Jekyll2021-05-10T03:38:19+01:00https://www.flyingcoloursmaths.co.uk/feed.xmlFlying Colours MathsFlying Colours Maths helps make sense of maths at A-level and beyond.A chalkboard challenge2021-05-10T00:00:00+01:002021-05-10T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/a-chalkboard-challenge<p>On <a href="https://www.reddit.com/r/math/comments/jlk0vb/given_one_hour_an_unlimited_amount_of_chalk_and/">reddit</a>, an interesting question:</p>
<blockquote>
<p>Given one hour, an unlimited amount of chalk, and an unlimited amount of blackboard space, how many (correct) digits of $\sqrt{10}$ could you find?</p>
</blockquote>
<p>(without any calculation aids, obviously).</p>
<p>At the moment, the idea of having a free hour to do anything, let alone work out huge bits of chalkboard arithmetic, seems like a crazy luxury. However, I can grab a few minutes here and there to try out a few different methods. I have four morally different ones I might attempt.</p>
<h3 id="method-1-the-long-division-method">Method 1: the long division method</h3>
<p>There are few things in maths, especially mental maths, that I dislike intensely. Despite @colinthemathmo’s <a href="http://www.solipsys.co.uk/new/SquareRootByLongDivision.html">excellent write-up</a> on the method, the long division way of finding square roots sadly leaves me cold. I could probably do it, in a pinch, but it’s not my idea of a fun hour. Let’s put that to one side.</p>
<p>(I’m going to treat “Figure out $n$ so that $n^2 \approx 10$, then use that to find $n^2 \approx 1000$, etc” as a variation on this.)</p>
<h3 id="method-2-taylor-series--binomial-expansion">Method 2: Taylor series / binomial expansion</h3>
<p>This is probably the method most of my mathematically-minded friends would try: use $(1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots$. Stick in $x=\frac{1}{9}$ and out pops $\frac{\sqrt{10}}{3}$.</p>
<p>It’s a pain, though. The numbers get big, quickly. And it’s also a pain because the Mathematical Ninja is likely to pay a visit to anyone who uses that form rather than $(a^2+b)^{1/2}= a + \frac{1}{2a}x - \frac{1}{8a^3}x^2 + \dots$. My first thought was $a=3$ and $b=1$, but I’ve since realised that $a=19$ and $b=-1$ converges much more quickly – two terms of that gives $\frac{721}{228}$, which is correct to better than one part in a million.</p>
<p>It may also be worth trying $(1-0.1)^{1/2}$, which gives $0.3\sqrt{10}$ and numbers that aren’t too horrible.</p>
<h3 id="method-3-babylonian">Method 3: Babylonian!</h3>
<p>I’d forgotten that the Babylonians had a method for this: let your initial guess be $x_0$; then $x_1 = \frac{x_0 + \frac{10}{x_0}}{2}$ is a better guess. This doesn’t converge super-quickly, and the fractions get ugly fast, but I still prefer it to long division.</p>
<h3 id="method-4-continued-fractions">Method 4: Continued fractions</h3>
<p>Of course I’d do it with continued fractions. $\sqrt{10}= [3; 6,6, 6,\dots]$, so it’s fairly easy to generate the convergents using the following process:</p>
<ul>
<li>$x_0 = 1$; $x_1 = 3$; $x_{n+2} = 6x_{n+1} + x_n$</li>
<li>$y_0 = 0$; $y_1 = 1$; $y_{n+2} = 6y_{n+1} + y_n$</li>
</ul>
<p>And $\frac{x_n}{y_n}$ is a good estimate for $\sqrt{10}$.</p>
<p>A few minutes’ work got me to $\frac{168,717}{53,353}$, which is good to ten or eleven decimal places.</p>
<hr />
<p>What about you? Do you have any better methods, either more efficient for chalkboard work or less likely to go wrong? I’d love to hear about them!</p>On reddit, an interesting question:Ask Uncle Colin: Couples2021-05-05T00:00:00+01:002021-05-05T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-couples<blockquote>
<p>Dear Uncle Colin,</p>
