Math Olympiad Question - Can You Spot the Trick?
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- Опубликовано: 30 дек 2023
- Diving into some of the most beautiful Mathematic Olympiad Problems ever set! In today's video, we look at a problem from INMO 1993 which uses geometry!
Apologies for not showing my face in this video! I came down with the flu and looked rather ill!
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For those of you that are new here, hi there 🌞 my name is Ellie and I'm a Cambridge Part III Mathematics Graduate and current Astrodynamics Software Engineer! This channel is where I nerd out about maths, physics, space and coding so if that sounds like something you're interested in, click the subscribe button to follow along ☺️
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Geometrical proves are so complex yet beautiful.
Really enjoyed the whole explanation! Keep it up!
Thank you so much!☺️
Can you please try to make a video on William Lowell Putnam Maths competition. Some of those questions are really interesting!
Yes definitely!!
@EllieSleightholm Can you do some AIME questions also..and hope you cam respond to my other question when you can. Thanks very much.
Very beautiful solution! I first thought of involving complex numbers but this is much more elegant and easy to explain.
Happy new year Ellie🎉😊❤
It’s really wow how I end up saying wow after every video of you
That’s not commonly happening with mathematics content
Great work 👏🏻✨✨✨
i don’t like math but i feel the joy ot gives when someone solves questions… your video give me some little motivation ❤
This is one of the most beautiful Mathematical proves Ive yet to see, good job to the ppl who got it in the IMO.
Please keep more problems like THIS comming up👍
You got it!
@@EllieSleightholmWhy did tou gloss over the important step.of HOW you deduce each of the angles is 120 since if you don't know what you said by memory you didn't show how to deduce the angle size.
This is like such a cute video! I was waiting on my Neural Network to finish training and watched this refreshing video! 😇
One word: Beautiful!
Ellie, happy new year to you! I enjoyed your explanation, And am wondering where else can, I follow you, you re likely active than this platform?? And thanks
Such an elegant proof.
hello! happy new year!! all the best.. greetings from Brazil
I am UR biggest fan🤗
a=1, b=3, c=5, and x=10. I did it a more random way and got a different answer. The triangle method is very elegant, and reassured me my result was valid.
I was wondering if you would react to an exam from Math 55 at Harvard. It's supposed to be one of the most difficult undergraduate math classes in the USA.
Hey you got new sub today
By the way, nice explanation!
Did it myself by same method , really proud of myself love from india
Another solution at 9:22 is {a=1, b=3, c=5, x=10}. As far as I can tell this is the only other solution (up to permutations of a,b,c)
Hallo Ellie mam! Describe your journey to Cambridge
I hope ❤
I used a different solution. Considering six sides as a to f. Since all angles are equal so 120 degrees, ad, be and cf must be parallel, which means a+b=d+e, b+c=e+f and c+d=a+f, combine this and the 123456 and the question is solved.
This could even have been done considering every side as a complex number with exponent angle being multiple of 60 and adding all of them to get zero.
Hey elli 😊
Hey Ellie! If you want to, then take a look at ISI UGA and UGB Exam, it's for highschool students in India and has better question in the UGB section compared to JEE Advanced, It'd be interesting to hear your thoughts on it!
Hey Ellie mam! ❤🎉🎉🎉
Do you use notability for note taking?
Ah you seem to have done otherwise - I'll be sure to check out the method soon
The picture threw me off for a second. I was like," Hey why does b look greater than x-a-b!?"
ellie good evenung ....as a maths lover i also want to be a mathematician like you ...and be cambridge graduate ..can you plzz make a video how you enterd cambridge means which exam we have to clear to be a student at cambridge, and what are the fees structure thereplzzzzzzzzzzzzzz didi
What do u use to write/type?
Preparing for jee but i do not have much intrest in Maths .
Could you get slightly sneaky with combinatorics and counting here instead?
Say, tessellate the plane with equilateral triangles and pick 6 around a vertex. Then, since the pattern is infinitely repeating, you just need to get crafty with permuting {1...6} for your edge labelling throughout the tessellation to match the question spec. Presto! A slightly troll answer which probably would get no points, but inside you'd burn with a passion of a thousand suns!🌞🕶😎
I did the same. This solution should get maximum points if explained properly.
I have some questions, can you please tell me, there are some question which i want to ask can you tell me how can i send it to you.
sure, what question? :)
Can you plz solve jee advanced paper 2023?...
Can this hexagon be constructed from equilateral triangles with side 1?
