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If you haven't watched the video yet, here is the link : https://www.facebook.com/watch/?v=373411481208054 if you don't speak Chinese, the question can be translated into the following : "given that both 4 sided shapes are squares, find the area of the shade". Teacher 柳 used the rotating method to rotate the Right triangle in order to calculate what is requested, but he is lucky that the triangle he is rotating is a right triangle, but what will happen if that triangle is not a right triangle? And this took me a while to prove : when you have 2 squares not overlapping, sharing the same vertex each with their corners, the two triangles adjacent to the 2 squares will have the same area. and so here's the proof: say the side of one of the squares is ℓ, and the other one is 𝑑, let's call the 2 triangles A and B, with their angles α and β. according to ∵ Area of a Triangle = 0.5 × a × b × sinθ then we know that A = 0.5 × ℓ × 𝑑 × sinα B = 0.5 × ℓ × 𝑑 × sinβ According to the diagram, 90° + α + 90° + β = 360° α + β = 180° ∴ sinα = sin(180° - β) = sin180°cosβ - cos180°sinβ = 0 × cosβ - (-1)sinβ = sinβ sinα = sinβ 0.5 × ℓ × 𝑑 × sinα = 0.5 × ℓ × 𝑑 × sinβ A = B (Proof complete) if the given situation is : w and 𝑑 and θ are given just like how the question given by him did we can just simply use A = 0.5 × w × 𝑑 × sinθ = B

What.... what just happened ?! About 1 month ago, an active VRChat World "MEROOM" has been taken down by the creator, MERU. Here is a link to a reddit post: https://www.reddit.com/r/VRchat/comments/qp81tv/meroom_no_longer_available_to_the_public/ According to the comments and replies, there was a thread on Twitter, written in Japanese, complaining about why the creator added a mahjong table {InfoProvidedBy:lastalliance}. Noted that the creator has asked the visitors not to inquire about the world {InfoProvidedBy:lastalliance}. MEROOM is not the only world that faced the same thread, take JP Hub for example, the creator got flooded with complaints and abuse, this caused him to delete the world. according to lastalliance, MERU stated that the biggest reason that he made MEROOM public was he hoped that maybe the visitors would get inspired to create their own worlds. MEROOM is gone for now but he said that he will continue making worlds to play with his friends. What does MEROOM look like when it was there ? the floor, the wall, the ceiling are in polished wood textures, though the support beams are darker. Sometimes the floor could be green. The room has a gigantic video player, a canvas for drawing, a steamboat set, a small wooden house, a few couches, a tabletop for games, a sauna bathroom, tons of pickups, and perhaps even more. Many VRChat citizens (including me) had a lot of precious memories there, and many others met other citizens and become friends. It is true that this world was so comfy that this world got featured at the "Hot" section in the Worlds tab, and so many newcomers would visit this masterpiece, this led it to become many citizens' first world where they met their VRChat friends/groups. Here are some photos taken at MEROOM before it was taken down: 14th June 2021 19th July 2021 Luckily, someone made a replica of MEROOM, so we can at least have a slight idea of MEROOM: https://vrchat.com/home/world/wrld_2b76e44d-acd4-4002-9b1a-90d5ffaea79b *COUGH COUGH..., EXCUSE YOU... Internet gangsters Often I will go world-hopping to look for ideas and inspirations. Whenever I see a world that has issues, I know that it is not exactly the creators are lazy or bad at it, it is also possible that they are bounded by the hardware/software/tools they have at that moment. I will only go "AW MAN~" when basic technical issues are found, for example, when there are no colliders for chairs, tables and couches, therefore I cannot stand on them to see from a higher angle. This can be easily fixed by adding colliders to it. But it is possible that the creator doesn't want to have colliders there for some reason, or maybe he/she is learning. Another common issue is UNITY CAN'T BAKE LIGHTS sometimes, that is why sometimes you can find weird shadows or unreasonable lightmap texels. The only thing the creator can do is adjust the parameters such that the machine/hardware he/she has can deal with the numbers and still has the best result. There is nothing much else the creator can do. Maybe there are gangsters out there saying "then get a better hardware then", well then, either you help him/her to get one or shut up. Some gangsters love to complain about worlds and do nothing to it, if they are just complaining about the world, it is still slightly toleratable, but abusing the creator is where they cross the line. I wish I could tell the gangsters: Did you create anything? No ? You didn't contribute, and yet you still complain about it. Yes ? Even if you had created worlds/avatars/models/prefabs or anything, this is his/her world after all, he/she gets to decide everything here, he/she didn't make this world just to satisfy only you. Who do you think you are ? If you don't like a feature here and you want to change it, go make your own world but with the features you wanted, go ahead, otherwise, don't complain, suggest instead, or else, shut up. If you saw an issue, don't complain about it, either you report it to him/her or come up with a solution and tell him/her, or else, stop being annoying. Of course, there are many creators out there trying to make this platform better and better. I would say I am lucky that I can also create worlds for M2007U and the VRChat citizens. It is possible that I may take long rests but I will never stop (until my physical form is not functioning to do so). To those who have been creating for a certain amount of time (including some of my VRC friends): My lord/lady/fellow friends, no matter you create worlds/avatars/models/prefabs/code or you are a content creator on VRChat, thanks for making this platform a better place, every day. Keep on the good work, I am excited and looking forward to experiencing what you have all created, and as a creator as well, I will keep on going as well. And I am sure, I am not the only one who is thrilled. OwO (eg: vannthedev, totobono4, Jiyue, meltingarmymen, NerdKoopa, Qhala, ImLexz, SpookyGhostBoo, Meru, Blue-kun, MLucifer, CyanLaser, PhaxeNor, Squid, Fiona, Vowgan, Varneon, Lakuza, Fins, and so much more that I'm unable to name all of them here...) and to those who are new to VRChat creations: NEWO saplings, thanks for deciding to contribute to this fascinating platform. This is a long journey, but I can assure you, it is worth it. Perhaps you are doing this for yourself, or maybe you are doing this to please others (not recommended thou O_O), or maybe this is your hobby, whatever reason it is, thanks for deciding to become a creator to make this platform progress. And who knows, maybe you would become a lord/lady as well. PS: to all the creators Sometimes, feedbacks and suggestions are important, they can encourage you to think from another perspective, and maybe you can learn something new and also make what you have created a better version of itself. But I have to mention that not all suggestions are mandatory to apply, you will still have to make decisions, whether to apply the suggestion, or alter them a little, or identify them as "not suitable suggestions" Welp, that's a lot to cover. One thing, this is not just only about VRChat, it is actually about the whole world, and the internet. Similar rules apply: experiencing an issue? report it, or suggest it, or come up with a solution or shut up, don't just complain, and never, ever, ever, abuse/cyberbully etc. Because you are not wearing their shoes. Anyway, Hail M2007U, creators, technology, peace, and harmony.

