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There are three people A, B, and C, and you meet them. You are given that A always tell the truth and B always lies, and C either always tells the truth or always lies. You don't know who is who. Please determine a strategy to find out if C is a truth-teller or a liar by asking the least amount of questions needed.

What number rhymes with angry and hungry? (Evan Chen wrote this for an April Fools Math Contest)

Alright guys, I am announcing a riddle and puzzle competition for LoC and CCMSL. I will be posting riddles or puzzles daily. Only 1 submission for a puzzle or a riddle for 1 contestant. You will have to submit the answer to me through PM. The fastest correct answer each day will win one point for that contestant. If there were no correct answers for a certain day, no points are given out. We will be posting puzzles until 10 points have been given out. The standings will be updated every day. The top 4 will advance to a sudden death stage, where speed does not matter (1 day time limit per puzzle/riddle) but a wrong answer will cause that contestant to be eliminated. The final person not eliminated will be the winner, and wins 50 trophies. Second place gets 25 trophies, and third gets 10 trophies. if in the sudden death stage there is a puzzle/riddle nobody solved, then that puzzle/riddle is not counted.

1. 180 [@StupidPersonRomania got correct] AOB can be 180 at most and 0 at least. 180-0 = 180 2. 203010 [@StupidPersonRomania got correct] 10ac = 1000a+10c+1000 ac = 100a+c+100 a(c-100)-(c-100)=200 [SFFT] (a-1)(c-100) = 200 (a,c) = (201, 101) 10ac = 203010 3. 103 [@StupidPersonRomania got correct] (6!+1)/(3!+1)=103. We claim this is smallest. Suppose that it is not the smallest, and for some a=x and b=y, (x!+1)/(y!+1) < 103 and is an integer. Then 103 > (x!+1)/(y!+1) = 1+ (x!-y!)/(y!+1)=1+y!*((y+1)*y*...*(x-1))/(y!+1) >= 1+y!, so 102>y!, and y = 1, 2, 3, or 4. y=1: (x!+1)/2 is integer, so x! is odd, so x = 1, but 1 is not larger than 1. y=2: (x!+1)/3 is integer, but x>2 so x!+1 is not multiple of 3. y=3: (x!+1)/7 is integer, and x>3. x=4 does not work, x=5 does not work, x=6 was our original solution. y=4: (x!+1)/25 is integer, and x>4. x=5 does not work, x=6 does not work, but x=7 is too large. Therefore, there is no solution less than 103, proving the claim.

Before we start looking at the problems, let me give a quick note. At early years of Math Olympiads, the geometry were mostly constructions, and 3D geometry locus problems. I will not be going through those type of problems. In the recent years, the style of olympiads has changed significantly. These new type of problems started around 1988. The first such 'new type' problem appeared on the 1988 USAMO, as problem 4. This problem was the first USAMO problem I ever solved in my life. Now, at that time, Internet was relatively new, and the usual exposition for Olympiad Geometry was "Geometry Revisited", which unfortunately did not contain the Incenter-Excenter Lemma I talked about in the last math stuff of the week post before the CCMSL Contest. The reason I did not prove it on the post, is because today, we are going to do the proof in the problem, because the 1988/4 is trivial by the I-E Lemma. 1988/4 USAMO Let D be the circumcenter of IAB In fact, all we need to prove is that DB=DI=DC in the I-E lemma, because that would mean that D is the circumcenter of BIC, and D is on the circumcircle of ABC, and the same goes for E and F by symmetry. Proof By the Definition of D, we note that DB=DC. So, we just have to prove DB=DI. angle(DBI)=angle(DBC)+angle(IBC)=angle(DAC)+angle(ABC)/2=angle(BAC)/2+angle(ABC)/2. angle(DIB)=angle(ABI)+angle(BAI)=angle(ABC)/2+angle(BAC)/2=angle(DBI). So, angle(DBI)=angle(DIB), meaning DB=DI, and so we are done. 1988/4 USAMO Formal Solution Write-up (My Original Write-Up when I first solved it) Let the three circumcenters be C', A', and B' respectively. We claim that A' is the midpoint of arc BC on the circumcircle of ABC. Let D be the midpoint of arc BC on the circumcircle of ABC. Then, we note that DB=DC. angle(DBI)=angle(DBC)+angle(IBC)=angle(DAC)+angle(ABC)/2=angle(BAC)/2+angle(ABC)/2. angle(DIB)=angle(ABI)+angle(BAI)=angle(ABC)/2+angle(BAC)/2=angle(DBI). So, angle(DBI)=angle(DIB), meaning DB=DI=DC, and D is indeed the circumcenter of IAB, so D=A', and A' is on the circumcircle of ABC. The results for B' and C' can be proven similarly. Q.E.D.

