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: Can You Solve This 6th Grade Geometry Problem From China? (1.7 million views) ua-cam.com/video/xnE_sO7PbBs/v-deo.html HARD Geometry Problem: Can You Solve The Horse Grazing Puzzle? (165,000 views) ua-cam.com/video/kaLiagYuYPc/v-deo.html Can You Solve A REALLY HARD Math Problem? The Circle Inscribed In A Parabola Puzzle (83,000 views) ua-cam.com/video/z2P8q4QC53s/v-deo.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
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
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.
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.
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.
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).
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)
*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.
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°.
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
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
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 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
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
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'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.
@@졸로-s7o 그게 한국 한정이니까 그러죠. 수능은 분명 누가누가 수학적 사고가 잘되나 하고 보는건데 현실은 누가누가 더 오랫동안 공부했나를 보는거잖아요. 다른 분들이 말했듯이 이 문제가 기출문제집에 나오고 그러니까 이런걸 풀어본 사람들은 쉽게 풀지만 처음 나왔을때는 정답률 10%잖아요. 같은 유형의 문제만 외워서 뭐해요. 완전히 새로운 유형을 만나면 쩔쩔매는데. 그렇게 대학을 가니까 대학이 취업을 위한게 목적이 되고 말았죠. 취업은 기업에서 그동안 해오던 일을 수행하기 위해 하는거지만 연구는 기존에 없던 새로운 것을 만들어 내는거잖아요. 당연히 자연대나 공대나 상대라면 새로운 수학 문제도 만들테고요. 그런데 문제 유형을 외워서 푼다면 그런 문제를 풀수 있을까요? 흔히 수학 괴물이라고 불리는 천재들이 아니라면 힘들겠죠.
@Beom Lim저도 님들처럼 딱 그렇게 생각했었음 ㅋㅋ 재수하면서 인강도 많이 봤고 특히 인강 강사들 하는 말이 전부 맞는 말인줄 알았음. 수학 고난이도 문제 풀려면 단순히 공식 외우는게 아니라 개념을 정확히 알아야 풀 수 있는 것이고, 평가원 문제, 특히 수능문제는 정말 교수들이 머리 꽁꽁 싸매서 만든 질 좋은 문제라고 생각했음. 암기가 아니라 정말 개념과 원리를 이해해야 풀 수 있는 아름다운 문제라고 ㅋㅋ 실제로 한국 교육이나 수능 까는 글이나 영상 있으면 공부도 못하는 빡대가리 새기들이 뭣도 모르면서 깐다고 욕도 했었음. 그런 시절을 지나 학부에서 컴퓨터 보안를 전공하고 지금은 대학원에서 인공지능 연구하면서 선형대 통계 해석학 최적화 등등 다양한 분아의 수학을 공부하고 있는 상황인데 지금 생각해보면 입시 교육이 주입식이였다는 생각이 듬. physical한 영역에서의 수학은 정해진 유형도 없고 답도 없음. 외국애들 논문 내는거 보면 정말 창의적인 논문들 많이 내는데 그런것들이 다 어렸을 때 부터 저 유튜버처럼 다양하게 생각하고 고민하는 것에서 부터 발전되는 거임. 한국인들이 창의성이 떨어지고 시키는 것만 잘한다, 한국에서 노벨상이 안나온다 등등의 말이 괜히 나온게 아니구나 라고 생각이 듬. 입시 교육에서 미분을 쓰기 전에 정말 이 상황에서 미분을 사용할 수 있는 조건이라는 생각은 하긴 함? 미분을 사용하기 위해서는 많은 전제조건들이 필요한데 솔직히 대충 그래프 있고 문제에 f' 이런거 있으면 기계적으로 미분 쓰지 않음?
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.
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.
@@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.
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!
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
Don’t need cosine law. Extend AO (O is apex of cone) to C such that BC is perpendicular to AC. Angle BOC is pi/3, so OC is 50/2=25, and BC is 25root(3). One can get AB now. h (or AH) is perpendicular to AB. Triangles AHO and ACB are similar. One can find AH, then BH.
@@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
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.
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
I find it pretty hard. I mean u dont have all the time in the world to solve it, if u can figure it out in 5 minutes or so you have my congratiolations
@@yuno7825 koreans can cuz they study math in a different way... they study math by memorizing, not understanding. imo this is not a good way to learn math.
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
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.
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.
In CSAT, Geometry is not the problem. Always, with no exception, the hardest part is differential equation. The reason why students should solve this at least in 4 mins comes from it. They must put at least 30mins in solving differential equation.
Korean SAT or Korean high school math doesn't deal with differential equation. The trickiest part is calculus(excluding differential "equation") related to analysis(like a transcendental functions graph). As that person said, ODE has specific way to solve. Most of tricky part in Korean SAT isn't solved by memorizing. It demands creativity. (Sorry for my poor English. I'm just pure Korean)
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..
