The Three Square Geometry Problem - Numberphile

[@strongeststrike6737]

‘This problem has not only 1, not only 2, but at least 54 solutions.’ That escalated quickly

[@nessfrommother2147]

Pythagorean Theorem: am I a joke to you?

[@GamerShen98]

This is extremely easy question if all sides are equal u can put X as length .. then in the beta angle just do Pythagoras as the base is 2X and the hight is x , then the hypotenuse =sqr of 5 and then shift sin/cos/tan give you the angle and u do the same for the third one and thats it

[@ZachAttack6089]

@@GamerShen98 Sine and cosine are irrational, so they don't give exact answers. So you'll end up with something like 90.000000003, which doesn't "prove" that it's 90. Also, this is geometry so I don't even know if you're allowed to use sine and cosine.

[@kamil.g.m]

@@ZachAttack6089 So are you saying tan(45) is not 1, but it's actually 1.0000000003?

[@ZachAttack6089]

@@kamil.g.m Not tan(45), but most values of sine and cosine and tangent are inexact, so they can't be used to "prove" anything. Besides, it's geometry not trigonometry. I don't think you're even allowed to use trigonometric functions.

[@jk844100]

Wow, this whole thing felt like the climax of a Phoenix Wright case except instead of finding a murderer you're finding an angle.

[@vincent-of-the-bog]

A CONJECTURE!

[@mads_in_zero]

[Mia Fey Voice] Phoenix, turn your thinking around! Don't ask what the angle has to be. Just assume what a 5th grader would know!

[@uuu12343]

Eq_NightGlider_ CONJECTURE! *slams table* I beg to disagree, Angle-o

[@cptaincrunch4438]

Angles are sharp

[@Lengo67]

It was Colonel Mustard in the dining room with the right triangle.

[@nameguy101]

"The size does not matter" - A renowned professional scientist Take that, society

[@LegendaryFartMaster]

This maybe be six years too late, but this comment is underrated.

[@rockifythis]

Thanks, that makes me feel better about myself

[@machineman8920]

don't say the s word

[@samisiddiqi5411]

Okay, Coomer.

[@migfed]

I like her style, she explains the problem as telling you a detective story, it thrills you! Greetings from Colombia.

[@ivpantev]

This video is brilliant not because of the problem, but because it shows you the thinking behind reaching a solution. Everyone who said to use trigonometry is not wrong per se, but simply taking a more complicated route to reach the same conclusion. The beauty here is that you can get where you want to be just by drawing a few lines rather than using advanced functions, i.e. you can solve a much more complicated problem with simpler tools and some creativity :)

[@nsq2487]

In the real world where I have a calculator and know about inverse trigonometry, I would use the so called "complex" method any day. Yes, the constructing lines solution was elegant, but not practical. Trigonometry was meant for problems like this and it would be the first thing to pop into the mind of any engineer. It is the most simple solution, find each of the angles, add em up , done. Just because it requires a calculator doesn't mean it's complex.

[@freddiehand6551]

@@nsq2487 In practical use, trigonometric methods are indeed the most efficient, and is thus used in engineering. However, this is only a maths problem intended for recreation, where we are not concerned about the speed of reaching the solution.

[@freddiehand6551]

@@nsq2487 in addition, you would have to use trigonometric identities to verify that the angles add to exactly 90 if you are only in possession of a pocket calculator, because it is possible that the sum is very close to 90 so that the calculator round the answer, but it is not exactly 90.

[@heronimousbrapson863]

I Pantev So much easier to use trigonometry. 5th graders are not normally taught trigonometry though. Otherwise, you could measure with a protractor. Cutting pieces of paper and putting them together seems more like art class.

[@keescanalfp5143]

@@heronimousbrapson863, A nice thing is that children aged up to 14-16 love seeing you folding, cutting, adding real material to illustrate no matter what. Even geometry.

[@numberphile]

There's a lot of links to go with this video (extra footage, associated video, brown paper, discuss on reddit, etc)... See the full video description for all these links.

[@randomguyontheinternet801]

.An underrated comment>

[@superj1e2z6]

I give this video 7 points.

[@zchelmerjoashgamboa7366]

nice reference.

[@NoriMori1992]

Ohohohoho!

[@composto]

Oh, nice one!

[@hblanco530]

Is that an imo reference?

[@dominikskorjanc]

Nicee😂

[@yugyfoog]

This problem has a very simple solution using complex numbers. (1+i)(2+i)(3+i) = 10i, which has an argument of 90 degrees. Of course I didn't know about complex numbers in fifth grade.

