VERY HARD South Korean Geometry Problem (CSAT Exam)

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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• HARD Geometry Problem:...
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• Can You Solve A Challe...
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[@MindYourDecisions]

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) kzbin.info/www/bejne/rp-okKaFbLWVeNU HARD Geometry Problem: Can You Solve The Horse Grazing Puzzle? (165,000 views) kzbin.info/www/bejne/oZKvmpSdjtqMhsU Can You Solve A REALLY HARD Math Problem? The Circle Inscribed In A Parabola Puzzle (83,000 views) kzbin.info/www/bejne/sGOzaaRqhqhoadU

[@jessstuart7495]

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

[@billy.7113]

Jess Stuart How long and how many pages did it take to get the right answer?

[@jessstuart7495]

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.

[@MindYourDecisions]

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!

[@peterz5731]

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

[@rossmarievivas4501]

the video: “what method did you use?” me: ennie meanie minie mo

[@buybotgt316]

Best comment so far haha😂

[@cyanide6954]

U would actually get it correct if u used that method

[@mohamedhusam8189]

@@cyanide6954 this method always gets the correct answer it proved itself many many times

[@normaldude2892]

What a comment 😂

[@tanishqkumar780]

*UNDERRATTED*

[@fahadmohamed5457]

I wouldve just guessed it and pray

[@pinklady7184]

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.

[@RobinClower]

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.

[@e1woqf]

Pink Lady: You cannot bribe God because God exists in your head only.

[@user-rb9nq7rm5n]

Logically you should pray first and then guess.

[@Vanessa24449]

fahad mohamed same

[@user-ko3kh8ci2p]

시행착오과정을 다 기록한다는 점이 독특하다 저러면 한 문제를 풀어도 밀도가 다를듯

[@user-gx6wf7su7n]

킹정

[@user-qi7rv5bq2f]

m/v가왜다름

[@user-wc4ii4hd1g]

@@user-qi7rv5bq2f 시행착오가 세기성질이 아닌가보지

[@ls-qq4iv]

그래서 원래 한국애들 유학가면 중고딩까진 우리가 훨씬 ㅈㄴ 더 잘하는데 점점 더 외국 애들이 잘해져서 나중엔 한국 애들이 더 못해짐

[@user-rr5tk2lc8b]

한국인이 외국 중고등학교에선 성적이 높은데 대학가선 적응 힘든 이유임

[@EastBurningRed]

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.

[@publiconions6313]

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

[@MotoRideswJohn]

I can't decide if I like the question more, or the solution. Brilliant!

[@massivetornado348]

Hardest part is you have less than a minute to solve this accurately

[@charlificate]

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.

[@massivetornado348]

Steven McCulloch In Korea you're expected to be able to solve quickly and accurately. This is highschool level stuff here.

[@stevenmcculloch5727]

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.

[@massivetornado348]

Steven McCulloch Why do you think Koreans students study outside of school as well

[@massivetornado348]

Steven McCulloch Korean education, especially math and English are harder than almost anywhere else in the world

[@biprashisdas]

Wonderful.... This problem and it’s solution are pieces of artwork :) Math is beautiful...

[@aaroneady7330]

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.

[@reldahr01]

I wouldn't be able to solve this. Congrats to anyone who figured this out. Great problem!

[@hannahmorris1835]

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.

[@leonhardfrommhold8463]

*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

[@palindromic283]

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+.

[@user-fz4wh1qo8u]

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.

[@mrfechu6281]

Larz B shortest line is not always a straight line. And this took me just 1 second.

[@vigspark]

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.

[@kdy5617]

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

[@redwren4182]

Math is certainly more fun when you have the time and the context to solve the problems!

[@jp-sn6si]

americans start hating math in the first grade.

[@11_amankumarmall14]

yeah i also usually hate maths but now i love to solve it

[@mercedesamgpetronas2439]

@@kdy5617 math before college in S.Korea is more like memorizing problems. Just solving thousands mindlessly.

[@antikertech157]

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.

[@takdudung]

> Be me > A Korean that just turned into our equivalent of 8th grade > Legit scared

[@supercool1312]

탁두둥 i wishnyou luck, you will probably need it based on this video

[@varunmanjunath9123]

nice, all the best! xP

[@juh4664]

>using comedy chevrons on YT >why?

[@jeanfrancois8145]

asdmvva 4chan user out of his natural element.

