Maths

We let \(a\), \(b\) and \(c\) represent the amount of time taken to paint a wall on their own by Amy, Bertie and Charlie respectively.

Since we have three unknowns, we require three equations; from the statement about the individual work, we obtain \(b=2a\) and \(a=c+2\).

For the final equation, we can use the hint. Amy painted \(\frac{2}{a}\) of the wall, Bertie painted \(\frac{2}{b}\) of the wall, and Charlie painted \(\frac{2}{c}\) of the wall; this gives \(\frac{2}{a}+\frac{2}{b}+\frac{2}{c} = 1\). Rearranging, this leads to: \(2bc+2ac+2ab = abc\)

Substituting the first two equations into the third equation and dividing through by \(2\), we obtain:

\( \displaystyle 2c \left(c+2\right) + c \left(c+2\right) + 2{\left(c+2\right)}^2 = c {\left(c+2\right)}^2\)

Rearranging and factorising then gives:

\( \displaystyle \left(c+2\right) \left[c\left(c+2\right) - 2c - c - 2\left(c+2\right)\right] = 0\)

Simplifying and factorising leads to:

\( \displaystyle \left(c+2\right)\left(c-4\right)\left(c+1\right) = 0\)

The only valid solution is then \(c = 4\), so \(a = 6\) and \(b = 12\).

As alluded to in the hint, there are a few ways to approach and solve this problem. Below is my favourite (and in my opinion, the most elegant) method.

First, instead of trying to deal with the two fractions separately, rewrite both fractions and then notice that it is easy to combine the two fractions over a common denominator.

\( \displaystyle \int \frac{1}{\sin^2{x}+\sec^2{x}} - \frac{1}{\cos^2{x}+\csc^2{x}} \, \mathrm{d}x\)

\( \displaystyle = \int \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}+1} - \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}+1} \, \mathrm{d}x\)

\( \displaystyle = \int \frac{\cos^2{x}-\sin^2{x}}{\sin^2{x}\cos^2{x}+1} \, \mathrm{d}x\)

Now, using the substitution \(u = \sin{x}\cos{x}\), we have:

\(\displaystyle \mathrm{d}x = \frac{\mathrm{d}u}{\cos^2{x} - \sin^2{x}}\)

And so, substituting into the result above, we obtain:

\( \displaystyle = \int \frac{\cos^2{x}-\sin^2{x}}{u^2+1} \cdot \frac{1}{\cos^2{x} - \sin^2{x}} \, \mathrm{d}u\)

\( \displaystyle = \int \frac{1}{u^2+1} \, \mathrm{d}u\)

Now, using the standard result for this integral (or the substitution \(u=\tan{w}\)):

\( \displaystyle =\arctan{u}+c \)

\( \displaystyle = \arctan\!{\left(\sin{x}\cos{x}\right)}+c\)

\(c\) is the arbitrary constant of integration.

If you did choose the "brute-force" route, you may end up with a different (yet equally valid) answer to the one given above. If you want a challenge, try using the substitution \(u = \cot{x}\).

Set \(r\) to be the radius of the circle. Notice that \(r\) is also equal to half of the side length of the larger square. We can now form a right angled triangle with side lengths \(r-12\), \(r-6\) and \(r\), since we know the dimensions of the rectangle. Using Pythagoras, we have:

\(\left(r-12\right)^2 + \left(r-6\right)^2 = r^2\), and solving this leads to \(r = 18 \pm 12\). Since the radius must be greater than 6 (otherwise the rectangles would not fit), we can conclude that \(r = 30\).