[@justpaulo]

Your solution is the most elegant, non-geometric, solution of all ! True that you don't know complex numbers in fifth grade, but you don't know trig either :) In fact it makes me think how good it would have been if, when I learned complex numbers, such a practical example was given to me.

[@foobargorch]

Beautiful!

[@timh.6872]

yugyfoog I feel bad for necrobumping, but that is a beautiful thing, and yet another reason why we should introduce complex numbers right after fractions alongside the irrationals. Not only does it make algebra easier (being an algebraically closed set), it's really stinking useful for plane geometry.

[@banjonpro5649]

can someone please expand on this solution?

[@dragonite7780]

Complex numbers have the property that if you multiply 2 complex numbers, their arguments add together and their modulus' multiply together. yugyfoog exploited the adding angles part.

[@keithwilson6060]

The best part of this video was her accent.

[@joshlewis8753]

@Paco Bulgarian

[@mariostar13]

@Paco Yep! Definitely Russian!

[@brenmcclure128]

Yes Dr. Jones

[@gmortimer20031]

@masonery123 Just for a fleeting moment, but one which extended to a point at infinity in a hidden dimension, I wished I was a single man!

[@ericastonehouse9979]

definitely lol

[@FlyingTurtleLP]

I really love vids like that. Will never regret that I subscribed this channel.

[@numberphile]

FlyingTurtle thanks - we're happy to have you as a subscriber

[@zacharygeorge4595]

Numberphile I love this channel! I'm a number nerd, so I learn everything about numbers as possible. I've watched like every single video!

[@aghaanantyab]

actually this channel is dedicated for psychopath. if you like this channel it means that you could potentially be a psychopath

[@aghaanantyab]

Thomas Anderson you're welcome :)

[@fishyeverything8530]

Me too 😄

[@Martial-Mat]

Jeez - if that's her idea of an easy solution that a 5th grader should be able to work out, she went to a different school to me! I don't think we touched algebra till 8th or 9th grade and cancelling of elements till a year or two after that. It was elegant though, and she's very pleasant to listen to.

[@whychoooseausername4763]

A lot of mathematically gifted kids were pushed into competitions for maths or physics. I did competitive maths physics and literature just for fun throughout middle school and high school. We were often asked difficult questions and asked to produce original proof.

[@travisbaskerfield]

Mat Broomfield. The deliberate dumbing down in American education?

[@mequable]

Bulgarian education was crazy back then. Now kids are lucky to be able to write properly when finishing 7th grade.

[@Martial-Mat]

Oh really? Why do you think that is pastichka?

[@GRBtutorials]

Well, here in Spain, we are introduced to algebra in the equivalent to 6th grade, and I personally knew a bit of it by the equivalent of 5th grade as well.

[@gnomee9447]

She is so great at explaining! I think even people, who didn't work with angles for long, will understand it :)

[@ItsClint]

This is brilliant. Loved both this and 'Pebbling a chessboard'. I'd love to see more videos from her. Keep up the good work, Numberphile!

[@headrockbeats]

What scares me is that we only learned this sort of stuff in 7th grade, in geometry class for _advanced_ students...

[@HoratioAccel]

+Headrock In the US Geometry is usually taken by Sophomores in _high school_. It's a sad world.

[@water7648]

+TheCycloneRanger im in 7th grade doin geometry but im da only 7th grader for it which is really sadddddddddddddd....

[@HoratioAccel]

Turtle cat For high school credit? If so, damn, I only knew a handful of people at my middle school who did that.

[@water7648]

not for high school credit. IM in middle school but they let me get into geometry since I passed algebra last year in six grade

[@HoratioAccel]

Turtle cat If it isn't for high school credit (and shows as such on your end of year transcript) then you are more than likely going to have to take it again in high school. I would check this if I were you, just in case.

[@gabcastel]

i love absolutely love her accent.

[@tomasvolejnik2809]

i hate it

[@FLomasterZ]

It's Bulgarian accent and it's very similar to Russian. ;)

[@iammaxhailme]

arctan(1/1) + arctan (1/2) + arctan (1/3) = 90 A bit easier, but less interesting than the video's method, of course.