[@joris8032]

Do you also study all day long till midnight?

[@yonglog1264]

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.

[@seungmoohan8152]

OMG.....so ture

[@juunnyy02]

현실이네 시부레

[@yK_Ki_JoE]

@@djmikecr9284 형님 한국인 이과라면 당연한거 아입니까

[@yK_Ki_JoE]

@@ys9018 아 수고하세요 저는 6학종..

[@yK_Ki_JoE]

@@user-tg9uz4ve8w 아따 01년생한테 6학년이라니 거 말이 너무 심한거 아니오! 수학 못해도 드립치는건 내 자유다 이말이야!!

[@baka0556]

외국인들 댓글1% / 수능뽕차서 열심히 번역기 돌리면서 댓글다는 한국인99%

[@user-kc7ww1tk8k]

쓸데없는 국뽕이다ㅋㅋㅋㅋ 오히려 독인데 그걸 자랑스럽게 여기는듯한

[@jjh4928]

@@user-kc7ww1tk8k 어떤점에서 독이라는거임

[@user-pp8bo4jz1c]

@@jjh4928 이러한 뽕은 필연적으로 다른 나라에 대한 무시와 차별로 이어지죠

[@user-kc7ww1tk8k]

@@jjh4928 이미 친 사람들은 모르겠지만 학생이면 개고생 해야되니까. 그리고 수능에 맞춰진 교육방식도 문제가 있고

[@user-ml9hn4ww1v]

@@user-pp8bo4jz1c 그게 필연적임? 문화절대주의랑 필연적인 연관성은 없는거 같은데

[@siddharthdikondwar8971]

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

[@LughSummerson]

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°.

[@arthurg.machado6803]

Lugh Summerson yeah , that is where that formula comes from .

[@johndavidalexander6646]

That’s what I did by that does equal 2(pi)/3... am I missing something?

[@GmanMilli]

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.

[@justinc2633]

@@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

[@ekstremumnoktas8836]

I always solve like that

[@Dinesh7219]

Your problem solving skills are exemplary & very absorbing .. I m an engr with 43 yrs experience, and still enjoy watching your solutions..

[@thizAHandle]

Sir

[@d4v1dc0fuse9]

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

[@user-oz8ze4ol2j]

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

[@user-mk5ln8iz7l]

맨날 수능 까는애들이 하는말이 "이런거 배워서 어따써먹음?" ㅇㅈㄹ하는데 애초에 수능은 대학교육에 필요한 수학 능력을 측정하는 시험임 이런 댓글보면 한숨만 나온다

[@user-rr5tk2lc8b]

@@user-mk5ln8iz7l 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.

[@user-rr5tk2lc8b]

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.

[@user-oz8ze4ol2j]

@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?

[@user-gr6fi8qi1z]

수능의 수학이 그수학인줄아는사람들이있네

[@utoronto5928]

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 :)

[@zachcioe5803]

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!

[@wizerdcloe]

zz 한국인들이 이걸 쉽게느끼는건 뇌리에 전개도면이 이미 박혀 있어서 그렇다.

[@MangChi_Hammer]

ㄹㅇ 이미 졸라본듯

[@user-nx4bd8ee2z]

ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ 한국인에게는 그닥 어렵지 않았다...

[@tsuminova]

@UCGBHl99d5_PHk_cl4ZyQlZg 지금은 2021년입니다

[@user-rz2yk9kt2x]

전개도면은 누구나알겠지만 내려가는 부분이 어디서 시작하는지 아는 사람은 별로 없을 것 같습니다

[@user-lw5um8ng7m]

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

[@skil6265]

우리는 전개도를 그려서 시각화하는게 좋은 방법이라는 것을 알고 있어서 그런가.. 이게 그렇게 어려운 아이디어인가? 솔직히 모르겠네요

[@tk0329]

저런 유형들 많이 분석돼서 중딩때 수업 잘 들었고 코사인 법칙만 배웠다면 쉽게 푸는 문제가 되었죠...

[@allisgood60]

ㄹㅇ ㅋㅋ 딱 중3문제인데

[@allisgood60]

@@floccinaucinihiliphilifica5265 ㅇ 맞음 나도 초등학교때 수학학원에서 이 문제 보고 벌벌떨었음 근데 이문제가 정답률이 10퍼센트라고요?

