Sangaku - Blog Posts

Sangaku Sunday Bonus - what about the other problems?

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.

Sangaku Saturday #14 - the grand finale!

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.

Sangaku Saturday #13

Last week, we uncovered this configuration which is also a solution to our "three circles in a triangle" problem, just not the one we were hoping for.

This is something that happens in all isosceles triangles. Draw the inscribed circle, with centre B, and the circle with centre C, tangent to the extended base (ON) and the side [SN] at the same point as the first circle is. Then it can be proved that the circle with centre A, whose diameter completes the height [SO] as our problem demands, is tangent to the circle with centre C.

But that's not what I'm going to concentrate on. Despite this plot twist, we are actually very close to getting what we want. What the above configuration means is that, returning to the initial scaled situation with SO = h = 1 and ON = b, we get

Knowing a solution to a degree 3 equation is extremely powerful, as we can factor the polynomial and leave a degree 2 equation, which has simple formulas for solutions. So, to finish off, can you:

1: prove that

2: solve the equation 2x²-(s-b)x-1 = 0, and deduce the general formulas for p, q and r that fit the configuration we are aiming for;

3: test the formulas for an equilateral triangle, in which s = 2b.

This last question is the one the sangaku tablet claims to solve.

Sangaku Saturday #12

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.

Sangaku Sunday #11

We're almost there! We have three relations between our unknowns, the radii p, q and r. Actually, let's write them in the general setting, with any height.

Set SO = h and ON = k (so the number b in the problem so far has been equal to k/h). Repeating what we've done in previous steps, and substituting q and r in the final equation so that we get an equation with just p (I've done it so you don't have to), this is what we're solving:

The plan is simple: get p with the last equation, deduce q then r with the first two. The execution of the plan... not so simple. That last equation is messy. Let's tidy it up a bit by noting that it is actually a polynomial equation of the variable x=squareroot(2p):

There are formulas for the solutions to an equation like this, but if we can avoid using them, we'll be happy.

Here's what I did - and you can do too: a numerical test. Let's take the simplest dimensions for a right triangle, h = 4 and k = 3. Replace in the last equation and notice an obvious solution. Deduce p, then q, then r. Jubilate - until you realise something is very, very wrong...

Sangaku Sunday #10

On the historical front, we previously established that mathematics didn't stop during the Edo period. Accountants and engineers were still in demand, but these weren't necessarily the people who were making sangaku tablets. The problems weren't always practical, and often, the solutions were incomplete, as they didn't say how the problems were solved.

There was another type of person who used mathematics at the time: people who regarded mathematics as a field in which all possibilities should be explored. Today, these would be called researchers, but in Edo-period Japan, they probably regarded mathematics more as an art form.

As in many other art forms (Hiroshige's Okazaki from The 53 Stations of the Tôkaidô series as an example), wasan mathematics organised into schools with masters and apprentices. This would have consequences on how mathematics advanced during this time, but besides that, wasan schools were on the look-out for promising talents. In this light, sangaku appear as an illustration of particular school's abilities with solved or unsolved problems to bait potential recruits, who would prove their worth by presenting their solutions.

Speaking which, we now continue to present our solution to the "three circles in a triangle" problem.

Recall that we are looking for two expressions of the length CN.

1: Knowing that ON = b and OQ = 2*sqrt(qr), it is immediate that QN is the subtraction of the two. Moreover, CQ = r, so by using Pythagoras's theorem in the right triangle CQN, we get

2: We get a second expression by using a cascade of right triangles to reach CN "from above". Working backwards, in the right triangle CRN, we known that CR = r, but RN is unknown, and we would need it to conclude with Pythagoras's theorem. We can get RN if we know SR, given that SN = SR+RN is known by using Pythagoras's theorem in the right triangle SON, with SO = 1 and ON = b. But again, in the right triangle CRS, we do not know CS, but (counter-but!) we could get CS by using the right triangle PCS, where PC and PS are both easy to calculate. We've reached a point where we can start calculating, so let's work forward from there.

Step 1: CPS. PCQO is a rectangle, so PC = OQ and PS = SO-OP = SO-CQ = 1-r, therefore

Step 2: CRS. Knowing CR = r, we deduce

At this point, we can note that 2r-4qr = 2r(1-2q) = 2r*2p, using the first relation between p and q obtained in the first post on this problem. So SR² = 1-4pr.

Step 3a: SON. Knowing SO = 1 and ON = b, we have SN² = 1+b².

Step 3b: CRN. From SN and SR, we deduce

so, using Pythagoras's theorem one more time:

Conclusion. At the end of this lengthy (but elementary) process, we can write CN² = CN² with different expressions either side, and get the final equation for our problem:

Note that 2*(p+q) = 1, and divide by 2 to get the announced result.

