VERY HARD South Korean Geometry Problem (CSAT Exam)
ฝัง
- เผยแพร่เมื่อ 31 ก.ค. 2024
- Thanks to Hyeong-jun (H. J.) for emailing me this problem! This is a challenging problem from the math section of the 1997 CSAT, a standardized test in South Korea. Can you figure it out? It took me several attempts, but it was really satisfying when I solved it (I did need to look up one key insight, which is the first thing presented in the solution). So give it a try, and don't peek at the rest of the solution too early!
0:00 Problem
1:58 My many attempts
3:28 Unwrapping a cone
5:14 Solution
Coordinate geometry solution and proof of uphill/downhill track
www.desmos.com/calculator/l18...
Note: pretty much every time I said or wrote "circular arc" I meant to say or write "circular sector." A "circular arc" is a portion of a circle's circumference; a "circular sector" is the enclosed region between the arc and two radii.
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Note: pretty much every time I said or wrote "circular arc" I meant to say or write "circular sector." A "circular arc" is a portion of a circle's circumference; a "circular sector" is the enclosed region between the arc and two radii. I was trying to avoid saying "circular segment" which is the enclosed region between the arc and a line segment between the endpoints of the radii. You can see the term "circular sector" used correctly in the following videos:
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th-cam.com/video/xnE_sO7PbBs/w-d-xo.html
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th-cam.com/video/kaLiagYuYPc/w-d-xo.html
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th-cam.com/video/z2P8q4QC53s/w-d-xo.html
I solved this problem using calculus of variations and some help from wolfram-alpha.
I used spherical coordinates, with the origin at the cone's peak. This has the advantage of making the phi coordinate (angle from the z axis) constant.
The differential arc length on the cone is...
L = sqrt((dr/dϑ)^2+(r*sin(φ))^2 )*dϑ where r is a function of theta.
You can use the Euler-Lagrange equation (dL/dr - d/dϑ[dL/dr'] =0 )to give you a differential equation, who's solution will minimize the integrated arc length (functional).
With the help of a computer-algebra system, I was able to reduce the differential equation to the form...
r'' - (2/r)*(r')^2 - r/9 =0.
This is where I used Wolfram-Alpha's General Differential Equation Solver for help, as I am pretty clueless when it comes to solving non-linear differential equations.
The general solution is...
r=C2*sec((9*C1+ϑ)/3)
And I used Wolfram-Alpha's systems of equations solver to solve for the integration constants given the boundary conditions (r=60,ϑ=0) and (r=50,ϑ=2pi)
C2 = 150*sqrt(3/91) = 27.235239
C1 = (2/3)*atan( (5*sqrt(3)-2*sqrt(91))/17 )= -0.36654527
This gives you the distance from the peak to the points on the track as a function of the angle around the mountain. Here is a plot of the radius.
www.wolframalpha.com/input/?i=plot+27.23523897009611*sec((x-3.298907465836848)%2F3)+from+x%3D0+to+2*pi&wal=header
And if you want to calculate the altitude below the peak, useful for calculating the elevation the track needs to be at, you can just multiply the radius function by sqrt(1-(20/60)^2) = 2*sqrt(2)/3 = 0.943
From this, you can calculate the grade of the track (rise over run) or (dz/dx).
The vertical distance delta_z = (-0.943)*delta_r. dz/dϑ = -0.943*(dr/dϑ). The horizontal distance delta_x = r*sin(φ)*delta_ϑ.
You can solve for dϑ/dx from the horizontal distance formula, and use dϑ/dx to change the vertical distance independant variable from ϑ to x.
(dz/dϑ)*(dϑ/dx) = dz/dx = -0.943*(dr/dϑ)*(dϑ/dx)
now substitute expressions for (dr/dϑ) and (dϑ/dx) into the right hand side of the previous equation...
dϑ/dx = 1/(r*sin(φ))
and
dr/dϑ = 9.078412*sec((ϑ-3.298907)/3)*tan((ϑ-3.298907)/3)
and get a track grade (slope, not a percentage)...
dz/dx = -0.333333*2^(3/2)*tan((ϑ-3.298907)/3)
www.wolframalpha.com/input/?i=plot+-0.3333333333333332*2%5E(3%2F2)*tan((x-3.298907465836848)%2F3)+from+x%3D0+to+2*pi
Jess Stuart
How long and how many pages did it take to get the right answer?
Bill Y.
Probably about 4 pages and atleast an hour of trying to solve the Differential Equation, then giving up and using Wolfram Alpha. It was a good excuse to get more practice doing calculus of variations stuff anyway. You would have to do it this way if your surface couldn't be unrolled to a nice flat surface.
Jess Stuart: Thank you for explaining this! I do enjoy when people find "easier" ways to solve the problem, but this is one case I really enjoyed seeing a "harder" way to solve it. Your comment will inspire many to study calculus of variations (and for me to brush up on it and learn more too). Thanks!