[@billykim7179]

@@floccinaucinihiliphilifica5265 그건 맞지 근데 이게 10퍼센트는 아닌거 같은데 어디 자룐진 모르겠지만 못해도 40퍼정도는 나올듯

[@user-rh8qi9sj5s]

@@billykim7179 어디자룐지 모르는게 아니라 당시 수능 정답률인데요ㅋㅋ

[@viktoriavadon2222]

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).

[@mab9316]

Bravo.

[@weijholtz]

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)

[@jrbleau]

Exactly how I conceived the solution, though I didn't bother doing the grunt work.

[@guesswho6038]

@@weijholtz If the choices are spaced enough to be resolved within your ruler accuracy then it's a great time saver.

[@mikkumi]

Brilliant

[@kylejohnson8462]

I would just take my 25% chance

[@toughguy1013]

that case i would choice 2 or 3, 4 so will be given 33% chance

[@Wyvern11]

You're genius.

[@youarenotlost]

Too easy in 2minutes clear

[@user-xv7dk3op8w]

And now.... we solve that kind of thing for practical qusetion

[@user-xv7dk3op8w]

Just basic question

[@glitchrang]

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!

[@andrekv]

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)

[@albertthorval4674]

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

[@Tehom1]

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.

[@khbye2411]

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?

[@Tehom1]

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.

[@sallylauper8222]

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?!?

[@Tehom1]

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.

[@trueriver1950]

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

[@cohgnh]

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.

[@wesleysuen4140]

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.

[@BruceLCM]

I love this question. It is distinct from those traditional textbook drills and sparks the imagination of the mathematical mind. Thank you very much.

[@sugoy092369]

Just remember this, this IS NOT HARDEST QUESTION in Korea SAT

[@classicloverautum7391]

그래두 자이스토리 29번인가 30번이었다네용 허헣

[@anmolbaloni]

Which is the hardest I want to take a crack so if you find that please post a link thank you

[@revesw]

@@anmolbaloni every problem numbered 21, 29, 30 in the korean math SAT is difficult

[@batmendbatbaatar4290]

Which is still kinda easy

[@frakus3068]

Bruh this is easy tho😑

[@JLvatron]

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.

[@claramagdalenasitorus3505]

I really like your creativity in solving math problems, you are very critical. I'm really amazed. greetings from me in Indonesia :)

[@rigelstar1130]

Hello. I.m a viewer of this channel too and i.m from Indonesia:)

[@shubhramishra8698]

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.

[@crappypoopycrap9800]

Still working on this problem, huh haha

[@trash9598]

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 😂

[@LibertyGunsBeerTrump]

You will never use this once in your life. I guarantee that.

[@angelspearl9974]

@@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

[@user-wy3rj2wf1q]

시부럴 우리나라 수학 30번 풀면기절하것네ㅋㅋㅋㅋㅋㄱㄱ

[@user-vr7vm5if4s]

ㄹㅇ ㅋㅋㅋㅋㅋㅋㅋㅋㅋ 요즘 수학 가형풀면 기절하겠누 ㅋㅋㅋㅋㅋㅋㅋㅋ 97수능이면 요즘 가형 27번쯤 될 듯

[@billykim7179]

ㄹㅇㅋㅋ

[@user-we1dd7kj3u]

@@user-vr7vm5if4s 17학년도 수능 가형 30번이면ㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋㅋ

[@user-we1dd7kj3u]

@이경훈 솔직히 근데 전체적인 수능 성적으로 따지면 언제가 더 어렵고 쉽고 비교 할 수 있나...? 97수능 문제 구경도 못해보긴 했지만.

[@as0325]

@@user-we1dd7kj3u 전반적으로 수준이 올라서 예전이랑은 비교하면 안됨 ㅇㅇ

[@henrytang2203]

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)

[@goretician02]

By the way, the hardest question does not provide multiple choices, which makes this question relatively easy

[@qqqquito]

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.

[@masteroogway2405]

You can find the central angle using this formula- x divided by 360 = radius of the base of the cone ÷ slant height of the cone (x= central angle of the sector)

[@programaths]

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.

[@user-fx9bl9fx6d]

I'm a liberal arts student who was passing by. I think i should keep going.

[@xxphoenixx8398]

SAME...