We can now calculate the red area as follows: \(60 \times 60 - 4 \times 6 \times 12 - \pi \cdot 30^2\), which gives \(3312-900\pi \approx 485 \textrm{m}^2\)

We proceed as follows:

\( \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left[\ln\left(\frac{\tan\!{\left(\sin{x}\right)}}{\tan\!{\left(\cos{x}\right)}}\right)\right] \)

\( \displaystyle = \frac{\mathrm{d}}{\mathrm{d}x}\left[\ln\left(\tan\!{\left(\sin{x}\right)}\right) - \ln\left(\tan\!{\left(\cos{x}\right)}\right)\right] \)

\( \displaystyle = \frac{\frac{\mathrm{d}}{\mathrm{d}x}\left[\tan\!{\left(\sin{x}\right)}\right]}{\tan\!{\left(\sin{x}\right)}}-\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left[\tan\!{\left(\cos{x}\right)}\right]}{\tan\!{\left(\cos{x}\right)}} \)

\( \displaystyle = \frac{\frac{\mathrm{d}}{\mathrm{d}x}\left[\sin{x}\right]\sec^2\!{\left(\sin{x}\right)}}{\tan\!{\left(\sin{x}\right)}}-\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left[\cos{x}\right]\sec^2\!{\left(\cos{x}\right)}}{\tan^2\!{\left(\cos{x}\right)}} \)

\( \displaystyle = \frac{\cos{x}\sec^2\!{\left(\sin{x}\right)}}{\tan\!{\left(\sin{x}\right)}}+\frac{\sin{x}\sec^2\!{\left(\cos{x}\right)}}{\tan\!{\left(\cos{x}\right)}} \)

\( \displaystyle = \frac{\cos{x}}{\sin\!{\left(\sin{x}\right)}\cos\!{\left(\sin{x}\right)}}+\frac{\sin{x}}{\sin\!{\left(\cos{x}\right)}\cos\!{\left(\cos{x}\right)}} \)

\( \displaystyle = \frac{2\cos{x}}{\sin\!{\left(2\sin{x}\right)}}+\frac{2\sin{x}}{\sin\!{\left(2\cos{x}\right)}} \)

\( \displaystyle = 2\cos{x}\csc{\left(2\sin{x}\right)}+2\sin{x}\csc{\left(2\cos{x}\right)}\)

Start by expanding the brackets; this results in:

\( \displaystyle \sqrt{\left(n^2+n\right)\left(n^2+5n+6\right)+1} = \sqrt{n^4+6n^3+11n^2+6n+1}\)

For this to be the result we are expecting, we need to show that \(n^4+6n^3+11n^2+6n+1\) is a perfect square. Because it is a quartic polynomial with terms in \(n^3\) and \(n\), we are therefore trying to show that:

\(\displaystyle n^4+6n^3+11n^2+6n+1 = {\left(an^2+bn+c\right)}^2\) where \(a\), \(b\) and \(c\) are constants to be found.

We can quickly deduce the values of \(a\) and \(c\). As the coefficients of \(n^4\) and \(n^0\) are \(1\), and all of the coefficients are positive, \(a = c = 1\).

By expanding the brackets, we can find the value of \(b\) by comparing coefficients:

\(\displaystyle n^4+6n^3+11n^2+6n+1 = n^4+\left(b+b\right)n^3+\left(1+1+b^2\right)n^2+\left(b+b\right)n+1\) which implies \(b = 3\).

Therefore we have the following, which is obviously an integer if \(n\) is an integer:

\( \displaystyle \sqrt{\left(n \times \left(n+1\right) \times \left(n+2\right) \times \left(n+3\right)\right) + 1}=n^2+3n+1\)

For the first part (as per the hint), we will write the sums together, but with the second sum as an imaginary component. When we compute this complex sum, the real component will give us the first sum, and the imaginary component will give us the second sum. This gives:

\(\displaystyle \sum_{n=0}^{\infty} \frac{\cos\!{\left(n\theta\right)}}{n!} + i\frac{\sin\!{\left(n\theta\right)}}{n!}\)

Using Euler's Identity, we have:

\(\displaystyle = \sum_{n=0}^{\infty} \frac{e^{in\theta}}{n!}\)