I thought about using coordinates to find the shortest distance btw the chord and the arc (on the unfolded cone), then I thought it's multiple choice it can't be that complicated and there has to be a "easier way." Then realize that I can simplify this into solve a triangle with two sides that are 50, 60 and 120 degree angle in btw. sadly I did make some calculation error when applying law of cosine. I like how you just used pythagorean theorem and set up 2 functions and solved for the part directly, I really made the calculation more complicate by trying to solve those angles
I wouldve just guessed it and pray
fahad mohamed you cannot bribe God for easy answers with all your flowery duas: God does not reward laziness by giving away easy answers to questions or even giving free scores in exams. God does not grant you easy answers to maths question: otherwise, that would be cheating and unfair.
In this case you should guess #3. It has the same denominator as #4, and the same numerator as #2, so it's the most likely to be correct.
Pink Lady: You cannot bribe God because God exists in your head only.
Logically you should pray first and then guess.
fahad mohamed same
the video: “what method did you use?”
me: ennie meanie minie mo
Best comment so far haha😂
U would actually get it correct if u used that method
@@cyanide6954 this method always gets the correct answer it proved itself many many times
What a comment 😂
*UNDERRATTED*
시행착오과정을 다 기록한다는 점이 독특하다
저러면 한 문제를 풀어도 밀도가 다를듯
킹정
m/v가왜다름
@@user-qi7rv5bq2f 시행착오가 세기성질이 아닌가보지
그래서 원래 한국애들 유학가면 중고딩까진 우리가 훨씬 ㅈㄴ 더 잘하는데 점점 더 외국 애들이 잘해져서 나중엔 한국 애들이 더 못해짐
한국인이 외국 중고등학교에선 성적이 높은데
대학가선 적응 힘든 이유임
Wonderful.... This problem and it’s solution are pieces of artwork :)
Math is beautiful...
I can't decide if I like the question more, or the solution. Brilliant!
Hardest part is you have less than a minute to solve this accurately
This is true, we have about 4 minutes. This question was worth 4 points out of total of 100 points, for which we are given 100 minutes to solve. We have about a minute per point.
Steven McCulloch In Korea you're expected to be able to solve quickly and accurately. This is highschool level stuff here.
MassiveTornado the concepts being applied are, but a good understanding of geometry that they don't teach in schools are needed to figure the entire problem out.
Steven McCulloch Why do you think Koreans students study outside of school as well
Steven McCulloch Korean education, especially math and English are harder than almost anywhere else in the world
I'd never have worked this out in time for the test either... I saw the unfolded cone immediately, but I forgot my cosine law and somehow couldn't get the Pythagorean bit to work. I used a bit of coordinate geometry (plus some basic calculus), and got the right answer. Took me over 30 minutes. Very clever question, and very tidy solution given here.
I’m Korean and this is why I hated math so much before college. I found out math is actually fun after I started working and use that in practice.
Isn’t math more difficult and complicated in college ? Or that’s true but still since you’re learning more things it’s becoming more interesting the whole lesson
Math is certainly more fun when you have the time and the context to solve the problems!
americans start hating math in the first grade.
yeah i also usually hate maths but now i love to solve it
@@kdy5617 math before college in S.Korea is more like memorizing problems. Just solving thousands mindlessly.
I wouldn't be able to solve this. Congrats to anyone who figured this out. Great problem!
Lol I wouldn’t be able to solve this either. All these people in the comments saying they could solve this quickly are crazy good at math haha.
*Hannah M* it’s not that hard, if you take out everything i calculator can do for you only a few steps are left
-understand it’s a circular cone
-put the points st their respective spots
-figure out the angle
-realize that the right angle line to the path is the highest point
That’s the only math involved in this problem, everything else he shows in this video is just boring calculator stuff
Larz B 20 sec? I doubt that any person that ever existed could solve it that fast. I don't even think Terence Tao could solve it that fast and his iq is 200+.
The important part in korean sat you have to solve this at least a minute to maximum 2 minute to solve this because there are like 10 more questions even harder than this. Also you don't use calculator.
Larz B shortest line is not always a straight line. And this took me just 1 second.
외국인들 댓글1% / 수능뽕차서 열심히 번역기 돌리면서 댓글다는 한국인99%
쓸데없는 국뽕이다ㅋㅋㅋㅋ 오히려 독인데 그걸 자랑스럽게 여기는듯한
@@user-kc7ww1tk8k 어떤점에서 독이라는거임
@@jjh4928 이러한 뽕은 필연적으로 다른 나라에 대한 무시와 차별로 이어지죠
@@jjh4928 이미 친 사람들은 모르겠지만 학생이면 개고생 해야되니까. 그리고 수능에 맞춰진 교육방식도 문제가 있고
@@user-pp8bo4jz1c 그게 필연적임? 문화절대주의랑 필연적인 연관성은 없는거 같은데
I unfolded the cone's lateral surface area into a sector and was able to figure out the straight line distance from A to B. Trying to find where the part that changes from going uphill to going downhill was tricky until I realized that it would be a point along the arc whose tangent is parallel to the line, and since the radius of a circle is always perpendicular to tangents of the circle, it would also be perpendicular to the line AB. Easy to solve from there.