<p>How big a group of people do you need so that the probability of two of them being married is more than 50%?</p>
<p>- Counting Out Unmarried People Leveraging Exponential Sums</p>
</blockquote>
<p>Hi, COUPLES, and thanks for your message!</p>
<p>Im going to model this as a graph theory problem, rather than a sociopolitical one about who’s allowed to marry whom. ((For the record, I think any consenting adults who want to be married should be allowed to be married.))</p>
<p>Our graph consists of the whole population under consideration. Let’s say there are $N$ people, each corresponding to a node, and $c$ marriages, each corresponding to an edge.</p>
<p>If we pick $n < N$ people at random from the graph, that subgraph contains $\frac{1}{2}n(n-1)$ edges. What’s the probability that none of them coincide with the $c$ married edges in the population graph?</p>
<p>Each edge of the subgraph has a $\left(1-\frac{1}{c}\right)$ probability of not being a marriage. Taking the subgraph as a whole, and assuming independence (which seems reasonable enough), that means $P(\text{no marriages})\approx \left(1-\frac{1}{c} \right)^{n(n-1)/2}$.</p>
<p>We want $P(\text{no marriages})<0.5$, so let’s take logs: $\ln\left(0.5\right) > \frac{1}{2}n(n-1) \ln\left(1-\frac{1}{c}\right)$, and we want to solve this for $n$.</p>
<p>Let’s assume $c$ is large, so $\ln\left(1 -\frac{1}{c}\right) \approx -\frac{1}{c}$.</p>
<p>This gives $\ln(0.5) > -\frac{n(n-1)}{2c}$, or $n(n-1) > 2c \ln(2)$.</p>
<p>We’ve assumed $c$ is large, so we can approximate $n(n-1)$ as $n^2$ (we could solve the quadratic if the Mathematical Ninja held a sword to our throat, of course) - but a decent approximation would be $n > \sqrt{2c\ln2}$, or $\sqrt{1.4c}$.</p>
<p>In the UK, there are approximately 16 million married couples, so about 4,500 to 5,000 people randomly picked from the population would likely contain a married couple.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, How big a group of people do you need so that the probability of two of them being married is more than 50%? - Counting Out Unmarried People Leveraging Exponential SumsDictionary of Mathematical Eponymy: The Dubins Path2021-05-03T00:00:00+01:002021-05-03T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/dictionary-of-mathematical-eponymy-the-dubins-path<h3 id="what-is-the-dubins-path">What is the Dubins path?</h3>
<p>Geometrically speaking, the shortest way to get from point A to point B is along a straight line. But what’s the shortest route if you have directional restrictions?</p>
<p>Suppose you have a car moving in a 2D plane and a restriction on how sharply it can turn, a maximum curvature. You’re in point A facing in direction A’, and need to arrive at point B from direction B’. What’s the shortest route? It’s along a Dubins path.</p>
<p>In general, the Dubins path consists of three segments – either a turn, a straight line and another turn, or three turns in alternating directions. (All turns are made at the maximum curvature.)</p>
<h3 id="why-is-it-important">Why is it important?</h3>
<p>I love the Dubins path – it’s an elegant solution to what looks like a fairly simple problem. It has an obvious application in path planning – if you’re designing an autonomous vehicle, you probably want it to follow a Dubins path at least some of the time.</p>
<h3 id="who-was-lester-dubins">Who was Lester Dubins?</h3>
<p>Lester Dubins (1920-2010) was born in New York and served in the USAF during the Second World War. He then worked on radar research for several years before undertaking graduate study at the University of Chicago. After receiving his PhD in 1955, he worked at the IAS in Princeton and at Carnegie Mellon before joining UC Berkeley for the rest of his career. (Famously, he won a court case in the early 1990s to allow him to come out of retirement.) He eventually retired on his own terms in 2004 and died six years later.</p>