Yes. You can even (rather easily) find the hexagon by tiling the plane using such triangles.
oh! you should to see the problems from the exam called ''ITA and IME''. They are military exams from Brazil and really hard! But.. its funny. Physics from ITA... wow.. so hard.
Hello, Ellie in mathematics there are some topic which doesn't feel. Please tell me how can I take the feel of that topic 😊
Hey there, I'm not quite sure what you mean, could you elaborate please? ☺️
@@EllieSleightholm There are some topics which are not understood easily. How can I understand that one very easily
@@sourabhsoni2930 i think finding out what type of learner you are helps massively. If you're a visual learner then try find youtube videos explaining certain concepts. if you prefer reading, then try find books on similar topics. For me, if i was learning a difficult topic I would break it down into the prerequisite topics and make sure I truly understood them before moving on! It's all about building a base foundation and building from there!
@@EllieSleightholm thank you so much Ellie 😊
@@EllieSleightholmwhat if you don't think k of the trick..can't youndolve.snotjer wah? I thought of making a swuare or recession gle around the hexagon..surely this would work also just mauve be a bit more complex? Mauve breaking the recession gle then I to triangles mauve..not sure if inwpjldve..hope you can dare feedback..
Nice, but I believe that Olympiad problems have changed over time, since this one is kind of an old one. To show what olympiad questions look nowadays it would be better to take a problem from past IMO(International Math Olympiad) questions.
Is the word “convex” needed? I’d have thought “all angles equal” implied “all angles 120°” and hence convex.
That's true because the sum of all interior angles must be 720° in both cases, convex and concave. So every angle must be 120°.
Hay Ellie mam ❤🎉🎉
❤👍
Hey Ellie today it's my birthday
Happy birthday!!
Please tell me whether I should take a laptop or a tablet for studies.
Only use paper and pen if you really want to learn mathematics!
Please mam or didi teach me olympiad AMC or ioqm i am from India. Top or hardest questions topics on geometry, algorithm,etc. I read this time and like blackpenredpen channel one short videos topic wise or etc.reasons
And board in front read so face emotions read because my English language some weakness😊😊😊
And Didi or mam tibees from read no reasons
I actually worked it out. Isometric grid, then you can pick the obvious coordinate basis, from there it works out fairly easily.
Wlog, start with 6 going right, then find some loop that takes you back where you started - "left" 6, and "up" 0. Each side travels along a nice line in the grid so that the coordinates change by integer amounts so you can get a couple equations if you want.
Hey your really a great person I admire the way you go about the maths word I have my personal question I will like to ask you if you don't mind
Why is x equal to 9?
Oh, I have already understand it, nice prove!
Клетки, клетки , клетки , как в метрополитене вагонетки
why x=9?
x must equal 9 for the sides of the hexagon to match the values in the question. i.e. each side of the large equilateral triangle will have sides:
1 + 6 + 2 = 9,
1 + 5 + 3 = 9,
2 + 4 + 3 = 9.
Hope that helps!
@@EllieSleightholm thank you
As someone pointed out in another comment, {a=1, b=3, c=5, x=10} is also possible.
@@EllieSleightholm 9 is the smallest possible value for x. The side with length 6 must be in one of the sets and adding the shortest side lengths of 1 and 2 totals 9. It is easy then to see that {1,6,2,4,3,5} is the only valid permutation for x=9 excluding rotations and reflection. This permutation when rotated is also the solution for x=12 which is the largest possible value for x.
The only side length combinations that would work for x=10 are 1+3+6,1+4+5 and 2+3+5. Only 1, 3 and 5 are repeated so these must be the vertices and thus {1,6,3,2,5,4} is the only solution for x=10 and x=11.
There can be no other solutions.
Another solution: take the first side to be the interval [0,1] in the complex plane, and then we require that
1+ Σa_k e^{ikπ/3}=0 where a_k are the numbers 2,3,4,5,6 in some order and k goes from 1 to 5. Then taking real and imaginary parts we get (2+a1-a2-2a3-a4+a5)=0 and (a1+a2-a4-a5)=0. Then subtracting these gives 1+a5=a2+a3. Adding the equations gives 1+a1=a3+a4. Set a3=2, then a5=a2+1 and a1=a4+1, so we can take a2=5,a5=6,a4=3,a1=4. So the sides ordered counterclockwise have lengths 1,4,5,2,3,6
Also nice video!
OMG you are smart & cute ^.^
The phrase "in some order" is redundant - the sides could not be in no order.
No, the phrase means that any order is acceptable.
No. There is no suggestion in 1,2,3,4,5,6 that a given side has to be adjacent to a given other side.@@Tommy_007
Learn my student time 😂,my mother teaches me