Think SMO is hard now? See SMO Junior 2011.

Ajouté : 2026-05-24

[00:00:04]Hello.

[00:00:05]While waiting for the hard smo review, I am going to take this time to actually talk about the OG hard SMO, which is one of the most infamous papers in SMO history. That is the 2011 SMO junior.

[00:00:20]Uh, and I would say both the round one and the round two are pretty infamous, but the round one is particularly obnoxious because it never really was ever so strange or so hard again after 2011. Which means that unless some of you have gone through 15 years of past year SMO juniors, you may not know that the SMO junior ever looked like this.

[00:00:46]I've mentioned a few of the weird points here before we start going through the paper which includes the fact that uh in a 35 question paper the gold cutoff mind you not the round two cutoff the gold cutoff was 12 points so really both round one and round two were exceedingly strange um there was a problem that was basically incorrect there were at least three answers in the SMO book that are not correct I was a student then so I have no idea what they did with is uh more precisely I was a student who was taking the SMO senior at that time. So I don't know what they did for junior. Uh all I remember was that that year a lot of my juniors were saying that how come SMO junior is even harder than SMO senior and open which honestly if I were to look at it now I semi agreed that it is actually completely ridiculous and honestly it's not a very good quality paper.

[00:01:42]uh apart from the mistaken problems uh it's not a very good quality paper because we want it to be hard because of creativity but this paper is just hard because it uses a lot of things that are from the SMO open level. Uh of course what this means is that if you are trying the paper I have put a link in the description below. Uh do not worry about how many you can do but at the same time uh it is useful for learning.

[00:02:08]Even if you are taking the SMO open this year, um you will still be able to learn a lot from this paper.

[00:02:14]So I'm going to definitely split it up into more than one part because it has 35 questions and as I said uh it is SMO open level. So I'm not going to make this into a 4hour video or something.

[00:02:27]But without further ado, we're just going to go through the problems. And if you want to try the whole paper on your own, you can. But you can also just pause the video, try each of the problems for a few minutes before I explain it. uh because each of the questions maybe with about five or six exceptions could theoretically be a SMO junior or SMO senior question. Um it is just that this collection of 35 questions is way too ridiculous.

[00:02:53]Okay, question one. Uh calculate the following sum. This is a question that the moment it appears as question one uh there will be some students who will panic and decide that never mind I will just calculate it manually and uh they will unfortunately have a pretty high chance of messing up and then there is a none of the above to sort of catch everything where if you messed up you would just think it's none of the above.

[00:03:21]So that is a pretty nasty way to start.

[00:03:25]Now the way to do this this is almost a smo open style question already is that it looks a bit like a geometric series.

[00:03:34]So you can call this thing s the usual thing we do with geometric series is that we use the common ratio which is in this case sort of like times half or time two.

[00:03:48]We multiply to everything so that the denominators will sort of run to the next thing. Uh this is also sort of likened to how our recurring decimals are converted to fractions and then we subtract.

[00:04:11]You get half one quarter 1/8 all the way to 1 / 2 to the 10 plus 10 over 2 to the 11 or rather - 10 / 2 to the 11 as the excess.

[00:04:32]So this is uh going to be multiplied by two and then we basically can go again if we do not already know what this sums to.

[00:04:54]Uh I can just multiply s by 1/2 again.

[00:05:01]Actually, I don't need to multiply s by 1/2 again, right? Because uh this is the half s already.

[00:05:06]So, if you were to just subtract them again, you get one minus this and then minus this plus this.

[00:05:26]Okay, I can cancel off a two down here and uh simplify it a little bit more and then multiply that by two again which is going to give you 59 over 256.

[00:05:55]So that is option D.

[00:06:06]This is going to be apart from our intro a more serious video as I'm always coming soon. So I'm just really going to talk through the questions. Uh not so many quips, not so many side notes. Also because uh the paper is hard enough that if I were to say too many other comments, this video and the next couple of parts are going to take forever.

[00:06:25]Okay, question number two. Uh, it is known that the roots of this equation are all integers. How many distinct roots does the equation have? So, of course, roots of the equation uh this is a power five.

[00:06:44]So, we do not have a formula to solve such equations.

[00:06:48]We do not even know that all of the roots are real numbers in general. We don't have such thing as a discriminant to test that.

[00:06:58]But we do know that uh just like a quadratic has two roots, a cubic has three roots, uh this can be said to have five roots. And so if it says they are all integers, uh what it means is that you should have five roots that are integers and I need to have a quick way of well I won't really want to factoriize this because that would take too long even to just test a factorization.

[00:07:34]But uh Vietas if I were to just call these numbers x1 x2 x3 x4 x5 vas still applies and I can still use the sum and the product of roots because that's not so ugly.

