The hardest problem in Korean SAT math history (from the 2019 Sunning exam) involves a function f(x) = g(x)/(x-a) where g(x) is a quartic function with leading coefficient -1. The key insight is to visualize f(x) as the slope of a line from the point (a, 0) to the point (x, g(x)). This geometric interpretation reveals that f(x) having a local maximum at x means the line from (a, 0) to (x, g(x)) is tangent to g(x) at that point. Given that f(x) has local maxima at both x = alpha and x = beta with the same value m, and that the number of local extrema of f(x) exceeds that of g(x), we can deduce that g(x) must have exactly one local extremum. Using the tangency conditions and the given beta - alpha = 6√3, the smallest possible value of m is 216.
The HARDEST Korean Math SAT question ever (Feat. DrZye)
Added:Okay. Hello. Welcome to the Korean SAT math experience presented by Z.
Now, disclaimer. Today's talk is about nice math questions. I will mostly be saying positive things about the math questions. That is not my opinion on the SAT system as a whole. Uh I I spent three years of time in the system myself. So, if anything, I would happily watch it burn. Now, anyways, I have translated the 2019 Sununning math in English and gave it to you to the Tu swap math club. Tus Swap Math Club. Uh, Sununning is the Korean SAT SAT because that's the one I took eight years ago in November of 2018. Um, it's called 2019, but it's taken in the year before.
Anyone watching this can try it yourself in z.orlds sununing. That's sunning in English.
That's why it's sunning. Uh you can also try any of the other past Sunning math exams because I've been translating most of them in the last few years.
I've also given the math club what I think to be the hardest question in Sunning math history as a challenge. It has a very beautiful solution. I'm going to talk about this problem first in today's talk. Now, uh let me put this here so that the people watching this on YouTube can have some time reading it.
And here today's talk will have either two parts or three parts. First part I'm going to solve that problem. Second part we're going to go through the entire SAT math exam I had to solve. And if we have time after that uh two swap once said what if we have to solve the problems in Korean too. And it's a great idea. The math club really likes a challenge like this. I've prepared a set of problems in Korean that I think makes sense as a decipher challenge. Everyone will turn on their mic. I'm gonna share an excgaly draw link with the problems and I'm gonna let you do your thing. Uh, sounds exciting. Okay, let's start. Uh, and by the way, you can stop me at any time and ask me any question. Uh, because first I'll give you some world building. Uh, what's the culture behind this exam?
Now, from now on, imagine that the world of Sunning is like a fictional fantasy, a fairy tale. In this fantasy world, there is a single deity, single deity called Kais. It's the institute that makes the exam every year. Uh in this world, every year, Kais bestows three holy scriptures down to us mortals. They are called the two mock exams. Two mock exams in June and September and then the real exam in November. These three tests are considered as part of the cannon.
They form the basis of what you need to study for the SAT. Uh I mean of course you need to study the school textbooks as a baseline and there are practice books designated by the government as well but anyway the prior kais exams are thought to hold absolute authority.
Other institutes give out mock exams too but these are not considered to be canon only the ones made by kais. Uh and it's that way because the quality of the exam questions is actually different. It's actually really hard to replicate Kais's style.
There are thousands of private tutors selling thousands of private exams to students but only few ever get close. I would say in this fantasy world uh the tutors are like wannabe oracles. Uh they try to earn trust trust from students give them hope prophesize what kais would do that year something like that.
But still tutors are not kais. Uh nothing they say is really canon.
highest exams are just like an untouchable platonic ideal at least for the characters in this fairy tale.
Uh, of course, this is a small fairy tale called SAT math in the much bigger world called the entire college entrance system because there are many ways to get to college with less focus on the SAT. Anyway, are you immersed in the world building now? Then let's get into the problem.
Uh, by the way, any questions up till now, you can ask me anything about the Korean college entrance exam or whatever.
What about uh let's see, let's look at the chat.
Yeah, this is the Korean version of College Board. Uh, I mean, Yep. I I don't know what College Board is exactly, but I think that would be the case. Anyway, this problem to me is >> Yeah. Yeah. This problem is wellmade in the sense that it rewards mathematical intuition and creativity. There are many ways to solve it and in every step of the solution there exists a way to save time and do less calculation if you have the right intuition. Uh now today I'm going to solve this problem with as little calculation as possible mostly using visual explanations. Of course, in the constraints of the actual exam, it would be realistically not feasible for a high school student to find every single time save that I'm going to explain. Uh because when I first tried this question while I was in high school, I believe it was the summer of 2018 around six months before I had I had to I had to take my SAT. Yeah. At the time, I had none of the intuition and I wrestled with the problem for a full day.
uh but today I will just take the role of someone visualizing the ideas behind the problem. Uh let's let's hear what three blue one brown has to say about visualization.
Uh three blue one brown says ask the question what does it look like?
I'm not saying every single piece of math has to be visualized. I'm not even necessarily saying visualizing it will always give you the best explanation.
But what's undeniably true is that there is a lot of lowhanging fruit simply asking the question, what does it look like? What's Okay, so I'm going to ask the question, what does it look like?
Every single step of the question and just solve it like that. Okay, let's go to the question.
Uh and I Okay.
Now, um I believe there are a few people here who have tried the question beforehand and uh I believe there are people here who got it correct. So, do you have h how did you like approach this problem at first? What did you think about it?
Um you can answer it like in the chat or like uh with your mic on anything is anything is okay and I will >> I saw g of x >> partic function and I saw that f ofx would be a rational function and at that point I was like oh this is going to be much harder than I anticipated and I sort of just gave up >> right right [laughter] you have this part uh x minus a fx equals gx wait my my thing is bugged Wait, wait.
Okay.
Why does it not follow my mouse? Okay.
Now, now it's following.
Let's follow.
Okay.
Now, this part, let's let's read it from the beginning. a function f(x) defined on x larger than a. So it's just a function and a quartic function gx with a leading coefficient of negative one.
So if you don't know, a quartic function is a function that kind of looks like this. We're doing a visual explanation here. So yeah, it kind of has ups and downs. It has like two ups and two downs. Or it could have like uh it could just go like that. This this bumps could differ in their size. But anyway, it can have at at most like these two bumps and at least it has like this one bump.
Anyway, it has a leading coefficient of minus one, which is why it goes like that. It goes down like this.
Yep, that's the part. That's the first part. And now it looks like x minus a time f(x) is g(x) for all x larger than a. uh for all that's just uh the entire domain. So f fx is just gx / x - a and that's why it's a rational function. fx is some sort of like cordic function divided by x - a.
Uh now here I'm going to ask what does it look like? [laughter] What does fx look like? But not in the usual sense like not in the usual sense of what does a graph of fx look like.
I'm going to instead ask what is the meaning of this expression visually.
What is the meaning of gx / xus a?
Now uh I could give you some time to think about what the meaning is.
Bam.
Now uh Valo says is val zero val zero says if g has a pole at a then f is continuous if not it has a pole right that is also correct like it this thing the shape of this thing matters whether gx has an xus a term in it or not.
Uh right. But it um unrelated to whether g(x) has that or not. This can be thought of as something relating to a line. Can you guess what it is? Something relating to a line.
Maybe it's better if I write this.
No.
This is what we call a slope. It's a slope from the point a comma 0 to the point x comma gx.
Right? Did you get that? Uh it's yeah x - a gx - 0. And that's that kind of gives us a nice intuition on what f(x) is. It's a always it's always a slope from the point a comma 0 to the point x comma gx. Uh I went in thinking f and g were polomials. So I assumed probably incorrectly that g equals f with another root at x= a. Yep. Uh that is an assumption you can make but unfortunately in this problem it turns out to be wrong. Uh let's kind of try to think about what this means. Now I'm gonna what? Okay, my my hands are shaky.
Wait, here's the point a comma 0. This is the x-axis. Now imagine gx looks like I don't know anything like this.
Let's just think about a single point x comma gx.
Now fx at x fx would be the slope of this line.
Now that is all we can get from the all we can get from this a. At this point we're going to go to b. Now b is where things get really interesting.
>> Where does that come from?
>> Where did that a zero come from? Because I feel like f of a is just going to be undefined, right? No.
>> Oh. Uh I'm what I'm saying is that f ofx is equal to the slope of this line given that this point is x gx.
>> Okay. I see. I see. I see. Okay. So you're taking the derivative basically at a specific point not in general.
Uh uh >> no it is in general because that's like the definition of a slope. This is like basically x2 - x1 and y2 - y1.
>> Oh I see now. I see now. Okay. Yeah I see now. Okay. Okay.
>> Okay.
>> All right. Yeah.