\(\displaystyle = \sum_{n=0}^{\infty} \frac{{\left(e^{i\theta}\right)}^n}{n!}\)

Now, this sum is in the form of the series expansion for \(e^x\), with \(x = e^{i\theta}\). We can therefore write the sum as:

\(\displaystyle = e^{e^{i\theta}}\)

Now, using Euler's Identity twice, we obtain:

\(\displaystyle = e^{\cos{\theta} + i\sin{\theta}}\)

\(\displaystyle = e^{\cos{\theta}}\cdot e^{i\sin{\theta}}\)

\(\displaystyle = e^{\cos{\theta}}\left(\cos\!{\left(\sin{\theta}\right)}+i\sin\!{\left(\sin{\theta}\right)}\right)\)

Taking real and imaginary components then gives the desired result.

For the second part, we start by applying the sine and cosine addition formulae:

\(\displaystyle \sum_{n=0}^{\infty} \frac{\sin\!{\left(n\theta\right)}\cos\!{\left(k\theta\right)}+\cos\!{\left(n\theta\right)}\sin\!{\left(k\theta\right)}+\cos\!{\left(n\theta\right)}\cos\!{\left(k\theta\right)}-\sin\!{\left(n\theta\right)}\sin\!{\left(k\theta\right)}}{n!}\)

Now since \(\sin\!{\left(k\theta\right)}\) and \(\cos\!{\left(k\theta\right)}\) are constants independent of \(n\), we can factor them out of the sum, and rewrite as:

\(\displaystyle = \cos\!{\left(k\theta\right)}\left(\sum_{n=0}^{\infty}\frac{\sin\!{\left(n\theta\right)}}{n!} + \frac{\cos\!{\left(n\theta\right)}}{n!}\right) + \sin\!{\left(k\theta\right)}\left(\sum_{n=0}^{\infty}\frac{\cos\!{\left(n\theta\right)}}{n!} - \frac{\sin\!{\left(n\theta\right)}}{n!}\right) \)

We can then apply the result from part one to obtain:

\(\displaystyle = \cos\!{\left(k\theta\right)}\left(e^{\cos{\theta}}\left(\cos\!{\left(\sin{\theta}\right)}+\sin\!{\left(\sin{\theta}\right)}\right)\right)+\sin\!{\left(k\theta\right)}\left(e^{\cos{\theta}}\left(\cos\!{\left(\sin{\theta}\right)}-\sin\!{\left(\sin{\theta}\right)}\right)\right) \)

\(\displaystyle = e^{\cos{\theta}}\left[\cos\!{\left(\sin{\theta}\right)}\cos\!{\left(k\theta\right)} -\sin\!{\left(\sin{\theta}\right)}\sin\!{\left(k\theta\right)} + \sin\!{\left(\sin{\theta}\right)}\cos\!{\left(k\theta\right)} + \cos\!{\left(\sin{\theta}\right)}\sin\!{\left(k\theta\right)}\right]\)

With this rearrangement, we can apply the sine and cosine addition formulae again to simplify the result to:

\(\displaystyle = e^{\cos{\theta}}\left[\cos\!{\left(\sin{\theta}+k\theta\right)}+\sin\!{\left(\sin{\theta}+k\theta\right)}\right]\)

We begin by calculating the area of the sector centred at \(P\). Taking \(\alpha\) to be the angle marked on the diagram in the hint, the area is given by \(\displaystyle \frac{l^2}{2}\left(2\alpha\right) = l^2\alpha\).

Next, we need the area of the two segments. Given their areas are the same, we can consider a single segment, and double the result. To calculate this area, we must first calculate the area of a sector centred at \(O\), and take away the area of the isosceles triangle.

The angle at \(O\) for the segment is \(\pi-2\alpha\), since it is an isosceles triangle. The area of the sector must therefore be \(\displaystyle \frac{1}{2}r^2\left(\pi-2\alpha\right)\), and the area of the isosceles triangle is \(\frac{1}{2}r^2\sin\!{\left(\pi-2\alpha\right)} = \frac{1}{2}r^2\sin\!{\left(2\alpha\right)}\).