Sangaku Saturday #9

As in every odd-numbered episode, we're going to set a problem - the next stage towards solving the "three circles and a triangle" sangaku. We are looking for one more equation between the radii p, q and r, it will be obtained with a similar method to the previous step... but the formulas will be a bit longer, so roll your sleeves up and don't be scared!

Here are the lengths we know:

SO = 1 , SN = b , SA = p , BO = q , CQ = r and

Here is also a list of known pairs of perpendicular lines:

(SO) and (ON) , (SO) and (PC) , (ON) and (CQ) , (SN) and (CR).

[P, Q and R are defined as the orthogonal projections of C onto the sides of SON.]

The equation we are looking for will come from getting two expressions for the square of the length CN.

You can work out how to do this by yourself if you feel like it, or check below the cut for the steps and to check your result. As always, details and a bit of history next week!

1: After working out the length QN, get a first expression of CN² by using Pythagoras's theorem in the right triangle CQN.

2: Proceed similarly in the cascade of right triangles CPS, CRS and CRN, to get a second expression of CN².

Conclude that

Sangaku Saturday #8

Having established that sangaku were, in part, a form of advertisement for the local mathematicians, we can look at the target demographic. Who were the mathematicians of the Edo period? What did they work on and how?

The obvious answer is that the people in the Edo period who used mathematics were the ones who needed mathematics. As far back as the time when the capital was in Kashihara, in the early 8th century, evidence of mathematical references has been uncovered (link to a Mainichi Shinbun article, with thanks to @todayintokyo for the hat tip). All kinds of government jobs - accounting, such as determining taxes, customs, or engineering... - needed some form of mathematics. Examples above: 8th-century luggage labels and coins at the Heijô-kyô Museum in Nara, and an Edo-period ruler used for surveying shown at Matsue's local history museum.

As such, reference books for practical mathematics have existed for a long time, and continued to be published to pass on knowledge to the next generation. But sangaku are different: they are problems, not handbooks.

More on that soon. Below the cut is the solution to our latest puzzle.

Recall that SON is a right triangle with SO = 1 and ON = b. These are set values, and our unknowns are the radii p, q and r of the circles with centres A, B and C. While these are unknown, we assume that this configuration is possible to get equations, which we can then solve.

1: The two circles with centres B and C are tangent to a same line, so we can just re-use the very first result from this series, so

2: Also recalling what we said in that first problem about tangent circles, we know that

Moreover, PA = AO - OP = AO - CQ = (p+2*q) - r. Thus, using Pythagoras's theorem in the right triangle APC, we get a new expression for PC:

since 2(p+q)=1 (the first relation). Equating the two expressions we now have of PC², we solve the equation for r:

again using the first relation to write 2q-1 = -2p.

It only remains to find a third equation for p to solve the problem.

Sangaku Sunday #7

We're back with a new problem from Miminashi-yamaguchi-jinja! This is going to be more ambitious than the first one, though it won't be much harder from a geometry standpoint - the main tool will still be Pythagoras's theorem. But we really need to set the stage for this one.

Consider an isosceles triangle, with two circles whose diameters are on the height from the apex, tangent to each other, and so that the top circle passes through the apex and the bottom circle is tangent to the base. We seek to draw one more circle on either side, which is tangent to the first two, and tangent to two sides of the triangle.

Details and first questions below the cut.

The triangle is given: it is an isosceles triangle SNN'. For the sake of simplicity, let's shrink or blow up the figure so that the height SO is equal to 1 (for a configuration with height h, we will just need to multiply all the lengths by h). The length of the base NN' is therefore the fixed parameter of the problem, and, as the figure is symmetric with respect to SO, we only need to set the length ON as our parameter: set ON = b. Hence, we are working in the right triangle SON.

The problem involves finding the three circles that fit the configuration in SON. Let these circles have respective centres A, B and C, and respective radii p, q and r. The radii are the unknowns of our problem, and we need to find three independent relations between them to solve. From the sketch, it looks like there should be only one solution.

The first relation is obvious: 2*(p+q) = 1, as the diameters of the first two circles form the height SO. This is also very easy to solve: if we have p, then q = 1/2 - p.

A second relation must start to involve r. For this, project the centre of the third circle onto SO and ON, calling these projections P and Q respectively. Now we get to two questions for you to munch on.

1: Prove that

2: Get the lengths AC and PA. Deduce another expression for PC, and prove that

With that, we just need another equation to find p, and we'll be done.

Sangaku Sunday #6

We are about to solve our first sangaku problem, as seen on the tablet shown above from Miminashi-yamaguchi-jinja in Kashihara.

