If we can construct a square which has the hypotenuse of the triangle as its side, and work out its area, then we can work out the length of the hypotenuse. A square has four sides, so we’ll need four triangles. There’s two ways we can arrange them: on the inside of the square, or around the outside. Either way is fine; I’ll start with inside: We know that the area of each triangle is $\frac{1}{2} \times1\times1 = \frac{1}{2}$, so the total area of the square is $4\times\frac{1}{2} = 2$. The side length of the square is the square root of this, $\sqrt2$.

Alternatively, we can position the triangles around the outside of the square: This makes a big outer square of side length $2$, whose total area is $2^2 = 4$. The area of the four triangles is $2$ as before, so the remaining area of the square in the middle is $4-2 = 2$. So again we find that the side length of the square, and hence the hypotenuse of the right-angled triangle, is $\sqrt2$.

Again, we can choose to put the four triangles inside the square or outside. I’ll start with inside: Unlike before, there’s a gap in the middle: a square of side length $a-b$ (if $a$ is the longer side). The area of each triangle is $ab/2$, and the total area of the outer square is $c^2$ so we have: \begin{align*} c^2 &= (a-b)^2 + 4\times ab/2\\ &= (a^2 - 2ab + b^2) + 2ab\\ &= a^2 + b^2\\ c &= \sqrt{a^2 + b^2} \end{align*} Alternatively, we could put the triangles around the outside of the square, like this: We get a bigger square around the outside, with side length $a+b$ (we know that the triangles must make a straight line where they meet each other because the angles will sum to $180^\circ$).

In this case we can actually solve the problem without using algebra (although I’ll cover the algebraic solution afterwards), just by rearranging the triangles inside the big square: The area not covered by the triangles was $c^2$ before the rearrangement. Now it is composed of two squares: one of side length $a$ and one of side length $b$, whose total area is $a^2 + b^2$. This must be the same as $c^2$ because all we’ve done is move the triangles around, hence \begin{align*} c^2 &= a^2 + b^2 \end{align*} as before.

Instead of rearranging, we could use algebra. The total area of the big square is $(a+b)^2$, which is composed of the smaller square of side length $c$, and the four triangles, so \begin{align*} (a+b)^2 &= c^2 + 4\times ab/2\\ a^2 + 2ab + b^2 &= c^2 + 2ab\\ a^2 + b^2 &= c^2 \end{align*}

The distance is the hypotenuse, $c$, of a right-angled triangle with catheti $a=4$ and $b=3$, giving \begin{align*} c^2 &= 4^2 + 3^2\\ &= 16 + 9\\ &= 25\\ c &= \sqrt{25}\\ &= 5. \end{align*} So the distance to the origin is $5$. (Normally when solving the equation $c^2 = 25$ we’d have to consider the solution $c=-5$ but distances are always positive, so we can ignore it here.)

Calling the width $w$, we have \begin{align*} 13^2 &= w^2 + 5^2\\ 169 &= w^2 + 25\\ w^2 &= 169 - 25\\ &= 144\\ w &= \sqrt{144}\\ &= 12 \end{align*} So the width is $12\,\mathrm{cm}$.

The width of this right-angled triangle is $1-(-2) = 3$ and the height is $3 - 1 = 2$, so the hypotenuse is $\sqrt{3^2 + 2^2} = \sqrt{13}$.

Just as in the last question, the distance will be the hypotenuse of a right-angled triangle. The width of this triangle will be $x_2 - x_1$ or $x_1 - x_2$, depending on which one is positive, but it doesn’t actually matter because we’re going to square the width, and $(x_2 - x_1)^2$ is always the same as $(x_1 - x_2)^2$ so we can just choose one of them. It’s conventional to use $x_2 - x_1$ since we think of it as the change when $x$ goes from $x_1$ to $x_2$ (see Section 7.1.3: Letters). Similarly, the height of the triangle is $y_2 - y_1$. This gives: \begin{align*} d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \end{align*} You might choose to write this in terms of the change in $x$, $\Delta x = x_2 - x_1$, and the change in $y$, $\Delta y = y_2 - y_1$: \[d = \sqrt{(\Delta x)^2 + (\Delta y)^2}.\]

The length of the bottom side is also $5\,\mathrm{cm}$, so the remaining width to the right of that is $4\,\mathrm{cm}$. Since the height $h$ is measured perpendicular to the base, we have a right-angled triangle with hypotenuse $5\,\mathrm{cm}$, and sides $4\,\mathrm{cm}$ and $h$. Therefore, $h = 3\,\mathrm{cm}$, and the area is base times height, which comes to $15\,\mathrm{cm^2}$.

As the vine wraps around the tree six times, it covers a horizontal distance of $6\times2\,\mathrm{m}$. Effectively the vine is going $12\,\mathrm{m}$ horizontally and $5\,\mathrm{m}$ up. The vine’s length is the length of the hypotenuse, $c$: \begin{align*} c^2 &= 12^2 + 5^2 \\ &= 169 \\ c &= 13 \end{align*} Therefore the vine is $13\,\mathrm{m}$ long.