>> Great.
Now okay. Let's go to B. For some real numbers alpha and beta, the function f(x) has a local maximum of m at both x= alpha and x= beta." Now, a local a local maximum is something like this.
This is a local maximum. It's the largest part like it's the largest point nearby.
uh usually like in for a continuous function like a local maximum happens when you increase and then decrease.
Now f(x) is a slope between this point and this point. Now what does it mean for fx to have a local maximum? What does it mean for fx to pro possibly like increase and then decrease? Imagine.
Uh, wait.
Yeah. Imagine this line is the line between a comma 0 and alpha comma g alpha. No, f ofx the slope of the line has to somehow increase while gx like x while x goes this way and then decrease while x goes this way. How can that happen? What do you think will happen if for 2 gx for that to be true? The slope has to increase and then decrease.
Uh the slope is kind of like the angle here. It's not exactly the same but like it's a tangent of the angle but anyway >> g of x now.
>> Yes. Yes. Uh okay. V 0 says gx also has a maximum nearby. And what did you say?
Uh I couldn't hear it very well.
>> f of x would have to equal g of x and then g of x have to be like concave down.
>> Yeah. uh it has to be like concave down which means something like this because this is the only way for f ofx for the angle of this to be whoops to be angle to be less than this in this area and then it also has to be less than this in this area. So the angle has to be kind of maximized in this point and then decrease again. So all this boils down to this line from a a comm 0 to the gx function has to be tangent to gx at that point and gx has to go down like this.
That's what it means by fx having a local maximum.
Now I'm gonna like assume you got that and move to the next part which is f(x) has a local maximum at both x= alpha and x= beta.
Now that would mean at at some alpha this line will be tangent to this alpha at this point and at some beta this line would also be tangent to gx at this point.
And now notice how it says it has a local maximum of m at both x= alpha and x= beta. The value in that local maximum is the same for both of these variable both of these both of these values.
Which means the slope the slope of the line from a comma 0 to alpha comma something and then from a comma 0 to beta comma something is equal.
What does it say about the three points a comma 0 alpha comma thing and the beta comma thing?
Yeah, they are aligned. Yes, that's correct. Now the only way for the slope to be equal is that if the the two points alpha comma g alpha and beta comma g beta are both on this line and we know that it has to be tangent like this in both points. So now we kind of know the shape of gx. That's what b tells us. And uh I I kind of skimmed over it, but the reason this line goes up like this like in the positive direction is because this M is larger than zero.
Now that's all we got from B and we are going to move on to C. The number of values of X where FX has a local extremum is greater than the number of values of X where GX has a local extremum. Now what does that mean visually? Uh f(x) having a local extremum that's either a local maximum or a local minimum.
So uh if it's a continuous function it means like it increases and then decreases or it decreases or and then increases one of the two. Now we already know that f uh has a local maximum here has a local maximum here. What happens to x on all other values? like let's kind of draw this gx thing like all all the way through and let's think about how this line behaves now. Bam. Okay.
At first the line kind of uh could go to negative infinity. If gx passes a comma zero then something weird happens. But anyway in our current drawing it's kind of it kind of goes like this. Zoom zoom.
keeps the angle keeps growing and growing and then here this is the first part where it changes direction it decreases uh and then uh it's kind of hard but something happens when this line is tangent to gx this line also has to be like tangent to gx at some point in here like at this point and then after passing this point fx this line has to go up again and then after this part the line kind of goes down again and down and down and down probably to negative infinity.
So I think what happens is that fx changes direction three times in this drawing. It starts like negative infinity or something. It goes up. It meets here.
It goes down for a slight bit. It meets here. And then it goes up again, meets here, and then it goes down for the rest of eternity.
So I think the number of values of x where fx has a local extremum is three. FX changes direction three times.
Okay.
Now this has to be greater than the number of values of x where gx has a local extremum. gx can only be one of two kind of like in two shapes. gx can change direction three times.
Uh wait wait wait a bit. Gx can change direction three times or it can change direction only once. It goes up and then down. So gx has either three local extrema or one local extrema. And because this is three, that means this has to be one.
Uh wait for a bit. My my web is lagging.
Yeah, this has to be one. Okay.
Now, uh do you have any questions in that part?
I probably have might have may have skipped over the part where gx uh passes through a comma zero. That is a that is a possibility that I have skipped. you kind of need to uh kind of need to realize that in that case something weird happens at this point also like uh and it's kind of it kind of becomes impossible for there to be two values like alpha and beta where fx has a local maximum uh in this drawing like this kind of has to be how gx looks for there to be two of these bumps where this line meets gx h.
Now I'm gonna assume that we now understand how gx looks. gx has to meet this line somehow in two places both tangent and gx cannot have more than one place where it changes direction. So like this part this part cannot go down too much like this part cannot be shaped like this. It cannot be shaped like this because it goes down and up here and that gives too many uh local extrema.
So anyway now we know that GS looks like this and that's basically all we get from A, B and C. And then the question asks given that beta minus alpha is equal to 6 <unk>3 compute the smallest possible value of m.
Now m is the slope of this line. m is the slope of this line. We know that alpha and beta are points where gx is tangent to the line. And we now know that this is 6 <unk>3.
And from here all we have to do is try to uh somehow make this situation into an equation and then solve that and get the smallest possible value of m. Let's let's try that. Let's try that. Okay. Um let's try that. Now gx I'm gonna do some trickery here because this all of this is pretty advanced, but I promise that I'm going to show you like the shortest path to the answer. Uh the most visual path. So, what I'm going to do is this. I could do something like minus x 4 + I don't know b x3 + c x2 plus dx + e and try to solve what these variables are. But this is too much.
That requires too much computation. So what I'm going to do instead is try to make it as simple as possible. Try to make this expression as simple as possible before uh putting in these numbers and trying to solve things.
I see that people are typing in the chat. Do you have any ideas?
I guess maybe you can look at the tangency of alpha and beta. And if you have a because you have M, maybe you can um maybe you can write a nice thing for F and therefore get something nice for G or something like that, >> right? You can get something nice for F because you have the tangency or something nice for G.
>> That is pretty much true. Yeah, you can get something nice for G using this condition. uh val zero says we can subtract alpha x so that alpha and beta become maximums and we can take the derivative and solve equal zero. I believe you are saying you can subtract the this line you can subract subtract this line maybe so that alpha and beta become maximums. I think what you're I think I understand what you're saying.
uh gx is some some quartic function with negative one as a leading coefficient. If you subtract m yep yep uh you could take the derivative of g and subtract m to get zeros at alpha and beta. Yeah, that's true. That's true. Uh just a quark. Yep.
Alpha and beta are no longer maximus after you make the subtraction. Uh I think what they're saying is this like imagine you take the uh sub you subtract this line and this function like imagine at each point you calculate this this distance. Yeah, this distance yeah becomes zero here and here.
>> That distance is going to look something like I mean technically the negative of that distance but whatever it's going to look something like this with alpha here and beta here.
And this is still a chordic function with a leading coefficient of negative one. And that means we can write it something like this. uh this line this is uh has a slope m and it passes through point a comma 0. So it's basically y = m to the mx - a and if you if we subtract gx from I mean if we subtract that from gx then we are supposed to get this function and we know that that has a negative like negative one coefficient and then there we all we know all the four solutions to the equation. Like it has alpha twice and it has beta twice.
Uh if you don't know that's like if you if it ever meets this thing in like this fashion, it has uh at least two solutions in alpha and beta. Yeah. Yep.
Yep. So we get this and we can just move this there and get Yeah. Get get that.
>> Should that be x - alpha cub and x - beta to the one or the other way around?
>> Uh wait, I'm sorry. Uh >> - alpha squar x - beta squared. We have x - alpha cubed and x - beta to the^ of one.
>> Oh right. If you have x - alpha to x - al cubed x - beta to the one, then it's going to look like this. I mean, it's going to look like this.
>> Yeah.
>> The shape of the graph differs by how many solutions you have at that point.
Now, this is like ideal enough for us to proceed, but there I I promise you that I'm going to do the one with the least calculations. So, here's what we can do further from this this equation. Uh we can simplify this even further which is that you know how it says beta minus alpha equals 6<unk>3 and it h we all we need to compute is m. Now what that means is the exact position of like the exact absolute position of a alpha and beta these things are not important in getting the solution at all. All that's important is that this difference is 6 <unk>3 and that difference is gonna gonna uh somehow decide how gx as a graph looks like and that's going to that's going to be enough to give us the smallest possible value of m. So we can put in alpha and beta whatever values we want given that they differ by 6 root3.