[@byunsungwoo]

ㅋㅋㅋㅋㅋㅋ문송쓰

[@jonathankim2386]

지나가던 문과생입니다. 계속 지나가겠습니다 이거 영어버전인가 ㅋㅋㅋㅋㅋ

[@asdzx-me6ku]

문붕잌ㅋㅋㅋㅋ

[@eternity4885]

아 역시 드립의민족

[@jackvictoryfankyola105]

thoes South Korean be like : why this guy upload this video it was too easy

[@user-ck8zt1xr4g]

e-z

[@electro_yellow9295]

Geometry dash

[@user-vr7vm5if4s]

Fact

[@user-vp5sd7sq2d]

umm.. it was hard but not hardest.

[@cashmoneycum]

ez 4 me

[@eroraf8637]

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.

[@monkeseeaction21987]

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.

[@skswkdvks]

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

[@acutepenguin3577]

푸아송 What grade is this for?

[@skswkdvks]

@@acutepenguin3577SAT for hight school grade 3 students. This is last grade in Korea.

[@joonkwon9303]

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.

[@jursamaj]

How much time you have for this one depends on how fast you are on the easier problems. :)

[@rosa578]

But isn’t the math section 100 minutes for 30 questions? 현재 한국 고등학생으로서 이렇게 알고 있습니다

[@billyhope809]

Guys, thats not even the hardest one in the test.

[@park_jong_in]

So freakin true. Students can solve this problem 3 years before they take KSAT

[@user-cw6hq8rj8s]

facts...

[@redteadev]

i saw this problem in math olympiad book

[@Linea_Arpolite_]

Too easy for KSAT.

[@redteadev]

SPayee yea. ksat is harder.

[@wilhelmmeyer7970]

It is an easy 2-dim task. roll the cone surface flat or draw it as a segment of a circle with the given radius from top to bottom. To do so, compute the angle wit hthe given data.Then mark points A and B. draw a straight line connecting them. Also draw a line 90 degrees from the line a to top ( top == circle center). the point C is the point, where it meets the line between point A and B. The distance between C and B is the solution, if I have not mixed something up.

[@santiagoarosam430]

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

[@doodelay]

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

[@fanaticalplel1003]

Me: sees this is what imma have to do for high school. Also me: aight imma head out

[@joeywild2011]

High school questions won’t get more difficult than this. This is an example of a high school question at the absolute highest level

[@fabiolindner]

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.

[@user-kj3op5ty2l]

true

[@brimstoneheo]

당시에는 삼각형의 높이 h를 60, 50과 싸인 120도를 이용한 삼각함수로 구하고 피타고라스 정리로 내려가는 길의 길이를 구했었습니다.

[@GermansEagle]

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.

[@gagadaddy8713]

Agree! try to solve some interesting Math Olympic question here!

[@GermansEagle]

kzbin.info/www/bejne/hZzQf4uvbMqlbpY

[@GermansEagle]

the question is literally: "what is the probability that this tetrahedron contains the spheres center?"...

[@aviralpatel2443]

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

[@rohangeorge712]

this one didnt need calculus

[@tjkim8171]

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.

[@MovvaTapaswipeace]

Is reply 1996 relatable for you?

[@chinmayh2745]

I was the one who made this question.

[@prateekpanwar646]

LOL

[@jjraga]

@@chinmayh2745 I was the one who made u

[@adityakuswaha2813]

@@jjraga 😂 😂 badass

[@coffeecup1196]

Comment before watching the video: I believe the answer is #4, 400 / sqrt(91). This took me 15-20 minutes because of having to refresh my memory of the law of sines and cosines (its been a while folks) and multiple calculation errors from doing it by hand. Edit after watching solution: I got the unwrapping part of the visualization pretty quick. My main problem was multiple instances and miscalculation at the law of cosines, getting to the end and realizing my answer was not an answer choice. Also, I used the law of sines to find the sine of the angle at point B, used a trig identity to turn sine into cosine, and then 50 cos(B) = the answer.

[@kennethhowell5291]

Great video! Thank you! Explained perfectly!

[@lostphantom1]

근데 진짜 너무 당연스럽게 전개도 펼쳐서 저 그림 그림 시작할때 4번의 잘못된 풀이를 했다는거가 저렇게 많은 풀이시도가 있을 수 있구나는 생각에 젤 놀라웠음

[@Cannongabang]

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!

[@tomlongland5399]

don't you find lagrange is more applicable to parabolic questions than straight line problems?

[@Jack_Callcott_AU]

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!

[@narasimharaobejawada2776]

How sir

[@Jack_Callcott_AU]

@@narasimharaobejawada2776 Thanks for answering. This was a few months ago. I will have to think about it again and get back to you.