[00:07:54]So the sum of roots even for degree 5 is still minus b / a so minus 3 over 1. And then the product of roots is going to be the last thing over a uh but with a odd degree is going to be times -1. So it would be positive 2011 squared 2011 if you were taking the smo junior in 2011 is a prime.

[00:08:30]Therefore, you would be able to know that 2011 squared doesn't have a lot of factors. In fact, it doesn't have enough to have five different factors. So, you need some ones and uh that will pad it.

[00:08:42]In addition, you need some negatives.

[00:08:45]So, it is by this paper's standards pretty easy to now tell that the two 2011s need to be plus and minus to cancel each other out. And then to have minus three, you just have three more negative 1's. Uh the three more negative 1's are going to give you the sum of -3.

[00:09:08]And it confirms that there are three distinct roots 2011, - 2011, and -1.

[00:09:29]Now question three. This is ridiculous that uh this is how a paper opens up, right? It is uh going to frighten nearly everyone. Uh so a fair dice uh by default is 1 to six.

[00:09:48]If shown three times the results of the first, second, and third row recorded as XYZ respectively and X + Y equals to Z.

[00:09:54]What is the probability that at least one of X, Y, and Z is two?

[00:10:04]Essentially, uh this is a conditional probability question.

[00:10:10]Given that X + Y equals to Z, what's the chance that at least one of them is two?

[00:10:16]The easiest way to do this is simply to list because the number of ways that X + Y equals to Z is limited enough and there isn't really anything special about one of them being two. It's just uh a random question about a random situation. So uh if we were to just list them if x y and z have to satisfy this it can be 1 one and two can be 1 2 and 2 1. I'll just sort of group them like this so that it's easier and faster to write. Now take note that 1 2 3 and 213 are different outcomes and they are both equally likely. So we don't count 112 and like 1 2 3 and 213 to be two different outcomes. They are equally likely. There are three different outcomes that equally likely because X and Y are the first and the second row. So reverse is a different outcome.

[00:11:10]Okay.

[00:11:14]Now there does come a point where it is easier to just list everything than to think too hard. uh do not get lazy and uh just like look for a clever solution to something when there's just three or four things to list. When there's just a few things to list, honestly, you'll be faster if you just list that. So there's a total of 1 2 3 four and five. There are 15 cases of which there is 1 2 3 4 5 6 7 8 which have at least one two. So the answer is literally just 8 over 50.

[00:12:03]Okay. Question four. Let x equal to this thing here.

[00:12:08]Which of the following is a perfect square?

[00:12:17]You can actually already do this partially by elimination because even without working through all of the details, you know that it cannot end with 50, right? A square if it ends with one zero, it will have two zeros at least. So cannot be x. Uh we also may know that the last two digits cannot be uh 75. So uh squares if they end in five, they'll end in 25, which does actually eliminate these two as well.

[00:12:46]But uh for the last two, we probably would need to do the whatever the question intended for us to do.

[00:12:58]So uh what I'm going to do is that whenever I see a bunch of zeros like this uh it is easier to convert x into a power of 10 power of 10 by the place values.

[00:13:12]Okay. So there's 50.

[00:13:15]This one has 2014 digits behind it.

[00:13:26]This has a further 2012. So it has 4026 digits behind it.

[00:13:36]So I can check both just to see how it looks. And uh there's one thing I'll do before that which is since 4026 looks very close to double of 2014 uh I'm going to let y be 10 to the 20113 so that x can now be described as y^2 + 10 y + 50.

[00:14:01]Now of course once I write that it should be pretty easy to tell that I don't need to test them both separately.

[00:14:07]I can just complete the square and conclude that x - 25 is the perfect square.

[00:14:16]Some of you may also have the interesting idea of well I can eliminate x + 75 if the last three digits are 125 uh then I can use some sort of uh mod 8 argument for example that is also possible because uh 5 mod 8 is not a perfect square but in this case uh we can use the value itself to determine what is the nearby square which would be 10^ the 2013 + 5 whole thing squared.

[00:14:53]Okay. Question five.

[00:14:55]Suppose that n1 to n 2011 are positive integers and then x and y are defined as these products.

[00:15:06]Um, this actually does look like one of those uh so-called uh primary school style rectangle method questions.

[00:15:17]Uh it would not be wrong to do that. Uh what I will do though is that I am just going to call this whole thing s the sum.

[00:15:31]Most of these can be said to be one or well actually not most all of these can be said to be s with one or two terms missing.

[00:15:41]So therefore if I were to write it in terms of s that will compress it a lot and then I can see what it gives me. I'm supposed to compare x and y basically and I don't really know how that's going to turn out until I try to expand it. So let's do that.

[00:16:00]So that's s and that is y.

[00:16:08]If I expand it, this is s^ 2 - n1 s - n1 s + n1 n 2011.

