In the sangaku series, we've solved two of the four problems on this tablet, the middle two, which I believe were the easiest to work on in terms of geometric arguments - we hardly ever used more than Pythagoras's theorem, though the second one needed some more advanced algebra to finish off.
Here's a quick look at the problems at each end of the tablet, and the main ideas I had to solve them.
On the far left, we have two circles tangent to one another (with centres A and B), inside a larger circle (with centre O) so that their diameters add up to the diameter of the largest. The radii of these three circles, respectively p, q and p+q, are known. The unknown is the radius r of the circle with centre C, which must be tangent to all three original circles (it has a twin on the right-hand side with the same radius).
This is quite quick to solve. Remember that tangent circles mean that the distances between centres is equal to the sum of the radii, e.g. AC = p+r, BC = q+r... Al-Kashi's theorem, which is a general version of Pythagoras's theorem, links the lengths of three sides of a triangle with one of the triangle's angles, and the triangles CAO and CAB have an angle in common, which yields the equation for r by isolating this angle in each application of Al-Kashi's theorem. The result is:
The problem on the far right seems to start in a similar fashion: two circles with fixed radii are offset by a fixed distance. A third circle has its diameter equal to the remainder of the diameter of one of the large circles: this radius can be calculated with little difficulty. What we want to do next is construct circles which are tangent to the two large ones, and the one previously constructed.
The radius of the circle with centre C1 can be obtained as above, but this method does not seem to extend to the subsequent circles, as O, D and C1 are no longer aligned, and there no longer appears to be a common angle in the triangles we want to work with. So I went for a parametric approach, understanding the curves that contain points that are equidistant from two circles. The red curve (which looks like a circle but isn't one) is the set of points at equal distance from the two largest circles, and we seek to intersect this with the set of points that are at equal distance from one large circle and the smaller one, the green curve. The intersection is equidistant from all three circles, so it is the centre of the circle we want to construct. Rotate and repeat for subsequent circles.
The general formulas are horrible and not worth showing, but this is another problem where I have been able to read the results on the tablet. The large circles have radii 61 and 72, and the offset is 23. The radii of the smaller circles, starting with the one in the middle and working outwards are:
17, 15.55, 12.292, 8.832 and 6.038 (I see 八, but I'll give the authors the benefit of the doubt as the top of the character 六 may have been erased by time)
The results with our exact formulas are:
17, 15.58, 12.795, 9.076 and 6.444
Rather close! As with the "three circles in a triangle", I do not know how the authors originally solved this problem.
Busy week done! Lots of work especially on Wednesday, and helping Vent d'Est organise their mah-jong tournament on Sunday to cap it off. We were on the boat in the foreground of this photo taken a few years ago - yep, still there today. A more flattering angle for the river, at a greener time of year, below.
When I visited Kashihara, looking to explore some deep Japanese history in the former province of Yamato, I expected to move around a bit, but there was actually enough in Kashihara itself to make for a busy day.
First up was this curious green round space in the middle of a residential area on the town map I'd picked up. It just seemed conspicuous to me, I decided to check it out.
This is Miminashi-yama, one of the Yamato Sanzan, or Three Main Mountains of Yamato. Though it stood out on the map and it does stand out in the plain around it, it's not huge, and it's a short climb to the top where a shrine awaited.
In that shrine, a sangaku geometry tablet is displayed. By chance, based on a whim, I had found one! Nearly six years on, I've finally solved it - it's not very difficult mathematically, it's just taken me this long to get on with it, having said that, even today I'm still figuring out extra things on it! - and will be presenting it at a conference tomorrow. I wouldn't have thought it at the time... I guess curiosity didn't kill the cat that day!
After visiting Karlsruhe Palace, I had a bit of time while waiting for a (packed) regional train South to wander around the station. There was quite a lot going on, as on top of the local traffic from S-Bahns to REs, busy intercity lines to Freiburg and Basel, Stuttgart and Munich, Mannheim and Frankfurt meet here. A chance to see my favourite German high-speed train: the Baureihe 403 ICE 3.
Entering service in 2000, over 10 years after the first InterCity Expresses, the ICE 3 was a revolution in European high-speed rail. These were the first 300 km/h-capable trains that weren't of a "power car & carriages" layout, using distributed traction (motors all along the unit) like Shinkansen trains. And a sleek shape to boot! They were designed by Alexander Neumeister, who also penned the 500 Series Shinkansen. Hmm, maybe that's why.