[English version] According to SinChew Daily and Goxuan, some students cried on the spot, vomitted, table flipped, or asked the teacher for a longer duration during the Primary School Leaving Examination (PSLE) held by the Ministry of Education of Singapore. One question goes like this: Helen and Ivan have the same number of coins. Helen has some number of 50cents and 64 20cents, with a total mass of 1.134kg Ivan has some number of 50cents and 104 20cents (1) Who has more money than the others ? And by how much ? (2) say each 50cent is 0.0027kg more than a 20cent, what is the mass of Ivan total coins Solution: (Desktop Site is recommended) In the beginning, it seems terrifying, that is one of the main objectives of the exam. But luckily, since the question is a "relative type question" (a question that asks what is the relationship between parameters) rather than an "absolute type question" (a question that asks what is the absolute value of each parameter), we can try to cancel out what both Helen and Ivan have in common, the rest is the difference and the answer. let's turn the question into something we can see, and simplify it (use smaller numbers and understand the concept first) let's say "🔴" is a 50cent and "⚪" is a 20cent so we should have something like: H: 🔴🔴🔴🔴 .... 🔴🔴|🔴🔴🔴🔴🔴🔴|⚪⚪ .... ⚪⚪ I: 🔴🔴🔴🔴 .... 🔴🔴|⚪⚪⚪⚪⚪⚪|⚪⚪ .... ⚪⚪ <------L part------->|<----M part--->|<----R part----> as you can see, the question is actually just asking for differences, (obvious for question 1 by indirect for question 2, don't worry, I'll explain) so what matters, is not the L part, is not the R part, instead, it should be the difference between Helen and Ivan, the M part. (1) When we want to find out who has more money, both Helen's L part and Ivan's L part will cancel each other and both Helen's R part and Ivan's R part will cancel each other what is left is the M part. so the total difference of cash should be = the width of M Part × the cash difference of each coin So let's go back to the original question. So the width of M Part should be 104 - 64 = 40, and the picture should originally look like this: H: 🔴🔴🔴🔴 .... 🔴🔴|🔴🔴🔴....🔴🔴🔴|⚪⚪ .... ⚪⚪ I: 🔴🔴🔴🔴 .... 🔴🔴|⚪⚪⚪....⚪⚪⚪|⚪⚪ .... ⚪⚪ ? | 40 | 64 <------L part------->|<------M part----->|<----R part----> so the difference is 40 x (50c - 20c) = 40 x 30c = 120c since we know that Helen is has less 20cents then Ivan, meaning Helen has more 50cents then Ivan, meaning Helen has more money than Ivan, and Helen has 120cents more than Ivan (question 1 solved) (2) we can also apply the same Idea here, instead of cash value, we are dealing with mass. so the total difference of total mass should be = the width of M Part × the mass difference of each coin so the difference of mass in the original question should be 40 x 0.0027kg = 0.108kg since we know that the total mass of Helen's coins is 1.134kg, then the total mass of Ivan's coins should be 1.134kg - 0.108kg = 1.026kg (question 2 solved) Often, a question that is a "relative type question" will allow you to cancel common values, and what matters is the difference. PS: I am not academically good at maths, I am like other students: failing my math exams all the time during high school. Here is the original link with no answers: https://goxuan.my/trending/entertainment-news/studentcryexam [中文版] 据星洲日报，新加坡举行小学毕业会考（PSLE），让部分应试的12岁学童当场大哭、呕吐、掀桌、或向监考老师乞求更多答题时间。 其中一题如下： 海伦（Helen）和伊凡（Ivan）拥有数量相同的铜板。 海伦有一些50分铜板和64枚20分铜板，铜板重量为1.134公斤。 伊凡有一些50分铜板和104枚20分铜板。 请问： 1. 谁的钱比较多？多出多少钱？ 2. 假设每枚50分铜板比20分铜板重0.0027公斤，伊凡的铜板最多有几公斤重？ 解题： （建议使用桌面型模式） 刚开始时，会觉得这一题很恐怖，但当然这是考试题目的其中一个目标。 不过幸运的是 这两个题目都是”相对性的题目“，这样的话，我们可以抵消海伦和伊凡共同有的东西，剩下有差别的就是答案。 我们可以把题目视觉化，同时做一些必要的简化 （使用比较小的数字，先掌握好概念） 假设"🔴" 是一枚50分铜板，则"⚪"是一枚20分铜板。 那么情况大致上会是这个样子： H: 🔴🔴🔴🔴 .... 🔴🔴|🔴🔴🔴🔴🔴🔴|⚪⚪ .... ⚪⚪ I: 🔴🔴🔴🔴 .... 🔴🔴|⚪⚪⚪⚪⚪⚪|⚪⚪ .... ⚪⚪ <------L 部分------->|<----M 部分--->|<----R 部分----> 题目问的是海伦和伊凡的不同之处， 由此可见，关键不是 L部分，也不是R部分，而是 两人不一样的M 部分 （1） 当我们在探讨”谁的钱比较多？“时， 海伦的 L部分 和 伊凡的 L部分 会互相抵消（因为它们都是一样的，不用比较） 海伦的 R部分 和 伊凡的 R部分 也会互相抵消 剩下的 就是 M部分 所以 两人相差的钱 = M部分的宽度 × 价钱差别 回到原先的题目的话， M部分的宽度应该是 104-64=40，那么示意图大致上应该是： H: 🔴🔴🔴🔴 .... 🔴🔴|🔴🔴🔴....🔴🔴🔴|⚪⚪ .... ⚪⚪ I: 🔴🔴🔴🔴 .... 🔴🔴|⚪⚪⚪....⚪⚪⚪|⚪⚪ .... ⚪⚪ ? | 40 | 64 <------L 部分------->|<------M 部分----->|<----R 部分----> 所以 两人相差的钱 = 40 x (50分 - 20分) = 40 x 30分 = 120分 既然海伦的 20分铜板比 伊凡的少； 也就是说海伦的 50分铜板比 伊凡的多， 也就是说海伦的钱比伊凡的多，多了 120分。(第一题解答完毕) （2） 我们也可以使用相同的概念。 但是这一题我们对付的不是 价钱，而是质量。 所以 两人铜板相差质量 = M部分的宽度 × 铜板质量差别 回到题目 两人铜板相差质量是 = 40 × 0.0027kg = 0.108kg 既然 50分铜板比20分铜板质量多，且海伦有的50分铜板比伊凡多， 海伦的铜板质量比伊凡的大。 所以伊凡的铜板质量因该是 1.134kg - 0.108kg = 1.026kg（第二题解答完毕） 在大部分时候，相对性的题目将会允许我们抵消一些东西，而剩下来的就是答案。 PS:我数学没有你想象中的好，我也和其他学生一样，在中学时期常常”烤烂“数学考试 原帖子链接在这里： https://goxuan.my/trending/entertainment-news/studentcryexam