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 can guess based on the options. I got the large angle 120 (2/3pi). Therefore angle at B is larger than (180-120)/2 =30, therefore, h>50/2 similarly h x in the range around 40-43. It is D.
if h < 30 -> x > 40, B and C < 40 for sure. As h > 25 -> x < sqrt (2500 - 625) < sqrt (2000) = sqrt (40000/20) = 200/sqrt(20) < 200/sqrt(19). therefore A is not OK.
The Korean SAT gives about an hour and a half for each test. The number of questions in the math test is 30, and depending on the attitude of the Ministry of Education, the questions on the CSAT can be easily found, the normal difficulty level, or the difficulty level of hell. If it's easy, it will take about a minute for each question, and if it's hell, you have to solve it in a minute. Do you know why I have to solve it in a minute? Because of the time. We have to solve it once and review it, but if it takes two to three minutes to solve one problem, we will run out of time to solve another problem. But if such a difficult problem comes up, there are three ways. First, let's put the asterisk and solve it later. Second, analyze the answers you've taken so far because you're confident in your skills. If you don't get the most number 3, you'll get number 3. The third is to just pray to God and take a number that seems to be the most correct answer.
Me: Mom I want to go mountain Mom: We already have mountain at home Mountain at home: P.S:Wait, this is just a boring mass-production joke with no creativity at all. Why are you pressing the like button 400 times in this comment?
I’m in middle school and by no means a genius so I’m looking at these comments that are saying that it was relatively simple to solve because of tangent and calculus and unfolding shapes and my brain is melting. I don’t really know why I’m watching this.
The central angle of a cone is equal to 360 times the proportion of the base's radius to the cone's surface length. So (20/60)*360=120 degrees. This is because the perimeter of the base gives the arc length of the sector, and this perimeter lies entirely on the perimeter of the circle defined by the unwrapped cone's vertex and radius.
This is easy for japanese students. They have to more difficult questions. It's why Japan have a lots of Nobel prize unlike S.korea. Another levels of brain
Yes, but the persistence mentioned should be applied to thousands of problems before this test, making every new problem easier. It is impossible to be persistent only during tests!
My procedure: 1) cut the cone along line AB up to the apex (call that C), this gives a sector of a circle, with radius 60 and arc length 2*20*pi. 2) A is at one corner, and B is 10 in from the other corner; the shortest path between them will be a line. 3) Construct a radial line segment of the sector which is perpendicular to the line AB, and call that intersection X. 4) From the arc length you have the interior angle of the segment, and then from the law of cosines, you have the full length AB. 5) Using AB, CA and CB, you can find the area of the triangle ABC using Heron's formula. 6) The area of the triangle will allow you to determine the height, which will be CX. 7) Using CX and CB, you can use Pythagoras to find XB. Simple.
jagmarz exactly..but at first I thought that the shortest distance is a curved line even after spending the 3d cone to 2d..really silly isn't it?.otherwise it's not that tough..but to get it right at first attempt ina short time is the real challenge
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.
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.
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.
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
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.🌹❤❤❤🌹.
I did technical drawing at high school so I went to the development of the cone straight away. I use the cosine rule to find the shortest distance. Hence the answer must be less than 47.5. With many of these tests, possible answers rule themselves out. Answer 2 ruled itself out. Using intuition the answer was number 4. However I wanted a calculated answer. In the last step in this problem I just applied the sin rule to find the missing angle and then calculated the distance the using the cosine rule. If this is an exam then taking a purist approach and calculating everything is not the best option.
Makes sense, but using intuition to solve math questions like this, because they give you multiple choices and just one makes sense feels like cheating. Doesn't feel like you're doing actual math
@@IDyn4m1CI Hi Vinicius Thanks for the comment. Please let me explain my rational. My intuition is based on many years doing and checking engineering drawings so it is a bit different from the average person. In an examination you have to use every tool available. The question was select the correct answer ....not calculate the answer from first principals. If this was the intention it would not have been multiple choice. Make your choice... be morally correct and fail the exam or use all your tools and pass. Exams don't reflect reality. Also from past experience the smartest person does not make the best employee. I will always employ the person with the best imagination. I am an engineer and we never do important calculations in a hurry unlike an exam. We do them slowly and check then recheck. For important structures we often employ an independent party to check our calculations and assumptions. Often in engineering I select a member size based on experience then do my calculations and code checks to prove it is correct. If they don't align then alarm bells ring. In this problem I did do a manual calculation to verify my assumption. Sorry for this long winder reply. Cheers John
@@johnspathonis1078 Thanks for the reply man. Although that still doesn't quite sit right with me. If the exam is meant to evaluate your math knowledge, than you should have to use every bit of knowledge you have to calculate the answer, and not make educated guesses based on intuition. I had to do a similar exam (ENEM, here in Brazil) to get selected to study in my current uni (I'm an undergrad physics student) and it sucks, you have to develop a whole new set of skills to pass that thing because you simply don't have time to read some of the questions or do the math required to solve them. The exam is divided in 2 days, in each day you have to answer 90 questions, and in the second day you also have to write a 24-30 line text about a selected theme, elaborating about it and proposing a solution. Some of the questions are so large you can't bother to read the text, and in the Math part of the test, there are questions ranging from easy (trivial), medium and hard difficulty, the hard ones you simply have to skip or make educated guesses, and then take your time after the exam to solve them to find out if your guess was correct.