Oh, that's a cool thought. I was hung up for a bit on that too, but i unstuck by imagining the more simple question of if the track came back to the same point A, then the downhill would be halfway... noticing that's exactly when the unrolled straight track makes a right angle with the radius.. which i guess is essentially the same thing as parallel to the tangent
Your problem solving skills are exemplary & very absorbing .. I m an engr with 43 yrs experience, and still enjoy watching your solutions..
Sir
Someone who is not Korean : Wow I made it!
Korean : Okay, you have 29 more problems to solve.
웃자고 쓴 댓글에 답글달며 싸우지마 새끼들아. 난 수학쫄보라 맨뒤부터 풀다가 앞에 풀고 그랬어.
OMG.....so ture
현실이네 시부레
@@djmikecr9284 형님 한국인 이과라면 당연한거 아입니까
@@ys9018 아 수고하세요 저는 6학종..
@@user-tg9uz4ve8w 아따 01년생한테 6학년이라니 거 말이 너무 심한거 아니오! 수학 못해도 드립치는건 내 자유다 이말이야!!
Great challenge! I solved it using variational calculus on the arc length integral in polar coordinates in the unwrapped cone. This approach gives you instantly two arc length integrals: one for the uphill length and the other for the downhill length. After that I checked the answer by using almost the same method as yours, the exception was that I began my consideration of length for non euclidean spaces to actually unwrap the cone, which led to the same answer.
Great video! Thank you! Explained perfectly!
very comprehensive and detailed explanation so that many Korean students may be able to understand the process of solving this problem easily!! -private instructor from Korea :)
I saw this about a year ago and was completely clueless. I was scrolling through your channel to find this specific video to give it another try. I didn’t know the law of cosines and spent about an hour finding equations for lines to fit intersection points on Desmos and was excited to finally answer this question correctly! I love this channel so much as it has helped me advance in several concepts in geometry and especially calculus. I stay up late at night on the channel because these puzzles are so fun and addicting!
I was an jee aspirant, and I had also studied engineering graphic so I was also managed to figure it out. But this was a very helpful video and too good to understand how good mathematics is
Hey man, thank you for this question! I had a great time watching this!!!
> Be me
> A Korean that just turned into our equivalent of 8th grade
> Legit scared
탁두둥 i wishnyou luck, you will probably need it based on this video
nice, all the best! xP
>using comedy chevrons on YT
>why?
asdmvva 4chan user out of his natural element.
Do you also study all day long till midnight?
Another way to find the angle _θ_ is to work out the circumference of the big circle, radius 60, which is 120π. The cone's base has a circumference of 40π, so the sector is exactly 1/3 of the circle and the angle is (360/3)° = 120°.
Lugh Summerson yeah , that is where that formula comes from .
That’s what I did by that does equal 2(pi)/3... am I missing something?
I find it easier to work with revolutions (easy to conceptualize) then convert to degrees or radians when needed. I wish calculators had a 3rd option of revolutions, not just radians and degrees.
@@johndavidalexander6646 (40pi/120pi)*360 or you could shorten it to (1/3)360, using this angle youll get the correct answer, in degrees not radians you either didnt divide both sides by pi or forgot to cancel one out
I always solve like that
zz 한국인들이 이걸 쉽게느끼는건 뇌리에 전개도면이 이미 박혀 있어서 그렇다.
ㄹㅇ 이미 졸라본듯
ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ 한국인에게는 그닥 어렵지 않았다...
@UCGBHl99d5_PHk_cl4ZyQlZg 지금은 2021년입니다
전개도면은 누구나알겠지만 내려가는 부분이 어디서 시작하는지 아는 사람은 별로 없을 것 같습니다
여러분 밑댓글중에 중3이 몇초만에 푼다 이런 댓글들 많던데 저는 진짜 개멍청해서 저거 전개도 펼치고 삼각형 만들어서 그 삼각형의 꼭짓점에서 수선 긋고 코사인 법칙쓴 뒤에 넓이 이용해서 수선길이 구하고 개그지같은 루트91로 된 숫자를 피타고라스로 풀어야 하는데 나름 어려운 문제같은데 저만 그런가요
I really like your creativity in solving math problems, you are very critical. I'm really amazed. greetings from me in Indonesia :)
Hello. I.m a viewer of this channel too and i.m from Indonesia:)
I had a different method, when seeing the four possible answers, I picked the 4th. Then I went to the end of the video and won
I would just take my 25% chance
that case i would choice 2 or 3, 4 so will be given 33% chance
You're genius.