It’ll be easier to understand if we draw a very small planet, such as Le Petit Prince lived on. When the ship is at position $1$, it’s completely visible. At $3$ you might be able to see the mast, but the base of the ship has disappeared below the horizon – the planet gets in the way of seeing it. Position $2$ is the critical point, beyond which the ship starts to disappear. So, what’s special about this point? If we draw a line of sight passing through position $2$, it will be tangent to the planet, and the important thing about tangents to circles is that they make a right angle with a line from the centre of the circle.

It’ll be neater to use letters for the distances, so let $r$ be the Earth’s radius, $d$ be the distance to the horizon, and $h$ be the height of the eyes above sea-level. \begin{align*} (r+h)^2 &= r^2 + d^2\\ d &= \sqrt{(r+h)^2 - r^2} \end{align*} If you have a calculator, you can put those values in now. Otherwise, it’ll help to expand and simplify: \begin{align*} d &= \sqrt{r^2 + 2rh + h^2 - r^2}\\ &= \sqrt{2rh + h^2} \end{align*} or use the difference of two squares to simplify: \begin{align*} d &= \sqrt{(r+h)^2 - r^2}\\ &= \sqrt{(r+h - r)(r+h + r)}\\ &= \sqrt{h(2r + h)} \end{align*} If $h = 1.5 = 3/2$ and $r = 6,\!400,\!000$ (remembering to convert it to metres), then $2rh + h^2$ is approximately $19$ million. Since $4000^2 = 16,\!000,\!000$ and $5000^2 = 25,\!000,\!000$, the distance is somewhere between $4$ and $5\,\mathrm{km}$. (Those with a calculator will say it’s about $4.382\,\mathrm{km}$).

In the tower case, $h = 50$, so $2rh + h^2$ is $640,\!002,\!500$. Noting that $25,\!000^2 = 625,\!000,\!000$ and $26,\!000^2 = 676,\!000,\!000$, the distance is somewhere between $25$ and $26\,\mathrm{km}$ ($25.298\,\mathrm{km}$ on the calculator). Or if you’re lazier, $20,\!000^2 = 400,\!000,\!000$ and $30,\!000^2 = 900,\!000,\!000$, so it’s somewhere between $20$ and $30\,\mathrm{km}$.

For most purposes, the $h^2$ term can be ignored because it’s much smaller than $2rh$, so a good rule of thumb for distance to the horizon is $d = \sqrt{2rh}$ or $d = 8\sqrt{h/5}$ if $d$ is measured in km and $h$ in m, since $r = \frac{8000^2}{10}\,\mathrm{m}$.

The base is a chord of the circle, so its perpendicular bisector passes through the centre. We know the distance from the chord to the other side is $h$, so the distance from the chord to the centre is $h - r$. We now have a right-angled triangle with hypotenuse $r$, and sides $h-r$ and $b/2$ (because we bisected the length $b$). Hence, \begin{align*} r^2 &= (h-r)^2 + \left(\frac{b}{2}\right)^2\\ &= h^2 - 2hr + r^2 + \frac{b^2}{4}\\ 2hr &= h^2 + \frac{b^2}{4}\\ r &= \frac{h^2 + b^2/4}{2h}\\ &= \frac{h}{2} + \frac{b^2}{8h}. \end{align*}

Let the distance between opposite corners be $c$. First we need to consider the length of the diagonal of a face of the cube. Let’s call this length $b$. $b$ is the hypotenuse of a right-angled triangle with both other sides of length $a$: Hence, $b^2=a^2+a^2=2a^2$. $c$ is the hypotenuse of another right-angled triangle with the other two sides of lengths $a$ and $b$: Therefore \begin{align*} c^2&=a^2+b^2\\ c&=\sqrt{a^2+2a^2}\\ &=\sqrt{3}a \end{align*} So the distance is $\sqrt{3}a$.

Since the straight line is tangent to the smaller circle, it’s perpendicular to a line from the centre. Calling the larger radius $R$ and the smaller radius $r$, this gives us a right-angled triangle with $R$ as the hypotenuse and lengths of $r$ and $4\,\mathrm{cm}\div2$ on the other sides. Using Pythagoras’ theorem, \begin{align*} R^2 &= r^2 + 2^2\\ R^2 - r^2 &= 4 \end{align*} The area of the shaded region is $\pi(R^2 - r^2)$ (see Question 10.8.11) which is $4\pi\,\mathrm{cm^2}$. There’s no need to actually know the radii, which is just as well, because we can’t work them out from the given information.