The ICE 3 would be the first example of Siemens's Velaro series, which would be an export hit: the Netherlands, Spain, China and Russia purchased this model. The type received a redesign in the late 2000s that I could only describe as "more beefy": the Velaro D was taken up by Germany, Turkey and Eurostar.
A quick post in response to Ferrari's second consecutive overall win at Le Mans, with a few sights from the manufacturer's home town, Maranello. Everything here is Ferrari: their road car factory with the classic entrance gate (above), the more modern F1 team base, the Fiorano test track, the Scuderia museum, the statues to founder Enzo Ferrari and the Prancing Horse, a park featuring Ferrari's most famous road car models...
Yep, everything is Ferrari here. Except, well, this store front apparently. It's rather gutsy to show off a Lambo badge here, but then again, historically, middle fingers to Ferrari was what Ferrucio Lamborghini was all about!
Having mentioned previously how mathematical schools were organised during the Edo period in Japan, we can briefly talk about how mathematicians of the time worked. This was a time of near-perfect isolation, but some information from the outside did reach Japanese scholars via the Dutch outpost near Nagasaki. In fact, a whole field of work became known as "Dutch studies" or rangaku.
One such example was Fujioka Yûichi (藤岡雄市, a.k.a. Arisada), a surveyor from Matsue. I have only been able to find extra information on him on Kotobank: lived 1820-1850, described first as a wasanka (practitioner of Japanese mathematics), who also worked in astronomy, geography and "Dutch studies". The Matsue City History Museum displays some of the tools he would have used in his day: ruler, compass and chain, and counting sticks to perform calculations on the fly.
No doubt that those who had access to European knowledge would have seen the calculus revolution that was going on at the time. Some instances of differential and integral calculus can be found in Japan, but the theory was never formalised, owing to the secretive and clannish culture of the day.
That said, let's have a look at where our "three circles in a triangle" problem stands.
The crucial step is to solve this equation,
and I suggested that we start with a test case, setting the sizes of the triangle SON as SO = h = 4 and ON = k = 3. Therefore, simply, the square root of h is 2, and h²+k² = 16+9 = 25 = 5², and our equation is
x = 1 is an obvious solution, because 32+64 = 96 = 48+48. This means we can deduce a solution to our problem:
Hooray! We did it!
What do you mean, "six"? The triangle is 4x3, that last radius makes the third circle way larger...
Okay, looking back at how the problem was formulated, one has to admit that this is a solution: the third circle is tangent to the first two, and to two sides of the triangle SNN' - you just need to extend the side NN' to see it.
But evidently, we're not done.
On the right-hand side of the Rance river, just before the fortified city of Saint Malo on the estuary, is the smaller town of Saint Servan. In fact, it technically isn't a town anymore, it was absorbed by Saint Malo in the 1960s. But for most of history, there was a stark contrast between the two, as Saint Malo fiercely proclaimed its autonomy several times. Hence the Solidor Tower.
Consisting of three tightly-bunched round towers and their connections, the Solidor was built in the 14th century by the Dukes of Brittany as a means to control the Rance estuary, against the rebellious Saint Malo if needed.
Like other fortresses, such as the Bastille in Paris or the towers at La Rochelle (another time maybe), its strategic value soon dwindled, and it seemed best-suited to serve as a prison or as storage during the late 18th-early 19th centuries. It has been an officially classified monument since 1886, and had housed a maritime-themed museum since 1970, though this appears to be in limbo and I can't find the tower's current function.
A walk along the coast on the West side of Saint Servan will reveal a bit more history: an old lifeboat station, a small tower in the sea that serves as a tide gauge... further up, a WWII memorial with the remains of concrete bunkers, and further along, a view of Saint Malo. It's a worthwhile detour for people visiting Saint Malo, especially if you're concerned that the city centre will be too crowded. But I think I remember parking here wasn't easy either; on a nice day, the locals who don't want the hassle of "intra-muros" would come here.
Close to impressive Haut-Barr castle, a one-hour hike from Saverne, sit two more ruins. All of these castles were built around the same time, late 10th to early 11th century, but despite being so close, they weren't owned by the same people.