Remembering that the donkey can graze exactly half of the area of the field, we therefore have the following equation for the area that the donkey can graze:

\( \displaystyle \frac{1}{2}\pi r^2 = l^2\alpha + r^2 \left(\pi-2\alpha\right) - r^2 \sin\!{\left(2\alpha\right)}\)

Dividing through by \(r^2\) (since \(r \gt 0\)), setting \(\displaystyle R = \frac{l}{r}\) and rearranging, we obtain:

\( \displaystyle 0 = R^2\alpha + \frac{\pi}{2}-2\alpha - 2\sin{\alpha}\cos{\alpha}\)

We can now eliminate \(\alpha\) by using one of the right-angled triangles on the diagram, with hypotenuse \(r\) and adjacent side \(\displaystyle \frac{l}{2}\).

By trigonometry and Pythagoras' Theorem, we have \(\displaystyle \cos \alpha = \frac{l}{2r}\) and \(\displaystyle \sin \alpha = \frac{\sqrt{r^2-\left(\frac{l}{2}\right)^2}}{r}\). Using \(\displaystyle R = \frac{l}{r}\), we can substitute these expressions into our equation to obtain:

\( \displaystyle 0 = R^2\arccos\!{\left(\frac{R}{2}\right)} + \frac{\pi}{2}-2\arccos\!{\left(\frac{R}{2}\right)} - R\sqrt{1-\left(\frac{R}{2}\right)^2}\)

This is a frankly horrendous equation, and not one that can be solved algebraically. We will need to use the Newton-Raphson method to calculate an approximation to the solution. We set \(f{\left(R\right)}\) to be equal to the right hand side of the above equation; then we can compute:

\( \displaystyle f'{\left(R\right)} = 2R\arccos\!{\left(\frac{R}{2}\right)} - \frac{R^2}{2\sqrt{1-\left(\frac{R}{2}\right)^2}} + \frac{1}{\sqrt{1-\left(\frac{R}{2}\right)^2}} - \sqrt{1-\left(\frac{R}{2}\right)^2} + \frac{R^2}{4\sqrt{1-\left(\frac{R}{2}\right)^2}}\)

Fortunately for us, this simplifies quite nicely. By taking out a factor of \(\displaystyle \frac{1}{\sqrt{1-\left(\frac{R}{2}\right)^2}}\) from everything but the first term, we are left with:

\( \displaystyle f'{\left(R\right)} = 2R\arccos\!{\left(\frac{R}{2}\right)} - \frac{1}{\sqrt{1-\left(\frac{R}{2}\right)^2}}\left[\frac{R^2}{2} - 1 + \left(1-\left(\frac{R}{2}\right)^2\right) - \frac{R^2}{4}\right] = 2R\arccos\!{\left(\frac{R}{2}\right)}\)

We can now implement the Newton-Raphson method by choosing an initial guess of \(R_0 = 1\), then use the following formula (noting that the denominator is not equal to zero for \(R_n \in \left(0, 2\right)\)):

\( \displaystyle R_{n+1} = R_n - \frac{R_n^2\arccos\!{\left(\frac{R_n}{2}\right)} + \frac{\pi}{2}-2\arccos\!{\left(\frac{R_n}{2}\right)} - R_n\sqrt{1-\left(\frac{R_n}{2}\right)^2}}{2R_n\arccos\!{\left(\frac{R_n}{2}\right)}}\)

This gives the following values (to 9 decimal places): \(R_1 = 1.163496672\), \(R_2 = 1.158730912\), \(R_3 = 1.158728473\) and \(R_4 = 1.158728473\).

We can therefore conclude that \(R = 1.158728473\) to 9 decimal places.