@@IDyn4m1CI Hi Vinicius. Agreed exams really suck. They do not reflect reality but I am unsure what the practical alternative is. I think in a way we are saying the same thing. I used the word intuition and you said educated guess. I think they are the same. They are both based on our knowledge and understanding of the subject matter. It has been interesting discussing this with you. I am in Australia and am currently in lockdown due to a Covid outbreak new me. Cheers and stay safe. John
Are you two just good at math from the beginning? The reason I chose med school was because I could never survive this exams. All we need is just read and read no calculations. Most of my friends who became specialist surgeon etc are rather average but they’re all diligent. I bet they’ll have a stroke trying to pass this exam. The irony is there was this junior who was a math freak but after 4 semester he quit and moved back to his habitat: engineering school. Last time I heard he was in Japan pursuing PhD in fields I can’t even remember, still in the realm of physics & math.
Why 4 is the answer? 2 is very different from the rest bcos of number composition. 3 and 4 have similar denominator so the answer is most likely from the two. 1 has switched 91 to 19 and the numerator is 200 which is half of number of 4 as opposed to two thirds of 3. Therefore, 4 is the answer.
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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.
I did mechanical drawing in school and that question is not too hard. If you think like me, you first mentally unfold the cone, flatten it and work your calculations thereupon.
There is one thing that you should know: there was way difficult CSAT mathematics problem . That test took place in 1997, and only 0.08% of korean students solved that problem. That problem was about set
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.
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.
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 ...
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.
UA-cam algorism somehow brings me here. I am Korean and I had this test 10 years before. One thing I can tell for sure is it's not even close to hard problem thing in that test at all. The test's problems can be divided into 3 types according to it's difficulty. (2 points / 3 points / 4 points) The hardest problem is 4 points. In my memory, that one is 3 points problem. The most difficult problem was to find maximum or minimum distance using derivation and integration from the 'shadow' of a sphere.
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!
I used the Theory of Dog racing, No 4 is my favorite number. Solved in a guess! 4/1 chance of being correct. normally Dog numbers 1,2,3 end up in a bump and run at the start which paves the way for the 4 to have a chance of winning. BS i'm sure!
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!
Yes - I think that's right. Sometimes I mix up segment and sector. (Segment is the part of the circle made by a chord). I guess "arc" would just be the line part of the circumference in this problem. Cool problem though.
You are correct. I pinned this note to the top, added to the video description, and edited my blog post. 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: Can You Solve This 6th Grade Geometry Problem From China? (1.7 million views) ua-cam.com/video/xnE_sO7PbBs/v-deo.html HARD Geometry Problem: Can You Solve The Horse Grazing Puzzle? (165,000 views) ua-cam.com/video/kaLiagYuYPc/v-deo.html Can You Solve A REALLY HARD Math Problem? The Circle Inscribed In A Parabola Puzzle (83,000 views) ua-cam.com/video/z2P8q4QC53s/v-deo.html
Pretty elaborate for a test question. I would suggest the same problem with the radius of the base circle being > 30 instead of 20. The angle theta is larger than pi inthis case. Then the answer is 60-10=50 because the shortest path will be infinitely close to the path going from A along the cut to the cone vertex, then go down along the same cut from the vertex to B.
근데 저 문제 원래 풀 대상이 고3인 부분에서 중딩들이 막막해도 정상아님? 그런거를 배워나가눈 거잖음 중1 2 때 고딩 수학을 끝내야하는 이 교육제도가 이상한거임 저건 막막한게 평범한거고 처음보자마자 알면 대단한거임 심지어 접하다의 개념도 잘모르는데 저걸 꼭대기에서부터위 거리고 생각하는게 말이나 되냐 일반인한테
I started my solve the same way, but deviated after drawing the path. The method I used did not involve the length of the whole path (though I DID calculate it at first). I divided the angle 2Pi/3 into Theta and 2Pi/3-Theta and established their cosines (h/50 and h/60). I then used the formula for the cosine of a difference (cos (A-B) = cos A cos B + sin A sin B). I solved for h (h = x*3*(sqrt 3)/8), then used it in the Pythagorean theorem (h^2 + x^2 = 2500). It was a bit more convoluted but arrived at the same answer.
Hi I am a graduate student studying quantum computing from Korea. I've seen this problem when I was a middle school student. It had become a very famous and typical, and most of students can solve this easily. Since I didnt know the calculus but geometry and trigonometry, it was relatively simple to find way to solve. If I met this problem for the first time during the college course, I might use calculus as you did. I think that as we learn more complete and complex theories, simple but insightful theories become hidden from our sight. Thanks for remind this problem to me, have a nice day
A to B straight line is not comprehensible. This has to be a arc? In order to find solution, he has made a straight line. For example, if there is no downhill (only uphill climbing), then he would have used an arc from A to B for shorter distance.