And if we do that time save, if we somehow get that intuition and are able to make a time save, then we can put in values here that's going to simplify this equation as much as possible. Uh what do you think? What do you think are the values we can put in? They are they have to be values that differ by uh 6 roo<unk>3.
Let a equal 0, alpha= 6 roo<unk>3, beta= 12<unk>3, says Sunny. Um, that's a good idea. But unfortunately, I don't think we know yet whether a and alpha differ by 6 root3. I don't think >> yeah I don't think that's >> right. I was acting under the assumption that it wouldn't matter. Um, then we can just let >> let's just let alpha be zero and beta behus 33.
>> Right. Right.
>> Yeah. Yeah. Plus 33.
>> Yeah. Yeah. Yeah. There are like there are two ways to do it. You can do 0 and 6<unk>3. But I think in the end what gives you the least amount of computations is negative -3<unk>3 and positive 3<unk>3 because they like cancel each other out and yeah they just become x - 27^ squared right -3 or 3 positive 3 or 3 uh yeah and then I guess we are trying to find out the largest value of m >> x^2us 27^ squ >> oh sorry sorry yeah that's true >> yeah and that is the simplest way we could write this uh I mean the simplest way I know that of writing this equation and then we could maybe differentiate this once and try to find the maximum value of m uh yeah we know like the only thing we have left is that the number of values of x where gx has a local extremum has to be one. And the way to uh compute that is to differentiate gx and find out where it's like max local maxima and stuff are and have the have just have it have uh less than three real solutions. Like the derivative of gx is a cubic function. If it has if this if this thing has three solutions then gx has uh three local m local extrema and if it has less than three like if it has one solution then gx has one local extrema. So we need this function g primex to have less than three solutions.
Okay, let's calculate G prime X. g primex minus this is just going to be m equals we are going to do a little trick here.
We do that and then we do x² derivative that's 2x.
No. Uh, we can put this here. Just plus m to this part. And let's move it down here so we can look at it better. Yeah.
Now it all comes down to have this thing uh have less than have this equal zero at less than three points.
Now this thing looks like this.
Uh I mean this this part uh wait it looks like this.
I think this part uh can be simplified.
I mean not simplified but if we move this over it simply becomes uh 4x * x^2 - 27. Oh hello hello Carrie. We are just uh almost at the end of solving the hardest problem in Korean SAT math history.
>> Hello Z. [laughter] >> Hello.
This part looks like this. Bam. Right.
This part looks like this. And we need it so that this the solution to this equation has less than three solutions.
So m has to be like what this or something. Yeah, it can it can be here.
It can be here. It can be here even because that gives us two solutions. But it cannot go any further below.
This is y equals m.
Okay. So I believe we are supposed to find the smallest possible value of m and m is larger than zero. So the smallest possible value of m is going to occur when this is m.
So now we have to find out the greatest this this local maxima of this graph and then just calculate them. Okay. How do you calculate uh the local maxima here?
Um you just differentiate it. Again I mean you there are techniques like about there like uh things about cubic functions that you can use to like derive it faster. What is happening?
Error. Error. Okay. Plus, wait, I I don't even need to do that.
This is This is just faster.
This is just faster.
And then we get 4 * 3x^2 - 27. That's that equals 0 at x = + -3.
This is going to be x = -3. We put that in here. That's going to give us m= m = 4 * -3 cubed minus 27 * - 3 which is 27 * 2 which is 54. 54 * 4 = 216.
And that is the answer to the hardest problem in SAT math, Korean SAT math history.
Bam. Any questions?
I think the really um at least unintuitive part is putting this as a slope, which is something that I that I think it it just makes sense only in this one specific scenario where you're like dividing by xus a. That's very >> right. Right. Yeah. I mean I I agree that that's the most unintuitive part like even in most uh solutions like there are um textbooks meant meant for students that have solutions to these questions even they like don't really solve it that way. They don't really try to solve it that way because probably the people who wrote the solutions thought it was too unintuitive. So they usually just like differentiate the both both sides of this to kind of get like things about FX kind of. Yeah, >> I would assume doing this the the the bashy way is not completely infeasible.
But like just doing this would be like like if you just want to improve your SAT score, this would be like it would be useless to you know present such a solution in a textbook because it doesn't tell you how to get more SAT grades.
>> Right. Right.
Yeah. I I think like because of this problem like this was like a completely new meta at the time. This was in the November of 2016 for the 2017 SAT. And this these kind of problems were never the meta at that point. Like this was just this just came in like and just just smashed all [laughter] all the competition for the hardest part hardest hardest problem in SAT math and then like it's still a legend. Uh nowadays the meta has changed a lot and SAT math do not have these kinds of extremely hard questions anymore.
But anyway, yeah, that's what that was it.
Okay, I deleted all that because it kind of takes up too much of my memory, I think.
Okay, >> it's just the storage, which means it's it's just the storage, which means that if you refresh, I believe it'll be gone.
But if you don't refresh, uh it's fine.
Oh, okay. I am currently uh wait I'm I think I'm currently sharing this screen between Yes. Sharing the X Excali draw between my Windows and my iPad. So I I don't think I can refresh.
>> Oh, I see. Yeah. Then then then you have to control memory, I guess.
>> Right. Right.
Okay. Now that we're done with that, let's return to the slideshow.
Okay, we had three blue, one brown. Now we're up to this. Yeah, we've seen the hardest problem. What does the full exam feel like? Okay, the exam I took the 2019 SAT. Now, I'm going to set the stage again, so you'll have full immersion. It's a cold day in 2018. Not as cold as the picture realistically, but we're considering the mental tool here that this fish has. uh you wake up a little before 6, eat breakfast while looking at your textbooks for the final time. Uh try not to think about the tests too much and go to the designated test sites every year. Sunning is held once in November. It is taken in one day for 8 to nine hours comprising the following subjects. Language arts, this is Korean.
Uh so that's the first one we take and then mathematics is the second one we take. And after that you have lunch and then you have all these other exams.
This one is optional. I didn't take this one. So up till here you can like pause. You can stop me at any point and ask me anything about SAT SAT stuff. Uh I'm fine with it.
>> We should we should have everybody take language arts and then Z take classical Chinese.
>> Right. Yeah, [laughter] >> everyone takes. By the way, I have to mention the language arts, the Korean language exam this year was the hardest it has ever been.
>> Oh, no.
>> I mean, it's not going to make it.
>> Yeah, [laughter] I believe more than 10,000 students gave up and left their exam site after the language arts exam. Anyway, another insane mental toll as you go into the math exam. Uh let's look at the okay let's go to the oh you get the exam papers like big big creamy papers and before starting there's a long tradition in swooning this part says it's a rudimentary way uh what's what's this yeah oh I'm sorry I for I read my script write the following text in the certifying statement section in the answer sheet this is a rudimentary way to check and compare prepare your handwriting in case any problems occur. Uh, this year's certifying statement was exceptionally moving. It said, "Never have I seen one as lovely as you."
Now, remember when I asked you to try the exam and I said, "I will give you a plus one point bonus if you write the phrase never have I seen one as lovely as you in your own handwriting and send it along the answers." Trust me, this is important. Yep. Uh we've got all these amazing handwriting from the solvers, even a long one from Sunny. Uh the phrase is from a poem by Kim Namu, the poet of po the poet of love. Rest in peace from a poem called Letter. Uh there were no English translations, so this one is my amateurish translation attempt.
Okay, I'll give you some time to read probably. Should I read it? Uh I'll give you some time. I think you can read the Korean version so we can get a sense of immersion.
>> Oh, right. I should read the Korean version. Okay. Um Clap clap clap clap clap clap clap.
Thank you.
Right, that was the letter and now like you are so touched, you're moved, you're crying opening this first page of the math exam. Okay, let's go. Time 100 minutes. Total score 100 points, 30 questions, 21 multiple choice questions, nine short answer questions. The answer to each short answer question is an integer from 1 to 999. Calculators are not allowed.
Bam. Let's go. Um, let's go to the test.
Okay, I have the test here to BAM. This is kind of the format. It looks like this is the BAM.
Uh, let's see.
I think I want to since there are people here who have tried the full exam, I kind of want to ask you how you felt about the exam as a whole. Like how what were your expectations going in? What were how did it actually feel? Do you think it lived up to the hype or no?
>> I mean, this is definitely harder than the standard like American SAT because I know like the standard American SAT you're expected to get full marks, >> right? like as in like the standard is you you should get full marks and this one is just I mean h I feel like it could be practical maybe to get full marks. I mean it's definitely possible.
I've later tried um the last question which I wasn't able to finish and it's not like terribly hard um but I couldn't do it. So that's Yeah.