[@usurper1091]

yeah

[@ptyamin6976]

you have to refer to these trancendental functipns which are not as elementary, and thus elegant, as simple geometry

[@stevenvanhulle7242]

I guess you used your calculator for this. That's not a way mathematicians do it. (Especially in inverse geometric functions) calculators make small errors, and you may end up with a result of 90.00 degrees, while in reality it could be 89.9999999943. In that case the conjecture doesn't hold. For instance, it's impossible to construct a regular nonagon with only compass and unmarked ruler. Yet there is a construction which comes real close, something like 99.999999999999%. Your calculator may incorrectly suggest that it's a valid construction for a perfect regular nonagon, but like I said, it's impossible.

[@shvoregavim9435]

Steven Van Hulle Just use the formula for sum of arctangents.

[@meJevin]

yeah, without a calculator you wouldn't be able to calculate this, so you would have to use an arctanget summation formula. you will get something like this if you do: arctan(1) + arctan(1/2) + arctan(1/3) = arctan(1) + arctan((1/2 + 1/3)/(1 - 1/2 * 1/3)) = arctan(1) + arctan((5/6)/(5/6)) = arctan(1) + arctan(1) = pi/4 + pi/4 = pi/2; (or 90 degrees if you prefer)

[@aboubacaramine8689]

I wanted to try to resolve the problem myself just for fun, so I paused the video and I spent like 10 min on it. I really like the fact that even though the basis of the method is the same, I still resolved it in another way than Pr. Stankova did. And it was so fun to do. Basically I multiplied the squares to a small grid, then I recreated a similar construction but with the diagonals of the first squares as the sides of the new squares. I had then new 90° angles and with the use of all the parrallel lines I could put together alpha, beta and gamma in one of the 90° corner and they fit perfectly. Proving then that their sum is 90°. I love how you can manipulate and distribuate angles using parallel lines. It's like the energy of the 2D geometric world x)

[@peterpowis4145]

I paused the video too to try and work it out for a minute or two. Came up with 90 as a guess

[@Flumphinator]

Speaking as an architect who makes a living working with geometry, this is brilliant. Also just another reason to love the number 6. More geometry videos!

[@slouch186]

is this really something they taught in bulgarian elementary school? pretty sure i was still struggling to memorize multiplication tables back then

[@yofoxjoke]

isnt multiplication taught in grade 1?

[@fatpanda3305]

I'm not sure about Bulgaria, but in Poland we were done with multiplication tables by grade 3. I'd expect a problem like the one in the video as an "extra difficult" question for 6th or 7th graders.

[@CraftQueenJr]

slouch in fith grade?! That was second and beginning of third! I learned protractors in 3rd!

[@CraftQueenJr]

slouch we did this type of stuff in third and fourth grade. I’m from Texas.

[@KucheKlizma]

Aye, bg here. Ack I remember this type of solution always looks so easy when someone else does it. Yet when I do it I have an triangle mona lisa on my list of paper and I'm trying to divide some angle by 0 to find my answer. -_-

[@lucasschuetz2111]

It started so simple, then grew to something so exciting. Love these videos! Easily my favorite channel, keep it up!

[@eldiospadre128]

Aw man, I haven't worked with geometry in years! This video took me back! Thanks for posting this it!

[@numberphile]

Diego Rivera you're welcome

[@Maverikmkd]

I was always amazed of all mathematical problems where you have to think "out of the box" in order to solve the problem. That is the most difficult skill to get when you participate math contests, and I was also amazed by all kids who had those kind of skills. For me, it was just magical and astonishing. I am still wondering, how to train kids to think like that?

[@Sirenhound]

I paused it until I figured it out. My solution was a little different (one of the other 54 I suppose): I chose the top left corner where all the lines radiate from; The 45 and the 18ish are already there, so to solve I need to show that the remainder is the 26ish. I mentally extended the diagonal (45) line down to be twice as long (terminating directly below C), and bring it up from that point to corner D. this constructs a similar triangle at sqrt(2) scale, seating the 26ish angle neatly in the desired position. Q.E.D.

