A fun probability puzzle with a neat geometric solution.

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Any further questions:
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Programs used:
► Cinema 4D
► Processing
Music by
► Lee: / @leemindme

[@MuPrimeMath]

Your geometric proof animations are so colorful and clean at the same time! Everything flows so nicely. I love it!

[@ThinkTwiceLtu]

Thank you:)

[@ZEEGA]

i want to see dirty and inaccurate animations with death grips music

[@johandaun874]

I didnt think twice when I saw your video getting notified

[@jamesmnguyen]

I didn't try to prove the solution but somehow my brain guessed the solution space of this problem is probably a semicircle.

[@MrSumkinFedor]

same here.

[@bonifacioteodorico7919]

I put my phone in a scuare and i prove it jajaja

[@Emeskey412]

James Nguyen Same

[@tofidu]

Please explain! Why it still counts when it's touching the line? It would be 90°, right? Bu, applying the formula, we'll get : pi/8 - pi/2 which is -3pi/8...

[@Fitzgibbon299]

@@tofidu The probability of the point actually being on the line isn't pi/2, but actually infinitesimal. Since we are looking at the probability in terms of area instead of in terms of the line, we are looking at all points within and arbitrarily close to the boundary of the semicircle, i.e. the area of the semicircle as a fraction of the area of the square.

[@SnydeX9]

I was able to intuit the solution and am feeling pretty smart right now, even though I probably shouldn't.

[@brandoncalvert8379]

why would you doubt yourself like that after admitting you did well? congrats on seeing the solution :) that is pretty smart

[@TiagoSantos-kj6tc]

Nice. What was the rationale? Would be nice to learn your process. :)

[@jakubstepo4309]

@@TiagoSantos-kj6tc I was able to solve this, too. I cannot speak for Snyde, but my reasoning was as follows: It is clear that both obtuse and acute angles are reachable. Now the input space of all P is continuous. Hence, if we consider the set of all P for which the angle is right, it divides the square in two areas; for all P in one of them, the angle will be acute, and obtuse in the other one. Per Thales’s theorem, such set is a circle with diameter AB, and the rest immediately follows.

[@PavanKumar-xv1hg]

Don't feel smart Feel confident.

[@amir3515]

High quality content, visually and audibly pleasing. Thank you!

[@ThinkTwiceLtu]

Thank you:)

[@jkid1134]

I paused when instructed and came up with pi/8. This was a surprisingly straightforward application of Thales theorem for a KZbin math puzzle :)

[@ollolol3303]

Here comes a challenging question: How about the possibility of α > x° for any given x?

[@kidze31]

is it an ellipse?

[@badrunnaimal-faraby309]

It's still a circular arc with a side of the square being the circle's chord that corresponds to the arc measure 2a for a90° (minor arc).

[@logicbreaker]

If a random internal point be taken inside a square, and it subtends an angle on one of its sides, what is the probability that the angle is greater than x? Let the probability be P. Let all angles be expressed in π radians. Let cotα = 2, and, π/2 > α > 0. Let cosθ = 2sinx - cosx. If x≥π, P = 0% If π≥x≥π/2, P = 25{(π-x)cosec^2(π-x) - cot(π-x)} % If π/2≥x≥2α, P = 25{xcosec^2x + 3cotx} % If 2α≥x≥π/4, P = 25{3cotx + (2 - cotx)(4cotx - 3)^(1/2) + (x-θ)cosec^(2)x} % If x≤π/4, P = 100 % I hope it’s correct. Please feel free to point out any mistakes.

[@yeast4529]

I could watch these videos forever. Combining interesting mathematics with great visuals and audio is about as good as it gets

[@ferrucciocastellini9266]

I have a harder question: what is the average angle that you get

[@mamamia12369]

45 right?

[@tofidu]

Gooood

[@ferrucciocastellini9266]

@@mamamia12369 not at all!

[@bamberghh1691]

90 degrees?