Chapter 1 when you are trying to simplify fractions, chances are you are having a hard time, you don't know what to factorize. Here are some of the techniques to know if a number is divisible by a simple factor, say from 2 to 9. Note that we are going to deal only with base-10. These tests can be divided into 3 types : Dividing Carries, Digit Sums & Complex Equations Type 1: Dividing Carries the main idea is that some factors can divide some power of ten perfectly, so we don't need to worry about how many that power of ten we have, instead, can we divide what is not that power of ten ? these factor numbers are: 2,4,5 and 8 [dividing by 2] since 10/2 = 5, we know that, no matter how many 10s we have, they always can be divided by 2, but it is unsure for the ones. Divisible test: if the ones can be divided by 2 (a.k.a. if the ones is 0,2,4,6 or 8), then the original number can be divided by 2. Example: 1356 --(look at ones)--> 6 (can be divided by 2) ----> divisible by 2 1357 --(look at ones)--> 7 (can't be divided by 2) ----> not divisible by 2 [dividing by 4] since 100/4 = 25, we know that, no matter how many 100s we have, they always can be divided by 4, but it is unsure for the tens and ones Divisible test: if the tens and ones can be divided by 4, then the original number can be divided by 4. Example: 1316 --(look at tens and ones)--> 16 (can be divided by 4) ----> divisible by 4 1357 --(look at tens and ones)--> 57 (can't be divided by 4) ----> not divisible by 4 [dividing by 5] // the simplest factor since 10/5 = 2, we know that, no matter how many 10s we have, they always can be divided by 5, but it is unsure for the ones Divisible test: if the ones can be divided by 5 (a.k.a. if the ones is 0 or 5) then the original number can be divided by 5. Example: 1355 --(look at ones)--> 5 (can be divided by 5) ----> divisible by 5 1370 --(look at ones)--> 0 (can be divided by 5) ----> divisible by 5 2689 --(look at ones)--> 9 (can't be divided by 5) ----> not divisible by 5 [dividing by 8] since 1000/8 = 125, we know that, no matter how many 1000s we have, they always can be divided by 8, but it is unsure for the hundreds, tens and ones Divisible test: if the hundred, tens and ones can be divided by 8, then the original number can be divided by 8. Example: 24000 --(look at hundred, tens and ones)--> 000 (can be divided by 8) ----> divisible by 8 24832 --(look at hundred, tens and ones)--> 832 (can be divided by 8) ----> divisible by 8 24257 --(look at hundred, tens and ones)--> 257 (can't be divided by 8) ----> not divisible by 8 Type 2: Digit sums the main idea is assuming a certain criterion then we can work out something from that. these factors numbers are : 3 and 9 [dividing by 3] say we call each digit d0 (ones),d1 (tens),d2 (hundreds),d3 (thousands), ... dn example : for the number 4869, d3=4, d2=8, d2=6, d0=9 let's set the criteria: sum of the digits is a multiple of 3 (divisible by 3) then formular1: d0 + d1 + d2 + d3 + d4 + ... = 3n we also know that 9d1 + 99d2 + 999d3 + 9999d4 + ... is a multiple of 3, say 3m then formular2: 9d1 + 99d2 + 999d3 + 9999d4 + ... = 3m when we add both formulas together, we would have d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... = 3n + 3m = 3 (n + m) (tada, a multiple of 3), where d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... is the original number we are testing Divisible test: if the sum of the digits is a multiple of 3, then the original number can be divided by3 Example: 1368 ----> 1+3+6+8 ----> 18 (divisible by 3) ----> divisible by 3 [dividing by 9] the proof is similar, but instead of setting the sum of the digits is a multiple of 3, this time the sum should be a multiple of 9, i.e. formular1: d0 + d1 + d2 + d3 + d4 + ... = 9n and for sure we know that formular2: 9d1 + 99d2 + 999d3 + 9999d4 + ... = 9m add both equations together we have d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... = 9n + 9m = 9 (n + m) (tada, a multiple of 9), where d0 + 10d1 + 100d2 + 1000d3 + 10000d4 + ... is the original number we are testing Divisible test: if the sum of the digits is a multiple of 9, then the original number can be divided by 9 Example: 1368 ----> 1+3+6+8 ----> 18 (divisible by 9) ----> divisible by 9 Type 3: Complex Equations These factors are combinations of previous factors OR individual factors which is totally different from the others, these factors are 6 and 7 [dividing by 6] since dividing by 6 can be treated as dividing by 2 then by 3 so it is actually the combination of 2 and 3 Divisible test: if the ones can be divided by 2 (a.k.a. if the ones is 0,2,4,6 or 8) AND if the sum of the digits is a multiple of 3, then the original number can be divided by 6. Example: 1368 --(look at ones)--> 8 (divisible by 2) ----> divisible by 2 1368 ----> 1+3+6+8 ----> 18 (divisible by 3) ----> divisible by 3 and finally, weirdest of all, the number 7 [dividing by 7] first let's say the number of tens we have is d1, and the digit of ones is d0, and the original number is K. example : if K = 8641969, then d0 = 9, and d1 = 864196 let's set a criterion : K is a multiple of 7 so we know that K = 10d1 + d0 = 7n, where n is an interger 10d1 + d0 - 21d0 = 7n - 21d0 10d1 - 20d0 = 7n - 21d0 10(d1 - 2d0) = 7(n - 3d0) since 10(d1 - 2d0) is a multiple