As mentioned in the hint, we can only reach step \(n\) from step \(n-1\) or step \(n-2\). The probability of reaching step \(n\) can therefore be written in terms of the probabilities of reaching steps \(n-1\) and \(n-2\):

\( \displaystyle p{\left(n\right)} = \frac{1}{2}p{\left(n-1\right)} + \frac{1}{2}p{\left(n-2\right)}\)

We can use induction to prove that the formula given in the question is true for all \(n\). Firstly, we must deal with the base cases, \(n = 1\) and \(n = 2\). Using the formula, we have:

\(p{\left(1\right)} = \frac{2-\left(-1\right)}{3\cdot1} = 1\) and \(p{\left(2\right)} = \frac{4-\left(1\right)}{3\cdot2} = \frac{1}{2}\)

We have verified that the formula is true for the base cases. Now, we can assume that the formula is true for \(n = k-1\) and \(n = k-2\), where \(k\) is some integer. Consider \(p{\left(k\right)}\); using the recursion and then substituting in our assumed formula, we have:

\( \displaystyle p{\left(k\right)} = \frac{1}{2}p{\left(k-1\right)} + \frac{1}{2}p{\left(k-2\right)}\)

\( \displaystyle p{\left(k\right)} = \frac{1}{2}\cdot \frac{2^{k-1}-{\left(-1\right)}^{k-1}}{3\cdot 2^{k-2}} + \frac{1}{2}\cdot \frac{2^{k-2}-{\left(-1\right)}^{k-2}}{3\cdot 2^{k-3}}\)

\( \displaystyle p{\left(k\right)} = \frac{\left(2^{k-1}-{\left(-1\right)}^{k-1}\right) + 2\left(2^{k-2}+{\left(-1\right)}^{k-1}\right)}{3\cdot 2^{k-1}}\)

\( \displaystyle p{\left(k\right)} = \frac{2\cdot 2^{k-1} + {\left(-1\right)}^{k-1}}{3\cdot 2^{k-1}}\)

\( \displaystyle p{\left(k\right)} = \frac{2^k - {\left(-1\right)}^k}{3\cdot 2^{k-1}}\)

Thus the formula is also true for \(n = k\). We conclude that by the principle of induction, the formula is true for all \(n \in \mathbb{N}\).

The solutions are given below in the same order as the questions above.

Start by letting \(a = \arccos{x}\), which rearranges to \(x = \cos{a}\). Now consider a right-angled triangle with hypotenuse of length \(1\), an angle of \(a\), and adjacent side of length \(x\). By applying Pythagoras' Theorem, we can deduce that the opposite side is of length \(\sqrt{1-x^2}\). Hence:

\( \displaystyle \sin\!{\left(\arccos{x}\right)} = \sin{a} = \sqrt{1-x^2}\,\) for \(x\in \left[-1, 1\right]\)

Next, let \(a = \arcsin{x}\), which rearranges to \(x = \sin{a}\). Now consider a right-angled triangle with hypotenuse of length \(1\), an angle of \(a\), and opposite side of length \(x\). By applying Pythagoras' Theorem, we can deduce that the adjacent side is of length \(\sqrt{1-x^2}\). Hence:

\( \displaystyle \tan\!{\left(\arcsin{x}\right)} = \tan{a} = \frac{x}{\sqrt{1-x^2}}\,\) for \(x\in \left(-1, 1\right)\)

Furthermore, by looking at the graphs for \(\sin\) and \(\cos\), you can see that \(\cos{x} = \sin\!{\left(\frac{\pi}{2}-x\right)}\). Therefore, we have:

\(\arcsin\!{\left(\cos{x}\right)} = \arcsin\!{\left(\sin\!{\left(\frac{\pi}{2}-x\right)}\right)} = \frac{\pi}{2}-x\,\) for \(x \in \left[0, \pi\right]\)

Recalling \(\sin\!{\left(\arccos{x}\right)} = \sqrt{1-x^2}\), we get \(y = \sqrt{1-x^2}\) by applying \(\arcsin\) to both sides of the first equation. However, this solution is only valid for \(x \in \left[0, 1\right]\), since \(\arcsin\) is an odd function, but \(\arccos\) is an even function. The graph of this function is therefore the first quadrant (i.e. top right quarter) of the unit circle.