[@davidgould9431]

Four years old (the video, not me) and people are still watching and commenting, which is great. Love it. Ignore the nay-sayers. I decided to have a stab at it and was pleased to find a geometric solution with minimal extra drawing. Presumably one of the 53, but here it is anyway: We know that we have an α at the top left in the isoceles triangle AEB. Let's try to construct a β on top of it. To do that, we need to go two units along AB, turn through 90° and go another one unit. Let's go two 'half diagonals' down to B. 45° takes us to BG and another 45° takes us to the diagonal of the middle square and one 'half diagonal' takes us exactly to its centre. I then realised that ED must also go through that centre point†, so we have AEB = α, BED = β and DEJ = γ by the 'Z' rule (EJ being parallel to AD) and AEJ = 90° because it's the corner of a square. QED † to prove it, draw a horizontal line through the centres of the squares: it's easy to show the triangles at top top left and bottom right are congruent, but I think "by symmetry" should be enough for a recreational maths posting :-)

[@jadengraner5004]

Solving this as a 5th grader and being one of 11 out of 260 kids to solve the infamous "Problem 6"..... much respect

[@tri5ford]

Approx 3:10: "Or some ugly angle nearby?" Priceless...!

[@PallasTurrets]

"Oh, it's a little bit obtuse" - my teachers about me in 5th grade

[@xovvo3950]

I am so glad to see anothervideo featuring Dr. Stankova. The problems she takes us through are always fascinating. Simple questions with complex implications. I can't wait to see the next one!

[@lhopitalified]

There's also an elegant solution via complex numbers: Let A = 1+i, B = 2+i, C = 3+i. Then the argument (angle) of the product ABC is the desired sum alpha + beta + gamma. Since ABC = 10i, the sum of the angles is pi/2.

[@aminsust1236]

ru

[@luksiv]

every single video of Numberphile leaves my mind blown of how someone can come up with these kind of ways of solving such complex problems... kudos for that !

[@lammatt]

wow... this is brilliant.

[@orbemsolis]

rilesthegiles I have completely forgotten what we were discussing

[@hellje]

matt lam жцжцжц

[@johngowers5403]

A method I found that doesn't require drawing any more lines on the diagram: Go to the diagram at 1:08 on the video. Look at the top left hand corner. It is divided into four angles by the three lines that meet that corner. The leftmost of these four angles is clearly equal to alpha (isosceles triangle). The topmost is equal to gamma ('Z' rule). Therefore, it suffices to show that the two angles in the middle sum to beta. For this, we spot a pair of similar triangles in the diagram. If we label the vertices along the top A,B,C,D and the vertices along the bottom W,X,Y,Z then we claim that the triangles AXY and AXZ are similar. Indeed, they share the angle

[@j0nthegreat]

how many squares do you need so the angles would add up to 180?

[@holdenew]

17 squares is 181.3. To get over 360, you'd need over 500.

[@me_hanics]

Holden Watson Source?

[@akanegally]

No source needed. Use your mind : arctan(1) + arctan(1/2) + arctan(1/3) + ... + arctan(1/n) >= PI It happens that n=17 squares

[@quinn7894]

2. That was the most easy question.

[@kipshiux333]

Professor Stankova is SO engaging, I really enjoy videos with her. Brady, can we have more of her, please?

[@irwinwinaris9800]

Just brilliant that math can be so simple yet so fascinating.

[@Jodabomb24]

Professor Stankova might be my favorite professor featured on this channel. She always presents the information clearly and in an interesting way; you can always see how much she loves what she's doing, and that's really important. I just love listening to her. And yes, her accent is fantastic :D

[@FirstLast-dd8ff]

This video was amazing; really liked the simple animations and the woman's voice felt nice to hear. Awesome!

[@numberphile]

First Last that is nice of you to say

[@SomeDude881]

That method was thinking right out of the box literally. This video impressed me for such an unique way to solve problems.

[@allluckyseven]

Beautiful problem. I wonder to how much would the angle tend to, if you kept adding squares to the right.

[@DerClaudius]

That was neat. The actual problem, but also the sparkles in her eyes, remembering one of the things that started her fascination and love for mathematics. I also remember multiple such situations with math, physics and programming in my youth.

[@MrV1604]

This is so awesome I cried .. I'm showing this to my little sister !

Let me just point out that I loved this professor. Her demeanor, her choice of words, the way she explained conjecture and of course her accent really makes it fun to listen to.

[@raykent3211]

what a brilliant teacher!

[@DeathlyTired]

I remember Prof. Stankova from the 'Pebbling a Chess Board' film. Again, she really is excellent; clear, concise & thorough.

[@DewZJ1997]

"The size does not matter" -Zvezdelina Stankova2014

[@manukrishnanms1317]

Time stamp please

[@xaviconde]

"Exaaaactly"!! 😂

[@EugenAugustinus]

It's really a very beautiful problem. I think the most simple proof can be given with complex numbers. You can write (1+i)(2+i)(3+i) = 10i (i^2 = -1).