Too easy
in 2minutes clear
And now.... we solve that kind of thing for practical qusetion
Just basic question
well to prove that from the line from I (center) to AB at H on a perpendicular line marks the change of "slope" is easy
all you need is to intercept AB with the cone's base, which you will create a bow, where AB is the string (I want to call it that ), extend IH to intercept with the arc, that's how I will interpret it. Love this, been years now
Normally, I struggle with starting problems like this. But halfway through the intro, I remembered the unwrapping trick, and I realized that the shortest path is just a partial chord. I’m so proud of myself for realizing that without any help.
I love this question. It is distinct from those traditional textbook drills and sparks the imagination of the mathematical mind. Thank you very much.
i feel like these will help me in the future someday so even if i dont know what he’s talking about, i still binge these 😂
You will never use this once in your life. I guarantee that.
@@LibertyGunsBeerTrump maybe, maybe not....
A lot of things that we eyeball can actually be done using math. We just find it easier to adjust to failure rather than doing some complex calculation and getting it right
Very informative! I probably would not have thought of flattening the cone, and I didn't know the Length of Circular Arc formula, so I would not have solved this or known how.
이분 채널 들어가 보니까 수학 영상 찍고 계시던데 이런 수학문제 푸시는걸 보니 대단하다. 저 문제 푸는 것만 해도 그당시엔 ㅈㄴ 어려운 문제였는데 저렇게 잘 푸시네!
Thanks for the little note about persistence. Usually, I take all the time I need to solve a problem, but today I was a little impatient and I played the video solution before even giving the problem a try. The persistence note made me pause the video and now I'll only watch the slution once I have an answer.
Still working on this problem, huh haha
I did it slightly differently, with a bit more trigonometry! It was the same with unwrapping the cone, finding that its angle is 120°, connecting AB with a straight segment, and realizing it changes from uphill to downhill at the closest point to the vertex, which is the perpendicular.
But I did not need the length of AB. Looking at the right triangle involving x, you see that x = 50 cos B.
Now how to figure out that angle B? Looking at the triangle between A, B, and the vertex of the cone/center of the circular sector, and using law of sines: sin A / 50 = sin B / 60. You also know A + B = 60°.
You substitute A = 60°-B, and expand sin A = sin(60°-B) = sin 60° cos B - cos 60° sin B = sqrt(3)/2 cos B - 1/2 sin B. Now this equals 5/6 sin B, from the law of sines. Rearrange and get 3 sqrt(3) cos B = 8 sin B, or 27 (cos B)^2 = 64 (sin B)^2. We want to solve for cos B, so we write (sin B)^2 = 1 - (cos B)^2 (trigonometric Pythagorean theorem). Rearrange again and get (cos B)^2 = 64/91, or cos B = 8/sqrt(91). Multiply by 50 and you get x = 400/sqrt(91).
Bravo.
yay! precisely the same solution path I came up with, however, as someone pointed out, "you only have ~10 min to solve this problem", I would have failed.
If I were smarter, since it is a multple choice question, I think it is possible to find the right alternative by constructing the path in the 2-d case and just measure the length of the decent part with a ruler. it will be (when I measure) ~42 length units and the correct alternative (400/sqrt(91)=41.9314 lu)
Exactly how I conceived the solution, though I didn't bother doing the grunt work.
@@weijholtz If the choices are spaced enough to be resolved within your ruler accuracy then it's a great time saver.
Brilliant
I think this is a very beautiful problem. For me it was difficult to understand the question but once you wrap your head around the shape of the cone and really understand it, the geometry behind it is wonderful and easier than it may seem.
true
The cone part is actually something very standard in Grade 9 Hong Kong maths. But you’re right, there’re so many other concepts got mixed into this single question.
Just remember this, this IS NOT HARDEST QUESTION in Korea SAT
그래두 자이스토리 29번인가 30번이었다네용 허헣
Which is the hardest I want to take a crack so if you find that please post a link
thank you
@@anmolbaloni every problem numbered 21, 29, 30 in the korean math SAT is difficult
Which is still kinda easy
Bruh this is easy tho😑
Love your videos, because I get better in English and maths at the same time ^^
The proof you showed is possibly the most natural one. This problem is very nice for motivating students towards Olympiad geometry.
Very well conceived analysis.
Salute the Professor
Your soft voice is soothing
And your explanation ( like first you introduced us to the possible mistakes i liked it :))
Thank You 😊
I'm a liberal arts student who was passing by. I think i should keep going.
SAME...
ㅋㅋㅋㅋㅋㅋ문송쓰
지나가던 문과생입니다. 계속 지나가겠습니다
이거 영어버전인가 ㅋㅋㅋㅋㅋ
문붕잌ㅋㅋㅋㅋ
아 역시 드립의민족
Very neat problem. The only hint I needed was to "unravel the cone" and the solution more or less jumped out afterwards. It's crazy how much the solution can sometimes just hinge on a single insight like that.
This problem is so instructive. I would like to know how to find the coordinates of the point between A and B which is at h from the cone ´s vertex and also how to find the equation of the line AB. Thank you very much.
