Puzzles

The exterior angles of any polygon add up to 360°.

The interior angles of our polygon are 178°, and so the exterior angles are 2°. There must be 180 sides (and so also 180 angles) on the polygon, as 180×2=360.

The triangle is equilateral.

To see why, add a copy of point \(E\) rotated by 90°. This is labelled \(F\) on the diagram below.

Angles \(BDE\) and \(CDF\) are both 75°. Therefore angles \(CDE\) and \(BDF\) are both 15°. This means that angle \(FDE\) is 60°.

Line \(AD\) is a line of symmetry of the diagram, so angles \(DFE\) and \(DEF\) are equal and both 60°. Therefore, triangle DEF is equilateral. This triangle is show in green in the diagram below.

Lines \(EF\), \(DF\) and \(BF\) are all equal length, so triangles \(BFE\) and \(BFD\) are isosceles. Angles \(BDF\) and \(FBD\) are both 15°. Angles \(FBE\) and \(FEB\) are equal, and the angles in triangle \(BED\) add to 180°: this means that angle \(FBE\) is 15°.

Angles \(FBE\) and \(FBD\) are both 15°, and so angle \(EBD\) is 30°. Angles \(EBD\) and \(ABE\) add to 90°, and so angle \(ABE\) is 60°.

By symmetry, angle \(BAE\) is also 60°. Angle \(BEA\) must therefore also be 60°, so triangle \(ABE\) is equilateral.

179

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so it is not possible

If the second hand moves continuously instead of moving every second, will there be a time when the hands of the clock are all 120° apart?

The interior angle of a regular hexagon is 120°. The interior angle of a regular dodecagon is 150°. Therefore angle CAM is 30°.

Now, consider the quadrilateral ACDE. This quadrilateral is symmetric (as the dodecagon is regular) so the angles CAE and DEA are equal. Hence:

The angles CAM and CAE are equal, so A, M and E lie on a straight line.

The vertices \(P_1\), \(P_2\), ..., \(P_n\) make up a regular \(n\)-gon and \(Q_1\), \(Q_2\), ..., \(Q_m\) make up a regular \(m\)-gon, with \(P_1=Q_1\) and \(P_2=Q_2\).

The vertices \(P_2\), \(Q_3\) and \(P_5\) lie on a straight line. What is the relationship between \(m\) and \(n\)?

Draw three more squares and add these lines (I have coloured the angles to make equal angles clearer):

Triangles \(ACE\), \(LDK\) and \(IKE\) are congruent, so angle \(KDL\) is equal to \(\beta\).

The congruence of these triangles tells us that angles \(DKL\) and \(EKI\) add up to a right angle, so angle \(EKD\) is also a right angle.

The congruence of the triangles also tells us that \(KD\) and \(KE\) are the same length and so angle \(EDK\) is the angle in an isosceles right-angled triangle. \(\alpha\) is also the angle in an isosceles right-angled triangle, so these two angles are equal.

Therefore \(\alpha+\beta+\gamma=90^\circ\).

The diagram shows three rhombuses with diagonals drawn on and three angles labelled.

What is the value of \(\alpha+\beta+\gamma\)?