[@beaclaster]

@@Jack_Callcott_AU have you thought about it again?

[@Jack_Callcott_AU]

@@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.

[@lilgooseboi7354]

@@Jack_Callcott_AU did you try again?

[@Reignspike]

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.

[@jameszenos4045]

Your soft voice is soothing And your explanation ( like first you introduced us to the possible mistakes i liked it :)) Thank You 😊

[@SelftaughtAnimator]

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..

[@PeterGeras]

When he started drawing the lines from the vertex to the radius... I also facepalmed.

[@anchalpandey9074]

Peter Geras 😂😂😂

[@lixxap7471]

Hey man, thank you for this question! I had a great time watching this!!!

[@user-xz8xx8sm3m]

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...

[@anushkasharma38]

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 ...

[@Motavi]

Love your videos, because I get better in English and maths at the same time ^^

[@Mayank-mf7xr]

"what method did you use ?" .....i couldnt do it man xd

[@ninepoints5932]

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.

[@erazorheader]

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.

[@idiot9359]

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

[@u.v.s.5583]

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.

[@user-skdewnvxk16]

- 외국댓글 중 - 한국인이 아닌 사람들: 마침내 내가 이 어려운 난제를 해결했어!! 한국인: 좋아 이제 앞으로 29문제가 남았군

[@soobark2]

ㅋㅋㅋㅋㅋㅋ 저문제 중3때 풀어본거 같은데

[@dana_0627]

@@soobark2 그러게 이거 중3 거 아니냐? ㅋㅋㅋㅋㅋㅋㅋㅋ

[@user-yp5tk5gn5c]

그 만큼 헬조선..

[@MoongTaeng2]

@@soobark2 블랙라벨에서 나온거같은데

[@liberationhomefront]

솔까 수능 이거 못 풀면... 3등급 이하 확정인듯. 곧 마주칠 수능은 범위 달라져서 어떻게 나올지 감은 안오지만 일단 내 시절이라면. 방정식도 그리 복잡하지 않고

[@devi1sdoz3n]

The easier way to get the circular arc angle is to get the circumference of the full circle (2x60xπ) and divide that by the circumference of the base (2x20xπ) - which gives you 120π/40π=3, which means the arc is one third of the circle, and consequently, the angle is 360/3=120. No need for the extra formula (which I have never seen before).

[@Kasei87]

I am so rusty at math! My first instinct was to draw the cone as a flat arc, but I then made so many little mistakes that I eventually started over. In my failed run, I did manage to find the length of the track but didn't know what to do with it. In my successful attempt, I used a coordinate system where the origin was the cone's vertex and A was the point (0,-60) to turn the track into a line segment and use the equation to find the shortest distance to a line from a point to get the length that completes the right triangle with the 50 and the length of the downhill segment, then used Pythagoras. I felt so silly when you did Pythagoras from the 10*sqrt(91) by making both line segments variables. In my defense, I did that as well but never thought to change my y (length of the uphill track) to 10*sqrt(91)-x. I was sooo close to doing it in the shortest way possible, too!

[@mathiaswehr6444]

Yes i solved i just thought where come these Numbers from. Then i saw the 20 as Radius at the Base and knew 4 couse 400 is 20^2

[@user-sp5vc7ws8h]

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

[@user-kb1bp6hh3u]

가연이 남친있노

[@zanti4132]

Let r = radius of the cone's base and s = slant height. In the video's problem, r/s = 1/3. Suppose we change the problem so that r/s = 1/2. Then the cone unwraps into a semicircle, and the straight line distance described in the video goes along the straight edge of the semicircle. So, the shortest distance wouldn't involve any circling at all - the path would be to go straight up the cone, passing the finishing point, to the top of the cone, then coming back down to the finishing point. Very strange and quite counter-intuitive, but apparently this is correct.

[@gabornagy4692]

Yes, I saw imediatly that one should unwrap the cone and trace a straight line but I failed to see where to lay point B and the AB length. The trick for me was the central angle measure. The rest seemed to be "trivial".

[@MR0MYSTERY87]

Watching the solution was like watching a neat magic trick 😊

[@Trebukeet]

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.

[@Andleoric]

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.

[@Macieks300]

@Andrés Rico That's not really rigorous. And where is the proof?

[@rmsgrey]

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...)

[@Trebukeet]

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.