of 7, then 10(d1 - 2d0) must somehow contain the factor 7, so d1-2d0 must contain the factor 7, i,e, d1 - 2d0 = 7m where m is an integer Divisible test: if the ((quantity of tens) minus the (double of ones)), is a multiple of 7, then the number K is divisible by 7, or if d1 - 2d0 = 7n the number is divisible by 7 Example: 8641969 -> 864196 - 18 = 864178 -> 86417 - 16 = 86401 -> 8640 - 2 = 8638 -> 863 - 16 = 847 -> 84 - 14 = 70 -> 7 - 0 = 7 (divisible by 7) so to gather up what we have so far, here's a table: [Added on 2021-10-22 and 2021-10-23] Chapter 2 Extra Numbers I have tried 2 new numbers : 13, and 11, turns out, we can use almost the same method to find the determinating equation, so let's get started: say the original number is K, d1 is the number of tens we have and d0 is the number of ones. note that d1 ≥ 1 and d1 ∈ W, d0 ∈ W as well. [dividing by 13] then let's set the criterion as : when K is a multiple of 13, say 13 n then we know that: K = 10d1 + d0 = 13n then we can also work out the following : 10d1 + d0 = 13n 10d1 + d0 - 91d0 = 13n - 91d0 10d1 - 90d0 = 13(n-7d0) 10(d1 - 9d0) = 13(n - 7d0) since 10(d1-9d0) is a multiple of 13, then the term 10(d1-9d0) must contain the factor 13, therefore: d1 - 9do = 13m, m ∈ W Divisible test: if the ((quantity of tens) minus the (nine times of ones)), is a multiple of 13, then the number K is divisible by 13, or if d1 - 9d0 = 13n the number is divisible by 13 Example: 59538232 -> 5953823 - 2x9 = 5953805 -> 595380 - 5x9 = 595335 -> 59533 - 5x9 = 59488 -> 5948 - 8x9 = 5876 -> 587 - 6x9 = 533 -> 53 - 3x9 = 26 (divisible by 13) [dividing by 11] surprisingly, this is much simpler. let's set the criterion as: when K is a multiple of 11, say 11 n then we know that: K = 10d1 + d0 = 11n then we can also work out the following : 10d1 + d0 = 11n 10d1 + d0 - 11d0 = 11n - 11d0 10d1 - 10d0 = 11(n-d0) 10(d1 - d0) = 11(n-d0) since 10(d1-d0) is a multiple of 11, then the term 10(d1-d0) must contain the factor 11, therefore: d1 - do = 11m, m ∈ W Divisible test: if the ((quantity of tens) minus the (quantity of ones)), is a multiple of 11, then the number K is divisible by 11, or if d1 - d0 = 11n the number is divisible by 11 Example: 54745394 -> 5474539 - 4 = 5474535 -> 547453 - 5 = 547448 -> 54744 - 8 = 54736 -> 5473 - 6 = 5467 -> 546 - 7 = 539 -> 53 - 9 = 44 (divisible by 11) Chapter 3: Similarly, we can use the same technique to come up with different algorithms to check if a number is divisible by something... therefore we an try to generalize them: say the number we want to test is "K", the number system base we are dealing with is "b", the numbers of bases we have is d1, and the numbers of ones we have is d0 and the number K will be divided by "p" where K,b,d1,d0,p ∈ W so we would have: K = d1b + d0 = pq, where q ∈ W by adding "- (βb + 1)d0" on both sides, where β ∈ W, we would have d1b + d0 - (βb + 1)d0 = pq - (βb + 1)d0 d1b - βbd0 = pq - (βb + 1)d0 from the examples from the previous parts, we know that (βb + 1) must be a multiple of p, so let (βb + 1) = pr, where r ∈ W β = (pr - 1)/b so d1b - βbd0 = pq - (βb + 1)d0 (d1 - βd0)b = pq - prd0 (d1 - (pr-1)/b)b = p(q-rd0) welp, this is still a draft though working is still in progress... I will add more details and content when more progressions are made. basically, we have the following formula... for now: βb +1 = pr I hope this helps you a lot, if you are a primary school maths teacher or a tuition teacher, why not share this with your students ? if you have other division tests, share it in the comment section below.

Starting from March 2021, I played VRChat until now, the avatar I use as my normal virtual form is a Kon, a short cute HalfHuman Half-fox character designed by Midori. Because of that, I am short to other users with average height, sometimes other players just walk to me and asked: "Why are you so smol ?". And I will reply to them: "Everything here is relative. Size is relative, time is relative. There is nothing which is the biggest, only bigger; there is nothing is the smallest, only smaller. When you say you are bigger, always remember that there will be someone or something bigger than you are; and since I am smaller than you are, of course, there is someone smaller than I am" Since length, location, time, mass, velocity, energy, temperature, etc are relative, we need a reference point. That's why we have units, That's why we need a reference. We are trying to describe properties around us relative to that unit or perspective. Still don't get it? Here are some more concrete examples: - I have a pencil which has a length of 8 times a centimeter, a chosen specific length by the society - My water is hotter than ice 28 degrees celsius. - That roller coaster has that certain amount of potential energy relative to that certain height. - I have Kinetic energy to the sun, but I have no Kinetic Energy to the Earth If you are a VRChat citizen, share what you have learned down in the comment section below.