The area $A$ of the triangle is $bh/2$ where $h$ is the height of the triangle: with $b_1 + b_2 = b$. From Pythagoras’ theorem, \[c^2 = h^2 + b_1^2\] and \[a^2 = h^2 + b_2^2.\] We need to find an expression for $h$ in terms of $a$, $b$, and $c$, so let’s try to eliminate $b_1$ and $b_2$: \begin{align*} a^2 &= h^2 + b_2^2\\ &= h^2 + (b - b_1)^2\\ &= h^2 + b^2 - 2bb_1 + b_1^2\\ &= h^2 + b^2 - 2bb_1 + (c^2 - h^2)\\ &= b^2 - 2bb_1 + c^2\\ b_1 &= \frac{b^2 + c^2 - a^2}{2b} \end{align*} Now we can solve for $h$: \begin{align*} h^2 &= c^2 - b_1^2\\ &= c^2 - \frac{(b^2 + c^2 - a^2)^2}{4b^2}\\ &= \frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{4b^2} \end{align*} And find $A$: \begin{align*} A &= \frac{bh}{2}\\ A^2 &= \frac{b^2h^2}{4}\\ &= \frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{16} \end{align*} Now we could expand this, and expand $s(s-a)(s-b)(s-c)$ and show that they’re the same, but that wouldn’t be very satisfying. Instead we can factorise it using the difference of two squares ($x^2 - y^2 = (x-y)(x+y)$, see Section 7.4.3: Expanding and Factorising): \begin{align*} A^2 &= \tfrac{1}{16}\left(4b^2c^2 - (b^2 + c^2 - a^2)^2\right)\\ &= \tfrac{1}{16}\left(2bc - (b^2 + c^2 - a^2)\right)\left(2bc + (b^2 + c^2 - a^2)\right)\\ &= \tfrac{1}{16}(a^2 - b^2 + 2bc - c^2)(b^2 + 2bc + c^2 - a^2)\\ &= \tfrac{1}{16}\left(a^2 - (b - c)^2\right)\left((b + c)^2 - a^2\right)\\ &= \tfrac{1}{16}\left(a - (b - c)\right)\left(a + (b-c)\right)\left((b + c) - a\right)\left((b + c) + a\right)\\ &= \frac{a - b + c}{2}\cdot\frac{a + b - c}{2}\cdot\frac{b + c - a}{2}\cdot\frac{a + b + c}{2}\\ &= (s - b)(s - c)(s - a)s\\ A &= \sqrt{s(s-a)(s-b)(s-c)} \end{align*}

Hopefully you’ve imagined the path looking something like this: It has to be composed of straight lines because the shortest distance between two points is a straight line. The only question is where on the river to stop and get water.

You could make a guess: maybe $10\,\mathrm{km}$ east of the start, maybe $9$. That looks about right. If you’ve got a calculator you could experiment with a few values and see what comes out the shortest. For $10\,\mathrm{km}$, the total distance would be \begin{align*} \sqrt{5^2 + 10^2} + \sqrt{(15-10)^2 + 3^2} &= \sqrt{125} + \sqrt{34}\\ &\approx 17.011 \end{align*} whereas if we choose $9\,\mathrm{km}$ then we get \begin{align*} \sqrt{5^2 + 9^2} + \sqrt{(15-9)^2 + 3^2} &= \sqrt{106} + \sqrt{45}\\ &\approx 17.004 \end{align*} If you choose lower values then the distance gets bigger. So the shortest distance seems to be about $17\,\mathrm{km}$.

You might make the guess that the two triangles in the diagram should be similar. Maybe that just feels intuitively right – when you drew the path you might have instinctively made them similar, or maybe you know a little physics and remember that light tends to travel the shortest path possible, and when it bounces off a mirror it leaves at the same angle it approaches at, so the triangles have to be similar. That would mean that we have to divide the $15\,\mathrm{km}$ in the ratio $5:3$, and reach the river $\frac{5\times15}{5+3}$ km east of the start. If we use this value ($9.375$) then the distance is exactly $17\,\mathrm{km}$.

You can take this a step further, and imagine the mirror image of the camp (with the bank of the river as a mirror). For any path we might think of taking, we can imagine the mirror image of the part of the path from the river to the camp. In this way, the distance to the camp must be the same as the distance to the ‘‘mirror-camp’’. The shortest distance to the mirror-camp is very easy to work out, as it’s just a straight line, the hypotenuse of a right-angled triangle with catheti $15$ and $5+3$: \begin{align*} \sqrt{15^2 + 8^2} &= \sqrt{225 + 64}\\ &= \sqrt{289}\\ &= 17 \end{align*}

We tried $n=1$ and $m=3$ in the table above, and got $a=8$, $b=6$, and $c=10$. These numbers have a common factor of $2$, and are multiples of the $3$, $4$, $5$ Pythagorean triple.

If you try other combinations of odd numbers for $n$ and $m$, you’ll find that all numbers in the triple will be even. We know that $b$ is always even because $b=2mn$, so if $a$ or $c$ are also even then the triple is non-primitive.

If $m$ and $n$ are both odd then $m^2$ and $n^2$ will also be odd, so $c = m^2 + n^2$ and $a = m^2 - n^2$ will both be even. Therefore, odd values for both $m$ and $n$ will produce a non-primitive Pythagorean triple.