https://lichess.org/swiss/qTZQSt0l You will need to join the CCMSL Lichess team. There is one round every day of 3+0 blitz starting tomorrow. In HKT it is 2:00pm. Please convert to your time zone so that you can be ready.

Hosted on lichess. 1. @StupidRomaniaGuy - @LLLhk 0 - 1 https://lichess.org/NtNUagQG/white#39 2. @MrChessGuyFromHk - @StupidRomaniaGuy 0 - 1 https://lichess.org/puxrywqK/black#30 3. @LLLhk - @StupidRomaniaGuy 1 - 0 (Note: Don't know why he resigned) https://lichess.org/a3lA1fRp/black#17 4. @StupidRomaniaGuy - @LLLhk 0 - 1 https://lichess.org/QdwmkE2n/white#32 5. @LLLhk - @StupidRomaniaGuy 1 - 0 https://lichess.org/OGTL4L5c/black#0 6. @StupidRomaniaGuy - @LLLhk 0.5 - 0.5 (I blundered a queen so I was losing, so I asked for a draw and he did not see the blundering, so we drew) https://lichess.org/TKqCNfZt/white 7. @LLLhk - @StupidRomaniaGuy 1 - 0 (He had bad internet connection here, so he lost a lot of time, but managed to get a winning position. With 20 secs left, he blundered a bishop so the game was a drawn position, and he knew he would lose on time) https://lichess.org/m0AVzmIT/black#44 8.@StupidRomaniaGuy - @LLLhk 0 - 1 (He wanted a time odds for this last game and I refused so he resigned) https://lichess.org/efcFsVdP/white

The reason I wasn't able to post a math trick on friday was because I was preparing for the AMT 2020 Math Competition (still going on). On saturday and sunday, we had the power round:http://attach.seedasdan.com/STEM/2020%20AMT%20PowerRound-problems.pdf, password=202008AMT. Now, my team had 4 members. However, one of them did not help, and another called Ross had Ross camp, so only had time to solve three problems(the hardest 3), and another called Daniel Chen also had Ross, and had to stay up very late in order to discuss the problems with me, and also only had time to solve three problems. But, Daniel helped us a lot in order to improve our LaTeX formatting and checking our solutions. In other, words, it was literally only the two of us and a bit of Ross's work. Here were our solutions: https://drive.google.com/file/d/1M-p_K2htOLwTPSLgb71K-4T42BVE3c6q/view?usp=sharing. Note that we used the ross formatting for LaTeX. Funniest moment: Daniel saying: "Imagine if I sent the pdf to submit through the wrong group chat" What it meant: I was the team leader, so i had to submit the pdf solutions. Now, I do not have Adobe Acrobat, so Daniel helped us cut the pdf into 21 parts (we had to submit 1 pdf per problem), and so he had to send it to me. We were using WeChat. Now, there was a team group chat and the competition group chat(consists of all the teams). The 'wrong' group chat he referred to was the competition group chat. Anyways, onto our math trick for this week... inspired by this year's power round... a proof technique called... INDUCTION! In the problems pdf, they explain principle of mathematical induction (PMI). In our solutions to 1c and 2a, we used PMI. 1a was some straight forward modular arithmetic, 1b was pigeonhole principle, 2b was Lifting the Exponent (we think that there should be a way easier solution), 2c was Strong Induction, and 2d was a combinatorial approach. For PMI, just read the problems pdf, in fact their explanation is nearly the same as what I would say. For STRONG INDUCTION, see below. In strong induction, the base case is still the same. However, the assumption is everything below. In other words, strong induction is S(0) implies S(1) S(0) AND S(1) implies S(2) S(0), S(1), AND S(2) implies S(3), and so on. Compare this to usual induction, which is: S(0) implies S(1) S(1) implies S(2) S(2) implies S(3) For an example see our solution to Q2(c).