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:
Can You Solve This 6th Grade Geometry Problem From China? (1.7 million views)
ua-cam.com/video/xnE_sO7PbBs/v-deo.html
HARD Geometry Problem: Can You Solve The Horse Grazing Puzzle? (165,000 views)
ua-cam.com/video/kaLiagYuYPc/v-deo.html
Can You Solve A REALLY HARD Math Problem? The Circle Inscribed In A Parabola Puzzle (83,000 views)
ua-cam.com/video/z2P8q4QC53s/v-deo.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
I can't decide if I like the question more, or the solution. Brilliant!
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*
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 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
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..
@@피즈-m2d 아따 01년생한테 6학년이라니 거 말이 너무 심한거 아니오! 수학 못해도 드립치는건 내 자유다 이말이야!!
시행착오과정을 다 기록한다는 점이 독특하다
저러면 한 문제를 풀어도 밀도가 다를듯
킹정
m/v가왜다름
@@뭐뭣 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
Your problem solving skills are exemplary & very absorbing .. I m an engr with 43 yrs experience, and still enjoy watching your solutions..
Sir
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.
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
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.
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😑
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
외국인들 댓글1% / 수능뽕차서 열심히 번역기 돌리면서 댓글다는 한국인99%
쓸데없는 국뽕이다ㅋㅋㅋㅋ 오히려 독인데 그걸 자랑스럽게 여기는듯한
@@홍인산삼 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
@@지원임-v3d 그게 필연적임? 문화절대주의랑 필연적인 연관성은 없는거 같은데
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
> 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?
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
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
Yea not that hard unless the time is very limited
시부럴 우리나라 수학 30번 풀면기절하것네ㅋㅋㅋㅋㅋㄱㄱ
ㄹㅇ ㅋㅋㅋㅋㅋㅋㅋㅋㅋ 요즘 수학 가형풀면 기절하겠누 ㅋㅋㅋㅋㅋㅋㅋㅋ 97수능이면 요즘 가형 27번쯤 될 듯
ㄹㅇㅋㅋ
@@구굴게이를외치다진짜 17학년도 수능 가형 30번이면ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ
@이경훈 솔직히 근데 전체적인 수능 성적으로 따지면 언제가 더 어렵고 쉽고 비교 할 수 있나...? 97수능 문제 구경도 못해보긴 했지만.
@@뾱뾱이-w8d 전반적으로 수준이 올라서 예전이랑은 비교하면 안됨 ㅇㅇ
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 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'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.
댓글과 영상에서 한국 입시교육의 문제점이 보이네. 저 와국인처럼 어떻게 문제를 해결헐 수 있을꺼 다각도로 접근해야지 전개도를 써야겠다는 주입식 지식으로 푸는 것이 무슨 의미가 있을까.. 실제 세상에서 쓰이는 수학은 입시처럼 유형화 되어 있지 않은데
맨날 수능 까는애들이 하는말이
"이런거 배워서 어따써먹음?" ㅇㅈㄹ하는데 애초에 수능은 대학교육에 필요한 수학 능력을 측정하는 시험임 이런 댓글보면 한숨만 나온다
@@졸로-s7o 그게 한국 한정이니까 그러죠.
수능은 분명 누가누가 수학적 사고가 잘되나 하고 보는건데 현실은 누가누가 더 오랫동안 공부했나를 보는거잖아요.
다른 분들이 말했듯이 이 문제가 기출문제집에 나오고 그러니까 이런걸 풀어본 사람들은 쉽게 풀지만
처음 나왔을때는 정답률 10%잖아요.
같은 유형의 문제만 외워서 뭐해요. 완전히 새로운 유형을 만나면 쩔쩔매는데.
그렇게 대학을 가니까 대학이 취업을 위한게 목적이 되고 말았죠. 취업은 기업에서 그동안 해오던 일을 수행하기 위해 하는거지만
연구는 기존에 없던 새로운 것을 만들어 내는거잖아요. 당연히 자연대나 공대나 상대라면 새로운 수학 문제도 만들테고요. 그런데 문제 유형을 외워서 푼다면 그런 문제를 풀수 있을까요? 흔히 수학 괴물이라고 불리는 천재들이 아니라면 힘들겠죠.
몇년전, 거의 10년전에 하버드 학생들이 한국 고등학교에 와서 고등문제를 못풀었던 적이 있었죠. 그리곤 고등학생들이 풀고서 하는 말이 '이런 문제를 풀어봐서 풀수 있었다.' 였고요. 우리나라 교육과 외국 교육의 차이를 나타내죠.
@Beom Lim저도 님들처럼 딱 그렇게 생각했었음 ㅋㅋ 재수하면서 인강도 많이 봤고 특히 인강 강사들 하는 말이 전부 맞는 말인줄 알았음.
