Math Olympiad Question - Can You Spot the 1 Line Solution?
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- čas přidán 9. 01. 2024
- Another beautiful Math Olympiad Question! Did you manage to solve it?
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No matter how you rotate the squares about their common center, there will always be a circle of diameter 1 that will fit inside of them. This circle will have an area of pi*(.5)^2, which is approximately equal to .7854, which is greater than .75.
love this - you could even do it as approximate pi as 3 find the area to be .75 exactly then since we rounded pi down this shows area is greater than .75 if we wanted an even easier computation
This looks like a really easy problem as per IMO standards! Ellie you should try and solve IMO 2023 paper. Only 6 students out of 618 participants were able to solve the 6th question of the paper completely! Give it a try! Btw love your videos from India 🇮🇳 ❤
Oooo I'll give it a go!!🚀
@@EllieSleightholmtry collaborating with The Math Sorcerer. Also help him with his style and hair. Maybe help him find a good gym trainer.
Very interesting explanation
Loved this series.. ❤
thank you for such a lovely comment! :)
Thanks PROF👍
Very relaxing.
Could you make a video on the UK space program? I think it would be amazing with your touch 🚀
To those familiar with geometry, this is a standard area enclosing problem. It’s actually how one can prove the lower bound for π i.e. 3 < π.
Excellent video!
Nice Explanation .
thank you!
That blew my mind! I watched the whole video although I was tempted to skip to the end - just like how I read my Harry Potter books!
In retrospect, the area of intersection is bound below due to symmetry and the inscribed circle happens to be an object in the set of lower bounds whose area is obviously less than the
infimum" and greater than 3/4. What would happen if we were asked to prove that the area of intersection > alpha, where --> area_of_inscribed_circle < alpha < infimum.... does an elegant on-line solution exist in that case? (alpha is an arbitrary positive real number {appropriately chosen})
There's a problem with the elegant solution right?..itndoesnt always work..youndont know if part of thr circle.will be outside the rotsting square as the square is rotated inside the other square..even though the four points where the circumference touches the non-rotating square will always be inside you don't know formsure if part of.thr circle.might "jut out" of the lines of the rotating square...see what I mean? That's true despite the symmetry of the square.
Correct me if there's a way to show otherwise
Thanks for your solutions.
How can we show that the min of z appears at 45 degree rotation situation by geometry method ?
I came up with the pi/4 solution within a couple of minutes. I'm not sure if (how quickly) I would have gotten it without being told that a one line solution exists. I think I would have calculated the area as a function of the angle of rotation.
1-4[(√2-1)/2]^2=1-(√2-1)^2=2√2-2=2(√2-1)=0.8284...>0.75
Here is the one-line solution where minimum overlap is at 45° rotation. In the above equation:
1 is the full area of the unrotated square.
[(√2-1)/2]^2 is the area of one of the four chopped-off corners of this square.
4[(√2-1)/2]^2 is the total area of the four chopped-off corners of this square.
love to see it 😎
@EllieSleightholm so he means he just used just then squares and triangle properties not asl circle right? That was my one line solution..isn't that equally valid as the circle
@@EllieSleightholmWAIT ELlie isnt there an issue with the legant solution..youndont actually know if the entire.circle will be within the intersection area..part of thr circle might "stick out" of the inner square and be in only the outer square..see what I mean? So this doesn't quite always work. I don't see someone always proving the circle is within both squares
@EllieSleightholm yea I'm pretty.sure I found a flaw.there in the elegant solution...hope.you can acknowledge and respond when you can.
Excellent work, Ellie, but please correct me if I am wrong. You have A = 2z, which seems to be dimensionally inconsistent. How can an area equal a constant multiplied by a length?
It's because there's a hidden 1 there, which is also a length: the side length of the squares. So it's more like A = 2(1·z), where 1 and z are both lengths.
Notice that in equation (1), x + y + z = 1, the 1 is the side length of the squares, and that gets carried on all the way to the end.
In particular, when in (2) you substitute z = 1 - x - y and take its square, you get some area z·z = 1·1 + x·x + y·y - 2(1·x) - 2(1·y) + 2(x·y). Notice how I explicitly put the 1 in the corresponding terms, which represents that side length 1.
@@nomechomath5644 Ok, yes I now see how Ellie's solution is dimensionally consistent. I had imagined the answer to my query would be quite straight-forward, and it was. I am in fact surprised I failed to see the 1 was a length! Thanks for that! :-)
I like watching your CZcams channel and "Looking Glass Universe" CZcams channel for motivation. "Looking Glass Universe" CZcamsr is also from Cambridge. Her story is inspirational because she was very bad in Mathematics and Physics until high school but still managed to get a PhD in Quantum Computing from Cambridge through hard work.
First comment :D
Haven't watched yet, but lemme try:
Because one square is a rotation of the other, the intersection contains both squares' circumcircle of area π/4, so its area > π/4 > 3/4.
EDIT: "incircle", not "circumcircle", as Farhan Awais pointed out
And that’s quite a bit of slack!
You mean incircle, but nice!
nice 😎
@fahrenheit2101 yea but as you rotate part of the incircle cpuld be within one of the squares bit outside the other..see what I mean
.like part of thecircle for example, could be on the bottom right corner inside the fixed square but outside the rotating square as it rotates.. even though they are congruent squares..see what I mean? I can picture it in my head.
@@leif1075 i disagree. Rotating the square keeps it the same size and keeps the centre of the square in the same place, so its incircle is surely invariant too, no?
Can you please teach me am really bad 😢😢😢😢 in math😮😮😮.
You're class study
thank you!!