One day I assigned a question to my students : If Aaron can walk 1.2 meters for each second, how long will it take for him to walk 25000cm. And the next lesson we discussed the question, and I asked them "how long the distance Aaron can walk within a second ?". I was expecting them to say "1.2 meters" since it is so easy and it is stated directly in the question, but instead, one of them said 250seconds". Of course, I have to correct them, but after the lesson, I think to myself, if what they said is a possible situation, what would this mean? Most of the time we would think that "time moves in one direction", this means that time is one dimensional. I can simply draw a time line. When I say "Aaron can walk 5 seconds in a second", This means that time can be more than one dimension. if we try to plot the statement onto a graph, what we have is a "time plane", in fact, it is possible that we can go to higher dimensions. This can also explain why time is relative... I guess. Let's set up a thinking experiment : there are 2 worlds with perpendicular timelines, A and B, Aaron is holding an hourglass, with the hourglass he walked 3 seconds along with world A, then walked 5 seconds along world B, Then he continued walked 4 seconds along with world A. How would the two worlds observe him? From the perspective of world A, At first, the 3 seconds seems normal, then the 5 seconds of sand in the hourglass suddenly flowed, then the hourglass flows normally for 4 seconds. From the perspective of world B, At first, 3 seconds of sand in the hourglass suddenly flowed, then the hourglass flowed normally for 5 seconds, then 4 seconds of sand suddenly flowed. From Aaron's perspective for the first 3 seconds, time is flowing normally for world A, but time is frozen for world B then for 5 seconds, time is frozen for world A, but time is flowing normally for world B then for 4 seconds, time is flowing normally for world A, but time is frozen for world B We can actually plot a graph like this: Let's Setup another experiment : Adam is holding a stopwatch and observing Celia from the perspective of world A Bob is holding a stopwatch and observing Celia from the perspective of world B Celia is holding a stopwatch and traveling along world A and B for some angle or proportion When Celia's stopwatch counted 5 seconds, Adams counted 4 and Bob's counted 3; When Celia's stopwatch counted 10 seconds, Adam's counted 7 as well as Bob's To Celia, she traveled 4 seconds along with world A in 5 seconds and traveled 3 seconds along world B in 5 seconds at first, then the other way around. To her, for the first 5 seconds, Adam slows down, and Bob is even slower for the second 5 seconds, Adams becomes slower, and Bob becomes slightly faster but still slower than Celia To Adom, At the beginning, Bob vanishes for 4 seconds, Celia is experiencing time faster, then for 3, Celia speeds up even faster To Bob, At the beginning, Adam vanishes for 3 seconds, Celia is experiencing time very fast then for 4 second, Celia is still "faster" but slows down a bit As you can see, Each of them experiences time at a relatively different rate. My 5 seconds, may not be your 5 seconds. Time is relative. In the thought experiment above, we are just dealing with 2D time, if you have nothing to do today or you are completely free today, think about this: What can we observe if time is 3D or higher ?

M2007U-E2016-A2017-S2017-M2021_SMT-20200618x01 You must have played with some questions which asked you to count the quantity of some sort of object or shapes, took this diagram for example, how many triangles can you find from this diagram Most of the people will just try to search all the triangles without any order, it is possible that they may overcount or miss some of the triangles. "Thankfully, there is a way to count the triangles according to an order", yup, I said this when I shared this question to my classmates when I was 14 from my big family Whatsapp group, though this question is forwarded by my father. ​ Please note that this article is going to discuss ONLY the triangles which are similar to this type Section 1 Let's look at the first part of the problem: a triangle (let's call this the General Yellow Triangle) with a bunch of lines connected from an angle to the base of the triangle Let's define that a "Y Line"(or a "Yellow Line") is a line connected from the Angle (or let's call it "Y Angle") to the very base of the triangle. ​ Let's define that a "Base Segment" is a line segment 1) that is not connected to the Y angle 2) associated EXACTLY with 2 yellow lines (no other yellow lines touching it) You can see each base segment is accompanied by a thin white rectangle. 3) a part of the base of the General Yellow Triangle Let's match all the Y lines to a base segment by coloring it with the same color, notice that ONLY 1 Y line doesn't have a base segment. ​ Key point : Numbers of Base Segment = Number of Y Lines - 1 = Y - 1 where "Y" is the Qty of Y Lines A base of a constructed triangle in the General Yellow triangle can be a combination of Base Segments Inside a base of a constructed triangle, Let's call the Base Segment on the left as "Base Head Segment" ( Orange Rectangle) , and the rest are "Base tail Segments" (White Rectangles) Let's assign Y - 1 base Segments (all Base Segments we have so far) with Y - 1 brackets, each Base Segment has its own brackets, each Base Segments can be a Base Head Segment or a Base Tail Segment at the same time (except only the one on the left which only can be the Base Head Segment) ​ The