>> Yeah. Yeah. V says it was hard mostly because the questions were so time expensive and so little time. Yeah, that's true. A lot of the part like Yeah, I think a lot of problem design goes into making it so that there are multiple ways to solve each problem and you can save time by doing something uh cool or intuitive or something creative or something like that.
Um I think it's also this is there's a surprising amount of statistics that when I like wrote up the solution document I realized there's a lot of statistics.
>> Yep. Uh that's true for only this era.
Like the meta has changed a lot throughout the entire ex existence of Sunning. Like there [snorts] there's like uh I have divided it into eras.
This is the website I made. Uh like this is the new meta. You can see that 2019 is kind of an old meta uh for for today.
Usually like in these days they have a completely separate section for statistics. So the main part of the exam handles only like uh arithmetic and basic calculus and then they have >> like I'm sorry >> there's no distinction between the type guy and type.
Oh, sorry. Uh, I'll go through that part just after this part. Okay. So, right now the meta is like uh there's there's a you can select between what type of elective things you want to do. Like there are three types you can choose from. There's probability, probability and statistics.
Uh if you choose that you need to solve these eight questions here. You can also choose calculus that handles like calculus of uh trigonometric functions and stuff. You can take that that's these four papers. And then you have geometry. If you pick geometry, you can do these four pages. Now this is a new meta. The meta is also going to change in 2027. The meta changes a lot. Anyway, in my era, all those three elective types were in one test. So, I had to take the three at the same time. Uh, probability and statistics, uh, calculus and geometry. So, this contains probability and statistics as much as it contains calculus as much as it contains geometry, which is probably why you felt like there are a real lot of statistics questions.
Yep. And then you also asked about what this type ka means. So ka is just the first letter in Korean alphabet.
>> Research. So I did do research but like it the distinction was removed. Yeah. In like the 15 reversion or >> uh yeah in this era there was ka for people who wanted to go into engineering and stuff and na which is for everyone else. Um so this one definitely harder has more covers more topics nowadays it's all in one exam you if you just if you want an engineering you just choose calculus within the three elective types and that's what you do today. Now if the meta changes again in 2027 the GA part will be completely gone everyone will only take uh arithmetic and calculus no advanced calculus in starting from next year. Yep.
>> See?
>> But anyway, that's what's happening right now. Uh let's wait, where's the where's the thing? Bam.
>> Okay, >> one brief question. So, I saw that you're um you said that it was like 100 points total. How are the points distributed? Because I assume you might go into this with like favoring some questions over others or something, >> right?
>> Oh, cool. So from 1 to 21 are the multiple choice questions. From 22 to 30 are the short answer questions. Each section is supposed to be in ascending order of difficulty. So 222 then 33 and then from here 444 all all the way up to here and again from 22 three and all the way up to four. And the hardest questions are supposed to be this one, this one and this one. At least in this meta. Again, the meta changes a lot. Nowadays, it's not the meta.
Anyway, that's how the points are distributed.
>> So, you know that they're ascending, but as a student, you don't go into this knowing the actual point distribution.
Exactly.
>> Uh, no, they are written here.
>> Oh, cool.
>> It's always fixed. Yeah, it's always fixed within the meta.
Now, yeah, usually you would just go from here to here, go on on and on and then if you get stuck, you will move on here and then try some of these. If you get stuck again, then you will look at harder questions and then try to solve them stuff. Yeah, I think what we should do is I don't think we have to go through every problem. I think we can because we're a math club, we only need to look at the mathematically interesting ones, right?
>> Let's start at problem one, which which actually I think problem one is really weird like I I don't understand what it means to take the sum of the components like like it it seems such a weird thing to ask for, >> right? Uh let's say okay, I've divid divided the problem into a few sections.
This is the easy part. And the easy part is supposed to be like basically memorizable. Basically the it's very fixed in the way they present themselves. And yeah, this type of question has just been the meta for problem number one for a few years. And you just see this and you just know what it means. But anyway, uh I think it makes more sense in Korean. I I don't think Yeah. what is the sum of all components is really a that big of a thing but yeah components is a concept that students learn it's just the these are components >> because yeah >> I would assume that the components are just like you you can break them up into vectors so you would the components would still be vectors so it would be weird but like I guess the only natural interpretation is to just take take them as treat them as scalers >> yeah And like because the these early parts are so uh should I say they are so optimized that people tend to develop like ways to even like solve this faster. You probably just add this first add this first and do minus one time plus 2 * 3 and that gives us kind of a faster solution like just a mere seconds or two but yeah like you're speed funny.
>> Yeah. Anyway, like these these are speedruns too. You just look at this and you just write 3x instead of this and then yeah, you just go like delete x and then you get five over three and stuff.
Yep. And then you have like internally dividing the line segment a in the ratio 2:1. Uh that's al so something very like a minus 2. This has to be divided into a ratio of 2 to one and this has to lie on the x- axis which means the y-coordinate has to be zero. So a satisfying this has to be four something like that.
>> Yep. It's basically a speedrun just to save time for later questions. You are correct. And even in this part like uh I believe the intention behind this expression is that you are supposed to re recognize that A and B complement are mutually exclusive means A is contained in B.
But if I were a student like solving this exam, I wouldn't do this. I just have a fixed strat for solving this kind of questions. I just do this. I just write a A complement B complement. A and B complement are mutually exclusive. I just write this part is zero. PA equals 1/3. This has to be 1/3. I write 1/3 here. P A C this B. This this has to be 16. And PB I add up these two. I do 1/2.
And that's faster than thinking about mathematical concepts. And it's also like safer because the more creativity you have to think about, the more uh faulty you could get, I guess. Yeah.
Anyway, that's what happens in the earlier questions.
I think the one I'm going to touch upon is this one. Parabola y^2= 12x.
>> Oh yeah.
>> Now, yeah. Parabola y^2 = 12x. Now a parabola is defined as a curve where in any point in on the parabola if you shoot a laser at this thing uh you're going to get you're going to get a what I I mean if you shoot a laser from the focus to here you're going to get a line that's per perpendicular to the x- axis. That's kind of a I don't know a definition of the parabola.
Yeah.
>> And you usually uh I don't know if this equation is that well known to us students but this thing is equal to 4 p where p is the coordinate of this focus.
So here you can just instantly know if you know this know this expression you just instantly know that the focus is at three.
>> I definitely don't know this equation and took me a little bit of time to derive it but it was possible.
>> Right. Right.
>> It was quite painful and I I is it the is it the norm to say y^2= 12x instead of x^2= 12 y? It's kind of weird because yeah, this is the this is in the geometry section and in the geometry section they weirdly use this more. I think it's because they want to have some like uh >> maybe something similar. Yeah, it looks similar to the to this. You know, this is the hyperbola and >> but like I would want it to look similar. [laughter] >> Yeah, I I don't know. But like so I ended up doing like a changing X and Y and it just confused me a lot more and >> yeah yeah >> this was >> also uh maybe this is also less known in the US but there's a thing called directric here uh at a coordinate that's just a negative of the focus and the directx has this uh feature of um being equidistant as the focus So if you are on any point on the parabola then these two lines have equal distance.
>> We were definitely taught this at some point. I just forgot the equation for it because it was in like uh >> right >> I think it would be equivalent to maybe 11th grade math >> or something like that.
>> Yeah.
>> Right. Right.
Okay. Anyway, so the fastest way to solve this would be PF equals 9. This is 9. This is also 9. So the x coordinate is this is my3 so this has to be six.
Yep. And then we could skip the next few questions. Uh from here it gets to medium and this is where you need to think a little bit. So for example here you have a function you have its inverse. What's the value of g prime of f of -1?
Now here um if you ever uh get if you ever get an expression like this what you have to think about is that f and g satisfy this f of gx is always x g of fx is also always x.
Now this kind of looks similar to this but there's a prime here. What if we d take the derivative of both sides? we get g prime fx times frimex equals 1. And that just gives g prime f negative 1 is just one over frime 1. And that's how you save time here and do just the derivative.
>> There's also a geometric interpretation where you take the inverse as reflection across the y= x line and you get the same result.
>> Yeah. Yeah, >> when I did this, I kind of just went, uh, I don't want to deal with this. And I just, um, computed G, computed G prime, computed F, computed F of negative 1, computed G prime of that.
>> Yeah. Yeah. I mean, that's the thing you can do like Yeah. As I said before, doing like thinking about the concepts and actually trying to do something creative could result in making a faulty uh because I know this is stuff that I am likely to get wrong, so I didn't want to.
>> Yeah. Yeah. Yeah.
And there's some probability stuff.