[@ruroruro]

It's 90 degrees. The average (expected) value is the integral of the probability density times the value at every point (see en.wikipedia.org/wiki/Expected_value#Absolutely_continuous_case ). The distribution of points in the square is uniform, so we need to find ∫∫ angle(x, y) * 1 dx dy from 0 to 1 for both variables (1 is the probability density). You can find the angle for the arbitrary point (x,y) using the law of cosines.(see en.wikipedia.org/wiki/Law_of_cosines ) c^2 = a^2 + b^2 - 2*a*b*cos(phi) therefore cos(phi) = (a^2 + b^2 - c^2)/(2*a*b) or, given c = 1, a = sqrt(x^2 + y^2), b = sqrt((1-x)^2 + y^2) cos(phi) = (x^2 + y^2 - x) / sqrt((x^2 + y^2)((1-x)^2 + y^2)) Take the arccos of the above and multiply by 180/pi and plug it into wolfram. www.wolframalpha.com/input/?i=Integrate%5BArcCos%5B%28x%5E2+%2B+y%5E2+-+x%29%2FSqrt%5B%28x%5E2+%2B+y%5E2%29*%28%281-x%29%5E2+%2B+y%5E2%29%5D%5D+*+180+%2F+Pi%2C+%7Bx%2C+0%2C+1%7D%2C+%7By%2C+0%2C+1%7D%5D

[@karthikthiagarajan2331]

I like your method of not having any voice overs. It is just the visuals and the math. Good job!

[@ThinkTwiceLtu]

Thank you 😊

[@cloudwolf3972]

Dude, I literally mentioned and recommended this channel to a friend today and we get a new video? Nice timing.

[@Aleksandar0100]

Not only are you awesome at animating but you give your channel a dope vibe with the music. Who said math couldn't be relaxing?

[@cardiomatica1091]

I don't know how someone can do such amazing videos. Please keep the great work. ThinkTwice, 3Blue1Brown and others math channels are not just incredible in content but also ethereal in design and animation.

[@imelse]

The longest video on your channel! And it's so much easier to understand everything and feel the beauty of math this way! Absolutely amazing. Mldc

[@ThinkTwiceLtu]

Thanks:)

[@aaronr.9644]

Pretty awesome but at 5:20 I would suggest that the purple region represents alpha

[@badrunnaimal-faraby309]

It actually does work because a line has a width and area of 0. For continuous distributions the probability of hitting any exact value is 0, so P(X>a) and P(X≥a) are the same.

[@SaNjA2659]

me: it's gotta be about 39% Think Twice, an intellectual: π / 8 me: *surprised pikachu face*

[@hamiltonianpathondodecahed5236]

its actually about π/8 = 39.27 %

[@hamiltonianpathondodecahed5236]

@@akshaj7011 me destroying the joke

[@mayabartolabac]

By my defense, they have no sense of good humor since they are uneducated of such and are raised with ifunny memes, so don't charge them with death =)

[@mathwizard8927]

Maths + Beautiful animation + Calm music = Think twice

[@vukasinkrsmanovic4568]

Video quality is on the whole another level and also explanations. I really love and apreciate the fact that you first prove theorems in explanings and than use it. Keep it up, you are amazing!

[@ThinkTwiceLtu]

Thank you:)

[@YellowBunny]

It took me about 10 seconds to find the solution, but seeing it all animated was still cool.

[@nataliarodriguez3740]

Same uwu

[@zakimoshi8686]

same

[@YellowBunny]

I didn't even remember this at all but after looking at the puzzle again I immediately thought of Thale's Theorem and the semicircle area which is also the solution shown here. So, I most likely solved it the same way.

[@alexandersanchez9138]

Thanks for the reminder to pause and figure out the problem beforehand. I was so mesmerized that I would have just watched without the thought even occurring to me!

[@amyshaw893]

Nailed it. I laready knew about thales' theorem, and knew from somewhere that pi/4 is the probability of being in a circle in its smallest square

[@reynandr.w.279]

Chill beats to solve mathematical problems to

[@alejrandom6592]

Was looking at this problem for like 10 min yesterday and had no clue. Today when I got up I Inmediately started to think about where the critical points are, that form 90⁰ angles. I thought about the semicircle and I was sure that every point inside would be greater than 90⁰, but didn't know how to prove that every point outside is less than 90⁰. Nice video!