Since \( \displaystyle \tan\!{\left(\arcsin{x}\right)} = \frac{x}{\sqrt{1-x^2}}\), we get \( \displaystyle y = \frac{x}{\sqrt{1-x^2}}\) by applying \(\arctan\) to both sides of the second equation. This solution is still valid for \(x\in \left(-1, 1\right)\). The graph of this function has asymptotes at \(x = \pm 1\), with \(\lim_{x \to -1} y = -\infty\) and \(\lim_{x \to 1} y = \infty\).

The final equation isn't quite as simple as \(y = \frac{\pi}{2} - x\), since there are actually an infinite number of possibilities for the function \(a\!\left(x\right)\) such that \(\cos{x} = \sin{\left(a\!\left(x\right)\right)}\). Recalling that \(\sin\) and \(\cos\) are \(2\pi\)-periodic, we have:

\(\cos{x} = \sin\!{\left(\frac{\pi}{2}+2k\pi - x \right)}\) for \(k \in \mathbb{Z}\)

Furthermore, by replacing \(x\) by \(-x\) and remembering \(\cos\) is an even function, we also obtain:

\(\cos{x} = \sin\!{\left(\frac{\pi}{2}+2k\pi + x \right)}\) for \(k \in \mathbb{Z}\)

Hence the full set of solutions is given by \(y = \frac{\pi}{2} + 2k\pi \pm x\) for \( k \in \mathbb{Z}\). The graph therefore consists of every straight line of that form, i.e. lines with a \(y\)-intercept in the set \(\left\{\dotsc, \frac{-3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, \dotsc \right\}\) and gradient either \(1\) or \(-1\).

Word based explanations are never the same as seeing it for yourself — I recommend going to the Desmos Graphing Calculator and viewing the equations there.

Taking the approach suggested by the hint, we will consider the probability of Albert's success on each attempt.

We are told that the probability that Albert succeeds on any one attempt is \(\frac{1}{3}\), so the probability that Albert succeeds on the first go is \(\frac{1}{3}\).

For Albert to succeed on the second attempt, each of Albert, Bernard and Charlie must have failed on the first attempt. The probability of this happening will be the product of the probabilities of each of them failing, i.e. \(\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{2}{27}\). Thus the probability that Albert succeeds on the second attempt will be: \(\frac{2}{27} \cdot \frac{1}{3}\).

Similarly to the second attempt, for Albert to succeed on the third attempt, each of Albert, Bernard and Charlie must have failed on both rounds of attempts. The probability of this happening will therefore be \(\frac{2}{27} \cdot \frac{2}{27} = {\frac{2}{27}}^2\). Therefore the probability that Albert succeeds on the third attempt is \({\frac{2}{27}}^2 \cdot \frac{1}{3}\).

This pattern will clearly continue, so the probability that Albert will succeed on the fourth attempt will be \({\frac{2}{27}}^3 \cdot \frac{1}{3}\), etc.

Now to find the overall probability that Albert is the successful one, we must add all of these probabilities together to obtain:

\(\displaystyle \frac{1}{3} + \frac{2}{27} \cdot \frac{1}{3} + {\frac{2}{27}}^2 \cdot \frac{1}{3} + ... = P\).

This is a geometric series with \(a = \frac{1}{3}\) and \(r = \frac{2}{27}\). As \(\left|r\right|<1\) we can find the sum of this infinite series using the formula \(S_{\infty}=\frac{a}{1-r}\).

Therefore \(P\), the probability that Albert is the successful one, is \(\frac{9}{25}\).