[@flam1ngicecream]

Instead of figuring Beta + 90 - Beta, you could also say that HE and HD have slopes that are opposite reciprocals of each other and are therefore perpendicular.

[@f1f1s]

Two sample problems from a Soviet quiz for second-graders (that is, 8-year-old children): 1. A digital clock displays hours and minutes (e.g. 13:08). Little Tim loves the digit “5”, and he enjoys watching it on the display. What is the longest possible continuous time span during which Tim can revel the sight of the number “5” on the display? 2. Theo picked two 3-digit numbers with equal sums of digits. He subtracted the smaller number from the larger one. What is the largest possible difference Theo could obtain? Now you get it why Soviet mathematicians emigrated to the USA and Europe after the fall of the Iron Curtain? Because their skills and wit were highly demanded in the West.

[@weekendresearcher]

The 1988 IMO Problem no 6 killer!

[@milosmitrovic1271]

8:35 Those two lines (EH, HD) were catheti or Cathetuses of that triangle, and the bottom line (ED) was a hypotenuse. One of my favorite solutions, very nice work. :)

[@SmileyMPV]

Omg pls why all the arctan ;-; guys this is geometry, not trigonometry. The difference is that geometry gives more of a pure proof while trigonometry is not that pure. You guys disappoint me :( Sure, arctan(a) + arctan(b) = arctan((a+b)/(1-ab)) and therefore arctan(1/2) + arctan(1/3) = 45 exactly, but you are using a formula that i bet nearly none of the people in this comment sections, who are posting this trigonomial 'proof' here know how to prove, on top of that, proving this formula takes more effort than proving the conjecture in this video. But if you are willing to take the time into noting that you are using a formula that has already been proven, then ok, I guess it's a proof, but it doesn't leave you with much understanding as to why the angles sum up to exactly 90 degrees. When you are proving a geometrical conjecture using methods described in this video (congruence of shapes, equal angles and sides, right angles, isosceles triangles) you get a more pure proof which leaves you with a better understanding as to the reason why the angles sum up to 90 degrees. On top of this all, who in the world knows about trigonometry in freaking fifth grade, i mean come on. If you walk up to the most clever fifth grader you know and tell them about your trigonomial 'proof', do you really think they understand? I think the geometrical proof will be way easier to get your head around as a fifth grader. And to those who are just saying 'arctan(1/1) + arctan(1/2) + arctan(1/3) = 90. There. Done', just go away and enjoy your oh so open minded life.

[@williamliu9584]

Hey Brady, I'm a Chinese student now in Grade nine. This problem came out in our math midterm exam. I just simply can't thank you enough. Thank you soooooooo much.

[@billy.7113]

Grade 9 in China or USA?

[@billy.7113]

Grade 9 in China or USA?

[@qbwkp]

"The size does noy matter:"

[@valeriobertoncello1809]

I was just looking for a comment like that ahahah it didn't take much time

[@JamBos11]

Only the proportion matters 😉

[@randomaccessfemale]

If you are worried about size, just change the metric. That's what I always say.

[@NoriMori1992]

"noy"

[@qorilla]

You can also collect them in the top left corner. The leftmost angle is 45°, the middle two add up to beta and the rightmost/topmost angle is gamma. Gamma because it's just a flipped version of gamma triangle up there, and beta because if you extend the lines downwards you can make a 2 by 1 triangle diagonally. Too bad I can't post a pic.

[@archimedesworld3202]

@jimpikles She said it was 5th grade. Here is why you DON'T jump to trig with sin and cosine etc... Its bad practice.. why bother explaining arithmetic to kids.. just teach them long division and forget about the reason it works or why. See the point? It's about understanding not about passing a test fast. Understanding is far superior to memorizing formulas. Creativity will get you further. The age of genius Einstein's and Euler is gone precisely because understanding has been thrown to the side for the sake of practical or commercial use.

[@S1rRender]

ArchimedesWorld The imagination is being reduced to almost non existence

[@angelmendez-rivera351]

Well, historically, schools were never meant to teach. The governments are perfectly fine with not having children learn as it would be inconvenient for them to have children learn only to overthrow them later. Private schools are not much better for the exact same reason. It is unfortunate.

[@taweller]

Professor Zvezdelina Stankova + that blue dress = Nice combo!

[@michaelbauers8800]

Where does someone get those mega tools she's using?

[@joeyhardin5903]

i just used trigonometry because i noticed for the second square the line comes in exactly half way through the square and on the third one it comes in exactly one third in. With a calculator, type in 45 + atan(1/2) + atan(1/3) and you'll get 90.

