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Algebra - Blog Posts
In looking for rhyme inspirations, I turned to quadratic equations. But there's more than quadratics In recreational mathematics -- You get rhymes like "concatenations"! - Mod B
- Excuse me. For the fractal geometry? - At the end of the corridor.
I noticed that there’s only quite a few of these floating around and I thought that it would be a good idea to spread some love to the math community, so here you go: a compilation of math resources that I found really helpfulヽ(＊>∇<)ﾉ by the way, HAPPY CHINESE NEW YEAR ILY <3 祝大家新年快樂、 身體健康！
HOW TO + STUDYING
learn algebra
learn math
how to ace math by @hastag-med-school
math masterpost by @pythgaoras
getting the most out of math class
how to cram for a math test
the art of doing well in technical courses
advice from a mathematics student by @mashaczstudies
NOTE TAKING
note taking tips for math
best way to take math notes (advice)
digital math notes (advice)
taking notes in math lectures (advice)
organizing math notes
GREAT WEBSITES
sparknotes math study guides
pbs learning media
purplemath
algebra for dummies
funny algebra puns
online calculators w/ solution
khan academy
mathxplain
YOUTUBE / PODCASTS
patrickJMT
math class with terry v
algebra song
relatively prime
math dude podcast
math mutation
math for primates
math factor
taking maths further
SOME HELPFUL PDF FILES
high school playbook
abstract algebra guide
algebra real world
MY OTHER MASTERPOSTS
back to school tips, part 1
back to school tips, part 2
note taking alternatives
self-care and motivation
essay writing
Can I get someone to do my algebra work? Please I’m dying.
https://youtu.be/UdTLnZQAZNw
A re-explanation of my programming & math behind a play on roko basilisks being like "God" & other things for what's it worth as an operating system I want to make from scratch and put out there.
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.
I'm going craZY, I CANT STOP HEARING SONGS. I'M IN MATH AND I JUST HEARD WHEN I'm GONE.