There's some equations. Another more pro counting stuff.
>> I guess 11 is >> geometry.
11.
>> Yeah, I was just mentioning 11 because this is I like I don't know how how obvious it is how to how to solve this.
Is this just like is this >> you just plug you just compute the what's it called discriminant of this formula it's going to be like a b ^2 - 4 ac >> like I guess it's obvious to the person that this is you you you just did directly take the discriminant change to sin square I I like how I like how it uses like quadratic quadratic stuff like twice which is funny.
>> Yeah. Yeah. I like that. Like that as well. These are some like pretty neat problems. I think Oh, this has to be less than zero. And then there's cosine and s. So you have to change cosine to sign >> for it to work as a quadratic equation again. And then you solve this. And then >> you kind of draw out how a sign looks.
And then you kind of try to deduce alpha and beta.
>> Bam.
Um, yeah.
14.
I I don't think I really have to explain it. Yeah. Uh, I believe Sunny got this one wrong because of the mistake, but it was just a mistake. [laughter] >> It was a simple mistake. So, I could like, but like if I was taking the question, I would probably try to solve this question like three times because I know I could get this one wrong. I [laughter] I could I know I can make a mistake here. Yeah.
>> I I I don't know why I thought it it it it worked that way and I I like I really don't know what happened [laughter] basically.
>> Right. Right. Like it looks simple. You kind of just do Yeah. This is equal to 2 to the minus F * G. This is equal to 2 to the minus 3 G. And this just means minus FG is greater than or equal to minus 3 G.
>> Yeah.
>> Take the take the negative away, you get that and you have to kind of you can't just divide it by g because you have to divide a case where g is positive and where g is negative and that just in the end it ends up giving us 1 3 4 5 I believe. Yeah.
>> Like I don't know why I believe that you know if you if you if you like like I kind of just believe that at two because f is zero you get zero which is less than or equal to everything which is just not true >> right.
>> Yeah. I mean, I believe this question, a lot of students got this question wrong for a problem at number 14. I believe it was a very mistakable question.
>> Yeah.
>> Now, what's that for that? Yep. Now, we're getting to the hard section.
>> Okay. This is where I think it really gets interesting. From here on, I'll try to at least mention all the solutions to the problem. Now, this one was unreasonably hard for number 16 at the time with the meta. I remember myself getting stuck here uh going going to number 17 and returning here all the way after I've looked at all the problems uh afterwards. So, I I just had to reset my mind. I was not solving this at at first, [snorts] right? Um, so >> I still can't believe I messed up you substitution that badly, [laughter] >> right?
>> I Yeah, I believe there are two solutions to this. Uh, not not as in two different solutions uh two different answers. I just mean yeah two ways to solve this. Um, one is noticing this and trying to do some sort of x equals 1 over u substitution.
It kind of works, but you have to be creative. You have to recognize that it kind of uh adds up to something. Okay.
If if we make the substitution, it ends up being something like this, I believe.
Yeah. If you make this substitution, then this is equal to this. But then this again this alone is not enough to solve it because like even if you try to divide this into one to two f and then plus half to 1 f and then only substituted this part that's going to give you something like 1 to 2 fx plus 1 /x^2 f 1 over x dx.
And this is not this because there's this two here. And I believe this is the first thing I tried in the tests while I was checking test and I got stuck.
>> I see.
>> Yeah. But what you can do is just let this be I. Let this be I and then have 3 I equals this plus this plus this and that just gives us 2 fx + 1x^2* f of 1 /x and then this just solves to this. Now that's one solution. I I went through it pretty fast, so I don't think everyone would have gotten it. But that's because I wanted to talk about this next solution because I think this the next one might be uh dare I say uh a more complete solution because what you can do is notice that this is this holds for all x larger than zero. Just plug in 1 /x 2x like just plug in 1 / t then you get 2 f 1 / t plus e^2 ft t equals what am I doing equals t + t ^2 we did this substitution and now this is where it kind of gets creative like look at this fx f1 overx x. This is f(x). This is f1 /x. This is just like a linear equation with with two variables and two equations. You can just solve this completely and just get what f(x) is. This is just like these are variables. You just like subtract this from this and then you just get f(x) and then you just compute f(x) dx.
I mean in fact you can like multiply x squar to the original uh equation and then you just get something that's way nicer >> right right that's true you just get 2x^2 and then this is one this is x this just one and then >> then the solving also be much easier >> time this times two subtract from that whatever something like that yeah that's going to just solve fx now uh the next One is a proof question. This is a long one. You are supposed to read through a proof and follow it. This is uh they have been pushing this for a while. I think this existed from the start of Sunn this type these type of questions.
They really like to test students whether they are able to read a long proof. I mean it's not that long but if to read a long proof and understand what the proof is saying.
It's a like pretty interesting proof like not that complicated. Of course, it has to be something that a high school student is familiar with. All the logic has to be familiar.
>> I think I've seen this exact question just like without the intermediate steps like just asking for the answer in like some competition math.
>> Oh, right. Right.
>> Definitely after 2019, >> right?
>> Yeah. If if they like remove this part and just ask ask it then I guess it will be like a more of a math competition test.
>> This would be like question 20 or something [snorts] >> right? [laughter] Yeah. Yeah. In this test and then here we have some sort of geometry with limits.
Now this one you are supposed to just do all this stuff. Yeah, like like the correct way to approach this is just to uh AB equals 1. You just all it's telling you all it's telling you to do is find the area of this part. Find the area of this part. All you do is actually just write out all the things tangent theta. This one is like one over cossine theta. And then the these two angles are the same. So this and this has a ratio of this and this which is like this is s over cosine. So it's actually one to sin theta. One to sin theta. And you just do this and get the length of this part.
Get the length of this part. Get the length of everything else. Get the length. Get the length. Just compute the area. All the things is what you're supposed to do. But after you do that, uh, sorry. Yeah.
>> No, I was just laughing at how this is like just I think completely implausible to do if you're actually like wanting to finish everything in time.
>> Uh, right. Oh, no. But, uh, I actually want to say it's very um it's very risky for students to go for intuitive solutions in a question like this.
>> That is for Yeah. Um I >> even for this question the only reason an intuitive solution that I'm going to show from now on is works is because this part only consists of division and multiplication. Like if anything here if anything like if there was an equation like I don't know this I don't know exactly what it's going to be maybe maybe this like well if it told us to calculate something like this if there's a subtraction or addition or anything then all the intuitive solutions kind of crumble down because then you kind of have to like go for I don't know uh >> I mean I would probably >> tailor expansions and stuff. Yeah. Oh yeah, because well yeah right but like I also think a question is not going to do that. [laughter] Um but yeah I it's definitely very worrying to to to do things intuitively when you have limits because >> Yep. Yeah.
>> Yeah. I I so I think like I want to say that this is usually where students use the time that they have saved in the first part because I think every even the like private tutors tell students to actually write all the all the things down because it's so easy to make a mistake if you want to if you try to like actually send theta to zero and try to draw what happens and stuff like it's so easy to make a mistake. So yeah, I think I solved it like this as well when I was first doing it in the exam site.
But yeah, anyway, after you solve it, there is a way to kind of intuitively check if the solution is right. What this one is telling us is find the area of here and here when theta approaches zero. Now when this angle approaches zero, this triangle is going to look like this.
something very thin.
Theta approaches zero. This is a right angle. This is also going to approach a right angle almost 90°.
This is supposed to be dividing this angle in half. So that is going to approach 45°.
Like this is almost going to be equal.
Now this is one.
Uh this thing is going to look like this.
And this thing is going to look like this. Now, uh because this this and this are equal and this and this are also equal, we're we're just going to say that they are equal. Now, uh this part um let's start from here. The length of this part approaches theta because this is one. This is theta. This is basically the arc length of like radius one angle theta. This is just almost theta. Which means the this is also theta. This is all theta. Uh the area of this sector is going to be half times r 2. That's going to be this this squar time theta. Now this r is 1us theta. This part is 1 minus theta. But that's basically one because theta is approaching zero. So r is basically one. So the area of this part is just half of theta. And the area of this part, this is basically a right triangle with length theta and theta. So this part has an area of half times theta* theta.
So that's going to be this. And now you just plug this here. Plug this here and you get uh what's it quarter theta^ 2 over half theta squared and you get a half.
And you are not supposed to solve like this. This is only supposed to be used as a way to check if your intuition is correct, if your solution is correct.
Anyway, that doesn't work. Like surely this isn't that bad. Um I I can see how it breaks down when you do add and subtract, but also um this is like an intuitive method that kind of just doesn't also just fails intuitively when you have add and subtract.