bracket can be stretched ONLY to the right-hand side to cover multiple Base Segments. Each bracket will construct a triangle with the width of Base Segments covered with the 2 Y lines connected with the base of the constructed triangle. PS : Each White Rectangle is a Base Segment Let's see how many triangles can be constructed with the brackets with a width of 1 Base Segment (1 Head 0 Tail) The Yellow brackets should cover 1 Base Segment. Notice that each bracket can construct a triangle with a base of 1 Base Segment. ​ let's jot that down as: Δ(1bS) = (Y - 1) where "Δ(WbS)" means Qty of Triangles with a base of W Base Segments where "Y" is the Qty of Yellow Lines, in this case, Y - 1 is the Qty of Base Segments Now tet's see how many triangles can be constructed with the brackets with a width of 2 Base Segments (1 Head 1 Tail) The Yellow brackets should cover 2 Base Segments. Notice that each bracket can construct a triangle with a base of 2 Base Segment, except 1 colored red because it doesn't have a Base Segment as it's Base Tail Segment. ​ let's jot that down as: Δ(2bS) = (Y - 1) - 1 Now tet's see how many triangles can be constructed with the brackets with a width of 3 Base Segments (1 Head 2 Tails) ​ The Yellow brackets should cover 3 Base Segments. Notice that each bracket can construct a triangle with a base of 3 Base Segments, except 2 colored red, because there are not enough base segments as their Base Tail Segments. ​ let's jot that down as: Δ(3bS) = (Y - 1) - 2 Now tet's see how many triangles can be constructed with the brackets with a width of 4 Base Segments (1 Head 3 Tails) The Yellow brackets should cover 4 Base Segments. Notice that each bracket can construct a triangle with a base of 4 Base Segments, except 3 colored red, because there are not enough base segments as their Base Tail Segments. ​ let's jot that down as: Δ(4bS) = (Y - 1) - 3 If we keep on going like this, we can get ​ Δ((Y-1)bS) = (Y - 1) - (Y - 2) = Y - 1 - Y + 2 = 1 If you look closer enough, you can know that: Δ( W bS) = (Y - 1) - (W - 1) = Y - 1 - W + 1 = Y - W If we collect all the notes we have, you can see the pattern ​ Δ( 1 bS) = (Y - 1) - (1 - 1) = Y - 1 Δ( 2 bS) = (Y - 1) - (2 - 1) = Y - 2 Δ( 3 bS) = (Y - 1) - (3 - 1) = Y - 3 Δ( 4 bS) = (Y - 1) - (4 - 1) = Y - 4 ​ ... and so on until ​ Δ( (Y-4) bS) = (Y - 1) - ((Y-4) - 1) = 4 Δ( (Y-3) bS) = (Y - 1) - ((Y-3) - 1) = 3 Δ( (Y-2) bS) = (Y - 1) - ((Y-2) - 1) = 2 Δ( (Y-1) bS) = (Y - 1) - ((Y-1) - 1) = 1 ​ these are all the triangles that can be constructed from the General Yellow triangle. If we sum that up, we have ​ Δ(all with Y angle) = 1 + 2 + 3 + 4 + ... + (Y-4) + (Y-3) + (Y-2) + (Y-1) = (1 + (Y-1))(Y-1)/2 = (Y)(Y-1)/2 Section 2 Let's look at the second part of the problem: a triangle (Let's call this the General Cyan Triangle) with a bunch of lines connected from side to side Note that the lines connected from side to side will not intersect each other. And these lines are at the opposite of a shared angle, let's call this angle the C angle (Colored Cyan) and the Lines as "White Lines". There are 2 lines that are adjacent to the C angle (colored Cyan), so obviously enough let's call these 2 lines the Cyan Lines. Let's define that a "Side Segment" is a line segment 1) that is connected to the C angle when extended or is within the Cyan Lines 2) associated EXACTLY with 2 White lines, except the one which is directly adjacent with the C angle, they only have 1 White Line Intersected You can see each side segment is accompanied by a thin white rectangle. 3) a part of the side of the triangle The situation is much simpler than the first part of the problem. If we try to match all the white lines to a Side Segment, Amazingly all of them are matched. So we can get Number of Side Segments = Number of White Lines Let's assign each segment with a bracket, this bracket will construct a triangle with the angle C angle, notice that all triangles in this huge triangle will have to contain the C angle. If we stretch all the brackets from their own "first" Side Segment to the "last" Side Segment, we can know that no brackets are red, meaning: ​ Number of Triangles it can make = Number of Side Segments = Number of White Lines Section 3 Now. let's combine the General Yellow Triangle and the General Cyan Triangle together. ​ Noticed that EACH CYAN LINE CAN CREATE A GENERAL YELLOW TRIANGLE WITH THE Y ANGLE ON THE TOP AND EACH YELLOW LINE CAN BE A WHITE LINE TO THOSE GENERAL CYAN TRIANGLES WITH THEIR C ANGLE COLORED We know that in a General Yellow Triangle, there are Y(Y-1)/2 triangles, then X cyan Lines can make XY(Y-1)/2 Triangles ​ Let's match each C angle to a Cyan Line, you will find that there is only 1 Cyan Line that is not matched, so that Line is unable to form a General Cyan Triangle ​ so we know that the Number of General Cyan Triangles = X - 1 where X is the Qty of Cyan Lines. ​ Now just focus on a complete General Cyan Triangle, let's match each Yellow line to the White Lines, noticed that 1 of the Yellow lines is not a White Line available to the General Cyan Triangle, in fact, intersecting the C angles So the Number of Available White Lines in these General Cyan Triangles = Y -1 since 1 General Cyan Triangle can make Y - 1 triangles, then X-1 General Cyan Triangles can make (X-1)(Y-1) Triangles Finally, the sum of XY(Y-1)/2 Triangles & (X-1)(Y-1) Triangles is the total triangles it can make from this mess... ​ so the Formula would be Qty of Triangle = XY(Y-1)/2 + (X-1)(Y-1) where X is the Qty of Cyan Lines (Lines opposite to the Y angle), Y is the Qty of Yellow Lines(Lines connecting with the Y angle) ​ CAUTION: This formula this only applicable to ONLY these pattern-like triangles, especially the Cyan lines must connect in a zig-zag way, you have to extremely be careful to use it since small changes of the shape may alter the situation. If the situation alters try to start with how many Y angles and C angles are there, and see if there are any new features. Now let's get back to the provided question we have, as you can see we have 4 Yellow lines and 3 Cyan Lines, So we know that X = 3, Y = 4, so plug that in, we get 24 And here's all of them ​ 1st row : 1st General Yellow Triangle 2nd row : 2nd General Yellow Triangle 3rd row : 3rd General Yellow Triangle 4th row, 1~3 : 1st General Cyan Triangle 4th row, 4~6 : 2nd General Cyan Triangle