<p>He’s not best-known for his work on paths: rather, he was a renowned probability theorist and wrote (with Leonard Jimmie Savage) the marvellously-titled <em>How To Gamble If You Must (Inequalities for Stochastic Processes)</em>. He’s also partly responsible for the Dubins-Spanier theorems on fair division.</p>What is the Dubins path?Ask Uncle Colin: Which roots have the greater sum?2021-04-28T00:00:00+01:002021-04-28T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-which-roots-have-the-greater-sum<blockquote>
<p>Dear Uncle Colin,</p>
<p>Which is larger, $\sqrt{3}+\sqrt{11}$ or $\sqrt{5}+\sqrt{8}$? No calculator!</p>
<p>Roots Are Difficult (I Calculated Anyway, LOL)</p>
</blockquote>
<p>Hi, RADICAL, and thanks for your message!</p>
<p>If the Mathematical Ninja was nearby – and who can say, they might be – I would probably work out $\sqrt{11}$, the only one of those roots I don’t know to a couple of decimal places. In fact, you can never be too careful: $1089 = 33^2$, so $\sqrt{1100} \approx 33 + \frac{11}{66}$, or $33.17$. That means $\sqrt{11}\approx 3.32$, which looks about right.</p>
<p>Then 1.73 + 3.32 = 5.05 and 2.24 + 2.83 = 5.07 – so, assuming I’ve rounded right, the second is larger. It’s a bit unsatisfactory, though!</p>
<p>Instead, let’s do it properly: we’ll square both. The first gives $14 + 2\sqrt{33}$, and the second gives $13 + 2\sqrt{40}$. I think it’s easier to see if we write those as $14 + \sqrt{132}$ and $12 + \sqrt{160}$.</p>
<p>Why’s that? I know that $11.5^2 = 132.25$ and that $12.5^2 = 156.25$, so $\sqrt{132}<11.5$ and $\sqrt{160} > 12.5$. That means $13 + 2\sqrt{40} < 24.5$ and $14 + 2\sqrt{33} > 24.5$, so the second is larger.</p>
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, Which is larger, $\sqrt{3}+\sqrt{11}$ or $\sqrt{5}+\sqrt{8}$? No calculator! Roots Are Difficult (I Calculated Anyway, LOL)A piece of cake2021-04-26T00:00:00+01:002021-04-26T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/a-piece-of-cake<p>At MathsJam, I was pointed at a puzzle from the New Scientist, which I’ll paraphrase as:</p>
<blockquote>
<p>You have a long, thin cake of length 1. Two candles are places at random ((Throughout, “random” means “at a point drawn from a uniform distribution on the cake’s length”.)) points on the top of the cake, and the cake is cut (perpendicular to its edges) at a third random point. What is the probability that the two resulting pieces of cake each have a candle on?</p>
</blockquote>
<p>In discussion, we came up with three ways, which I’ll present in reverse order of complexity. Spoilers below the line.</p>
<hr />
<h3 id="in-three-dimensions-my-way">In three dimensions (my way)</h3>
<p>Let the first candle be at position $x$ and the second at position $y$. The probability of the cut dividing the candles is $|x-y|$, which I think of as a height above the $xy$-plane.</p>
<p>The resulting 3D graph forms a pair of tetrahedra, each with a base of area $\frac{1}{2}$ and height 1; the volume of each is therefore $\frac{1}{6}$ and the probability is $\frac{1}{3}$.</p>
<h3 id="a-single-integral">A single integral</h3>
<p>Suppose the cut takes place at position $x$. The probability the candles lie on either side of the cut is $2x(1-x)$, so the total probability is $\int_0^1 2x - 2x^2 \dx = \left[ x^2 - \frac{2}{3}x^3\right]_0^1$, which is again $\frac{1}{3}$.</p>
<h3 id="a-simple-and-logical-approach">A simple and logical approach</h3>
<p>Philipp pointed out that the three points are in an order from left to right, and the probability of any given one of them being in the middle is $\frac{1}{3}$.</p>