[00:16:19]This will have pretty much the same things, but it stops one term earlier. So great, I am able to confirm straight away that X is bigger.

[00:16:40]So I wasn't going to make too many remarks, but here's something pretty funny. Uh if you got to this point, congratulations. you are literally above average because apparently the average score for this year's uh in this 2011 round one was slightly under five points.

[00:16:57]And bear in mind that slightly under five points for a paper with 10 multiple choice questions is truly ridiculous.

[00:17:09]Why do I think that the cutoff is so low? It is probably just sheer panic. Uh I took the 2009 paper as uh year one for SMO junior which was probably the second worst SMO junior paper. Uh and I just remember people sleeping, people uh like mourning and generally looking pretty sad even in N US high. So it's just panic setting in uh I'm sure that any of you taking this paper now, if you're not doing it in a panicked mode, you can get more than five questions. Uh remember we are only at question five out of 35.

[00:17:46]Okay, question six. Uh, here is an eggshaped curve.

[00:17:52]ABCD is a circle and then there is an arc Ae, another ark cf, another arc EF and we are asked what is the area of this well egg.

[00:18:04]The area of the egg apparently involves pi and it involves uh square root two somewhere. I think I can understand roughly why both of those will appear. So let's try to chop it up like how we normally handle circle like areas into pieces that we know how to deal with.

[00:18:27]So firstly looking at this I can see that the half of it is just a normal semicircle.

[00:18:40]It's radius one. So this half of it I don't have to worry. I can just say that that is pi / 2.

[00:18:53]Then I have got each of those arcs. The arcs tell me that I basically have a few different circles and I took a few pieces of them and Frankenstein them together into this egg. So accordingly the only thing that I know how to find the area of to do with circles is a circle, a semicircle, a quarter circle or just one slice of a circle. So I should do that. And if I were to for example just chop up this part, I do know how to find that area.

[00:19:28]Likewise, if I were to just chop up this bit, I also know how to find that area.

[00:19:36]And finally, of course, I also know how to find this part because that's just a quarter circle.

[00:19:41]All of this can be combined together to give me the shape that I want.

[00:19:48]Okay. So, uh by symmetry x are symmetric and this is certainly symmetric. You know that this is going to be 45 4590 which means that uh it is a 1/8 slice of the circle which in turn has radius AC which is two.

[00:20:12]Therefore, if let's say I to take this portion, it is 1/8 of pi * 2^ 2. And there's also another 1/8 of p<unk> * 2 down here.

[00:20:28]And then I need to subtract off the triangle which has been counted twice.

[00:20:35]Uh this is also the radius of the original circle which has just size one.

[00:20:40]Therefore the area of that is just one which you're going to minus of at this point. uh some of my students who have seen this question correctly point out that does that mean that we can just say that the answer is a which is correct because if at this point you see that there is a minus one then that must be it right because the other last thing also has to do with pi that is completely correct uh but just for completeness sake we're going to get the last bit of it uh which is this uh quarter circle now for the quarter circle I need its radius and I know that a d is <unk>2. I know a f is two.

[00:21:26]Therefore, I know that this small radius is 2 minus <unk>2.

[00:21:30]And so the area of the quarter circle is going to be 2 minus <unk>2 squared uh 14 pi.

[00:21:49]I'll leave you to expand all of the things out. So<unk> / 2 +/<unk> +/ pi + 2 - <unk>2 2 * 1/4 pi - 1 and you will be able to confirm that option A is indeed actually correct.

[00:22:14]Okay. Question seven. Um the following annulus is cut into 14 regions.

[00:22:23]Each region is painted with one color.

[00:22:28]Do you all know what an annulus is?

[00:22:33]So uh an annulus uh is referring to a donut shape basically. So uh take note that this middle portion uh is not actually colored at all.

[00:22:45]uh you can count what the 14 regions are, right? So let's say this loop here has uh seven and then the outer weird shaped loop also has seven. So that's already 14, which confirms that the middle doesn't count.

[00:22:57]Uh what is the minimum number of colors needed to paint the annulus? So that no any no this is not even the mistake I'm talking about but uh there are a few of these such that no two adjacent regions share the same colors.

[00:23:15]The smo junior should not be testing you on this because this is a very very famous but it's almost a university level kind of statement that there is the four color theorem.

[00:23:32]The four color theorem just tells you that if you want to do this for any normal like um division into regions which is actually usually described as like let's say a map. You want to give every country or every city or every part a different color such that the adjacent ones are different colors so you don't confuse them. Uh four colors are enough is basically the four color theorem. Um you technically don't need this to solve the question but uh it is really useful to know that so that you can try to see whether you can do it with four colors or three colors you can already eliminate uh five six and seven.

[00:24:15]So let's try to do it with three colors and if I can't do it with three colors then I know that it will definitely be um four colors is the answer.