[@gosselinkfinecarpentry9786]

You make it so that a carpenter like me can understand and so that I can use it in my field. I used to do a lot of circular and elliptical work.

[@Green_Eclipse]

While it is easy to guess the answer I made an algebraic proof. Make square with side length 1 for simplicity with bottom left corner at (0,0). Pick a point in the square (a,b). Draw vector from bottom left corner to (a,b). This is vector [a,b]. Draw a vector from bottom right corner to (a,b). This is vector [a-1,b]. Using dot product we get this equation: a^2-a+b^2=|v1|•|v2|•cos(angle) Note: cos(90°)=0 cos(more than 90°) < 0 Therefore a^2-a+b^2

[@98perova]

I'been suscribed to your channel for a while, and I gotta say i really love your videos. They're really well made and beatifully animated, plus i love that your proofs focus on de visual and geomtric side of theorems. Keep up the good work! Love from Argentina.

[@ThinkTwiceLtu]

Thanks for the support:)

[@saurabhsharma7123]

Cute explanation, cute background tune!

[@hugopettersson3213]

I thought, where is the angle 90° and knew that all space below it would have a bigger angle

[@kikivoorburg]

I finally managed to guess at the answer! It felt super satisfying to realise that it would be the area of a semicircle. Keep up the great videos!

[@omerresnikoff3565]

I love the animation, but I'm pretty sure Thales's theorem stated that two triangles were similar iff their corresponding angles were congruent

[@AvelinoTiago]

That's only one Thales theorem. Try to study harder. Next it, u can see these videos...

[@nandish18]

Wow we need more such videos based on geometry!! BTW loved ur vid ❤️❤️

[@Nostra.Damus14]

Should we subtract all the points in the semicircle cause that will be 90 ? P = (π/8) - (points in semicircle)

[@Slavaver]

This question bothers me to.

[@IQuick143cz]

Well yes and no, without going into any advanced mathematics, the probability of a point being exactly 90° ie. on the semicircle is 0. Think about it like having a darts board (the square) and you wanna hit a single line (the semicircle) perfectly with a single point dart but the line is infinitely thin, meaning it's physically impossible to hit it. It's not entirely impossible, because certain points still lie on the line, but there are infinitely more points outside for you to hit. Another way to think about it is in terms of an area. The inner semicircle has some area and the outline is just a line, and lines don't have any area hence they have no probability. So, while your idea of subtracting the points on the semicircle is correct, there's no need to as the probability you're subtracting is 0. Relevant (and much much more precise than this comment) studying material is Measure Theory. en.wikipedia.org/wiki/Measure_(mathematics)

[@IQuick143cz]

@@That_One_Guy... There's a problem with units. A probability is just an unit-less number but your length/area would have a unit of 1/m which doesn't really make much sense.

[@That_One_Guy...]

@@IQuick143cz nevermind im just being stupid today, the probability value would be more than 1 if it's like that

[@That_One_Guy...]

@@IQuick143cz well thinking again, it seems plausible that the chance would be near zero, i was thinking that because all points on circle causing the angle to be 90°, intuitively the chance would be the arc length of semi circle

[@mystmuffin3600]

this was in my watch later for 2 years! Glad I finally got to watching it :)

[@agugyin]

I love the smooth music and animations! :) It is such a pleasant way to learn math!

[@davidb2885]

Well, it took me one second, so that felt pretty good, but experience is an advantage I guess. And it's great that the video brings that fun to really everybody, cause that's what we all want to reach, right.