<hr />
<p>I love it when there’s an elegant solution! Did you tackle it a different way?</p>At MathsJam, I was pointed at a puzzle from the New Scientist, which I’ll paraphrase as:Ask Uncle Colin: A Geometric Subset2021-04-21T00:00:00+01:002021-04-21T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-geometric-subset<blockquote>
<p>Dear Uncle Colin,</p>
<p>I was working on a MAT question that asked about finding a subset, $S$, of the 2D-plane, and a point $P$ such that no point in $S$ was the closest to $P$. I had no idea where to start!</p>
<p>Omitted Point, Euclidean Norm</p>
</blockquote>
<p>Hello, OPEN, and thanks for your message!</p>
<p>My first answer wasn’t the one the MAT answer suggested, but I stand by it: if $S$ is the empty set, it’s definitely a subset of the 2D-plane ((every point in the empty set is also a member of the set of points in the real plane, vacuously)). And since there are no points in $S$ to be nearest to $P$, this satisfies the condition.</p>
<p>But they had something a bit trickier in mind.</p>
<p>According to the answer sheet, a valid answer would be something like:</p>
<blockquote>
<p>$S$ is the set of points such that $x^2+y^2 < 1$, and $P$ is a distance greater than 1 from the origin.</p>
</blockquote>
<p>Let’s unpack that a bit. The set $S$ is an open disc, a filled circle centred on the origin, with a radius of 1, but such that the circumference is not part of the set. $P$ is outside of the circle. So why is there not a point in $S$ that’s closest to $P$?</p>
<p>Suppose (without loss of generality) that $P$ is on the positive $x$-axis. Any candidate closest point would also have to be on the positive $x$-axis, with an $x$-coordinate smaller than 1 to be a member of $S$.</p>
<p>If someone claims “The point $(X,0)$, is the closest member of $S$ to $P$”, you can immediately say “No, it isn’t – the point $\left(\frac{1+X}{2},0\right)$, midway between $P$ and the circumference, is inside of $S$ and closer to $P$ than your point.” Contradiction. Boom.</p>
<p>It’s a bit subtle (which is why I prefer my empty set method), and I’m not sure it’s a fair thing to ask an A-level student to come up with on the fly in an exam, but it’s interesting enough to write up.</p>
<p>I hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I was working on a MAT question that asked about finding a subset, $S$, of the 2D-plane, and a point $P$ such that no point in $S$ was the closest to $P$. I had no idea where to start! Omitted Point, Euclidean NormA pretty puzzle2021-04-19T00:00:00+01:002021-04-19T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/a-pretty-puzzle<p>I heard it from @benjamin_leis, and he says he heard it from @CMonMattTHINK, and I love it:</p>
<blockquote>
<p>The number of integer solutions to $x^2 + xy + y^2 = a$ appears to be a multiple of six for all $a \in \mathbb{Z_+}$ . Why?</p>
</blockquote>
<p>How good a puzzle is this? I started my swimming class on Thursday and decided to swim non-stop until I solved it. Forty lengths later they kicked me out of the pool.</p>
<p>As always, spoilers below the line.</p>
<hr />
<p>This is a very pretty puzzle. Some observations to start with:</p>
<ul>
<li>The curve describes an ellipse</li>
<li>There are two lines of symmetry to the ellipse, $x=y$ and $x=-y$.</li>
<li>The extreme points of the ellipse are on the lines $x+2y=0$ and $2x+y=0$.</li>
</ul>
<h3 id="how-i-did-it">How I did it</h3>
<p>I think it’s useful to break my solution into two bits, and acknowledge that the “correct”, underlying answer came after a lot of head-scratching and missteps. The key bit of reasoning was:</p>
<blockquote>