[@rmsgrey]

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)

[@lakai0307]

It's pretty cool I could only figure this one out after some good thinking because I sometimes write plugins/scripts for a 3d program I use. One way of writing a shortest path tool between vertices for polygon geometry is by looking at the uvs, basically an unwrapped 2d version of the 3d model and connecting the vertices with a straight line(for many reasons it's not a good way of doing it though). The second part reminded me of the closest points on two skew lines/curves in 3d space. The shortest line that connects these points basically shows two right angles to both skew lines. It's funny how this math really applies to some real world useful things. Eventhough I would consider myself very bad at math, not really knowing any formulas or many of the rules to solve equations, I use vector math, trigonometry etc on a daily basis to solving these little puzzles to writing my scripts and it's pretty fun because the solutions can many times be solved by thinking visually and being creative.

[@markp7262]

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.

[@pavelkotsev1542]

Nice! Yes, a bit more convoluted, but more systematic, straightforward and relentlessly working solution :)

[@panulli4]

Admit it, Presh: By „slightly editing the email“ you mean adding in a sentence about how they like to watch your videos.

[@m.g.6081]

panulli4 He most probably meant editing the Korean dude's English, since Korea is the Asian country that is least able use it.

[@GortigeGort]

Mitko Gospodinov dude the joke flew right over your head holy fook

[@noemigonczol7454]

@@m.g.6081 whoosh

[@discovaria9507]

probably fixing grammar and spelling?

[@yasinsaad6068]

i guess he made the problem easier by editing it

[@user-cw5vv3yz5h]

불면증있는데 마침 딱 좋은 영상 떴네 역시 유튜브 알고리즘

[@bgggeometry6082]

At the end you could’ve used geometric mean because the perpendicular line made 3 right triangles. And by that law, they are all similar so all you would need to do is set up two proportions.

[@mokadoggo5177]

considering sqrt(91) is close to 9~10, you can guess this by drawing too, takes about several seconds

[@chinareds54]

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.

[@dhk1126]

chinareds54 Actually, there are 5 choices in the test.

[@유형준1116]

Dong-hyeon Kim 그런데 인터넷으로 찾아보니 같은 선지가 두개있더라구요, 오타였던건지...

[@dhk1126]

劉형준 엥 저도 확인해봤는데 그렇네요..

[@skatastic57]

Hell I'd guess between 3&4 just because they have the same denominator.

[@heyhey97777]

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.

[@ekxo1126]

I don't know why but I understood everything in seconds. That's the first time but it is so satisfying and good

[@kimarnina]

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*

[@amritenduhait6239]

The proof you showed is possibly the most natural one. This problem is very nice for motivating students towards Olympiad geometry.

[@mikefochtman7164]

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.

[@martinschopflocher9908]

An interesting special case of this nice problem is when both points A and B are at the same height (at the bottom of the cone, for example). It’s wrong to think that the track for the sightseeing train has to go through the base of the cone. The track goes uphill from A and in the middle of the track goes downhill to reach B (coincident with A in this case) in order to achieve the shortest distance on his turn around the hill. Do you figure it out...(The answer then is approximately 52).

[@jandemeyer105]

That's what could be expected since higher the radius is smaller so the circumference is smaller but you have to go upside as well, that's an addition. So going as high as possible for as small a radius as possible competes with staying as low as possible to avoid having to make the distance up. That's why there is an optimal (shortest) path.

[@anshul_saxena]

30 * sqrt(3)

[@jagmarz]

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.

[@justarandomcatwithmoustache]

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

[@user-zk7ju1ft6c]

We should solve that at 3~4 minutes😭

[@redwren4182]

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.

[@sadraderhami2628]

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.🌹❤❤❤🌹.

[@johnspathonis1078]

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.

[@IDyn4m1CI]

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

[@johnspathonis1078]

@@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

[@IDyn4m1CI]

@@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.

[@johnspathonis1078]

@@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

[@iganpparamarta8813]

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.

[@Rr_9991]

I am also a Korean student, and that type of question isn't really hard anymore since it has been over 20 years

[@redteadev]

YH C true lol

[@user-yi1bq7nf3p]

RIGHT... It's not actually really hard question.

[@yuno7825]

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

[@capjus]

Lol

[@achaemenid]

@@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.

[@supersonicgamerguru]

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.

[@aidenkoenig5379]

Loved the problem, I got to the part about going up hill, but briliant solution for going down hill.