수학 고난이도 문제 풀려면 단순히 공식 외우는게 아니라 개념을 정확히 알아야 풀 수 있는 것이고, 평가원 문제, 특히 수능문제는 정말 교수들이 머리 꽁꽁 싸매서 만든 질 좋은 문제라고 생각했음. 암기가 아니라 정말 개념과 원리를 이해해야 풀 수 있는 아름다운 문제라고 ㅋㅋ 실제로 한국 교육이나 수능 까는 글이나 영상 있으면 공부도 못하는 빡대가리 새기들이 뭣도 모르면서 깐다고 욕도 했었음.
그런 시절을 지나 학부에서 컴퓨터 보안를 전공하고 지금은 대학원에서 인공지능 연구하면서 선형대 통계 해석학 최적화 등등 다양한 분아의 수학을 공부하고 있는 상황인데 지금 생각해보면 입시 교육이 주입식이였다는 생각이 듬. physical한 영역에서의 수학은 정해진 유형도 없고 답도 없음. 외국애들 논문 내는거 보면 정말 창의적인 논문들 많이 내는데 그런것들이 다 어렸을 때 부터 저 유튜버처럼 다양하게 생각하고 고민하는 것에서 부터 발전되는 거임. 한국인들이 창의성이 떨어지고 시키는 것만 잘한다, 한국에서 노벨상이 안나온다 등등의 말이 괜히 나온게 아니구나 라고 생각이 듬.
입시 교육에서 미분을 쓰기 전에 정말 이 상황에서 미분을 사용할 수 있는 조건이라는 생각은 하긴 함? 미분을 사용하기 위해서는 많은 전제조건들이 필요한데 솔직히 대충 그래프 있고 문제에 f' 이런거 있으면 기계적으로 미분 쓰지 않음?
수능의 수학이 그수학인줄아는사람들이있네
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!
Wonderful.... This problem and it’s solution are pieces of artwork :)
Math is beautiful...
우리는 전개도를 그려서 시각화하는게 좋은 방법이라는 것을 알고 있어서 그런가.. 이게 그렇게 어려운 아이디어인가? 솔직히 모르겠네요
저런 유형들 많이 분석돼서 중딩때 수업 잘 들었고 코사인 법칙만 배웠다면 쉽게 푸는 문제가 되었죠...
ㄹㅇ ㅋㅋ 딱 중3문제인데
@@부동산관심 ㅇ 맞음 나도 초등학교때 수학학원에서 이 문제 보고 벌벌떨었음 근데 이문제가 정답률이 10퍼센트라고요?
@@부동산관심 그건 맞지 근데 이게 10퍼센트는 아닌거 같은데 어디 자룐진 모르겠지만 못해도 40퍼정도는 나올듯
@@바르고고운말 어디자룐지 모르는게 아니라 당시 수능 정답률인데요ㅋㅋ
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
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.
- 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
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?
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!
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'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
Don’t need cosine law.
Extend AO (O is apex of cone) to C such that BC is perpendicular to AC. Angle BOC is pi/3, so OC is 50/2=25, and BC is 25root(3). One can get AB now. h (or AH) is perpendicular to AB.
Triangles AHO and ACB are similar. One can find AH, then BH.
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
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
I love this question. It is distinct from those traditional textbook drills and sparks the imagination of the mathematical mind. Thank you very much.
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.
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.
By the way, the hardest question does not provide multiple choices, which makes this question relatively easy
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
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
I am also a Korean student, and that type of question isn't really hard anymore since it has been over 20 years
YH C true lol
RIGHT... It's not actually really hard question.
I find it pretty hard. I mean u dont have all the time in the world to solve it, if u can figure it out in 5 minutes or so you have my congratiolations
Lol
@@yuno7825 koreans can cuz they study math in a different way... they study math by memorizing, not understanding. imo this is not a good way to learn math.
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
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!
ua-cam.com/video/OkmNXy7er84/v-deo.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
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.
In CSAT, Geometry is not the problem. Always, with no exception, the hardest part is differential equation. The reason why students should solve this at least in 4 mins comes from it. They must put at least 30mins in solving differential equation.
well but there are ways u can follow to solve ODE, so it's easier for some ppl
It really looked like a calculus problem. Should be no problem for students in differential equation.
Diff Eqs in high school sounds unusual
@@pierrecurie welcome to Asia
Korean SAT or Korean high school math doesn't deal with differential equation. The trickiest part is calculus(excluding differential "equation") related to analysis(like a transcendental functions graph). As that person said, ODE has specific way to solve. Most of tricky part in Korean SAT isn't solved by memorizing. It demands creativity. (Sorry for my poor English. I'm just pure Korean)
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 😂😂😂
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 can guess based on the options. I got the large angle 120 (2/3pi). Therefore angle at B is larger than (180-120)/2 =30, therefore, h>50/2 similarly h x in the range around 40-43. It is D.
if h < 30 -> x > 40, B and C < 40 for sure. As h > 25 -> x < sqrt (2500 - 625) < sqrt (2000) = sqrt (40000/20) = 200/sqrt(20) < 200/sqrt(19). therefore A is not OK.