[@damianmatma708]

Additional question: What is the probability that |

[@Haguatchi]

Here's a different but related question: I pick a number between 0 and 1 uniformly at random. What's the probability I pick the number 0.5? Certainly it must be possible to pick 0.5; it's between 0 and 1, but there are an infinite number of other options I can pick. I can pick 1/sqrt(2), or 0.355602, or 1/3, or any out of a literally uncountable number of options, so the probability I pick 0.5 must equal 0. While it's possible I choose 0.5, this happens with probability 0. Thats the same thing that happens here with the arc, and more generally what happens with what are called continuous distributions; there are events that are possible, but they occur "almost never", which means they occur with probability 0. While it's possible to pick a point such that

[@tofidu]

Well here s my answer:its semi perimetr of circle/2 which is 2*pi*1/4 which is pi/2 okay now i need help too

[@tofidu]

ITS MORE THAN A A>90 PROBABILITY THAT'S IMPOSSIBLE

[@victoriam6569]

I was searching for this question. And the answer makes me confused even more... It seems wrong to equal "almost never" and "never"... Isn't there an infinitly small number, which is greater than 0 anyway?

[@victoriam6569]

Sorry, I've learnt that this is a paradox of probability theory -- that some possible event can still have a probability of zero... It still sounds weird though :)))))

[@shaileshkaranjkar1906]

What an amazing problem!!!!

[@tizweber8871]

A visual proof of the law of the Parallelogram law would be awesome !

[@soutrikdas8403]

How do you even do these sick animation like it gets me calm and think any the problem really !

[@alxjones]

You can reach the same conclusion as Thales' theorem using analytic geometry. Suppose that P = (x,y) and APB is a right angle. Then, the length of the hypotenuse of the right triangle APB is 1 by definition. The left leg of the triangle has squared length x^2 + y^2, while the right leg has squared length (1-x)^2 + y^2, and so Pythagorean theorem gives [x^2 + y^2] + [(1-x)^2 + y^2] = 1 -> (x - 1/2)^2 + y^2 = (1/2)^2 This, together with x and y between 0 and 1, gives a semicircle of radius 1/2 centered at (1/2, 0). The rest of the proof follows.

[@YuTe3712]

My progression of thought: 1) "Hm. First, plot obvious points where the resulting angle is obtuse (or not). Near D and C are acute. Center of square is... exactly 90 degrees (interesting). Points near AB seems to be obtuse. So, the answer is 50% ("below" the center towards AB)?" 2) "No, points near AD and BC are acute, even at their midpoints, so has to be less than 50%." 3) "What's the barrier between acute and obtuse look like (where its 90 degrees)? Lots of points near bottom seem to be obtuse, but obviously not all of them..." 4) "Aha! What does moving the point from center of square around while preserving 90 degrees draw! That should give me the barrier!" 5) _Much fiddling with fingers pointing at the square on screen to imagine the movement later..._ 6) "Oh. It's a semi-circle. It draws a semi-circle. Ohhh! So it's (area of square) - (area of circle / 2)!" 7) A moment to remember the formula for the area of a circle, and Pi/8 it is! Took about a good 5 minutes of deliberation. That was a fun puzzle, thank you for the video!

[@urvishbhavsar8910]

Exactly accurate to me as well..

[@JuanIgnacioAlmenaraOrtiz]

I discused the problem with my father and we got the solution together. He didn't know anything about probabilistics.

[@RomanSpartan3]

I was able to solve this problem using the fact that the x and y coordinates of the point P had to be such that the side lengths AP^2 + PB^2 < AB^2 = 1. This that you eventually get x^2+y^2 - x = 0. Then you can rewrite y as a function of x and discard the negative solutions (since the point has to have an x value between 0 and 1) to get y = sqrt(x-x^2). This is a nasty derivative, but integrating from 0 to 1 with respect to x gets you pi/8! I think it is very interesting that there are different ways to get to the same answer. Also, if I had plotted out the equation I integrated, I could have realized that it is precisely a half-circle and finished solving the problem just by calculating the area of a half-circle of radius 0.5. Its very interesting how everything is related!

[@prajwal6416]

We love your videos... Please upload more😍😍

[@matejvedak]

What software do you use to make animations? They are amazing

[@galen5224]

I instead used the fact that, on the boundary where the angle is right, the slope of the segment AP is the opposite inverse of the slope of the segment PB. Then, you can show that distance from any point on the boundary to the midpoint of the base is a constant (the radius of the circle).