<p>If $(X,Y)$ is an integer solution, fixing $y=Y$ gives a quadratic $x^2 + Yx + Y^2-a=0$. > I could solve that explicitly, but I don’t need to; $Y^2-a$ is an integer, so the second solution is also an integer; the solutions sum to $-Y$, so the other solution is $x=-X-Y$, which (for convenience) I’ll call $Z$.</p>
</blockquote>
<p>That means, if $(X,Y)$ is an integer solution, so is $(Z,Y)$. Applying similar reasoning to the resulting point, fixing $x=Z$ gives a solution at $(Z,X)$.</p>
<p>We can carry on around to get $(Y,X)$, $(Y,Z)$ and $(X,Z)$ before returning to $(X,Y)$, giving six algebraically distinct solutions.</p>
<p>BUT! Rotational symmetry also gives a solution at $(-X, -Y)$ - so changing the signs on all of the points we found before gives a further six solutions, a total of 12.</p>
<p>However, they aren’t all necessarily distinct! If any pair of $X$, $Y$ and $Z$ are equal, then each point we’ve found appears twice, leaving us with six solutions (in fact, two lie on the minor axis and the remaining four at the extreme points).</p>
<h3 id="underlying-group-structure">Underlying group structure</h3>
<p>I wasn’t <em>happy</em> with my answer until I managed to shoehorn a group structure onto the solutions. The points can be classified by three pieces of information:</p>
<ul>
<li>Which letter is missing (isomorphic to $\mathbb{Z}_3$, with $X \sim 0$, $Y \sim 1$ and $Z \sim 2$)</li>
<li>Whether the signs are + or - (isomorphic to $\mathbb{Z}_2$)</li>
<li>Whether the second letter immediately follows the first (isomorphic to $\mathbb{Z}_2$ as well). By convention, $Y$ follows $X$, $Z$ follows $Y$ and $X$ follows $Z$ in a pleasing cycle.</li>
</ul>
<p>And this gives a natural way to combine any pair of integer solutions!</p>
<p>The point $(X,Z)$ has $Y$ missing, positive signs and is out of order.</p>
<p>The point $(-Y,-X)$ has letter $Z$, negative signs and is out of order.</p>
<p>Combining them together, $Y + Z \sim (1 + 2)$, which is $0 \sim X$; the sign is negative, and the letters must be in order, giving $(-Y, -Z)$.</p>
<p>This is a group, because it’s isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_2$, and has order 12.</p>
<p>(In the degenerate case when a pair of variables is equal, each point coincides with exactly one other.)</p>
<h3 id="even-more-underlying">Even more underlying</h3>
<p>After reading around a bit on Diophanitine equations - by which I mean scanning a few paragraphs in a huff - I realised that the ellipse equation can be rewritten as:</p>
<ul>
<li>$2x^2 + 2xy + y^2 = 2a$</li>
<li>$x^2 + x^2 + 2xy + y^2 + y^2 = 2a$</li>
<li>$x^2 + (x+y)^2 + y^2 = 2a$</li>
</ul>
<p>Or, equivalently, $x^2 + (-x-y)^2 + y^2 = 2a$. Why that way?</p>
<p>To tell the truth, that’s still a bit nebulous. However, it has the lovely pair of properties:</p>
<ul>
<li>by symmetry, if $(X,Y)$ is a solution, any permutation of two elements of ${ X, Y, -X-Y }$ is also a solution</li>
<li>by a different symmetry, if $(X,Y)$ is a solution, any permutation of two elements of ${-X, -Y, X+Y}$ is also a solution.</li>
</ul>
<p>If $X$, $Y$ and $-X-Y$ are all different, then there are six solutions from the first set and six from the second; if two are the same, then there are three from each.</p>
<p>In either case, the number of solutions is a multiple of six.</p>
<hr />
<p>Can I stop swimming now? I need to put my computer in rice.</p>I heard it from @benjamin_leis, and he says he heard it from @CMonMattTHINK, and I love it:Ask Uncle Colin: A classical mixture2021-04-14T00:00:00+01:002021-04-14T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-classical-mixture<blockquote>
<p>Dear Uncle Colin,</p>