The Korean SAT gives about an hour and a half for each test. The number of questions in the math test is 30, and depending on the attitude of the Ministry of Education, the questions on the CSAT can be easily found, the normal difficulty level, or the difficulty level of hell. If it's easy, it will take about a minute for each question, and if it's hell, you have to solve it in a minute. Do you know why I have to solve it in a minute? Because of the time. We have to solve it once and review it, but if it takes two to three minutes to solve one problem, we will run out of time to solve another problem.
But if such a difficult problem comes up, there are three ways. First, let's put the asterisk and solve it later. Second, analyze the answers you've taken so far because you're confident in your skills. If you don't get the most number 3, you'll get number 3. The third is to just pray to God and take a number that seems to be the most correct answer.
This is a very reasonable problem to solve in 2-3 min... just the calculations are a bit cumbersome
definitely NOT A HARD problem
@@Mike-rw1jw Altough there are problems that are harder this one is difficult.
Me: Mom I want to go mountain
Mom: We already have mountain at home
Mountain at home:
P.S:Wait, this is just a boring mass-production joke with no creativity at all. Why are you pressing the like button 400 times in this comment?
@Helloo There lmao
ㅋㅋㅋㅋㅋㅋㅋ
? 저만 이해 못한겅가요?
혹시 이거 왜 웃긴지? 알려주실분!
@궁예 아 ㅋㅋㅋㅋㅋ
감사합니다 ㅋㅋㅋ
UNDERRATED
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
I’m in middle school and by no means a genius so I’m looking at these comments that are saying that it was relatively simple to solve because of tangent and calculus and unfolding shapes and my brain is melting. I don’t really know why I’m watching this.
근데 진짜 너무 당연스럽게 전개도 펼쳐서 저 그림 그림
시작할때 4번의 잘못된 풀이를 했다는거가 저렇게 많은 풀이시도가 있을 수 있구나는 생각에 젤 놀라웠음
수능치는데 유튜브가 나보고 유튜브 보지말고 공부하라고 영상 추천해주네
유튜브 알고리즘 똑똑하누
머염쿤UA-cam 난 수능도 안보는데 왜 추천해주누 ㅋㅋㅋㅋ
어떻게 사람이 문과냐
I completely agree man you guys are nuts. In a good way.
@@ab-bo5hq 문들문들;
Eeney meeney miney Mo was invented for a reason
This is that reason
And it freaking worked for me .........question done in 1st attempt ....
The central angle of a cone is equal to 360 times the proportion of the base's radius to the cone's surface length.
So (20/60)*360=120 degrees.
This is because the perimeter of the base gives the arc length of the sector, and this perimeter lies entirely on the perimeter of the circle defined by the unwrapped cone's vertex and radius.
기출 그림이 왜있나 보러왔는데 수능 문제구나..
박상현 22 ㅋㅋㅋㅋㅋㅋ
박상현 hahahahah Kim jong un
This is easy for japanese students. They have to more difficult questions. It's why Japan have a lots of Nobel prize unlike S.korea. Another levels of brain
@@guozhou8114 中国人だとおもいますが、確かに認めざるを得ない部分もありますね.でも、動画に関係ない話で他国を批判することはお止めください.
김도현 translate please. I dont know so much. And why can you use complexed japanese language?
challenging problems often require persistence
and you got about an hour and a half to solve like 20 of these.....
nah, 45~50min. it was on my test actually.
Yes, but the persistence mentioned should be applied to thousands of problems before this test, making every new problem easier. It is impossible to be persistent only during tests!
My procedure: 1) cut the cone along line AB up to the apex (call that C), this gives a sector of a circle, with radius 60 and arc length 2*20*pi. 2) A is at one corner, and B is 10 in from the other corner; the shortest path between them will be a line. 3) Construct a radial line segment of the sector which is perpendicular to the line AB, and call that intersection X. 4) From the arc length you have the interior angle of the segment, and then from the law of cosines, you have the full length AB. 5) Using AB, CA and CB, you can find the area of the triangle ABC using Heron's formula. 6) The area of the triangle will allow you to determine the height, which will be CX. 7) Using CX and CB, you can use Pythagoras to find XB. Simple.
jagmarz exactly..but at first I thought that the shortest distance is a curved line even after spending the 3d cone to 2d..really silly isn't it?.otherwise it's not that tough..but to get it right at first attempt ina short time is the real challenge
We should solve that at 3~4 minutes😭
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.
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
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.
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
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.🌹❤❤❤🌹.
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
I did technical drawing at high school so I went to the development of the cone straight away. I use the cosine rule to find the shortest distance. Hence the answer must be less than 47.5. With many of these tests, possible answers rule themselves out. Answer 2 ruled itself out. Using intuition the answer was number 4. However I wanted a calculated answer. In the last step in this problem I just applied the sin rule to find the missing angle and then calculated the distance the using the cosine rule. If this is an exam then taking a purist approach and calculating everything is not the best option.