[@LBDRP]

Most beautiful explanation ever

[@ThinkTwiceLtu]

Thanks 😊

[@shivamkasera6139]

You are an artist!

[@rcb3921]

13 poor souls couldn't stand the delightfully distressed sound of the turntable static.

[@user-kl5ni4zd8p]

It was easy, but still thanks for the food for thought!

[@math2693]

This is astonishing

[@philipyao5989]

I got the answer almost immediately but I never stopped to wonder about Thale’s theorem. Great video!

[@jesusthroughmary]

This is so elegant.

[@madhavanand756]

First thing I did in morning watching your videos 😊 Indeed Brilliant

[@wolfie6175]

It was very intuitive . My mind immediately clicked to a circle. After that it was only a few seconds to find the solution space.

[@mandaparajosue]

We've seen that P(Alpha < π/2) = π/8. For which angles x we have P(alpha < x) = x?

[@zForce4]

I was going to give up so I take a sneal peak for a hint. And I facepalmed my self. The theorem (I'm lazy to scroll back but me and my teachers call it the right triangle in a circle theorem) I immediately have a picture of a semicircle in a unit square in my head immediately.

[@Taekwonjoe]

I did everything right except I forgot the extra 1/2 (for half the area of the circle) and now I'm kicking myself. Glad I recognised that Thales's theorem would be handy.

[@rhysphillips5120]

Nice solution.

[@Mayur7Garg]

Brilliant animation! This can also be an example of how we cannot completely comprehend infinity. If you calculate the probability of alpha being 90, it would come out to be 0 for the similar scenario. Since that would be equal to area of semicircular arc by area of square and area of a curve or a line is by definition 0. Hence alpha being 90 is easily possible but with probability 0 . 😅

[@italotobia478]

P {alpha > 90} = area of semicircle of radius (1/2) divided by area of the square

[@petemagnuson7357]

My guess, if p is within a half-circle who's diameter is the side AB, it will be obtuse.

[@ersin486]

This is amazing

[@heraklesg7774]

wow, beautiful solution!

[@NoNameAtAll2]

I wonder whether Bertrand paradox would make a good video What do you think about that?

[@ThinkTwiceLtu]

Yes it's a great topic for a video and It is on my "potential videos" list but I'm still not sure if I'm going to make a video on it. Thanks for the suggestion.

[@tofidu]

Ahem. why don't you consider that it would be 90° when it touches the line?

[@mattgsm]

what's the average angle of all of the points?

[@1llum1nate]

I'm bad at geometry so i solved it using algebra only. So we want to know what line do all the points such that the corner APB is right form. If we take corner PAB to be alpha, then the equation of line PA is y(PA)=tg (alpha) * x. The line PB is perpendicular to it, so its equation is y(PB) = - (1/tg(alpha)) x + k. Given that is passes through (1,0), its y(PB) = (1/tg(alpha))(1-x). We can solve the equation y(PA)=y(PB) to find the intersection point The point of intersection is then x = cos^2 (alpha), y = cos(alpha)*sin(alpha) (after some simple trigonometry). It then follows x + sin^2(alpha) = 1, sin^2 (alpha) = 1-x, y = sqrt(sin^2(alpha)*cos^2(alpha)) = sqrt(x(1-x)) or y^2 = x - x^2. This can be transformed as x^2 - x + y^2 = 0, or (x-0.5)^2 + y^2 = 0.5^2 which is clearly a circle. And the rest follows. All that additional stuff just cause i suck at geometry :(

[@mohitkulkarni943]

Great video. Audio was very pleasing. Keep posting :)

[@ThinkTwiceLtu]

Thanks:)

[@pasunurusaivineeth3739]

1:03 How can you call that a seemingly unrelated theorem, when I literally got the solution using the same method?! 😅

[@georgehiggins1320]

I'd heard of Thales theorem before so I figured it out fairly quick. I couldn't remember the proof for Thales theorem though.