<p>The puzzle asks: you have two glasses with equal volumes of water and juice. You take a tablespoon of water and mix it into the juice; you take a tablespoon of the juice-water mix and mix it into the water. Is there more water in the juice, or juice in the water? Apparently the amounts are the same, but I don’t see how they can be!</p>
<p>Combining Liquids And Stirring Strangely - It’s Confusing!</p>
</blockquote>
<p>Hi, CLASSIC, and thanks for your message!</p>
<p>There are many ways to approach this puzzle, and I’m going to leave what I think is the simplest one until last.</p>
<h3 id="algebra">Algebra</h3>
<p>Initially, one glass contains $V$ of water; the other contains $V$ of juice.</p>
<p>Take a volume $v$ of water and add it to the juice; there is now $V - v$ of water in one glass; in the other, there is $V$ of juice and $v$ of water.</p>
<p>Now the tricky bit: we’re going to move a total of $v$ of the mixture back into the first glass, but it could be all juice, all water, or any mixture in between. So, let’s say the $v$ is made up of $t$ of juice and $v-t$ of water.</p>
<p>After moving it, the first glass has $(V - v) + (v-t)$, or $V-t$ of water and $t$ of juice.</p>
<p>The second glass has $V-t$, of juice, and $v - (v-t)$ or $t$ of water.</p>
<p>There’s as much juice in the water as there is water in the juice.</p>
<h3 id="cards">Cards</h3>
<p>The algebra is all well and good, but it’s not exactly intuitive. Instead, I like to think about things you can count, like cards of different colours.</p>
<p>Imagine you have a number of red cards and a number of blue cards. You take a certain number of red cards and mix them with the blue ones; you take the same number from the mixed pack and put them back in the red pack.</p>
<p>Now, there’s the same number of cards in each pack, and there’s still the same number of red and blue cards split between them. For every red card in the blue pack, a blue card must have gone to the red pack, and vice versa: there are as many red cards in the blue pack as vice versa.</p>
<h3 id="intuitively">Intuitively</h3>
<p>Everything in the juice glass that isn’t juice, is water.</p>
<p>Everything in the water glass that isn’t water, is juice.</p>
<p>There’s the same amount of water and juice altogether, so the “isn’ts” must be the same.</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, The puzzle asks: you have two glasses with equal volumes of water and juice. You take a tablespoon of water and mix it into the juice; you take a tablespoon of the juice-water mix and mix it into the water. Is there more water in the juice, or juice in the water? Apparently the amounts are the same, but I don’t see how they can be! Combining Liquids And Stirring Strangely - It’s Confusing!Circles in Seattle2021-04-12T00:00:00+01:002021-04-12T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/circles-in-seattle<p>A day off ill, and the chance to look muddle-headedly at a tweet from @trianglemanscd long ago:</p>
<blockquote class="twitter-tweet"><p lang="en" dir="ltr">OK Geometers, I'm gonna need you to problem solve for me. <a href="https://t.co/00njFhiBEw">pic.twitter.com/00njFhiBEw</a></p>— Christopher Danielson (<span class="citation" data-cites="Trianglemancsd">@Trianglemancsd</span>) <a href="https://twitter.com/Trianglemancsd/status/1056955508995883010?ref_src=twsrc%5Etfw">October 29, 2018</a></blockquote>
<script async="" src="https://platform.twitter.com/widgets.js" charset="utf-8"></script>
<p>I’m a geometer! And I like problem-solving!</p>
<p>There are <em>loads</em> of valid approaches here, and I’m going to talk of three.</p>
<h3 id="a-really-boring-approach">A really boring approach</h3>
<p>Step 1: go to Google maps Step 2: find the scale of the map Step 3: use that to calibrate the distances.</p>