Makes sense, but using intuition to solve math questions like this, because they give you multiple choices and just one makes sense feels like cheating. Doesn't feel like you're doing actual math
@@IDyn4m1CI Hi Vinicius Thanks for the comment. Please let me explain my rational. My intuition is based on many years doing and checking engineering drawings so it is a bit different from the average person. In an examination you have to use every tool available. The question was select the correct answer ....not calculate the answer from first principals. If this was the intention it would not have been multiple choice. Make your choice... be morally correct and fail the exam or use all your tools and pass. Exams don't reflect reality. Also from past experience the smartest person does not make the best employee. I will always employ the person with the best imagination. I am an engineer and we never do important calculations in a hurry unlike an exam. We do them slowly and check then recheck. For important structures we often employ an independent party to check our calculations and assumptions. Often in engineering I select a member size based on experience then do my calculations and code checks to prove it is correct. If they don't align then alarm bells ring. In this problem I did do a manual calculation to verify my assumption. Sorry for this long winder reply. Cheers John
@@johnspathonis1078 Thanks for the reply man. Although that still doesn't quite sit right with me. If the exam is meant to evaluate your math knowledge, than you should have to use every bit of knowledge you have to calculate the answer, and not make educated guesses based on intuition. I had to do a similar exam (ENEM, here in Brazil) to get selected to study in my current uni (I'm an undergrad physics student) and it sucks, you have to develop a whole new set of skills to pass that thing because you simply don't have time to read some of the questions or do the math required to solve them. The exam is divided in 2 days, in each day you have to answer 90 questions, and in the second day you also have to write a 24-30 line text about a selected theme, elaborating about it and proposing a solution. Some of the questions are so large you can't bother to read the text, and in the Math part of the test, there are questions ranging from easy (trivial), medium and hard difficulty, the hard ones you simply have to skip or make educated guesses, and then take your time after the exam to solve them to find out if your guess was correct.
@@IDyn4m1CI Hi Vinicius. Agreed exams really suck. They do not reflect reality but I am unsure what the practical alternative is. I think in a way we are saying the same thing. I used the word intuition and you said educated guess. I think they are the same. They are both based on our knowledge and understanding of the subject matter. It has been interesting discussing this with you. I am in Australia and am currently in lockdown due to a Covid outbreak new me. Cheers and stay safe. John
Are you two just good at math from the beginning? The reason I chose med school was because I could never survive this exams. All we need is just read and read no calculations. Most of my friends who became specialist surgeon etc are rather average but they’re all diligent. I bet they’ll have a stroke trying to pass this exam.
The irony is there was this junior who was a math freak but after 4 semester he quit and moved back to his habitat: engineering school. Last time I heard he was in Japan pursuing PhD in fields I can’t even remember, still in the realm of physics & math.
Imagine the horror when u have spent so much of your time and your answer doesnot match with any in the mcqs...
N○OO◦◦oo....༼;´༎ຶ ༎ຶ༽
Just chose the one nearest to ur answer
Why 4 is the answer?
2 is very different from the rest bcos of number composition.
3 and 4 have similar denominator so the answer is most likely from the two.
1 has switched 91 to 19 and the numerator is 200 which is half of number of 4 as opposed to two thirds of 3.
Therefore, 4 is the answer.
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.
Likes and dislikes don't matter to the success of a video. Advertisers only care about the number of views. Liking is only to help the algorithms build up a profile of you.
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.
I did mechanical drawing in school and that question is not too hard. If you think like me, you first mentally unfold the cone, flatten it and work your calculations thereupon.
There is one thing that you should know: there was way difficult CSAT mathematics problem .
That test took place in 1997, and only 0.08% of korean students solved that problem. That problem was about set
이게 우리나라에서 0.08% 맞췄다구요???
ㅎㄷㄷ. 지금보면 쉬운문제중 1개인데
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
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.
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 ...
당시에는 삼각형의 높이 h를 60, 50과 싸인 120도를 이용한 삼각함수로 구하고 피타고라스 정리로 내려가는 길의 길이를 구했었습니다.
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.
UA-cam algorism somehow brings me here. I am Korean and I had this test 10 years before. One thing I can tell for sure is it's not even close to hard problem thing in that test at all. The test's problems can be divided into 3 types according to it's difficulty. (2 points / 3 points / 4 points) The hardest problem is 4 points. In my memory, that one is 3 points problem. The most difficult problem was to find maximum or minimum distance using derivation and integration from the 'shadow' of a sphere.
@Pete you must be kidding
Algorithm... m8
리빙포인트) 이 문제는 "중2 삼각형의 성질" 단원 쎈에도 수록된 문제이다
어억... 루트가 중3에 나올텐데 중2 문제집에 이게 나온다구요..?
중2때 블랙라벨에서 본 것 같기도 하네요
@@누리꾼-e9y 저도 왜인지 모르겠는데 2015 개정교육과정 중2 쎈 Cstep에 있습니다;;
@@누리꾼-e9y 음 예전수학이랑 지금수학이랑 배우는 것이 달라지지않았을까요
지금 곧 고1인데 요즘은 중학교 3학년 때 삼각비 파트 나옴 이런 문제는 중3 때 난이도 상짜리로 나올 수도 있을 듯
어림도 없지
바로 ‘문과’
씹 ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ
@이마넌 ?