[00:24:25]Okay. So I will start with let's say any of the random regions I can see and I will use 1 2 3 to represent three colors.

[00:24:35]If this is one and two then this one touches both of them so it should be three. And I'm going to continue reasoning like that. This touches two and three should be one. This touches one and two so it should be three. And I will just continue like that.

[00:24:50]and you realize that it will loop around.

[00:24:57]Is this going to be okay? No, it is not okay because now when I get to this guy, it has no option.

[00:25:04]So, this clearly shows that having uh three colors is not enough.

[00:25:14]Now, however, having four colors is possible. uh the four color theorem guarantees that. But like I said, you don't need the four color theorem because uh I can just now that I have a fourth color, I can for instance just use the fourth color down here. And this is color number three to finish off. So that will confirm that uh indeed you can do it with four colors.

[00:25:44]It will still be four colors. even if you had to shade the inner region. Okay, so the fourc coloror theorem does guarantee that uh it may not always be easy to do. uh but it is nonetheless possible and you can search out there's an interesting story about the fourcolor theorem being the first thing that was proven by a computer not AI but uh decades ago supercomputer went through a bunch of cases you can check it out it's pretty cool as the first example basically of a computer-based proof of a big theorem okay question Now uh question eight is essentially asking you about uh forat's little theorem. Again uh what is this doing in the smo junior? I don't know. Uh but if you know format's little theorem, you'll be able to do this very fast. If you don't know, then you will need to keep on testing different stuff and it will probably take us a little while before we can get there.

[00:26:52]The reason why format little the works quickly is that if you look at the exponents which are 4, 6, 10, and 12, those are one less than the things in the options 5 7 11 and 13.

[00:27:07]So for men's little theorem will immediately tell you that 2 to the 4 is 1 5 3 to the 6 is 1 7 5 to the 10 is 1 11 and 7 to the 12 is 1 30. All of these are immediately going to be confirmed and therefore n would be divisible by 5 7 11 and 13. So the answer is none of the above.

[00:27:37]You could check these one by one by saying okay 2 to the 4 - 1 is 15. Uh therefore it is divisible by five and then 3 to the 6 - 1 you either calculate it or you factoriize it with a difference of squares, difference of cubes and then you realize that it is divisible by seven. And you could but obviously this is going to be a lot quicker.

[00:28:07]Question nine. Uh, how many ways can you place a white knight and a black knight on a 8 by8 chess board such that they do not attack each other?

[00:28:20]Firstly, we know that uh by assumption if you are told that a black knight and a white knight are the two pieces then uh the order does matter, right? you swap them around that is a different thing.

[00:28:38]So my do not attack each other is also something that's pretty difficult to count. Uh it is a lot easier to look for. They do attack each other and subtract that from the total.

[00:28:51]So total is going to be just 64 * 63.

[00:29:00]So this is the total number.

[00:29:05]Now intuitively speaking already uh I know that if this is the answer I can eliminate A and B because there is no way that uh more than half of the time uh the white and the black knight do attack each other. That cannot be to count the number of ways that they attack each other. Um, for anyone that plays chess, you will know that the knights on the rim, their future is dim.

[00:29:31]They do not have as many possible squares that they attack. Um, the knight in the center has more possible moves.

[00:29:37]In fact, this is also demonstrated in the diagram here. So, I do not want to go and like say that if the white knight is here, if the white knight is there, if the white knight is here, that's going to take forever.

[00:29:52]Instead I am going to focus on this white knight and black knight. The way that knights move is uh L-shaped, right? Two steps in one direction, one step in the other direction. So I can just look for 2x3 rectangles because 2x3 rectangles are going to be like a proxy for the knights in that L-shaped distance apart.

[00:30:20]Now the number of 2x3 rectangles before that don't forget that uh we can also change the orientation but 2x3 rectangles you can have uh total of let's say seven horizontal positions and six vertical positions across.

[00:30:51]So, a total of 42 different rectangles.

[00:30:56]And then there's two orientations, meaning 2x3 or 3x two.

[00:31:08]And in each of these rectangles, the white knight can go into any of the four corners. And then the black knight will just occupy the diagonally opposite one.

[00:31:24]So there are four choices per rectangle.

[00:31:29]Therefore, you just take the 4032 and you subtract off the 42 * 2 * 4.

[00:31:44]This would give you the 3696 which is option C.

[00:32:02]Finally for the multiple choice question 10. In the set 1 6 7 9, which of the numbers appears the last digit of n to the n for infinitely many positive integers n.

[00:32:19]In other words, uh if you use something to the power of itself, can you make it these things not just once but infinitely many times?

[00:32:34]I don't think it's very hard to see that one is going to work, right? Because if I just use one to the 1, 11 to the 11, 21, and so on, these are definitely going to give me one. So that's easy to see.