<p>Of course I <em>could</em> do that. But that’s geography, not maths.</p>
<h3 id="slightly-nicer">Slightly nicer</h3>
<p>A more mathematical approach would be to centre a circle on each of the the landmarks with a radius proportional to the given distance to the mystery hotel. Slide the scale until the circles all intersect, or close enough. <a href="/images/Screenshot-2020-09-17-at-12.11.05.png"><img src="/images/Screenshot-2020-09-17-at-12.11.05.png" alt="" /></a></p>
<p><a href="/images/Screenshot-2020-09-17-at-12.11.23.png"><img src="/images/Screenshot-2020-09-17-at-12.11.23.png" alt="" /></a></p>
<p>This is <em>becoming</em> more mathematical, but still a bit boring: my best effort gives a location somewhere between Madison and Union, where there are half a dozen hotels.</p>
<h3 id="the-proper-way">The proper way</h3>
<p>But as a mathematician, fudging $k$ to get an answer feels a bit dirty. The circles don’t intersect properly because the distances are rounded. Surely we can do it another way?</p>
<p>Of course we can.</p>
<p>For example, the mystery hotel is marked as 0.3m from both Pike Place Market and Columbia Center. Or rather, both of those landmarks are between 0.25m and 0.35m away.</p>
<p>Or rather rather, the <em>ratio</em> of the distances from our hotel to those landmarks is between $\frac{5}{7}$ and $\frac{7}{5}$. What does that look like?</p>
<p>Well, like this:</p>
<p><a href="/images/Screenshot-2020-09-17-at-12.17.41.png"><img src="/images/Screenshot-2020-09-17-at-12.17.41.png" alt="" /></a></p>
<p>A large circle around each landmark is excluded - and I can do the same with any pair I choose: <a href="/images/Screenshot-2020-09-19-at-08.52.02.png"><img src="/images/Screenshot-2020-09-19-at-08.52.02.png" alt="" /></a></p>
<p>Several observations here: first, the orange region, centred on A and G, seems to disagree with the other pairs I picked. I don’t know why that is - possibly the distance is measured from a different part of CenturyLink Field. Secondly, the others converge on a point 2.5 blocks from both Union and I-5 ((Oh good grief, sudden flashback to the one time I drove on that)) - which could reasonably <a href="https://www.google.com/maps/@47.6077527,-122.3361587,17.12z">be the Executive Hotel Pacific or the West Seattle</a>. (There are several other plausible options nearby).</p>
<h3 id="where-was-it">Where was it?</h3>
<p>It was, it turns out, neither of those, but the Kimpton Alexis, several blocks further from the freeway – astonishingly, bang on where the bulk of the circles coincide in the “slightly nicer way”!</p>
<p>I don’t know what the take-home message here is, only that it was a nice thing to play with, and I’d be curious to hear what other methods you might try!</p>
<p>* Accessibility features on images are currently unavailable owing to a glitch on the website. I hope to find and address this before it goes live, but who can predict?</p>A day off ill, and the chance to look muddle-headedly at a tweet from @trianglemanscd long ago:Ask Uncle Colin: A Couple of Tangents2021-04-07T00:00:00+01:002021-04-07T00:00:00+01:00https://www.flyingcoloursmaths.co.uk/ask-uncle-colin-a-couple-of-tangents<blockquote>
<p>Dear Uncle Colin,</p>
<p>I’ve got two circles ($x^2 + y^2 = 3^2$ and $(x-10)^2 + y^2 =5^2$) and I want to find the equations of the common tangents. I’ve been stuck for ages!</p>
<p>- Tackling A New Geometry Exercise, Need Tuition</p>
</blockquote>
<p>Hi, TANGENT, and thanks for your message!</p>
<p>I can see two fairly nice ways to approach this.</p>
<h3 id="with-algebra">With algebra!</h3>
<p>Suppose the equation of a tangent line is $y = m(x-x_0)$ for some fixed values of $m$ and $x_0$. ((Other equations are available.))</p>