@이마넌 내년은 모르겠지만 2019 수학 나형은 기백 안배움
문프
어림도 없지
삼각함수
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!
I used the Theory of Dog racing, No 4 is my favorite number. Solved in a guess! 4/1 chance of being correct. normally Dog numbers 1,2,3 end up in a bump and run at the start which paves the way for the 4 to have a chance of winning. BS i'm sure!
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?
Imagine to live or die you must answer this question, I’m dying for sure bruh 😔
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:)
“What method did you use?” Guessing.
😄
I used exactly this methode in one shot
What the hell! All that for a single MCQ? You must be kidding me, right? But you solved it brilliantly I must say.
Shouldn't it be called a circular sector instead of circular arc?
I agree.
Yes - I think that's right. Sometimes I mix up segment and sector. (Segment is the part of the circle made by a chord). I guess "arc" would just be the line part of the circumference in this problem. Cool problem though.
Corey Derochie yep
You're right.
You are correct. I pinned this note to the top, added to the video description, and edited my blog post.
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:
Can You Solve This 6th Grade Geometry Problem From China? (1.7 million views)
ua-cam.com/video/xnE_sO7PbBs/v-deo.html
HARD Geometry Problem: Can You Solve The Horse Grazing Puzzle? (165,000 views)
ua-cam.com/video/kaLiagYuYPc/v-deo.html
Can You Solve A REALLY HARD Math Problem? The Circle Inscribed In A Parabola Puzzle (83,000 views)
ua-cam.com/video/z2P8q4QC53s/v-deo.html
Pretty elaborate for a test question. I would suggest the same problem with the radius of the base circle being > 30 instead of 20. The angle theta is larger than pi inthis case. Then the answer is 60-10=50 because the shortest path will be infinitely close to the path going from A along the cut to the cone vertex, then go down along the same cut from the vertex to B.
I couldn't solve this even if you gave me 5 years
너가 머리가 안좋아서 그럼ㅋㅋㅋㅋ
@@user-oe9sg5rn5v 진짜 역겹네
@@user-oe9sg5rn5v 이런애들 특) 영어독해 저사람보다 못함
@@YUUUUUUUUUUUUUUUUUUUUUUUN 아니 저사람은 원어민이자너ㅋㅋ
저거 쉬워보인다는 친구들
중학교때 저 문제 처음 만났을때 생각해보셈 ㅈㄴ 막막했을걸
ㅇㅈ 푸는법만 알면 풀 수 있는 저 유형의 기본문제니까 쉽다고 하는거지 저 유형에서도 조그만 변형들어가도 여기서 쉽다하는 애들 대부분 못풀듯
근데 저 문제 원래 풀 대상이 고3인 부분에서 중딩들이 막막해도 정상아님? 그런거를 배워나가눈 거잖음 중1 2 때 고딩 수학을 끝내야하는 이 교육제도가 이상한거임 저건 막막한게 평범한거고 처음보자마자 알면 대단한거임 심지어 접하다의 개념도 잘모르는데 저걸 꼭대기에서부터위 거리고 생각하는게 말이나 되냐 일반인한테
@ᄎᄀᄀᄂᄂ 다른 답글들이 뭐라함
@ᄎᄀᄀᄂᄂ 답글이 아니라 댓글
@@catmom2 미안하지만 중딩은 아니야
"what method did you use ?" .....i couldnt do it man xd
I started my solve the same way, but deviated after drawing the path. The method I used did not involve the length of the whole path (though I DID calculate it at first). I divided the angle 2Pi/3 into Theta and 2Pi/3-Theta and established their cosines (h/50 and h/60). I then used the formula for the cosine of a difference (cos (A-B) = cos A cos B + sin A sin B). I solved for h (h = x*3*(sqrt 3)/8), then used it in the Pythagorean theorem (h^2 + x^2 = 2500). It was a bit more convoluted but arrived at the same answer.
Nice! Yes, a bit more convoluted, but more systematic, straightforward and relentlessly working solution :)
Hi I am a graduate student studying quantum computing from Korea. I've seen this problem when I was a middle school student. It had become a very famous and typical, and most of students can solve this easily. Since I didnt know the calculus but geometry and trigonometry, it was relatively simple to find way to solve.
If I met this problem for the first time during the college course, I might use calculus as you did. I think that as we learn more complete and complex theories, simple but insightful theories become hidden from our sight. Thanks for remind this problem to me, have a nice day
불면증있는데 마침 딱 좋은 영상 떴네 역시 유튜브 알고리즘
I never realized how awesome Trigonometry was.
A to B straight line is not comprehensible. This has to be a arc? In order to find solution, he has made a straight line. For example, if there is no downhill (only uphill climbing), then he would have used an arc from A to B for shorter distance.