Presh's solution also basically solved a side question that caught my interest.
I assumed that the question was asking about paths with winding number 1 corresponding to the diagram. Winding number 0 would be trivial (no downward path so its length = 0), winding number -1 just reflects the problem, and other winding numbers are clearly not giving the shortest path. But I still wondered about other winding numbers for cones of arbitrary aspect ratio.
From the solution, the answer is fairly obvious: For winding number N, unroll the cone N times. If, as in the given problem, the angle at the apex is less than 180 degrees, ie N theta < pi, the solution is essentially the same: find the target point on the far side of the Nth unrolled section, draw a line, find the downwards path essentially the same way. Notice that the greater that N is, the closer the path passes to the apex - it goes up high so it can circle the cone with smaller circles.
But if the angle at the apex is greater than 180 degrees, we can no longer just zip across to the target point. The unrolled area is now concave, and when the angle surpasses 360 it will get even worse, giving a solid helix with multiple arms at the "same" point.
So if the angle > 180 degrees, the path goes directly to the apex, then circles around it at an infinitesimal radius until it aligns with the target point on the appropriate arm of the helix, and then goes straight down to the target point. Then the answer is always 60 - 10, or 50; adjust in the obvious way for cones of different sizes.
May I know what you mean by the angle at the apex being greater than 180 degrees? Does it mean like if the 'net' of the cone turns out to be a sector that has a central angle (subtended by an arc/the circumference of the base of the cone) which is a concave angle?
OK, what I said was elliptical, but I think you get what I mean.
But more carefully now: We unrolled one cone, and we mapped the apex of the cone to a particular point on the plane. I called that point the apex also. The circular sector subtends a particular angle at that point - I called that angle theta. Then I multiplied theta by N, the number of windings. Then I called N times theta the angle at the apex. That's the angle which we test for being greater than 180 degrees.
So yes, sounds like you got it.
Whoa! I didn't think about "unrolling N times." It's a strange concept (strange for 3D to 2D anyway). Can I unroll something a fraction/ negative number/ irrational number/ complex number Number of times?!?
Well, unrolling a negative number of times is the easy one. It just corresponding to negative winding numbers, which are just going around the cone the opposite way. So we'd just unroll the cone the other way.
Zero makes sense in this picture too: when you unroll the cone zero times, you just have the line segment from the base of the cone to the apex. Zero area, zero angle at the point corresponding to the apex, and the path corresponds to the solution of zero windings.
Fractional - we can say that it corresponds to placing the target point a fraction of the way around the circle at that latitude of the cone. That should cover irrational too.
Complex winding numbers would get a lot messier. We'd like to say such a path "winds around in it the imaginary direction", but that doesn't make sense on a real-valued cone. Maybe we embed the cone and the whole problem in complex 3d space, but then it's not a cone any more. We could generalize the cone - like, the base of it is not a circle but any solution to x^2 + y^2 = c and similarly for other latitudes, so x^2 + y^2 = c * z. Then in imaginary space it has this hyperbolic cross-section, and there are solutions that make the path length zero. That's if we're looking to minimize its absolute value - if we're looking to minimize its natural value, we can find paths of negative infinite length, which is a fair bit shorter than 400/sqrt(91). Like I said, it gets messy.
The cone with a vertex angle V >180 is geometrically identical to one with vertex angle (180-V). After you unroll it you can't tell the difference.
To see this in practice make a sector out of paper and draw on it the railway then roll it in to a cone.
Roll it into a cone by curving the paper the other way: all that is different is that the track is on the inner surface instead of the outer: the shape in 3D is the mirror image but all the lengths are the same
I never really understood the utility of the law of sines and cosines until this problem! They're useful for when you can't use the pythagorean theorem to find the side of a non-right triangle! Awesome
I had a very similar method to you. I just used the cosine rule to get cos(B) in triangle A-B-Vertex. The answer is x = 50cos(B)
Fascinating. Every math test problem always easier than it looks but takes so much time in tests that they're almost not worth the time to get the marks. Tests are a fallacy. Great job with the explanation.
Man i tried with variational principles.. got a truly hard lagrange equation to solve! :) will try again hahah
your visual method, looking at it, looks so simple, yet so satisfying!
don't you find lagrange is more applicable to parabolic questions than straight line problems?
Really love these types of questions! Could you go for some nice olympic questions that don't need calculus? Like that probability one that 3Blue1Brown did.