<p>This line meets the first circle when $x^2 + m^2(x-x_0)^2 = 3^2$; if the line is tangent, the equation has a repeated root.</p>
<p>Rearrange to get $x^2 (1+m^2) - 2m^2x_0x + (m^2x_0^2 - 9) = 0$; this has equal roots if $4m^4x_0^2 = 4(1+m^2)(m^2 x_0^2 - 9)$.</p>
<p>Expanding that and dividing by 4 gives $m^4 x_0^2 = m^2x_0^2 - 9 + m^4x_0^2 - 9m^2$</p>
<p>There’s an $m^2 x_0^2$ on each side, so $m^2(x_0^2 - 9) = 9$, or $m^2 = \frac{9}{x_0^2 - 9}$.</p>
<p>So, this is the relationship between $m$ and $x_0$ for any line that touches the first circle. What about the second one?</p>
<p>We can do a similar analysis. Substitute to get $(x-10)^2 + m^2(x-x_0)^2 = 5^2$, and rearrange to get $x^2 (1+m^2) - (2m^2 x_0 + 20)x + (75 + m^2x_0^2)= 0$.</p>
<p>Next, this needs to have repeated roots, so $(2m^2x_0 + 20)^2 = 4(1+m^2)(75 + m^2x_0^2)$.</p>
<p>Expanding and dividing by 4: $m^4 x_0^2 + 20m^2 x_0 x + 100 = 75 + m^2x_0^2 + 75m^2 + m^4x_0^2$.</p>
<p>Again, the $m^4$ term vanishes to leave us, this time, with $m^2(x_0^2-20x_0 + 75) = 25$.</p>
<p>We have two relationships now:</p>
<ul>
<li>$m^2 = \frac{9}{x_0^2 - 9}$</li>
<li>$m^2 = \frac{25}{x_0^2 - 20x_0 + 75}$</li>
</ul>
<p>We can combine those to give $\frac{9}{x_0^2 - 9} = \frac{25}{x_0^2 - 20x_0 + 75}$, or $9(x_0^2 - 20x_0 + 75) = 25(x_0^2 - 9)$.</p>
<p>This rearranges to $0 = 16x_0^2 + 180x_0 - 900$, or $4x_0^2 + 45x_0 - 225 = 0$…</p>
<p><strong>whoosh</strong></p>
<p>“$(4x_0-15)(x_0+15)$. You’re welcome.”</p>
<p><strong>whoosh</strong></p>
<p>So $x_0 = \frac{15}{4}$ or $-15$, and $m^2 = \frac{16}{9}$ or $\frac{1}{24}$.</p>
<p>The tangent lines are $y = \pm \frac{4}{3}\left( x - \frac{15}{4}\right)$ and $y = \pm \frac{1}{\sqrt{24}}\left( x + 15 \right)$.</p>
<p>The Mathematical Ninja, I couldn’t help noticing, wrinkled their face in passing.</p>
<h3 id="with-geometry">With geometry:</h3>
<p>We have a circle with radius 3, centred at the origin.</p>
<p>We have a circle with radius 5, centred at point P, (-10, 0).</p>
<p>Suppose we have a line that’s tangent to the first circle at T and the second circle at U, and that crosses the X-axis at X.</p>
<p>Triangle OTX and triangle PUX are similar, with a scale factor of 3:5.</p>
<p>Now, there are two classes of tangent: inner tangents (that cross between the circles) and outer tangents (that cross outside). Let’s look at those in turn.</p>
<p>If the tangents cross the axis outside the circles, we know that the difference between the hypotenuses of the triangles is 10. That means the longer hypotenuse is 25 units long and the shorter one 15 – and these must meet the axis at (-15,0).</p>
<p>If the shorter hypotenuse is 15 and its shorter leg is 3, its longer leg is $\sqrt{225-9} = 6\sqrt{6}$. The gradient of the tangent is then $\pm \frac{3}{6\sqrt{6}} = \pm \frac{1}{\sqrt{24}}$, so the equation of the line is $y = \pm \frac{1}{\sqrt{24}}(x+15)$. (The $\pm$ is because the tangent could be above or below the axis).</p>
<p>Looking at the inner tangents, the sum of the hypotenuses is 10, so their lengths are $\frac{15}{4}$ and $\frac{25}{4}$, and they cross at $\left(\frac{15}{4},0\right)$.</p>
<p>The shorter hypotenuse is $\frac{15}{4}$ long, and one of the legs is 3, so the other is $\frac{9}{4}$. The gradient of the tangent is $\pm\frac{9}{3/4} = \pm\frac{4}{3}$, and the equation of the line is $y = \frac{4}{3}\left(x - \frac{15}{4}\right)$.</p>
<hr />
<p>Hope that helps!</p>
<p>- Uncle Colin</p>Dear Uncle Colin, I’ve got two circles ($x^2 + y^2 = 3^2$ and $(x-10)^2 + y^2 =5^2$) and I want to find the equations of the common tangents. I’ve been stuck for ages! - Tackling A New Geometry Exercise, Need Tuition