https://www.chess.com/tournament/chess-club-for-maths-and-science-lovers In first place we have @mathsoccerchess (HE CHEATED!!!!!) In second place we have me, @LLLhk! In third place we have @Saint_Anne, who unfortunately is not in our club. The "prizes" will be received by Aug 1

This will be helpful for CCMSL Contest 1 problems 1, 8, and 13. The orthocenter of ABC, usually denoted by H, is the intersection of the perpendiculars (or altitudes) from A to BC, B to CA, and C to AB. The triangle formed by the feet of these altitudes is called the orthic triangle. The centroid, usually denoted by G, is the intersection the medians, which are the lines joining each vertex to the midpoint of the opposite side. The triangle formed by the midpoints is called the medial triangle. Next, the incenter, usually denoted by I , is the intersection of the angle bisectors of the angles of ABC. It is also the center of a circle (the incircle) tangent to all three sides. The radius of the incircle is called the inradius, denoted by r. Finally, the circumcenter, usually denoted by O, is the center of the unique circle (the circumcircle) passing through the vertices of ABC. The radius of this circumcircle is called the circumradius, denoted by R. Excenter There are three excenters, one for each vertex. Consider the angle external bisector to angle B and C. Then, these two bisectors meet at the point known as the A-excenter, denoted by I_A. You will use this in problem 8. In fact, there is a circle with center Excenter that is tangent to all of the sides, when extended. This circle has radius exradius, denoted by r_A for A-Excenter, r_B for B-Excenter, and r_C for C-Excenter. The Incenter-Excenter Lemma The problems for CCMSL Contest 1 will be posted on 1st Aug. Look at Problem 8. This section will be a hint for that problem. However, you will have to do some more research to solve problem 8. BTW this problem is stolen from a contest. However, I bet most of you don't have access to it since you need a membership. The Incenter-Excenter Lemma says: A, I (Incenter), D (mentioned in the problem), I_A(Excenter) is collinear, and that DB=DC=DI=DI_A (length). Similarly, you can say the same for B, I, E, I_B, EC, EA, EI, EI_B, and then same for C, I, F, I_C, FA, FB, FI, FI_C.

These clouds form at the crests of waves created by moist air crossing a mountain barrier; they are consequently lens-shaped. Lenticular clouds tend to form one above the other like a stack of pancakes or a hovering spacecraft, and they are therefore sometimes mistakenly reported as UFOs.

i'm not bad at maths, but i'm not elite in maths.. so can u guys plz teach me math tricks?? hehe

Many people from America who do good in Math Olympiad actually watch anime. (I watched Hikaru No Go 3 times) What about in other countries, what hobbies do people do good in Math Olympiads have in common? I guess in China it would be um... I don't really know, actually.