Agree! try to solve some interesting Math Olympic question here!
th-cam.com/video/OkmNXy7er84/w-d-xo.html
the question is literally: "what is the probability that this tetrahedron contains the spheres center?"...
if ur talking of that tetrahedral one I agree that was too difficult for me i spent 4 hrs working on it but finally had to look towards the answer
this one didnt need calculus
I don't know why but I understood everything in seconds. That's the first time but it is so satisfying and good
Good for you. I only understood it when he unraveled the cone into a circular arc, then it became easy from there. I didn’t even realize that there’ll be a downhill at first until I realized that the line should be parabolic for it to have no downhill. Facepalm*
Loved the problem, I got to the part about going up hill, but briliant solution for going down hill.
thoes South Korean be like : why this guy upload this video it was too easy
e-z
Geometry dash
Fact
umm.. it was hard but not hardest.
ez 4 me
Me: sees this is what imma have to do for high school.
Also me: aight imma head out
High school questions won’t get more difficult than this. This is an example of a high school question at the absolute highest level
Wow! Nicely solved, amazing work!
I figured out the right method but got the wrong answer. Once I watched the video, however, I was able to track down my mistake. Great video!
Guys, thats not even the hardest one in the test.
So freakin true. Students can solve this problem 3 years before they take KSAT
facts...
i saw this problem in math olympiad book
Too easy for KSAT.
SPayee yea. ksat is harder.
I was able to solve until the part, AB = 10√91.. using the same method as you.. but got stuck after that.. So I started watching your video for solution and was blown away that you used the same method as me.. when you dropped that perpendicular from vertex.. I facepalmed and paused the video.. It was easy to solve after that..
When he started drawing the lines from the vertex to the radius... I also facepalmed.
Peter Geras 😂😂😂
I wouldn't even notice that there is a downhill part!!! Still unbelievable
Amazing problem. Interesting way to solve the problem.
- 외국댓글 중 -
한국인이 아닌 사람들: 마침내 내가 이 어려운 난제를 해결했어!!
한국인: 좋아 이제 앞으로 29문제가 남았군
ㅋㅋㅋㅋㅋㅋ 저문제 중3때 풀어본거 같은데
@@soobark2 그러게 이거 중3 거 아니냐? ㅋㅋㅋㅋㅋㅋㅋㅋ
그 만큼 헬조선..
@@soobark2 블랙라벨에서 나온거같은데
솔까 수능 이거 못 풀면... 3등급 이하 확정인듯.
곧 마주칠 수능은 범위 달라져서 어떻게 나올지 감은 안오지만 일단 내 시절이라면. 방정식도 그리 복잡하지 않고
By the way, the hardest question does not provide multiple choices, which makes this question relatively easy
It was so interesting to see how you solved this and I'm definitely not one of the people who thinks this is easy but I guessed 20^2 is 400, might as well go with choice 4. I'm definitely one of those people who got through high school with good guessing skills.
This is one of the best problems I have seen
Watching the solution was like watching a neat magic trick 😊
The key to this problem is knowing that the side surface of a cone can be unwrapped into a sector. If you know this, the solution comes straightforward.
I never realized how awesome Trigonometry was.
I feel like I should make a plug here for estimation, since I didn't find any comments that do. Your full solution is wonderful and well-explained. Had I myself been faced with the problem, I'm far too rusty on my equations and would have utterly failed...
Which is why I would have estimated instead, in seconds instead of minutes -- important in a timed test. In fact, I did so at the start of the video and ended up with the correct answer. 1 & 2 seemed too big, 3 seemed too small, so I picked 4.
In addition to allowing me better than 25% chance of choosing the correct answer in less than 1/10 the time, estimation is great for knowing whether or not to spend more time to finish a solution properly. If I was a business person trying to decide whether or not to build the track, and I couldn't solve the problem completely myself, knowing that it will be between some x & y amount of track lets me approximate the price (& thus the chances of profit) before getting other people involved.
Obviously, you were asked to actually solve the problem and it really is a great solution. But I feel that estimation is a valuable tool that doesn't get enough credit. I do grant that it takes some time to develop the skill, but I think it's worth it.
I was the one of high school students who took the CSAT in Korea in 1997. I guess I randomly picked a number for the Q.
Is reply 1996 relatable for you?
I was the one who made this question.
LOL
@@chinmayh2745 I was the one who made u
@@jjraga 😂 😂 badass
근데 진짜 너무 당연스럽게 전개도 펼쳐서 저 그림 그림
시작할때 4번의 잘못된 풀이를 했다는거가 저렇게 많은 풀이시도가 있을 수 있구나는 생각에 젤 놀라웠음
bravo, well done presh!!
?? Bit confused but nice to see it again! It was a fun problem
I solved this problem at my math academy... for the weekly test... I actually didn't know that it was a korean sat geometry question!!
Sorry for my bad eng..
(Korean student)
You know what. in Korean SAT, we can use only 3~5 minutes each of problems. So we have to solve that in a 5 minutes. That's crazy
푸아송 What grade is this for?
@@acutepenguin3577SAT for hight school grade 3 students. This is last grade in Korea.
It isn't going to be that hard since lots of Korean students get familiarized with similar types of problems throughout their math courses at Hak-won prior to taking the CSAT.
How much time you have for this one depends on how fast you are on the easier problems. :)
But isn’t the math section 100 minutes for 30 questions?