[@Seth4All]

Brilliant solution. Very elegant.

[@numberphile]

TheMathKid glad you liked it

[@christianrodriguez823]

These are some of the most brilliant solutions, the ones that are rooted in basic geometry and algebra but require such outside-the-box thinking. As an aspiring teacher, I believe these are the kinds of questions we should be giving to our children in school and I hope to give my future students some problems like these to open their minds and challenge them to think differently about math.

[@bonez565]

Alright, I like this kind of Patreon reward.

[@nikolaalfredi3025]

Hey, you can take line segment at an angle -beta ( according to cartesian plane)... and after rotating it by -gamma then translate it over the line segment making angle -gamma . If the new line passes through the vertex of the square then beta + gamma = 45. How to rotate and translate... what you can do is just take the ratio of sides of the square for angle beta , then using scale draw same conditions on the line segment making angle gamma. I bet that the new line will pass through vertex of square... and all you have done is rotation and translation without a compass, only pencil paper and scale. :)

[@jujumas8708]

This is the math people should learn at school, not the one that is filled with doing the same things over and over again.

[@dr.mikelitoris]

Juju Mas it is what we learn at school though

[@AllieAndPeach]

some people need repitition to really learn and remember a concept. nothing wrong with that.

[@BonziFedoraINC]

Sam Harper ah someone that gets it

[@Drachenbauer]

actually i already played around with that stuff at constructing geometric shapes for graphics of wind-kites, where i like to use a squared snapping-grid. So i already knew, that the sum of the two crooked number angles is 45°. I also like to use it to turn my kite-graphics into an isometric 3D-looking flight view: 1. with theese rules i can rotate the kite 45° and make it snap to the grid again (it´s bigger, twice the area, now) 2. i squeeze it along the vertical axis of the snapping-grid to half hight(now former 45°lines became the angle of the seccond line in the puzzle-starting-setup) 3. sometimes i move points, that are not on the center-axis of the kite, a little bit up (the further away from the axis, the more up) to make it´s wings look a bit tilted up.

[@willdeary630]

0:19 "Size does not matter", you heard it here folks.

[@Kourindouinc]

As Zvezdelina Stankova stated, there's several ways to solve this problem, and several of them uses trigonometry. Another alternate solution, although it is similar to the video, has an alternate way of proving that EHD is a right angle. Let's assume the side of a square is of x. The slope of EH is x/2x or 1/2. The slope of HD is -2x/x or -2. If two slopes are opposite to each other, meaning one is x and the other is -1/x _(Where x here would be 1/2),_ the two segments are perpendicular, meaning EHD is exactly 90 degrees. Proving this theorem is as easy as the step they used instead. There's a lot of shortcuts that one can do while doing this, although a fifth grader may not have access to all of them.

[@Xalnop]

Great great great great video! I love it.

[@numberphile]

Xalnop thank you

[@chadtindale2095]

Proving space using Negative Space. I've never seen that applied to angles, but it's blowing my mind. And I love it.

[@Opaqu.e]

I feel really stupid now that I resorted to trig the first time I tried to solve this!

[@mr.champion7304]

you're not alone :D

[@silvaindupertuis]

Wonderful ! But I would suggest another way to prove (or show…) that the triangle EHCX is an isosceles right triangle, without any calculation on the angles. It suffices to observe that if we rotate the rectangle HFEX of 90° around H, we get the rectangle CHID, and rotating the diagonal HE moves it to HD, which proves that HA and HE are of same length and the angle EHD is right (90°).

[@bombasticbrian.]

Math is a beautiful subject

[@karrieadler2808]

I actually came up with a somewhat simpler solution, It's very close to this though. I start with the single square and draw the line to the corner to create that angle, I draw 2 more squares above it (3 high in total) and draw a line from that point down to the same corner. then if you imagine the original 45 degree line as a side to a square and draw the rest of the square, and then draw another square towards the point at the top of the stack of smaller squares, it lines up with point at the top (showing that the middle segment is 2 squares apart) and all the angles fit into the 90 degree corner yay, (It's a lot simpler to show visually than to explain)

[@ButzPunk]

Her accent sounds so familiar to me... is she Bulgarian? (I might be completely wrong)

[@nandvinchhi1217]

ya bulgarian olympiad winner

[@DiDi_DZB]

really enjoyed this talk and i think the teacher is quite amazing too. Watched all three parts and the discussion was fun, light and interesting!