[00:32:49]Um it is also quite easy to see that uh six is going to work because 6 to the power of anything ends in six.

[00:33:00]6 square is really 36 * 6 is 21 6.

[00:33:02]Basically it always ends in six. So therefore 6 to the 6 16 to the 16 26 26 and so on. All of these are also going to be good enough for us to get six there.

[00:33:21]Now where I need to get seven or nine based on the answer options I know that at least one of them is possible if not both right because uh already I have uh confirmed that one and six do work.

[00:33:37]So I'm going to just check all the possible last digits that make sense to me which is those ending in three, those ending in seven again and those ending in nine because one we've already done ending in five obviously gives you five and even gives you even. So is it going to be possible to get seven or nine?

[00:34:03]Well, let's check each of these. So three to the something the last digit pattern is 3 971 repeating.

[00:34:14]Am I able to get nine from here? And can I get seven from here?

[00:34:22]Well, uh 9 basically requires that my exponent is 2 mod 4 and seven here is n is congrent to 3 mod 4.

[00:34:38]Now clearly I will not be able to get n to be 2 mod 4 if it ends in three.

[00:34:44]But uh I can get n to be 3 mod 4 if it ends in three very easily. Right? I can just use uh 3 23 43 and so on and that will confirm to me that 7 is also possible.

[00:35:01]Let's check how uh seven and I go to see if we can make nine appear as the last digit.

[00:35:16]So seven looks like this. 9 is just going to be 9 1 is just repeating.

[00:35:22]So immediately I see that oh 9 to the some odd power because it repeats after two steps. So 9 to the some odd power is going to be good enough and certainly uh it is very possible because any time where n ends in nine it is odd. So therefore 9 is also possible.

[00:35:48]Uh, and this implies that the answer is E.

[00:36:02]So, this brings us to the end of the multiple choice section. I'm going to continue on a bit more, but as you can see, uh, this is not a very normal junior paper to say the least.

[00:36:15]And the weirdness continues as we look at the next few problems.

[00:36:27]Now, traditionally the questions right after the multiple choice section uh when you start on the short answer ones are sort of like relatively easy.

[00:36:36]uh by the standards of 2011 this would be relatively easy but by the standards of uh what we have now this is definitely quite a hard first short answer question.

[00:36:49]So uh we are given suppose x / a + y over b + z / c= to roo<unk>2 and ax plus b y + c over z = 0. Find x square a square + y square b square + z^2 over c².

[00:37:04]The rule of thumb for when you are given these sorts of algebraic questions.

[00:37:13]Well, two rules of thumb. The first is that if there are a lot of repeated things, try to substitute it away because otherwise it is just going to actively make it look more scary than it really is. Right? So x over a y over b z over c. And then here is just the flip over reciprocal a over x b over y c over z. The thing to find is just a square of those. So I'm going to get rid of them by a substitution.

[00:37:49]So that by doing the substitution now I have got a nicer first equation.

[00:37:54]I've got a nicer second equation.

[00:37:58]And the thing to find is certainly also nicer. So not so scary. Whenever you can uh do a substitution like that uh it's never going to be wrong, right? It may turn out that I didn't need to do that, but it's kind of like uh if let's say that you have some random uh power socket on the wall, nothing is plugged in, you can switch it off, right? I mean, it technically may not waste electricity. It may not harm anyone, but why not? You can just switch it off. So, uh you can always just do this even if it wasn't fully necessary.

[00:38:30]And now we ask ourselves, how do I get p + q + r 2? Which brings us to the second rule of thumb for algebraic manipulation questions. If you cannot make that thing uh show up immediately, just make it appear inside what you get.

[00:38:50]Even if it comes with a lot of other junk, try to make it appear first.

[00:38:56]So, how do I make this appear? I square the thing that has PQ and R inside.

[00:39:02]So, it appears, but it appears with some other stuff. Never mind.

[00:39:08]Roo<unk>2 squ is two.

[00:39:11]I assume that the other thing is going to help me to clean up the additional junk.

[00:39:18]and the additional junk is indeed going to get cleaned up because if you add up these fractions, uh you conclude that this pq + qr + p r equals to zero which means 2 * of that is also zero which means that your answer is just two.

[00:39:51]Okay, question 12.

[00:39:55]Now, if question 11 is slightly difficult for the first one out of the multiple choice section, question 12 is downright ridiculous for the second question to come out of the multiple choice section.

[00:40:06]And this is going to continue to uh be a trend uh in this 2011 paper as well as some of the old papers that uh there are just very weird difficulty distributions to the paper.

[00:40:24]There are two parts to this question or three parts to this question depending on how we want to view it. But the first part is definitely that I need to do something with x.

[00:40:37]Uh there's no way that I'm ever going to substitute this into here. It is way too horrible.

[00:40:43]And so what I'm going to do is that I am going to uh try to simplify the nested set.