현재 한국 고등학생으로서 이렇게 알고 있습니다
불면증있는데 마침 딱 좋은 영상 떴네 역시 유튜브 알고리즘
Böyle yerlerde çoğunluk Türkler olsa keşke
-an excellent channel thank you for your efforts
I saw immediately that you should 'unwrap' this cone. But was stumped by the starting and end points being at different distances from vertex. If it started and ended at the same point, I saw that by symmetry the inflection point from uphill to downhill would be half way, but couldn't figure out this case. Interestingly, this shows that the shortest trip around the mountain from A back to A, goes up hill halfway and downhill the other half of the trip and that the shortest distance is not just laterally straight around the base of the cone.
Great video and an interesting problem.
I am a high school student in Korea who took this 2020 CSAT. Anyone who solved many problems with spatial shapes would have thought of drawing a floor plan. After that, we found the minimum point of the distance and solved it easily
가연이 남친있노
시부럴 우리나라 수학 30번 풀면기절하것네ㅋㅋㅋㅋㅋㄱㄱ
ㄹㅇ ㅋㅋㅋㅋㅋㅋㅋㅋㅋ 요즘 수학 가형풀면 기절하겠누 ㅋㅋㅋㅋㅋㅋㅋㅋ 97수능이면 요즘 가형 27번쯤 될 듯
ㄹㅇㅋㅋ
@@user-vr7vm5if4s 17학년도 수능 가형 30번이면ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ
@이경훈 솔직히 근데 전체적인 수능 성적으로 따지면 언제가 더 어렵고 쉽고 비교 할 수 있나...? 97수능 문제 구경도 못해보긴 했지만.
@@user-we1dd7kj3u 전반적으로 수준이 올라서 예전이랑은 비교하면 안됨 ㅇㅇ
Hi. Im sadra derhami from iran. Im15 and I solved this problem. This is the best math problem I have ever seen. Thank you for this incredible question.🌹❤❤❤🌹.
It feels so counterintuitive that the shortest path from A to B actually goes above B in altitude and then back down to B. Your explanation makes perfect sense though!
첨에 되게 쉽다고 생각하고 바로 전개도 그리고 호길이구해서 호의 각 구하고 A부터 B까지 선분 긋고 길이 코사인법칙으로 구한 뒤, 그대로 5분동안 이게 뭐지 싶었네요 ㅋㅋㅋㅋ 도저히 내려가는 부분 찾는 방법이 안떠올라서 영상보다가 꼭짓점으로부터의 거리 듣고 나서야 유레카 외치고 풀었네요... 발상의 중요성을 깨닫고 갑니다...
I saw this type of question in my class 9th reference book ....not exactly this question but this type of question.
And I am form india ....there are some similarities between us ...
Is it true for all 'unfoldable' shapes that the shortest path on the solid surface is equal to the shortest path on the unfolded surface? I felt uncomfortable making that assumption.
Yes, if you can unfold the shape without "stretching" the surface.
Edit: The shortest path between 2 points on a continuous surface is always along a "geodesic".
A geodesic on a flat surface is always a straight line. To visualize what a geodesic on a curved surface would be like, you can imagine an ant walking on said surface always going "forward" without ever turning even a bit to the left or to the right. The trajectory of the ant is a geodesic.
@Andrés Rico
That's not really rigorous. And where is the proof?
If you think about what "unfolding" means, then it's trivial that the shortest path when folded is also the shortest path unfolded (provided none of the cut lines cross the path). For something flexible but non-stretchy like paper, when you curve or fold a sheet of it, you don't change the distances between points on it along the surface of the paper - otherwise you're not folding/unfolding it; you're stretching it.
If you draw any line on a sheet of paper, no matter how you deform the paper, the line will still have the same length (except in degenerate cases - if you tear the paper into non-measurable fragments, then you can reassemble them to give a different length...)
I saw a problem once that involved unfolding a rectangular box, and there were multiple ways to unfold it that greatly changed the shortest path.
Yeah, in this problem the way the path is specified makes it clear how to unfold the cone to avoid cutting the shortest path. In general, the shortest path on the surface is going to be the shortest path when unfolded correctly, but using the wrong net will give you the wrong path - figuring out the best net to use is part of the problem in such cases (either figuring out a best net and then calculating, or trying a range of nets that you are confident will include a best one - since you can cut across faces, not just along edges, there are an infinite number of possible nets, most of which are extremely silly)
I got up with an solution in very 5 mins.
I love it
I couldn't solve this problem my self. But I learned lot of thing about cones. Thank you very much sir.
solution :
step 1 : divide the cone in half .
step 2 : use integration on x(pi)r on limit 10 ,20
step 3 : try to copy others
step 4 : lay down and cry
1. Make an actual model out of the sheet the question is on.
2. Take a rubber band.
3. Mark the points A and B by pins.
4. Connect them by the rubber band.
5. Take a marble and let it roll from the highest point of the trajectory towards B.
6. Measure the path the marble took.
There you go.