[@freeshavaacadooo1095]

That was a nice neat little puzzle. I knew exactly where you were going with it when you said obtuse cause the first thing that came to my mind was Thale's Theorem and lines with 90-degree relations. Nevertheless, a nice little geometric puzzle indeed.

[@BigDBrian]

If P lies on the semicircle that goes through A, B and the centre of the square, then thanks to Thales we know that the angle is 90 degrees. My intuition says the angle is smaller than 90 degrees if and only of P is outside this semicircle, and likewise greater than 90 if it's inside. The area of the semicircle is ½pi*r² and r=½ if we take the sides of the square to be 1 unit. So the probability equals the area which is pi/8. time to watch the video now =p

[@BigDBrian]

hot damn I actually got it right. My initial guess was a quarter, but I figured the shape of one quadrant was off and it just popped onto my head that a circle may work because it always ends up with 90 degrees

[@gnikola2013]

As always, very cool

[@ericpham6192]

It is a trick to insert zero into mathematics. Actually the probability of angle to become obtuse is improbable unless time and space change in unequal term like the clock on INE dimension is out of syn with another clock then there is great chance the angle is obtuse.

[@anirudh67]

But if we consider drawing the two lines from all the four edges won't the probability higher??

[@AntoCharles]

5:43 transition was clean af 👌

[@ThinkTwiceLtu]

:))

[@nightish_one6007]

I actually did it! I solved it! But when P is right on the edge of the circle, where a=90°, the condition of a>90° isn't satisfied, so shouldn't the %chance be less then pi/8? I also wonder how this problem would look in a non-eucludian space, like an ecliptic space.

[@Joffrerap]

Landing on the line is probability exactly 0

[@mathyland4632]

Before watching the answer, I’m guessing it’s pi/8 because it’s got to be inside the semicircle with diameter AB. If it’s on that semicircle, it’s a right triangle. So for the angle to be obtuse, it has to be inside that circle with area pi/8.

[@mathyland4632]

Yay I was right

[@trelligan42]

Nice presentation. I haven't seen Thales' Theorem in years.

[@sumitmamoria]

Nicely done. What kind of tools do you use to make such beautiful animations ?

[@ThinkTwiceLtu]

Thanks, I use cinema4d and processing.

[@youneverknow5555]

So is it 8/pi for angle acute

[@satyamburnwal391]

Marvelous!

[@anilagarwal4983]

Really nice

[@yinq5384]

0:37 It's just the area of the half circle.

[@lailaalbuquerque873]

THIS IS SO COOL

[@azzazrachid1908]

Please , can you tell us the software used to move shapes and angles . Thank you so much New subscriber. 👏

[@shaileshkaranjkar1906]

How do you find the probability that angle alpha is exactly 90 degrees, because using the above method it comes out to be zero!!! Which I don't think is possible

[@joesiu4972]

very cool and interesting stuff

[@alirezaei2572]

Thank you.very good

[@ThinkTwiceLtu]

:)

[@jlpsinde]

Just perfect and amazing. That's why I support few creators as Veritasium, Science Asylum, 3blue1brown, and you are one of them!

[@ThinkTwiceLtu]

Thank you very much!

[@manuelb__r]

Wonderful channel

[@farisakmal2722]

Yay new video

[@kaziaburousan166]

Wow... ingenious 😍😍

[@Psyduc]

Here's a not so neat solution: Let the four vertices of the square be A(0,0) B(0,1) C(1,1) D(1,0) , then the problem is equivalent to choosing O(x,y) where 0

[@ethanwinters1519]

Cool visual math proof AND some banger lofi too?! Hell yeah, this is my kinda channel

[@khenanjamesnarisma5050]

I kinda get how the answer came to be. But why are the points P in which the angle become 90 not included in calculating the probability?

[@micalobia1515]

I'd imagine because it's similar to how the area is *considerably* more filling than a single line so that in comparison it's negligible if that makes sense.

[@ramk4004]

Thank you

[@ThinkTwiceLtu]

:)

[@KIGBYT]

Nice explaination 👍👍

[@disasterarea9341]

good choice of music, nice q, though quick to solve for me