What is AM-GM? AM-GM is an inequality. It states that AM≥GM for positive variables and equality holds if and only if they are all equal. What is AM? AM is the "mean" you are used to, the Arithmetic Mean. This is just adding the n variables and divide by n. It is called that because when there are 2 variables, the variables and the AM form an Arithmetic sequence. What is GM? GM is not the "mean" you are used to. This, the Geometric Mean, is just multiplying the n variables and taking the whole thing to the nth root. It is called that because when there are 2 variables, the variables and the GM form an Geometric sequence. Example: (15)^(1/3)≤3 This is true since 3=(27)^(1/3). But, another way to look at it is that (1*3*5)^(1/3)≥(1+3+5)/3=3 How do we prove AM-GM? See this article I wrote (a few months ago for no reason): I cannot attach files, so why not see screenshots I made of it? xD Example : a^7+b^7+c^7 ≥ a^4*b^3+b^4*c^3+c^4*a^3 (a^7+a^7+a^7+a^7+b^7+b^7+b^7)/7≥(a^28*b^21)^(1/7)=a^4*b^3. So, (4a^7+3b^7)≥a^4*b^3, and similarly we can write two other inequalities with b and c, c and a. Summing them up gives the desired result. Exercises: Chapter 1.3:https://web.evanchen.cc/handouts/Ineq/en.pdf THE END Next time, we will be looking at some geometry!

Also known as night-shining clouds, noctilucent clouds are Earth's highest clouds and are found in the mesosphere, roughly 50 miles above the surface (just for comparison, most are found in the troposphere, under 10 miles). They are mostly observed at high latitudes and during astronomical twilight (1.5-2 hrs before sunrise or after sunset). For information, you can visit the wiki site here: https://en.wikipedia.org/wiki/Noctilucent_cloud#:~:text=Noctilucent%20clouds%2C%20or%20night%20shining,%22night%20shining%22%20in%20Latin.

If you could only eat one thing for the rest of your life, what would it be?

These pouch like protrusions composed of ice are typically found at the bottom of cumulonimbus clouds and can stretch hundreds of miles in every direction. They are typically harbingers of violent storm weather. For more info on these, you can check out the wiki page here: https://en.wikipedia.org/wiki/Mammatus_cloud

Hi guys, Just wanted to share with you these awesome clouds, which occur in the stratosphere at polar latitudes during the wintertime. For more information, you can visit the wiki page here: https://en.wikipedia.org/wiki/Polar_stratospheric_cloud

For the first Math Trick of the Week (MTW), which will be uploaded every 23:30 Wednesday GMT time (as near that time as I can make), is going to be on... SFFT. WHAT DOES SFFT STAND FOR? SFFT stands for: Simon's Favorite Factoring Trick, which was popularized by AoPS (Art of Porblem Solving), in one of their videos. WHAT IS SFFT? To explain it, let us look at an example: "Solve for all integer pairs (x,y) in this: xy+x+y=0". At first glance, this problem seems hard. Then, we might observe this solution: (0,0). But, is that the only one? If we can find another, how can we prove there are only these pairs? Method 1 (Works in general, but slow) xy+y=-x (x+1)y=-x y=-x/(x+1) So, when is y an integer given x is an integer? Well, after a bit of deliberation, it is clear that only x=0 and x=-2 work, so the pairs are (0, 0) and (-2, -2). (Took me 1:34 mins using this method) Method 2 (Fast, SFFT) xy+y=-x (x+1)y=-x (x+1)y-1=-(x+1) (x+1)(y+1)=1 So, (x+1, y+1)=(1, 1) or (-1, -1). So, (x,y)=(0, 0) or (-2, -2) What is SFFT in general? axy+bx+cy=d xy+(b/a)x+(c/a)y=d/a (y+b/a)x+(c/a)y=d/a The next step, is to make y+b/a. (y+b/a)x+(c/a)y+bc/(a^2)=d/a+bc/(a^2) (y+b/a)x+(y+b/a)(c/a)=d/a+bc/(a^2)=(da+bc)/(a^2) (y+b/a)(x+c/a)=(da+bc)/(a^2) (ay+b)(ax+c)=da+bc. You now find factors of the right hand side, and then you can find (x, y). Conclusion SFFT is a powerful trick that solves diophantine equations of the form axy+bx+cy=d. I hope you understand this trick now. THE END