First, we should conclude our discussion: what are sangaku for? There's the religious function, as an offering, and this offering was put on display for all to see, though not all fully understood the problems and their solutions. But a few people would understand, and these would have been the mathematicians of the time. When they visited a new town, they would typically stop at a temple or shrine for some prayers, and they would see the sangaku, a sample of what the local mathematicians were capable of. Whether the problems were solved or open, the visitor could take up the challenges and find the authors to discuss.

And this is where everything lined up: the local school of mathematics would have someone new to talk to, possibly to impress or be impressed by, and maybe even recruit. With the Japanese-style mathematics of the time, called wasan, being considered something of an art form, there would be masters and apprentices, and the sangaku was therefore a means to perpetuate the art.

Now, what about that configuration of circles, second from right on the tablet?

Recall that we had a formula for the radii of three circles which are pairwise tangent and all tangent to the same line. Calling the radii p, q, r, s and t for the circles of centres A, B, C, D and E respectively, we have

for the circles with centres A, B and C (our previous problem), and adapting this formula to two other systems of three circles, we get

for the circles with centres A, C and D, and

for the circles with centres B, C and E. Add these together, and use the first relation on the right-hand side, we get a rather elegant relation between all five radii:

Of course, we can get formulas for s and t,

r having been calculated previously using just p and q, which were our starting radii.

For example, setting p=4 and q=3, we get, approximately, r=0.86, s=0.4 and t=0.37 (this is the configuration shown in the figure, not necessarily the one on the tablet - I will be able to make remarks about that on another example).

Sangaku Saturday #5

This is it, we've reached our first full sangaku problem! (At least one that I've seen on a real-life tablet.)

We set up the tools in previous posts, so I'll let you work this one out by yourselves: given two circles which are tangent to each other and tangent to a same line, what can you say about the radii of the three smaller, light-coloured circles?

Sangaku Saturday #4

In the previous info post, we went over the debate on the religious aspect of sangaku, and the fact that the absence of prayers on these tablets was more puzzling to some than the mathematics. As such, the tablets are not ema prayer tablets, but donations, which usually don't feature prayers on them. Case in point, some consecrated sake and French wine seen at Meiji-jingû in 2016.

Beyond wishing for good fortune and health, such donations serve two very worldly purposes: to contribute to the life and prestige of the shrine or temple (having a famous contributor makes the shrine famous by association), and to advertise the donor in return, as their name is on display. See this large torii at Fushimi Inari Taisha paid for by TV Asahi (テレビ朝日).

With that in mind, Meijizen's cynical comment from 1673 that sangaku aim "to celebrate the mathematical genius of their authors" may not far from the truth. The authors of sangaku are looking to gain notoriety through the publicity that the shrine or temple provides. But was the bemused Meijizen the target audience?

More on that in a couple of weeks. Below the cut is the solution to last week's problem.

The solution to the first problem (below the cut in this post) is the key. Name K, L and M the intersections of the three circles with the horizontal line. Then, by using that previous result,

Indeed, as in that problem, we can construct three right triangles, ABH, ACI and BCJ and apply Pythagoras's theorem in each.

Now, it suffices to note that KL = KM + LM, so

or, dividing by 2*squareroot(pqr), we get the desired result:

Inverting and squaring this yields the formula for r:

This gives us the means to construct this figure on paper using a compass and a marked ruler. Having chosen two radii p and q and constructed the two large circles (remember that AB=p+q) and a line tangent to both, placing M and C is done after calculating the lengths IK=CM=r and IC=KM=2*sqrt(pr).

Sangaku Saturday #3

Another problem this week, adding to the configuration we looked at previously.

Specifically, given two circles tangent to each other and tangent to a same line - these circles have respective centres A and B, and respective radii p and q -, we want to construct the circle tangent to both of the original circles, and tangent to the line beneath them.

Can you prove that the radius of this third circle, denoted r, satisfies

and deduce a formula for r as a function of p and q?

Help below the cut, answers next week.

Hint. Name K, L and M the intersections of the circles with the line below, and use the previous result on each pair of circles to get the lengths KL, KM and LM. One of these lengths is the sum of the two others.

Sangaku Sunday #2

As the tags in a reblog by @todayintokyo indicated, I waffled about what we'll do in this series in the first post without really defining its main object!

Sangaku are wooden tablets on display at Shintô shrines or Buddhist temples in Japan, featuring geometry problems and their solutions, usually without proof. They started appearing in the Edo period, a particular time for the Japanese people and Japanese scientists. The votive role of these tablets has been debated as far back as the Edo period, as indicated by Meijizen who wrote in 1673:

"There appears to be a trend these days, of mathematical problems on display at shrines. If they were true votive tablets (ema), they should contain a prayer of some sort. Lacking that, one wonders what they are for, other than to celebrate the mathematical genius of their authors. Their meaning eludes me."

