problem is taken from the UKMT Mathematical Challenges.

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