[00:40:52]The nested set is pretty clearly not something I want to have in here.

[00:40:59]And so uh I talked about this in my SMO junior formulas video. Incidentally 12 formulas and this is question 12. So the standard way to try to simplify the nested set. If you see something like this, guess that it looks like root a plus root b or minus if there's a minus in between.

[00:41:23]Square both sides.

[00:41:30]Match the nice part with the nice part and the not nice part with the not nice part. So root ab is 4<unk>3 meaning a b is 48 and then a and b. The whole point is that you're trying to simplify it. So a and b should be nice. They should be integers. So you just try to factoriize 48 and you will see that it's 16 and three.

[00:42:02]So that is roo<unk> 16 which is just four and roo<unk>3. So this is step one.

[00:42:09]Uh step two is uh simplify x which is now that uh using the conjugate I can get rid of the fraction because 4 + roo<unk>3 4 - roo<unk>3 is 4^ 2 - 3 is 13.

[00:42:33]So it is just 4 min - <unk>3.

[00:42:38]Great. That's a lot better than where we started. But we are clearly not done yet.

[00:42:46]And so this can be done in a couple of ways. One is that you just substitute in and expand and you just stomach it and hope that everything goes fine.

[00:43:01]that is technically okay. Uh but the other way that we can do this is that I can do a long division first so that the x^ 4 - 6x - 2x2 + 18 x + 23 uh I don't want to have to take power four. If I do a long division and simplify, the furthest I'll end up with is x².

[00:43:29]But hang on, before we do that, uh I'm going to do something that is often a good idea.

[00:43:36]A good idea is that when you see a set, the square root is actually more annoying than x.

[00:43:45]So it is fine if I were to rearrange it.

[00:43:49]square both sides and basically make myself an identity which is x^2 - 8 x + 13 equals to zero with no square roots inside.

[00:44:04]And that implies that the denominator is two.

[00:44:08]I'm happy with that.

[00:44:16]Okay. So now what do I do with the numerator?

[00:44:22]You see the idea is that now that I know that the denominator is two, I don't need to care about the denominator. So what I want to do is to write my numerator as x^2 - 8 x + 13 times something plus a remainder because x^2 - 8 x + 13 is zero.

[00:44:46]So equals to whatever that thing is.

[00:44:49]This is my target.

[00:44:52]To achieve that target uh I need to basically do long division on x^2 - 8 x + 13.

[00:45:05]I'm not sure if I left enough space for this. So I'm going to write a little bit smaller.

[00:45:18]Uh this is the point at which uh if you feel like pausing the video, you can quickly ask yourself what is 150 minutes divided by 35 questions. Uh and realize that uh although I am going at the fastest speed that is reasonable. I am only actually on pace for that.

[00:45:34]Basically I'm only actually at that rate which is really frightening. But of course uh nobody would be anywhere near finishing everything. So um it is not really that anyone is going to have to work at that speed but in theory uh pretty scary. Okay. So anyway long division works as you take the highest power available and divide it by the highest power here.

[00:46:00]So we subtract we bring this down. We repeat.

[00:46:12]You bring it down and then you repeat again. Oh, this is actually just uh + one and the remainder is 10.

[00:46:30]which tells you that the long division produced this identity.

[00:46:38]The numerator is x2 - 8 x + 13 * x2 + 2x + 1 + 10.

[00:46:48]So that implies that my answer is 10 / 2 which is equal to 5.

[00:46:59]Now, if this appeared at the SMO open today, some of you would not be very surprised, right? Uh SMO Drew near question 12.

[00:47:14]Okay, question 13. Another problem that actually has been in the SMO open not too long ago.

[00:47:22]uh let a1 equals to 3 and a n + 1 = to roo<unk>3 a n minus 1 / a n + roo<unk>3 for all positive integers and find a 2011.

[00:47:33]This is a sequence question.

[00:47:41]When you have a sequence question, the idea is that you should just test the initial values of the sequence and see how it looks like. This is true for junior, senior or open. So if we suffer a little bit, uh hopefully because this a 2011 is the final answer, it means that it's not going to get worse and worse and worse, right? uh whole number must come back around again at some point. So if a whole number must come around at some point then this sequence cannot just get more and more and more square roots. So that's the only reason why I dare to try to put things in.

[00:48:19]Okay. So what does this mean? This means that if n equals to 1, a2 equals to 3<unk>3 - 1 / 3 + roo<unk>3. Okay. I need to simplify that uh probably using the conjugate just to make it as le as uh well let's just say as maintainable as possible. I don't want to have to deal with a giant mess every time. Uh this is 10 <unk>3 minus 12.

[00:48:54]Already this doesn't seem very good.

[00:49:04]Never mind. Uh, let's just put it in again.