THE FIRST EVER PROBLEM I SOLVED MYSELF!!!❤💙❤💙❤
It's pretty easy to show that the perpendicular segment to AB marks the highest point of the track. Simply make an arc with radius h. It's easy to prove that the arc is tangent to AB. An arc on the 2D plane is just an equal altitude line on the cone. So the segment that leads up to that tangential point to the arc is going uphill, and the the other segment is downhill.
Cant believe i got it right in the very first attempt! It was really fun to solve it ,i m a jee aspirant nd i must say its a beautiful geometry problem with no hard mathematics involved yet it is so confusing to get to it
Imagine to live or die you must answer this question, I’m dying for sure bruh 😔
Not too hard, not too easy. I like it!
What a beautiful solution!
The flat development of the surface of the conical mountain is a circular sector with a radius equal to the generatrix of the cone (vertex “V”) and subtends an arc of length equal to the perimeter of the base. In this sector, the shortest possible train path AB is the line that joins the outer end A of the radius on the left and point B, which is 10 units away from point C, the outer end of the radius on the right. The ascending and descending sections of the train route are separated by the radius perpendicular to the line AB, which it cuts at point D. Moving in the direction A→B, to the left of D the points of the route move away from the subtended arc (which represents the base of the mountain) and that tells us that the train is going up; to the right the opposite happens and that tells us that the train descends. With these premises, the descending path can be calculated:
Angle AVC=α, circular sector opening.- 2x20π/2x60π=1/3 ⇒ α=360º/3=120º
Angles VAC=VCA=β=180º-(120º/2)-90º=30º ⇒ AC=60√3 → Projection of BC on AC = 5√3 → Projection of AB on AC = (60√3)-(5√3)=55√3 → Height of point B above AC = 10/2=5
Length AB, train run.- (AB)² = 5²+(55√3)² ⇒ AB=10√91
Angle BAC=γ ⇒ Angle VAD=δ =β-γ=30º-γ → Upstroke length = 60cosδ=60cos(30º-γ)=60(cos30º cosγ+sin30º sinγ)= =60[(√3/2)(55√3/10√91)+(1/2)(5/10 √91)] = 60(17/2√91) = 510/√91
Down stroke length = AB-(510/√91) = (10√91)-(510/√91) = 400/√91
I solved it (aren't I great), but I used a more complicated approach with the cosine rule and Heron's formula. It was fun!
How sir
@@narasimharaobejawada2776 Thanks for answering. This was a few months ago. I will have to think about it again and get back to you.
@@Jack_Callcott_AU have you thought about it again?
@@beaclaster I'm sorry to say I haven't.
It was very hard for me the first time, so I was loathe to try again. I will have to sit down with pen and paper and try again. I'm sure I'll see the way through. Thanks for the reply.
@@Jack_Callcott_AU did you try again?
Admit it, Presh: By „slightly editing the email“ you mean adding in a sentence about how they like to watch your videos.
panulli4 He most probably meant editing the Korean dude's English, since Korea is the Asian country that is least able use it.
Mitko Gospodinov dude the joke flew right over your head holy fook
@@m.g.6081 whoosh
probably fixing grammar and spelling?
i guess he made the problem easier by editing it
Extra credit:
Let O be the center of the circle and M where the perpendicular to AB through O intersect AB, then OM is the height of OAB relative to AB.
The height intersect it's relative base only if and only if neither of the adjacent angle to that base are obtuse.
The arc angle is obtuse, hence neither of the adjacent angles can be obtuse. Therefore, for this cone, the path will always go uphill and downhill if A and B are not lying on O.
General case:
if the arc angle is obtuse, see previous demonstration, else, it depends on the angles at the base if they are obtuse or not.
For the step 5, I just used the cosine theorem once again, because the downhill part can be easily found as 50*cos(ABO), where O is the Cone's vertex)
Based solely on the answer choices, I'd guess between 3 and 4. The other choices would be obvious immediately upon calculating the radical correctly.
chinareds54 Actually, there are 5 choices in the test.
Dong-hyeon Kim 그런데 인터넷으로 찾아보니 같은 선지가 두개있더라구요, 오타였던건지...
劉형준 엥 저도 확인해봤는데 그렇네요..
Hell I'd guess between 3&4 just because they have the same denominator.
I guessed between B and C not only because they have the same numerator but also that they both are the most common multiple choice answers. ’Course I chose C because it is the MOST common one.
Piers Morgan will be *totally stumped* by this true maths problem.
And Karren Brady didn't even understand the problem. 😉
And by its solution, if given on a silver plate.
So far, for this video,
👍 : 👎 = 810 : 10 = 81 : 1
For last video,
👍 : 👎 = 566 : 442 = 1.28 : 1
Presh, you are back on the right track.
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This was really an incredible question.
this problem is great! it took me a while to solve but I got there in the end XD