[@TheDiggster13]

Well suddenly, my solution involving the sine and cosine rules, and pythagoras' theorem seems horribly inelegant!

[@AustinGarrett777]

Conveniently, the slope of each line segment is given, which allows you to find the relative sizes of each side, and calculate the angles through trigonometry.

[@ben1996123]

okso if there are infinitely many squares then the sum of the angles diverges the sum of the angles formed by n squares is i/2 (log(gamma(1+i))-log(gamma(1-i))+log(gamma(1+n-i))-log(gamma(1+n+i))) which is cool

[@willericson7793]

I bet it also diverges when you only use the angles going to the prime numbered squares. It feels like the harmonic series, but I don't have the math to prove it.

[@EPICGUYDUDE]

Ben? Ben?! It's you!

[@PhillipH-san]

ben1996123 I like your profile picture. I also like your maths.

[@XBrainstoneX]

Here. I posted this proof in the other comment section about this topic. It shows that you can indeed use the harmonic series as minorant for the series. So the statement about the prime numbers from will ericson is correct. "Here is a proof idea. I'm not quite sure if it's the most elegant, but I'm pretty sure it works. Also, you need first semester university calculus: We are looking for sum(1,inf)(arctan(1/n)). Define f(x) := arctan(x) - ln(1+x). Because f '(x) = 1/(1+x²) -1/(1+x) we can conclude that f '(x) >= 0 for all x in [0,1]. With the mean value theorem we get that f(x) >= 0 for all x in [0,1].* Thus arctan(1/n) >= ln(1+ 1/n) for all natural n. Because ln(1+ 1/n) = 1/n*ln((1+1/n)^n)), and ln((1+1/n)^n) -> ln(e) = 1 for n to infinity, there is a N from where ln((1+1/n)^n) > 1/2. Therefore we get the harmonic sum as divergent minorant for sum( 1/n*ln((1+1/n)^n) ) and for sum(arctan(1/n)) aswell." *( In-detail-explanation for *: Let x be out of [0,1]. Then there exists a z out of (0,1) so that f(x) - f(0) = f '(z)*(x-0) => f(x) = f '(z)*x >= 0 )

[@nullpoint3346]

Row or grid?

[@JohnSmith-ry3rp]

In my opinion, the most logical way to go about this involves basic trigonometry. If the squares are all equal in length, then you can get the answer by using the tan^-1 function. If you look at it, the inverse tangent ratios for the three angles are tan^-1(1/1), tan^-1(1/2) and tan^-1(1/3). If you add them up you get: tan^-1(1/1) + tan^-1(1/2) + tan^-1(1/3) = 90 Try it yourself it works.

[@raoulhery]

Can I use Trigonometrics here?

[@Fun_maths]

You get arctan(1)+arctan(2)+arctan(3) Not clear that it sums to 90 degrees

[@Pingstery]

Came up with 2 solutions just after the problem was explained, this indeed is a brilliant problem that allows kids to apply what they've learned in their own ways, instead of forcing a teacher's decided approach. I wish this was shown to me in my geometry class! Definitely going to forward it to my old teacher, hehe.

[@naor5683]

WOW we should all be thankful to the greeks for trigonometry xD

[@zolv]

Welcome back! I couldn't wait for next videos with You after "pebbling a chessboard" videos!!!

[@akanegally]

Another way to demonstrate it arctan(1)+arctan(1/2)+arctan(1/3) = 90°

[@PeterAuto1]

but therefore you need the exact values of arctan(1/2) and arctan(1/3)

[@Shadowmere29]

+Peter Auto no you don't need the exact values. Just use the tangent angle sum formula

[@reetasingh1679]

+Peter Auto You can use the tan(x+y) identity... then you will get (1/2+1/3)/(1-1/6) which is equal to 1, therefore sum of the two angles is 45

[@sanyamahuja7896]

only if fifth graders knew trigonometry

[@Shadowmere29]

Lol

[@jorgesampaio5635]

You don't need such a complex construction. All you need is to prove that beta + gamma = alpha = 45 degrees. Note first that angle DEJ = gamma and angle CEJ = betta. Now all is needed is to prove that angle BEC = gamma. This is immediate if you extend line EB to intersect with the extension of line JC at, say, point M (below segment BC). Then note that triangle EMC is rectangle and proportional to triangle EAD, because EM = 3xMC. QED.

