Puzzles

Consider a line which bisects an equilateral triangle with sides of length 1. Let the sides of the smaller triangle created be \(a\) and \(b\) as shown below:

The area of the equilateral triangle is:

The area of the smaller triangle is:

This is half the area of the equilateral triangle, so:

The length (\(L\)) of the line bisecting the equilateral triangle is:

To find the minimum, we set the derivative to 0:

For this value of \(a\), \(b\) and \(L\) also equal to \(\frac{1}{\sqrt{2}}\).

What if the line does not need to be straight?

(i) If a circle could be drawn, then its centre would be equidistant from each pair of points. But the locus of points equidistant from B and C is parallel to the locus of points equidistant from A and D.

Therefore it is impossible to place the centre of the circle, so no circle can be drawn.

(ii)

(iii)

Let A, B, C and D be any four points. When is it possible to draw

(i) A circle

(ii) An equilateral triangle

(iii) A square

through the four points?

Could this be done with other numbers of friends?

Call the length of a side of the square \(x\). CAB and CEF are similar triangles, therefore:

Consider a tetrahedron with vertices at \((0,0,0)\), \((a,0,0)\), \((0,b,0)\) and \((0,0,c)\). A cube is placed in the tetrahedron so that three of its faces are in the xy, xz and yz planes and one of its vertices lies on the other face.

What is the length of a side of the cube?

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 \(\alpha+\beta+\gamma\)?

The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.

The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation \(y=2x-4\) and ON has equation \(y=\frac{5}{2}-\frac{1}{2}x\). These lines intersect at \((\frac{13}{5},\frac{6}{5})\), these are the co-ordinates of N. By the same method, the co-ordinates of P are \((-\frac{8}{5},-\frac{1}{5})\).

By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is 9.8.

Let \(S\) be the area of the large square, \(T\) be the area of one of the large triangles, \(U\) be one of the red overlaps and V be the uncovered blue square. We can write $$S=4T-4U+V$$ as the area of the square is the total of the four triangles, take away the overlaps as they have been double counted, add the blue square as it has been missed.

We know that 4U=V, so

Therefore one of the triangles covers one quarter of the square.

Five congruent triangles are drawn in a regular pentagon. The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large pentagon does each triangle take up?

\(n\) congruent triangles are drawn in a regular \(n\) sided polygon. The total area which the triangles overlap is equal to the area they don't cover. What proportion of the area of the large \(n\) sided polygon does each triangle take up?

The sides of A4 paper are in the ratio \(1:\sqrt{2}\). Let the width of the paper be 1 unit. This means that the height of the paper is \(\sqrt{2}\) units.

Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).

\(BF=DF=\sqrt{2}-1\) so by Pythagoras' Theorem again, \(BD=2-\sqrt{2}\). \(CD=1-(\sqrt{2}-1)=2-\sqrt{2}\). Hence, \(CD=BD\) and so the shape is a kite.

Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a cyclic quadrilateral.