M2007U-E2016-A2017-S2017-M2021_SMT-20200425x01 Let's draw Right Triangle with a Height of Th, Width with Tw, and install a rectangle inside it. In this article, we are going to prove the largest rectangle we can make in this triangle. Let's say the Height of the Triangle is U, and the width of the triangle is V Let's try to find the relation between H and W ∵ W // X, H // Y ∴ α = β ∴ tanα = tanβ Y / W = H / X H = XY / W H = (V - W)(U - H) / W HW = (V - W)(U - H) HW = VU - VH - UW + HW 0 = VU - VH - UW VH = VU - UW H = U - UW/V ∴ the area of the rectangle would be : A = WH A = W(U - UW/V) A = (UW) - (UW²)/V A = (U)W - (U/V)W² so we can treat W as an input and the area A would be the output, both U and V are fixed constants To make sure that the area of the rectangle has a maximum area, we just need to differentiate the equation twice and see if the result is negative or not. dA/dW = U - 2(U/V)W d²A/dW² = 0 - 2(U/V) d²A/dW² = - 2(U/V) since U and V are positive, then U/V > 0, then - 2(U/V) < 0, so the Area of the Rectangle A has a maximum value. When A is the maximum value, dA/dW = 0 ∴ when dA/dW = U - 2(U/V)W = 0 1 - 2(1/V)W = 0 1 = 2(1/V)W 1/2 = W/V W = (1/2)V surprisingly, when W is half of the width of the triangle, the area of the rectangle is at its maximum, we and also substitute when A = Amax & W = (1/2)V to find that H = U - UW/V H = U - U((1/2)V)/V H = U - (UV)/(2V) H = U - (1/2)U H = (1/2)U Did you know ? This proof is inspired by the main task of filling up a Triangular region in Minecraft by using the command /fill in order to achieve the maximum efficiency

M2007U-E2016-A2017-S2017-M2021_SMT-20200103x01 Teachers, Textbooks, Articles, the Internet always say that any real number is forbidden to be divided by 0. Some of us cannot understand why, some of us just accept it as a fact because we want to get good grades. In this article, we are going to discuss why it is forbidden. ​ One critical reason that it is a mathematical error is the definition of "Division" Definition 1 : To find the Quantity of How many times do we need to minus a certain number from the product until it reaches 0. Example : 20 / 4 = ? ​ 20 - 4 = 16 16 - 4 = 12 12 - 4 = 8 8 - 4 = 4 4 - 4 = 0 ​ it takes 5 times to minus 4 from the product 20, so 20/4 = 5 What if : 20 / 0 = ? ​ 20 - 0 = 20 20 - 0 = 20 20 - 0 = 20 ... and so on ​ as you can see, if we keep going on doing this infinitely, we are not even closer to 19 or something smaller than 20 Definition 2 : To find the Quantity of How many elements inside a group where all elements in all groups add up to the product For example: if we need to put 20 apples into 4 bags and all bags must have the same amount of apples, how many apples does each bag has? ​ 20 / 4 = 5 Classically, 5 apples each What if: we need to put 20 apples into 0 bags and all bags must have the same amount of apples. ​ At first, we don't have any bags since there are 0 bags, then how come we can put any apples inside them? Therefore this is a logical error. Definition 3 : To find the other factor of the product when one factor is given. The problem is, there are no numbers that can multiply with 0 and end up with a number that is not 0 since any number that multiplies with 0 ends up being 0 By using different definitions, you will end up with different answers But other than using definitions to explain, are there any other ways to prove that dividing a number by 0 is not a valid move? And the answers are obviously yes. Proof 1 : Prove by Contradiction Actually a lot of math pranks uses this technique to false proof something, but here is the general idea. Usually in algebra we "move" terms by using Addition, Subtraction, Multiplycation or Division or any other legitimate ways to arrange what is on both sides of an equation. for example : ? - 8 = 2, we can add 8 on both sides, then we will have ? - 8 + 8 = 2 + 8, tidy it up then we can have ? = 10. Let's just assume that 1) we are allowed to divide by 0, 2) and let A and B be any real numbers which are not equal, since A x 0 = 0 and B x 0 = 0, then we can conclude A x 0 = B x 0, but if we divide both sides by 0, we will have A = B, which contradicts with what we have assume earlier. So dividing a number by 0 is not a legitimate move. Proof 2 : Prove by Contradiction again First we all know that N x 0 = 0 no matter what N is. But let's just assume that a number, N, when divided by 0, is K, and we will have N/0 = K This is where we can end up with a lot of weird things... If we were to multiply 0 on both sides, we will have N/0 x 0 = K x 0 N = K x 0 This means "there is a number , multiply by 0, is not 0, is N instead", which does make any sense. If we continue, we can divide K on both sides, and we will have N / K = K x 0 / K N / K = 0 seriously... what the heck is that ? O_O