>> Yeah. Um, also this is kind of in the on the easier side of these kinds of questions. So if it gets harder than this, then sometimes your intuition is like kind of gets confused >> complicated shapes >> and it's definitely safer to just do it than >> Yeah.
Okay. 19. I'm just going to do the first half of 19 which is there's a tetrahedrin ABCD whose face BCD is an equilateral triangle with side length 12. Let H be the perpendicular foot from point A to plane BCD. So we know that BCD is an equilateral triangle.
>> Wait, no perfect opportunity. Can you construct a perfect equilateral triangle in Excel?
>> You can't. Perfect. Uh, square.
Um, turn it 30°.
I mean, I need to copy it first. Turn it 30°. Uh, >> can we just have this thing >> straight edge and compass constructions?
>> I don't think so. Wait, I need the thing to be even. But >> because you like >> I just do >> you just make a circle and then like make >> you can't take the intersection that easily. I'm I'm not sure if compost straight edge would be good.
>> Okay, first uh the shape of the circle is not promising. Um view mode where where where's the thing?
Oh, disable zen mode. Just do that. I don't think it's going to work. I don't think it's going to give me >> anything useful about points.
>> So, if you take like if you take a line segment and then you make uh two circles, one at each side of it, um uh sorry, centered at each side of it. Um >> I I I think yeah, Tusoft just wants the uh compass straight edge construction, but then you kind of have to take the intersection of the two circles, which you >> you don't have to you don't Sorry. Oh.
Oh, you're saying, "Oh, Excaladraw doesn't allow you to do this."
>> Yeah.
>> Yeah.
>> Is that not legal in the rules of construction? We can't just like use that point.
>> Yeah.
>> Uh, unfortunately, for all rules of PowerPoint construction, I don't think we have allowed that.
>> Oh, no.
>> So, I think the way to do it is to rotate this thing the square by 30 degrees. That's going to give us uh something like this. We know this is one. So this is a half. This is roo<unk>3 over 2.
And this is another half. So this whole thing is uh 1 +<unk>3 / 2. And now we do uh subtract 1/2 from this length which is this. Uh oh wait, how do I snap it to that? I I don't think it's possible yet.
I have to look more into it. Okay.
>> Okay. This is this is possible because um Excalador has a snapped object. If you right click the canvas, um, it has a it has a like it's it's kind of like a total grid option, but this snapped objects.
>> Oh, right. Right. In that case, we do have we do have root three.
>> I actually did this construction when I was making the graph for this thing.
>> Right. Okay. This is root three given that this is one.
This is one. This is root three. Bam.
Anyway, uh back to the triangle. I'm just going to eyeball the triangle with a with a pencil tool. B C D.
>> Yay.
[laughter] >> Just as Kais intended. Now uh H is a point inside triangle BCD such that triangle CDH has three times the area of triangle okay let's look let's H triangle CDH has three times the area of triangle BCH so this is like 3 A this is a triangle DBH has 2 times the area of triangle BCH dBH this is 2 AH= three. Okay, let's ignore that part. Uh, all I want to ask is this. What do you think is the best way to determine the position of this point h given that this is an equilateral triangle with side length 12.
This is a to 2 a to 3 a. Interesting. By the position I mean like for example the distance from this point to each of the three sides. Bam. Bam. Bam. Does anyone have an idea on how to calculate these lengths?
>> Wouldn't it just be factors of one two and like like if a is Yeah, I think like just the height of it would have to be like ratios of one to two to three, >> right? Right. Because these are all triangles with the same base.
>> Yeah.
>> Let's write K 2 K 3 K. Okay. Now the next question is how do we get K? How do you think we can get K?
Um, I don't know if I I probably missed some details, but I think we know the area of the entire triangle. So, that means that we can just like figure out the whole area and then divide it all by six and start multiplying by.
>> That's That's correct. Yep. The entire triangle. It's an equilateral triangle.
So, uh, what was it like something like this?
Yeah. And then that is equal to I mean six of that is equal to this little triangle because a + 2 a + 3 a equals 6 a. So this part has to equal the sixth of the entire triangle. And that's how you get K. And after that it's just geometry after geometry and you can get the line segments AQ somehow. Bam bam bam.
>> There is a separate approach where you can extend BHC and DH um to meet at the other side. So for instance if you extend BH to meet at let's say I don't know the choose your own letter then you can get um C to that point. Um so if you extend uh BH to let's say E then you can get CE to E D is equal to the ratio between the areas of BCH and BHD and therefore you would be able to get CE and CE to ED >> right >> because C CE to ED would be equal to the area of BCH and the area of BHD. And this allows you to this allows you to get, you know, all of these uh all of these proportions and you can solve similarly.
>> Right. Right. And yeah, I've read through the solution by Sunny and there's a theorem that I didn't know the name of, but yeah, there it was a great solution.
>> A lot of area manipulation like the theorem theorem itself is also just fuddling with areas.
>> Right. Right.
Okay. That was problem 19. Okay, we're going toward the end of the multiple choice section. 20. Now, this here is where I where I think it gets like harder. I'd say harder.
You usually say killer problems in Korean SAT. Killer problems. This one is like uh semikiller. This one would be uh killer. This is also an easy problem for a killer, but anyway. Uh, oh, by the way, did I mention like this this test was not that hard as the Korean SAT math? In terms of Korean SAT math, this exam was slightly easier than usual.
Slightly easier than usual.
Uh, anyway, back to the question.
Consider all lines passing through some point that are tangent to the curve y= sinx.
Now I think we are supposed to draw sinx kind of [snorts] now the point minus that's here and we are supposed to draw tangent lines like this.
Um bam.
Whoops. Bam. Bam.
Something like that.
And let a n be the nth number in this list like all the x-coordinates of the points of tangency. So this is a1, this is a2, this is a3 and so on.
Pretty okay pretty interesting. Now which statements are true?
Tangent of a n equals something something. Now the only thing we know about a n is that it's tangent like this. So there's an equation I use when there's the the there are these kinds of tangency. Uh what I use is that the rates of change there are two kinds of rates of change. Uh there's the insta in instant rate of change and there's like a I don't know how you call it in English. There's like an interval rate of change. Anyway, what I'm what I'm what I'm supposed to say is the slope between this point and this point is equal to the instantaneous rate of change at this point. Just the slope of the tangent line to this point. And this function is y equals sinx. So the the the the the prime of that function is just going to be cossine x.
That's the instantaneous rate of change.
Right? Average rate of change. Okay.
Right. So what I'm going to do is write down the slope the equation for the slope between these two points which is um x - x then y minus y. That's the slope of this line which is equal to the slope of the line at this point which is cosine a1.
Now this minus0 doesn't need you need you don't need that. Put this here put this below here you get s over cosine which is tangent and that gives this equation. So this is true. Yep. This is similar to problem three from before. Yeah, this also uses the rate of change thing, slope thing.
It's often used in like these sorts of situations when you have a line tangent to a curve and you have two points on a line.
Now after this part you can kind of think about what these things do and then stuff but I think the best way to solve uh intuitively solve this part is just to draw this again draw this again which mean I mean like draw tangent draw the tangent curve like this kind of bad kind of bad drawing but anyway this is tangent x and we're going to draw a line which is x + pi / 2. This is pi / 2. Now the fact that this equation holds for all a n and means that these are just a n. These coordinates are just a n a1 a2 a3 and so on.
Now, uh, drawing this kind of gets hard.
So, let's just open up Desmos.
Desmos. Okay. Um, y = tangent x and then y = x + pi / 2. And cheater cheater.
Okay. Uh, let's see. So, this this is how the thing looks like. This is a1. This is a2. This is a3. This is a4 and so on. I mean the x coordinates. Now what it's asking us first tangent a n + 2 minus tangent a n is it larger than 2 pi.
For example tangent a n + 2 like the difference between the x coordinates of these two. Is it larger than 2 pi? Now maximum maximum cheating. Okay. Bam. Now observe how this is 2 pi because tangent has a uh what it it repeats every pi.
>> Period.
>> Right. Yeah. Period. Period is pi. This is 2 pi. This has to be more to the right than this. So, so this is true.
Now, then let's look at this one. A n plus1 plus a n plus2 is larger than this. Now, what you have to do here is that um I mean you can just do it as just just like this. But you can also move these around and have a n + 1 minus a n is it greater than a n + 3 minus a n + 2?
And this is basically asking is this larger than this?
Is this larger than this? Now you see uh the tangent kind of gets steeper and steeper as you go this way.