[00:49:12]So, this time roo<unk>3. I'll multiply just to the top.

[00:49:18]Okay, actually I can already simplify this down. This is 5 minus 2<unk>3. So, that was uh less bad than I thought.

[00:49:27]and then uh minus one. So I'm doing simplification on the fly. Uh otherwise the working is going to look very long and ugly over this plus roo<unk>3.

[00:49:41]So in fact this plus roo<unk>3 would mean that I add 3<unk>3 to the numerator. So it's 8<unk>3 - 6 over um 3 again.

[00:50:00]Okay. So, I move this to the top and I cancel off a two.

[00:50:11]That is not very good.

[00:50:20]So, at this point, I would say that I'm pretty fed up already. Um I can actually continue but uh it is the theoretically possible way of solving it like a SMO junior question is to just continue but uh I just want to fast forward to well so what's really going on here and how can anyone do it uh and the answer unfortunately involves us doing a little bit of trigonometry.

[00:50:54]This is what I meant about uh the fact that it's technically some of these questions you can do it for junior but uh the method will then be just brute force.

[00:51:04]Does this look like trigle to you?

[00:51:09]If it does then you would recognize it looks like the tangent identity.

[00:51:21]So I'm not going to continue with that.

[00:51:22]Uh it can work but is pretty dreadful.

[00:51:26]Uh this looks like a tangent identity.

[00:51:28]uh it is not perfectly so but uh if I were to take the thing down here and let all of my ASB be tangents of some angle and if I were to simplify the fraction Yeah.

[00:52:02]By dividing the top and bottom by square<unk> 3.

[00:52:18]You'll realize that this is just nothing more than a tangent of 30°.

[00:52:25]uh and theta n combined together based on this identity.

[00:52:31]Um that does explain why all of the stuff are so horrible. It explains why all the stuff are so horrible because your initial value is just the tangent inverse of three whatever that is. So if the initial value is something ugly then um tangent of whatever that isus 30° would still be horrible and - 60° would still be horrible. But it does imply that uh if you subtract 30° six times because the period of tangent is 180° it is just itself.

[00:53:10]So the original sequence has a period of 6.

[00:53:17]Hence if you were to notice that this is 2011 uh 2011 is congrent to 1 6. So at the end of all of this it repeats every six terms. So the answer is still just the three that we started with.

[00:53:35][snorts] These are the first three questions. I'm sure you can understand why some people who uh after doing a multiple choice looked at this just decided that they should just go to sleep and just give up on smo forever.

[00:53:52]But uh it is a good example if we were to put this aside as junior. It's a good example for senior and open of like when you see something that looks vaguely familiar like you just don't know why then ask yourself what does it look like? It doesn't matter how ridiculous it is. If it looks like, let's say, calculus, maybe it's calculus. If it looks like, let's say, uh, trigle in a sequence question, maybe it's trigger. If it looks like, uh, some kind of, uh, log identity, maybe it's something to do with logs, right? Do not disregard even if it seems that this question has nothing to do with that.

[00:54:29]Whatever you feel like it is could well turn out to be the intended trick.

[00:54:35]So unfortunately if you continue here you're going to see that you end up with a pretty ugly A4 A5 A6 before A7 turns out to repeat and almost everyone is going to have given out this before then unless they knew the trigger idea.

[00:54:49]If I are teaching a class right now, uh some student would say, "Then I'll just guess three," which is probably not a bad idea if you don't know what to do to just guess that if the sequence looks too weird, maybe it just repeats.

[00:55:04]Okay, question 14.

[00:55:09]This continues to be pretty bad.

[00:55:14]It's a very famous question.

[00:55:15]Unfortunately, it's a famous question that's actually at like a easy IMO level famous easy IMO level of like the 1980s kind of famous.

[00:55:27]So, uh you can do this by just doing a lot of uh combining equations together, add and subtract things and then something good's going to happen. But, uh that is not the official solution.

[00:55:39]And honestly, that's also going to take quite long. So, I'm going to just get straight to the idea, which again is a more SMO senior style idea.

[00:55:51]I just said to look out for stuff that seem familiar. Does this look familiar in any way?

[00:55:58]Uh especially considering that uh these are squares. Does this look familiar in any way?

[00:56:14]If you look at it and you thought of the cosine rule, then congratulations to you for being probably one of only 10 people who thought of that without having seen it before. If you have seen the question before then you know uh it is the cosine rule because a squ b 2 and a 5 squ if I were to have a triangle which has sides a b and five and an angle theta here.

[00:56:49]This is the equation.

[00:56:53]So therefore uh if you want this to match you should let cosine theta be half i.e. theta equals to 120° in a triangle.

[00:57:07]You have three of those and a b and c are repeating which means that what you would actually have is this. You have a 578 triangle.

