Puzzles

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?

(ii) Differentiating \(y=x^2\) with respect to \(x\) \(\frac{dy}{dx}=2x\). Let \(g=\frac{dy}{dx}\). By the chain rule:

So \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{1}{x}\)

(iii) By the same method, \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{2}{x}\)

(iv) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=\frac{n-1}{x}\)

(v) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=1\)

(vi) \(\frac{d}{dy}\left(\frac{dy}{dx}\right)=-\tan(x)\)

What is

when \(y=f(x)\)?

Let \(p_1\), \(p_2\), ..., \(p_6\) be the probabilities of getting 1 to 6 on one die and \(q_1\), ..., \(q_6\) on the other. The probability of getting a total of 2 is \(p_1q_1\) and the probabilty of getting a total of 12 is \(p_6q_6\). Therefore \(p_1q_1=p_6q_6\).

If \(p_1\geq p_6\) then \(q_1\leq q_6\) (and vice-versa) as otherwise the above equality could not hole. Therefore:

The probability of rolling a total of 7 is \(p_1q_6+p_2q_5+...+p_6q_1\). This is larger than \(p_1q_6+p_6q_1\), which is larger than (or equal to) \(p_1q_1+q_6p_6\), which is larger than \(p_1q_1\).

Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.

Can two \(n\)-sided dice be weighted so that the probability of each of the numbers 2, 3, ..., 2\(n\) is the same?

Can a \(n\)-sided die and a \(m\)-sided die be weighted so that the probability of each of the numbers 2, 3, ..., \(n+m\) is the same?

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 problem can be expressed using the following probability tree:

The probability that the card turned over from C is an Ace of Spades is:

(i) Let \(a,b\in S\). Then \(\exists \alpha,\beta\in \mathbb{N}\) such that \(a=3\alpha+1\) and \(b=3\beta+1\).

This is a multiple of three plus one, so \(a\times b\in S\).

(ii) No, as \(36\times 22=4\times 253\) and 36,22,4 and 253 are all irreducible.

Try the task again with \(S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}\).

\(2009=7\times 7\times 41\), so there are four possible sets of dimensions of the cuboid:

In the first three cuboids, there is a face with an area of 2009 units (\(2009\times 1\), \(7\times 287\) and \(41\times 49\) respectively) and so 2009 stickers will not be enough. Therefore the cuboid has dimensions \(7\times 7\times 41\) and a surface area of 1246, leaving 763 stickers left over

For which numbers \(n\) can a cuboid be made with \(n\) unit cube such that \(n\) unit square stickers can cover the faces of the cuboid?

Cosine can be drawn the same way as sine but starting B at the top of the circle.

Tangent can be drawn by giving the following instructions:

A. Stand on the spot.

B. Walk around A in a circle, holding this string to keep you the same distance away.

C. Make a straight line with A and B, staying on the line tangent to the circle through B's starting point.

D. Walk in a straight line perpendicular to C's line.

E. Stay in line with C and D.

Could people be used to draw graphs of secant, cosecant and cotangent?