I feel the debate on their religious role is overrated. If you look at some food offerings at shrines today, I don't think you'll find a prayer on the bottle of tea or pack of rice, as the prayer is made at the time of offering. It likely is the same for sangaku tablets, which went on display with other offerings. But, as Meijizen hinted, they did have another purpose.

Until we expand on that, below the cut is the solution of last weekend's problem.

Place the point H on the line between A and C1 so that the distance between A and C1 is equal to r2. As the lines (AC1) and (BC2) are both perpendicular to the line (AB), they are parallel, and since AH=BC2=r2, HABC2 is a parallelogram with two right angles: it's a rectangle.

So the length we want, AB, is equal to HC2. The triangle HC1C2 has a right angle at the vertex H, so we can use Pythagoras's theorem:

HC1² + HC2² = C1C2²

In this equality, two lengths are known: C1C2=r1+r2, and

HC1 = AC1-AH = r1-r2 (assuming r1>r2, if not just switch the roles of r1 and r2)

Thus, HC2² = (r1+r2)²-(r1-r2)² = 4 r1 r2 after expanding both expressions (e.g. (r1+r2)² = (r1+r2)x(r1+r2) = r1² + 2 r1 r2 + r2²).

Taking the square root yields the result.

Sangaku Saturday #1

In this series, I'll try to put something up every Saturday or Sunday (depending on my plans and/or your time zone), either with a puzzle, a solution or some historical info on sangaku. If people find this interesting (and editing maths on this probably will be), we can work step by step to solving full sangaku problems.

Let's start with a puzzle, one of the most basic tools of the genre. Two circles, with radii r1 and r2, are tangent to each other and tangent to a same line.

This means that the distance between C1 and C2 is equal to r1+r2, the distance between C1 and A is r1, and the distance between C2 and B is r2. Moreover, the lines (AC1) and (AB) are perpendicular, and the lines (BC2) and (AB) are perpendicular.

Can you prove that the distance between A and B is equal to

???

Hints below the cut, questions possible in the reply section.

Take your time, I intend to show the solution next weekend.

Hints. Place the point H on the line between A and C1 so that the distance between A and H is equal to r2. Then the triangle HC1C2 is a right triangle with two sides of known length. A famous theorem gets the third length, which is equal to AB.

Hello! I just saw your post about the conference. I know it's very niche, but I'd love to hear / read more about your sangaku presentation. I actually went back to Konnō Hachiman-gū this afternoon, hoping to see more examples, but no such luck. (I cannot decipher them, of course, but I taught English at a faculty of engineering, and my students could. Sometimes. )

I'll put together something about the shrine, but どうぞお先に。Nudge nudge hint hint.

Hi, thanks for the message!

The presentation was in two main parts: first the historical context of the Edo period and function of sangaku in developing mathematics during that time, and second a closer look at Kashihara Miminashi Yamaguchi-jinja's example with a modern solution. I can't read the sangaku in full, but I have been able to pick out the parts with numbers and compare some of their results with the formulas.

I can probably put together a mini-series at some point. Which parts would you want to hear more about? (That's a general question btw: anyone can reply and add the conversation of course.)

Here we are: Miminashi-yama

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!

Everything is ready for Tuesday! How this particular configuration works, as well as the one below, will be covered - we can talk about it on here too afterwards if anyone's interested.

C'est avec grand plaisir que je présenterai le mardi 16 avril à la Maison Universitaire France-Japon de Strasbourg une conférence sur la géométrie pendant la période d'Edo, avec en support le sangaku de Kashihara. Entre grande Histoire et petits calculs. Lien vers les détails 4月16日(火)、ストラスブール市の日仏大学会館に江戸時代の算額についてコンファレンスをします。楽しみにしています！ Looking forward to giving a conference on Edo-period geometry on 16 April at Strasbourg's French-Japanese Institute. Expect a few posts about Kashihara around then. Has it really been 6 years?...

C'est avec grand plaisir que je présenterai le mardi 16 avril à la Maison Universitaire France-Japon de Strasbourg une conférence sur la géométrie pendant la période d'Edo, avec en support le sangaku de Kashihara. Entre grande Histoire et petits calculs. Lien vers les détails 4月16日(火)、ストラスブール市の日仏大学会館に江戸時代の算額についてコンファレンスをします。楽しみにしています！ Looking forward to giving a conference on Edo-period geometry on 16 April at Strasbourg's French-Japanese Institute. Expect a few posts about Kashihara around then. Has it really been 6 years?...