Yeah, this seems totally viable. Yeah, true. I I remember I vibed this question out using this using this drawing. It just because the tangent gets like steeper and steeper as you go to this side. This just approaches just just approaches pi more and more. the difference between this part and this part like the difference between this part and this part this little part is just gonna get closer and closer. It's going to reach zero. Yeah.
And I think that is enough intuition to just say that this is correct. Uh of course you can do like calculus and stuff to check that it's correct but yeah I think that's basically why it's correct.
Okay, number 21.
Now, I believe this one uh was hard for some people them.
Uh the first one seems plausible and the others fall out.
Yeah, true.
Okay, let's go. f a function fx differentiable on the set of all real numbers satisfies the following. What is the value of f1?
Huh? And this is a kind of a new question. This I've never seen this exact equation before before the the SAT.
But when you get f and fprime at the same time, usually what you want to do is integrate because when you differentiate f, that's why you get fprime. So if you can find a way to integrate this and remove fprime, then you're going to be left with a much more simpler equation.
So, oh by the way, this is not a set symbol. It's just like a >> par. It looks like you want to put the two on the right side so that then the left side just integrates naturally.
Wait, sorry. Just kidding. Uh, >> right. Okay, you can do that too. But I think the constant this constant will not matter that much when you integrate because >> yeah, imagine you are.
Imagine you are uh yeah taking the derivative of f2 the^ of three. The curly brackets are just supposed to be parenthesis. It's how you write things in Korea. You you just write them as parenthesis. Yep. Just grouping.
Uh yeah, Korea uses all the all the all the all kinds of things to group things.
You just use this one first and then this one and then this one.
China has this. What? That's the ordering is different.
>> That is what I learned as a child as well. Um >> Okay. Okay. That's an interesting difference. This is Korea.
[laughter] >> Interesting.
>> I mean I I think these these notations are equally bad.
Right.
>> Yeah. Ideally, if you type set your math well, then it'll be readable regardless, but not everybody does that here. Okay, let's integrate this.
Integrate both sides. You know how if you try to take the derivative of this, you get 3 f(x) squared and then frimex. And this part is basically just this. You just have to multiply uh what's it?
Multiply 2 over three like this. And then you know that this is the anti-derivative one anti-derivative of this. And this part is very similar. Uh, but you also need this two in front of the X. Wait, what just happened? My Excali draw bugged. Where did Oh my Where did it go? Where did it go? Okay, it's it's fine. I have it saved.
Bam.
Okay. So this becomes like this. And then the right side becomes uh this is what it would be if this 2x didn't exist. But because this exists, you have to multiply this with an additional half because that way when you take the derivative, this three is going to cancel out with this and this two is going to cancel out with this.
And then you will going to be left with exactly this.
And then of course since we are taking the anti-derivative, we need to put a plus c there.
Uh we can like do something better. We can multiply both sides by three and also multiply both sides by two.
This becomes four. Uh this tech technically becomes 6 C but C was just a random var variable we picked. So it doesn't matter. This is the uh what we get by integrating both sides. And now we know this this and we need to get this.
And here is where it where it gets kind of creative. You need to play around with numbers. This is like basically number theory at this point. You how do you get f of negative1 like realize how you if you put negative1 here this becomes negative 1.
This also becomes -1 because -1 is a fixed point of the line 2x + 1. 2x + 1 looking like this.
Yeah, it passes through like -1 comma1.
>> That should be under the x axis.
>> Oh, sorry. Negative 1 comma negative 1.
What am I doing? Bam.
And that gives us like what is it 3 f to the negative 1 I I mean fative1 to the 3 equals c. So now we just have to get c and that's going to give us the solution. Just a quirk sees it. Just a quirk. What did you see?
that >> um I see that if you take um six f of six and kind of like reverse the 2x + one three times then you get a negative 1/8.
>> Right. Right. Right. Uh or in other words, if you put negative - 1/8 in here and take 2x + one of it three times, you get six. Yeah. It's kind of it's weird.
You >> wait. Say that again.
>> You do what?
>> You put negative 1/8. Like imagine this is 2x + 1. This is x. You start fromgative8.
You you put it in this function. You re you you plug that amount again. So you're doing this.
You are kind of doing this and you >> second entry six. Oh, you arrive at six.
Wait. So what are the intermediate?
Let me show you. Let me show you. 4 f to the 8th cubed is equal to putgative8 there. You get -/4 + 1 3/4.
Okay. And then we're going to put 3/4 in that again. F 3/4 cubed is equal to put 3/4 in x 3 over 2 + 1 5 over two and then we put 5 over 2 inside here again then we get * 2 + 1 f6 cubed plus c and that's how we can get C because let's say this is one this is two we know that this is 4 - C 4 - C plug it in here 4 * 4 - C so 4 * 4 - C is equal to this and we get this two so we get two equations with two unknowns so we can get C. Uh, let's just do 16 - 4 C and then move the C aside. 5 C and then plug this in here. So, 16 * 4 - 5 * 4 C = 2 + C 21 C = uh 62, right? C equals Wait, am I wrong? Did I do something wrong? Because I don't think this is the >> Yeah, there's definitely Right.
>> Wait, where where did I make a mistake?
16 - 5 C.
Uh uh uh uh wait.
Oh, f6 is equal to two. I was supposed to plug in eight here because this was f6 cubed, right? This was actually F6 cubed. This had to be 8 + C. Now we get 21 C = 56. C is equal to uh this is 7 * 3 7 * 8 and that is C. And remember how F - 1 - 1 was the fixed point of this 2x + 1 function.
We plug in minus one here. We just get 3 F - 1 cubed equals C. And then this is just 38 9/8 we get uh two and then 9 to the third root and that's going to be 2 * 9 I mean what is it? This is 3 to the power of 2/3.
So that's equal to something like this.
>> Yeah.
>> And yeah, that is this. So it's number four.
Uh Sunny says if you let f(x)= g(x) + 1, you get 4 g(x) cubed equals g2x cubed + c." Right?
So fx [snorts] equals g. Wait, do you do it do it at this point?
>> Yeah, I do. I do it here.
>> Well, you get * g of x + 1 cubed equals g of 2x + 2.
>> Um, but then you can just rewrite x + one as x, >> right?
And then that's basically 4 gx cubed equals g 2x.
>> And then if you plug it to - 1 over 8 and 6, you get 7 over 8 and 7, which is more obviously solvable.
>> Uh right.
Yeah, that makes sense. That makes sense.
Um, anyway, it's it all has to do with this image, I think. I mean, of course, you're not supposed to draw this to solve it. You're just supposed to plug these in. But yeah, uh, they meet nicely. Minus one goes back to itself nicely. Cool. Cool problem. Okay, that was problem 21.
Hardest problem in the multiple choice section. And then we have all these uh only a few nine questions left. What if we had other values would we be able to solve it? Not within the definition like not with just the method we did and I don't know if this solves as a differential equation nicely into anything and I don't know if we can actually get f but it is differentiable on the set of all real numbers. So, I don't know. I think maybe it becomes something nice. It's meant to be something nice. Maybe. Yeah.
>> So, you think snow, you can manipulate the differentiable function pretty much arbitrarily >> in a way that still satisfies the first constraint?
Uh yeah I don't know for sure because it's like one value of x is effect affecting another value of x another value of x to be something >> clearly if you if you like if you take let's say my g and then you just say you know I I take like let's say a log base 2 before my g and then I uh subtract by the necessary constant uh then it's really just equivalent to finding a periodic function and it just so happens that I have the periodic function uh at at like two different at like two different places which correspond to the same period and therefore I can solve but if it's not on the same period I wouldn't be able to solve for C like at all.
>> Right.
Right. Okay. I kind of get it. I kind of Yeah, I I get the argument.
H I do believe like it was not considered while making the problem because that's not what students are meant to do. It's not even in the textbook the part differential equations and stuff. Yeah, I I don't know.
Let's go to the short answer questions.
And the first part is really easy. It's easier than the multiple choice questions. This part is also pretty easy up until 28. 28 has something interesting. So I'll just uh let's take a look.
Consider this ellipse. Now this is a square of the coordinate of this point.
33 that's a square of the coordinate of this point. And 49 minus 33 which is 16 is supposed to be the square of the coordinate of the focus. So this is four.
Now four point P on the circle. Bam. So it's a po circle with a center on Y3 and with a radius of two radius of two.
Let q be the point where a positive y with a positive y-coordinate where line frime P meets the ellipse. And now it says compute the greatest possible value of PQ + FQ which is this plus this.