[@Chriib]

45 degrees + 2/3 of 45 degrees + 1/3 of 45 degrees. That was my first thought. It is not the right approach but the answer is correct.

[@RomeoUW]

This is pretty awesome and really understandable due to the way it's explained. Thanks for sharing with us this beautiful knowledge !

[@robinbuster1323]

hypotenuses is correct. Singular: υποτείνουσα Plural: υποτείνουσες

[@DeltaHedra]

What I find more amazing about this problem is that the initial red lines which join the corners of the rectangles, where they touch the uprights of the squares are the points of a parabola.

[@GaryFerrao]

OMG!~ The most revered brown paper is cut up!~ :-o :D I guess this is the first time the brown paper has been desecrated and destructed, or is it? :P

[@divyanshbhandari8870]

Vsause music starts....

[@FLomasterZ]

Zvezdelina Entcheva Stankova (Bulgarian: Звезделина Енчева Станкова; born 15 September 1969) is a professor of mathematics at Mills College and a teaching professor at the University of California, Berkeley, the founder of the Berkeley Math Circle, and an expert in the combinatorial enumeration of permutations with forbidden patterns.

[@cursedswordsman]

USING ARCTAN IS NOT A PROOF UNLESS YOU USE A FORMULA TO SHOW IT ANALYTICALLY. ARRRGH

[@stellarwinda]

cursedswordsman Indeed, it seems half of the people here think that taking arctan on their calculator is fine.

[@paulbin]

cursedswordsman You are wrong my Little friend

[@liberphilosophus7481]

Hue hue arctan :P

[@dominantwolf4593]

Just use regular SOH-CAH-TOA on each angle and add. Assuming you don't have to prove your unit circle rules true which isn't that hard either.

[@dominantwolf4593]

+Harry Potter granted I'm a chemist so I believe integral are like addition lol

[@justpaulo]

My solution involves only drawing a single line. (5:49) Let's focus on the left top corner (the common vertice from where all the red lines start). It's easy to define Alfa & Beta or Alfa & Gamma over there. All it's left is to prove that the remaining angle in the middle of the others is Gamma or Beta respectively. Let's use the Alfa & Beta that are already in the top left corner and prove that the remaining angle is Gamma. For that just draw a line perpendicular to the red line that defines Alfa and crosses the red line that defines Beta where the latter intersects the vertical edge that separates the first 2 squares. That new line defines a right angle triangle that fills the empty angle (which we want to prove it is Gamma). This new triangle can be very easily proven to be a scaled down version of the Gamma triangle with the sides that define the 90" angle having a 3 to 1 ratio (1 unit = 1/4 diagonal of a square) . So the remaining angle in the top left corner must be Gamma, and therefore Alfa+Beta+Gamma=90".

[@shield543]

Awesome video :), Oh and first?

[@billzorbas3889]

Young girls fishing

[@girgakos]

you can also say that tanβ = 1/2 and tanγ = 1/3. Then you say that tan(β + γ) = (tanβ + tanγ)/(1 - tanβ*tanγ) => tan(β + γ) = 1 => tan(β + γ) = tan45° => β + γ = 45°. So α + β + γ = 90°

[@rhythml6229]

Secondary school level okay... even this way of solution

[@kg583]

You don't need to find the angles of the large triangle; it is already known that HD is the same as BE because they both go from a top corner and then two squares over to the bottom, and therefore have the same angle. Therefore beta is guaranteed to be angle HDJ and alpha EDH such that they all add up to 90. Loved your proof for it but it was a tad long-winded.

[@opcn18]

That was beautiful.

[@DjVortex-w]

That was an amazingly complicated way of explaining the six-square solution.

[@Michael_Arnold]

And so, so slow!

[@Michael_Arnold]

And so, so slow!

[@mullergyula4174]

I guess they wanted to make it easy for all audiences.

[@kiefac]

This is much simpler once you learn trig functions, but I guess 5th graders wouldn't have learned those yet.

[@wafikiri_]

There is a much simpler proof. No need to draw three other squares and a slanted big triangle. Just get a diagonal DK in the third square (new labeled vertex K) CDJK. Then all the figure has a 180º rotational symmetry, and the square with vertices A, B and E has diagonal symmetries, of course. Then, at vertex E, we have: angle JED = angle EDA = gamma; angle DEC= angle ECA = beta; and angle BEA = angle EBA = alpha, the three of them composing angle AEJ = 90º. And the said diagonal DK has not even been used, except to show the 180º symmetry. Update: typo correction.