[00:57:23]And you have this point somewhere inside. Uh this is technically has a name. This is your format point. Uh something that also appeared in our recent SMO open format point. Uh where all of these are 120°.

[00:57:41]This is your A, B and C.

[00:57:45]And now the question is okay so how do I actually find a + b + c whole thing squared from here? That's still not simple at all.

[00:57:59]Uh but at least it kind of uh vaguely explains what is the reason for having uh those weird equations. Now we need to figure out how do I make use of A, B and C.

[00:58:23]And the answer is that since we are already in the trigo world, we continue on by using area.

[00:58:34]area is half a b sin 120° and then bc and ca are also there.

[00:58:45]This is the area of the whole triangle.

[00:58:50]It is also equal to a value by heron's formula as the semiparimeter is 10.

[00:59:04]So 10 10 - 5 10 - 7 10 - 8. Uh this does evaluate to something reasonably nice.

[00:59:18]Which means that if you were to ask what does this give me? Uh there's no square roots inside. Uh sin 120° is roo<unk>3 over2.

[00:59:29]So it is just that a b plus bc plus ca is 40.

[00:59:37]Okay. Now that is enough for me to complete the question because if a b + bc plus c is 40, I can now solve this like algebra.

[00:59:47]uh I can solve this like algebra by adding these three equations and subtract the 40 and it will give me that 2 * of the sum of squares is equal to 52 + 7 + 8 2 - 40 uh which is equal to 98 8.

[01:00:19]So the sum of squares is equal to 49.

[01:00:23]And the answer would then be a square + b square + c² + 2 * of this equals to 129.

[01:00:54]Okay. So for this part of the video we will end on question 15 which is a classic question but it's a classic SMO senior question again. Uh let PX be a polomial of degree 2010. Suppose PN= to N / 1 + N for all n= to 0 2010. Find P of 2012.

[01:01:23]The answer is supposed to be an integer apparently.

[01:01:28]So if the answer is supposed to be an integer, uh it cannot be that the answer is 2012 over 2013.

[01:01:41]So then what is it?

[01:01:45]Well, px is a polomial. So it is not uh any function. I cannot let px be just x over 1 + x.

[01:01:55]The usual way that we handle these is by trying to use the polomial to turn the pattern into a statement of here are the roots of some polomial and then use the factor theorem.

[01:02:14]So uh I'm going to do that and say that px = to x over 1 + x for this bunch of values means that if I were to multiply 1 plus x over and subtract x that is going to be a new polomial that has this bunch of things as roots.

[01:02:48]this new polomial since the original polomial had degree 20110.

[01:02:56]Uh if the original polomial has degree 2010 then we know that uh your 20 11 roots of a new 2011 degree polomial almost perfectly locks it down.

[01:03:19]It almost perfectly locks it down because by the factor theorem, it tells you that this thing is going to have the factors x x -1 x - 2 until x - 2010 but with some constant waiting.

[01:03:54]So how do I determine that constant?

[01:03:59]It is going to be not possible to use the polomial because I don't know what is it yet. I definitely cannot just assume the constant is one because well I mean among other things if the constant is one and then you put in 2012 we are in deep deep trouble.

[01:04:16]So the way that our remainder and factor theorem normally works is that you substitute something that will cancel away the polomial since you don't know what is it yet.

[01:04:31]I will let x equal to1 and by letting x be1 you are just going to end up with uh the left hand side p of negative 1. I don't know what is it.

[01:04:43]It's okay. It got cancel off.

[01:04:51]So the entire thing is fine.

[01:04:56]If I were to now put in 2012, which is the thing I'm supposed to find, then something pretty funny happens.

[01:05:19]Okay. So, 2012, oops, not PX. Uh, 2013 P of 2012 minus 2012.

[01:05:32]Sorry, give me a second.

[01:05:44]Yeah. So 2013 P of 2012 minus 2012 is equal to C * all of this oh I forgot to say that what is C right so C is originally this horrible looking thing C is -1 over 2011 factorial negative because there are 2011 minus signs which is odd but now if you put it into here it just gives gives you negative 2012 because it cancels away all this other stuff. And so, uh, the funny thing is that it gives you P of 2012 is zero, which is pretty cool, but also admittedly, uh, not something you're going to guess despite how weird it may look.

[01:06:36]Okay, so this takes us up to question 15 uh which is uh amazingly still not yet having seen the hardest questions of this paper. Uh for again for everyone who is just seeing this if somehow anyone is doing last minute preparation for the SMO junior this is not the typical standard of the SMO junior today. Uh please go and look at the past recent years ones and it's a lot less scary than this. Uh but for anyone who is also doing senior open um this paper actually has a lot that you can learn from. So I hope it's useful not just for those of you who are taking junior and got frightened off but if you're taking senior or open these questions do appear in senior and open today. So thanks for watching and you can stay tuned for part two in a few days time.