M2007U-E2016-A2017-S2017-M2021_SMT-20160929x02 Let's draw a circle with a radius of r, and install a rectangle inside it. From the video, the 2 angles α and β have a total of 180 degrees. In this article, we are going to prove that the largest rectangle it can have inside is actually a square. by using the cosine rule, we can know that h² = 2r² - 2r²cosα and w² = 2r² - 2r²cosβ w² = 2r² - 2r²cos(180° - α) w² = 2r² - 2r²(cos180° cosα - sin180°sinα) w² = 2r² - 2r²((-1) cosα - (0)sinα) w² = 2r² + 2r²(cosα) therefore, to get the area of the rectangle, we can have h²w² = (2r² - 2r²cosα)(2r² + 2r²cosα) h²w² = (2r²)² - (2r²cosα)² h²w² = 4r⁴ - 4r⁴cos²α hw = (4r⁴ - 4r⁴cos²α)^(1/2) = A to make sure that the area has a maximum value, we just need to differentiate this equation twice and see if the result is negative or not. dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ d(4r⁴ - 4r⁴cos²α)/dα dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ d(-4r⁴(cosα)²)/dα dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ (-8r⁴cosα) ⋅ d(cosα)/dα dA/dα = (1/2)(4r⁴ - 4r⁴cos²α)^(-1/2) ⋅ (-8r⁴cosα) ⋅ (-1)sinα dA/dα = [(-8r⁴cosα) ⋅ (-1)sinα] / [2 ⋅ (4r⁴ - 4r⁴cos²α)^(1/2)] dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ ((4r⁴)(1 - cos²α))^(1/2)] dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ (4r⁴sin²α)^(1/2)] dA/dα = [8r⁴cosα ⋅ sinα] / [2 ⋅ 2r²sinα] dA/dα = 2r²cosα d²A/dα² = d(2r²cosα) /dα d²A/dα² = 2r²(1) ⋅ d(cosα) /dα d²A/dα² = 2r²(1) ⋅ (-1)sinα d²A/dα² = -2r²sinα since we are dealing with 0° ≤ α ≤ 180°, then sinα > 0, then -2r²sinα < 0, therefore d²A/dα² < 0, meaning the Area of the rectangle has a valid maximum value. If we plot a graph for A and α, when A reaches the maximum value, dA/dα = 0 from this we can know that when dA/dα = 0 = 2r²cosα cosα = 0 α = 90° and when α = 90°, then β = 90° the whole shape inside the circle is a square. Do you know ? This proof is inspired by the main task of filling up a circular region in Minecraft by using the command /fill in order to achieve the maximum efficiency

M2007U-E2016-A2017-S2017-M2021_SMT-20160929x01 Let's draw a circle with a radius of r, and install a polygon inside it. ​ As the number of the sides of the polygon increases, the polygon is approaching a circle, but the polygon will never be a circle. cutting the polygon into several equal triangles with one of their tips at the center of the circle, let's call the angle between 2 radii as θ. as the side of the polygon increases, the angle θ becomes smaller and smaller, the number of sides also indicates the number of triangles having one of its vertices as the center of the circle we have. Therefore θ = 360° / n where n is the number of sides we have / the number of triangles we have since we know that the area of a triangle is Δ = (1/2)ab sinθ in this case, a = b = r, so Δ = (1/2) ⋅ r² ⋅ sinθ since a single triangle has an area like that, then the whole polygon should have n times of that area, so we will have nΔ = (n/2) ⋅ r² ⋅ sinθ nΔ = (n/2) ⋅ r² ⋅ sin(360° / n) as n increases, the polygon will approximately become a circle, and therefore the area would be nΔ = (n/2) ⋅ r² ⋅ sin(360° / n) = πr² π = (n/2) ⋅ sin(360° / n) we can also write the equation with a variable approaching 0 instead of infinity by setting n = 1/x, when n→∞, then x→0+ therefore π = (n/2) ⋅ sin(360° / n) π = (1/2x) ⋅ sin(360° ⋅ x) π = sin(360° ⋅ x) / 2x

M2007U-E2016-A2017-S2017-M2021_20180508x01 Let's draw a circle with a radius of R and mark out all the angles with a factor of 5 degrees around the circle as the diagram shown on the right. now let's mark out all different radii with a length of (R x the sine of the angle the radius is pointing to) from the center. you'll find that the tips of those radii are likely to form a circle, the target of this article is to prove that the tips of the radius are really forming a circle. Proofing first, let's name the important parts of the setup: R is the length of the radius of the big circle, R = a + b θ is the angle which the radius is pointing to, Rsinθ is the length of the dynamic radii you've drawn ​the angle p is the one attached to the "likely circle" ​ to prove the "likely circle" is a circle, we just need to prove the angle p is a right angle. first we need to define the range of θ, α, and β in this case 0° ≤ θ ≤ 180° 0° ≤ α ≤ 90° 0° ≤ β ≤ 90° from the diagram, we know that ∵ cosα = a / Rsinθ ..... equation (1) ∴ a = R ⋅ sinθ ⋅ cosα ∵ tanα = h / a ..... equation (2) ∴ h = a ⋅ tanα ∵ α + θ = 90° ..... equation (3) ∴ sinα = cosθ ∴ cosα = sinθ starting with tanβ = h/b tanβ = (a ⋅ tanα) / b ..... from equation (2) tanβ = (a ⋅ sinα) / (b ⋅ cosα) sinβ / cosβ = (R ⋅ sinθ ⋅ cosα ⋅ sinα) / (b ⋅ cosα) ..... from equation (1) cosβ / sinβ = (b ⋅ cosα) / (R ⋅ sinθ ⋅ cosα ⋅ sinα) b = (cosβ)(R ⋅ sinθ ⋅ cosα ⋅ sinα) / (sinβ ⋅ cosα) b = (cosβ/sinβ) ⋅ R ⋅ sinθ ⋅ sinα ..... equation (4) from R = a + b R = R ⋅ sinθ ⋅ cosα + (cosβ/sinβ) ⋅ R ⋅ sinθ ⋅ sinα ..... from equation (1) and equation (4) 1 = sinθ ⋅ cosα + (cosβ/sinβ) ⋅ sinθ ⋅ sinα 1 = sinθ ⋅ sinθ + (cosβ ⋅ sinθ ⋅ sinα)/sinβ 1 = sin²θ + (cosβ ⋅ sinθ ⋅ sinα)/sinβ 1 - sin²θ = (cosβ ⋅ sinθ ⋅ sinα)/sinβ cos²θ = (cosβ ⋅ sinθ ⋅ cosθ)/sinβ ..... equation (3) cosθ = (cosβ ⋅ sinθ)/sinβ cosθ / sinθ = cosβ / sinβ tanθ = tanβ θ = β from α + β + p = 180° α + θ + p = 180° 90° + p = 180° p = 90° ..... (proofed) Usage by measuring the length of Rsinθ and R and a simple calculator, we can calculate the value of sinθ approximately：​ the length of Rsinθ / the length of R = the value of sinθ