While Haut-Barr was under the control of the Bishop of Strasbourg, the two Geroldseck castles, the Petit and the Grand, were built by the Geroldseck family, in charge of protecting the lands of the Abbey of Marmoutier. At the time, Alsace was part of the Holy Roman Empire and divided into many largely independent pieces, so these castles facing each other were on a border of sorts. However, the male Geroldseck line went extinct at the end of the 12th century, and the land was co-owned by so many people that no-one was maintaining the castle. The last stand came in 1471, when a group of disgruntled knights used it as their base. The Imperial bailiff laid siege, won and the castle was left as a ruin after that.
While Haut-Barr castle gets a lot of visitors, owing to the possibility of driving there, the Grand Geroldseck is worth the extra walk and brief climb from its neighbour. As well as the dungeon, lots of walls and rooms are still present, making it an interesting place to explore. The remaining walls continue to receive restoration work - there seem to be a few differences between my first visit with @teamroquette and my second this summer, for example, I don't remember seeing the little garden a few years ago.
All that's left to say is: "OI YOU!... YES, YOU! Have a good time."
We are only a few steps of algebra away from solving the "three circles in a triangle" problem we set in episode 7. This method will also yield general formulas for the solutions (first with height 1 and base b; for any height h and half-base k, set b=k/h and multiply the results by h).
Before we do that, it's worth noting what the sangaku tablet says. Now I don't read classical Japanese (the tablet dates back to 1854 according to wasan.jp), but I can read numbers, and fishing for these in the text at least allows me to understand the result. The authors of the sangaku consider an equilateral triangle whose sides measure 60: boxed text on the right: 三角面六尺, sankaku-men roku shaku (probably rosshaku), in which 尺, shaku, is the ten marker. In their writing of numbers, each level has its own marker: 尺 shaku for ten, 寸 sun for units, 分 fun for tenths and 厘 rin for hundredths (毛 mô for thousandths also appear, which I will ignore for brevity). Their results are as follows:
甲径三尺八寸八分六厘: diameter of the top (甲 kou) circle 38.86
乙径一尺六寸四分二厘: diameter of the side (乙 otsu) circle 16.42
反径一尺二寸四分二厘: diameter of the bottom (反 han) circle 12.42
I repeat that I don't know classical Japanese (or much modern Japanese for that matter), so my readings may be off, not to mention that these are the only parts of the tablet that I understand, but the results seem clear enough. Let's see how they hold up to our final proof.
1: to prove the equality
simply expand the expression on the right, taking into account that
(s+b)(s-b) = s²-b² = 1+b²-b² = 1.
2: the equation 2x²-(s-b)x-1 = 0 can be solved via the discriminant
As this is positive (which isn't obvious as s>b, but it can be proved), the solutions of the equation are
x+ is clearly positive, while it can be proved the x- is negative. Given that x is defined as the square root of 2p in the set-up of the equation, x- is discarded. This yields the formulas for the solution of the geometry problem we've been looking for:
3: in the equilateral triangle, s=2b. Moreover, the height is fixed at 1, so b can be determined exactly: by Pythagoras's theorem in SON,
Replacing b with this value in the formulas for p, q and r, we get
Now we can compare our results with the tablet, all we need to do is multiply these by the height of the equilateral triangle whose sides measure 60. The height is obtained with the same Pythagoras's theorem as above, this time knowing SN = 60 and ON = 30, and we get h = SO = 30*sqrt(3). Bearing in mind that p, q and r are radii, while the tablet gives the diameters, here are our results:
diameter of the top circle: 2hp = 45*sqrt(3)/2 = 38.97 approx.
diameter of the side circle: 2hr = 10*sqrt(3) = 17.32 approx.
diameter of the bottom circle: 2hq = 15*sqrt(3)/2 = 12.99 approx.
We notice that the sangaku is off by up to nearly a whole unit. Whether they used the same geometric reasoning as us isn't clear (I can't read the rest of the tablet and I don't know if the method is even described), but if they did, the difference could be explained by some approximations they may have used, such as the square root of 3. Bear in mind they didn't have calculators in Edo period Japan.
With that, thank you very much for following the Sangaku Weekends series, hoping that you found at least some of it interesting.
I know that a Doge is something else in Venetian history, and I knew it back in 2015 too (though I didn't look into what it was exactly).
But at the time, the meme was in full swing, and I couldn't resist picturing this sign. I don't know if it's a good hotel though, the reviews aren't rave...
ありがとうかぼす!
Landscapes, travel, memories... with extra info.Nerdier than the Instagram with the same username.60x Pedantle Gold medallistEnglish / Français / 下手の日本語
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