It looks daunting but uh you have what you have to recognize is that an ellipse has a nice feature where any point if you connect it to the two folky fosai if you connect it to the two folky the length of frime q plus fq is always equal to the this major axis of the of of the ellipse which is 14. So this is simply 14 minus frime p. And just as just a quirk said maximizing this is equal to minimizing this because of the negative here. Yeah. So we need to me minimize this length.
And how do we minimize this length? I hello.
We can minimize this length by thinking about this triangle connecting this point to the center of the circle and then connecting the P p to the center of the circle. Now we know that this length plus this length is always larger or equal to this length because of a triangle inequality.
And this is the same as this uh because it's the radius. So the smallest frime p can get is when frime p is on a line that extends to the center of the circle. If p is here then this length is smallest.
So all we have to do now is get this length and that is easy because we know the coordinates of this and this. So this just becomes a right triangle with length four and three which gives us five here. And we know that the the radius is two. So this is two. This is three. This is three. 14 minus 3 is 11.
And that's how you solve question number 28.
Bam. So the context here for cinepher is that we are solving the exam Korean SAT math exam I took I actually took in 2018 and we only have two problems left the two hardest problems and after that we're going to have a fun session where we solve Korean math questions in Korean. I've prepared a lot for it.
Okay 29 is fun and 30 is daunting. 29.
I'm gonna uh I'm gonna leave Yeah, I'm gonna leave the hardest for the last 29.
We have a triangle ABC on the XY plane with an area of nine.
Bam.
Okay. Let P, Q, and R be points that move freely on line segments A, B, B, C, and CA, respectively. The set of all points x that satisfy ax has to satisfy quarter of a p plus a r. Now p q and r are points on here here and here respectively. Now I yep sunny sniffer I believe it is solvable without any any dictionary. I believe I have like a few questions prepared and I've selected them very carefully. So I think it's possible.
Okay. Um there are multiple ways to do this but I'll talk about the way I did it in the actual exam. So AP plus AR.
Let's just think about like this. This thing looks weird. I was playing around with what does this half and a quarter mean for a while and I I thought maybe it's maybe it's supposed to be something like this and what I did was this equation is the same as doing this right and I think the con I thought the concept behind it is that you are supposed to take this median I mean the mean this mean and then you are supposed to take another mean.
Let's say M is the midpoint of P and R.
Then we know that two uh ap + a r over two is equal to I mean the vector a plus a r over two is equal to the vector am because this is the mean and this just becomes am.
Now this is another kind of mean thing.
So let x be the the midpoint of m and q.
Then this entire thing just becomes ax.
So that's what I thought this expression is entailing. We are supposed to take the midpoint of p and r and then we're supposed to take the midpoint of m and q and we need to find out where that point x can lie. And we need to take the area of the region.
So for example, if P happens to lie here and R happens to lie here, then the mean of that is just going to be here. Okay, let's think about all the places M can go first. M is the mean of P and R. Now let's say P goes all the way up here and all go R goes all the way down here.
Then M has to lie here because that's the middle point. If P is here and R is here, then M happens to be here. Like if P is here and R is here, then M happens to be here. And uh what you have to realize is that M can travel anywhere in this region if P and R moves freely on these two line segments.
That is because for any point P uh like let's say this is P and let's say R moves from here to here smoothly. What would happen to M? It would surely move from the midpoint of these two points to the midpoint of these two points freely.
Like it it's going to move in a line like this. And these lines when you add up all of those lines for every P on this part of the line segment then it becomes this entire diamond. It fills up this entire diamond. Now the way I just explained it is maybe not the most intuitive way of explaining it. I just use this way because that is the way I first thought of when I looked at this.
But anyway, that's how I got that M has M can be any point lying on this diamond and X can be any point lying on uh a little hexagon like this. I'm not going to explain why this happens, but I'll just say yeah X on in the end it happens to lie on a hexagon hexagon like this. And that's because this diamond gets shrinkedked in half towards Q and moves with Q. So yeah, the geometric intuition behind it is that this diamond when Q is all the way here, this diamond shrinks uh half the way from this original size to the direction of Q, which becomes something like this. And then if Q is all the way here, the diamond becomes something like this. And this diamond moves freely from this from here to here and swipes over all this area. So in the end uh X can lie within this diamond shape which is kind of like uh imagine we divide this up into tiny triangles. Then this this hexagon consists of one two three four five six seven eight nine 10 that >> triangles uh out of the entire one two three four five six 11 uh uh 16 triangles. So the area of this part is 10 / 16 um multiplied by the total area 9 which is 2 * 8 2 * 5 45 over 8 computes p + q 45 + 8 that's what you get okay we have only just one question left [sighs] what do you think do I need to go through the full solution or should I just lay down the intuition behind some parts? What do you think?
>> When do you want to sleep? I think is the question.
>> Okay. Wait, what time is it for you people right now? If it's okay for you, if it's not like time to sleep for you, then >> it's fine for me, but I'm just worried about you.
>> It's 11 a.m. here. So, >> Right. Right. Okay. Okay, then I'll go through this one fast and only lay out the fun parts, which is the one at the start. And uh okay, let's let's do this.
There's a little hill here, and I'm traveling on the hill.
Imagine I go like this in this hill. What's my height going to look like?
in respect with respect to time as I travel. Number one is it going to look like this? Uh number two.
Number three. Pretty easy question. What do you think? What? Number one. Number two. Number three. If I go like this, what is the height according to time going to look like?
>> Number three.
It is number three because I go up and down once, then I go back, I go up and down once again, I go forward, I go up and down again, up and down, up and down, up and down. That's going to be the basic idea behind this problem. Let's look at this. FX is a cubic function. Leading coefficient of oh my gosh, six pi. But it looks like this or this, whatever.
GX. Oh my gosh. One over. No. Uh I would say FX is really uh is an eyesore here. I kind of want to write it as something like this. I want to kind of write write out as a composite function of this and f. So I'm going to do f ofx first and then I'm going to put that value inside this function. 1 / 2 plus sinx.
Now 1 over 2 plus sin x. How does that look like? At least we can draw this, right? We can't draw this because we don't know what fx looks like. But at least we can draw this. Sinx goes this up and down, up and down. 2 plus sinx, just this but two up, right?
Starting from two up and down, up and down. And then 1 over 2 plus sin x.
That's going to like reciprocal of two.
Like this is three. This is one. It's going to go I guess down up down up down up. And then this is probably going to be 1 over two. This is then uh 1 over three. This is probably one. Yep. That's how 1 / 2 plus sin x looks like.
Now what am I doing? Let us list all values of alpha uh for which gx has a local maximum or a local minimum at x equals alpha. Okay, so gx has supposed to have a local maximum or a local minimum. Like at this point we can try to like take the derivative of this and find out where it's local maximum or local minimum. But what we can also do is think about f as this trajectory. It goes up, down, and then up. Forward, backward, and then forward. Right? What if we traveled forward, then backward, then forward on this hill?
When would GX have a local maximum or a local minimum?
Now, I'm going to cut to the chase because I said this is going to be fast.
The first thing I did is that I realized that the when this has a local maximum or local minimum is the same as when sinx has a local maximum or local minimum because this this is just a bloat. This bloat is not needed. It it goes up and down. This goes down and up. It has local maximum local minimum exactly at the same parts as this thing sinx.
We can just ask what if we travel in this trajectory in these hills, right? So that's what we're doing. And and then all we need to do is look at all these constraints and pick up all the things we need to pick up and do apply logic.
And that's going to solve the question.
I'm I kind of want to leave it as kind of open for people on YouTube to try to because if I lay out all the solutions then I guess they have nothing to go for. So yeah, anyway that's how you solve 30 and that's how you solve the entire exam that I took as my SAT.
Yay.
Okay.
Now, uh you can go sleep, but there's something really interesting after this.
[laughter] You know what's after this? There is if we have time, we have the Korean decipher challenge, right? We have the Korean decipher challenge. And uh bam bam bam again I'm going to read this again. Two swap once said what if we have to solve the problems in Korean too and it's a great idea. The math club really likes challenge like this. I prepared a set of problems in Korean that I think makes sense as a deciphered challenge.
Everyone will turn on their mic. I'm going to share an excellent draw link with the problems and I'm going to let you do your thing.
Sounds exciting.
>> Yes.
>> Says no.
>> Yes. [laughter] >> Okay, then I'm going to prepare.
>> I got to get my >> Okay.
I need to reset this thing.
Uh, open a new thing.
I guess this would be a a good time to ask Z how how did you do originally on this? Did you get full marks or >> uh it's a secret it's open to interpretation?
>> Okay, I see now.
[clears throat] Okay, I'm going to share I'm going to export the link. I have it open.
I'm going to send it to the Discord chat right now.
Let's see. Okay.
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