The Three Square Geometry Problem – Numberphile

The Three Square Geometry Problem – Numberphile

So what are we talking about today? About my favourite problem in geometry when I was a fifth grader and it is etched in my memory to the point of being able to recall it at any time so we have three squares of the
same sizes next to each other the size does not matter so what we will do you is pick one of the top corners let’s say the top left
corner and connect it to the bottom right corners of each of those three squares. Next, we
will take this angle here at the bottom let’s call
it alpha the next one at the bottom let’s call it well beta and and that’s incorrect yeah this is upside down and finally gamma alright so the question is asking what is the sum of these three angle so it’s a
simple enough problem the interesting part about it is that it
was given to me as a fifth grader so I had to a attack it as
fifth-grader in fifth grade we learn how to use a protractor so I measure the first angle it’s about 45
degrees and in fact then you can stop and think
about it but of course it has to be 45 since the triangle we’re looking at is
right and iscoclese this guy here so it should not be a surprise to
get a 45 degree angle there and we go about measuring the others
there perhaps we run a little bit out of luck because the second angle looks like it’s pointing to 25 and a little bit maybe 26 and the third angle Wow where is it going perhaps 15 20 so about 18 so when we add this up 89 could that be it what do you think it seems a bit arbitrary i don’t know Yeah and also there were all of these mistakes in our measurement and why would they be asking a fifth
grader how much this is and ending up with such a random angle? well if you try to make another
measurement perhaps you will come up with 91.3 and so you would say oh it’s a little
bit obtuse but maybe it’s a little bit acute so
what would be conjecture that people will leap to if they believe this is a beautiful
problem I would guess that the correct answer they would think the correct answer is ninety exactly ninety but how would we know really is it ninety? Or some ugly angle nearby? we cheat a little bit we say okay that
problem was given to a fifth grader it has got to be a nice answer we will
believe for a moment it’s ninety and this is what mathematicians call a
conjecture and then put all of our effort into
proving that it is ninety and we are kind of a third of the way to proving it because we know that one
of the angles for sure is forty five so what should be left for the other two angles
in order to make ninety? well obviously forty five. so somehow we
need to add up the other two angles beta and gamma and prove that they add up to up to forty five another experiment which a fifth grader might try and you can
also do at home is cut up the whole thing with scissors and eliminate some of the error that is coming from the protractor when you are
rounding the degrees okay so let’s try to put everything together if we can it does look like this is a ninety degree let’s check well yep it does look like 90 degrees okay so now we truly believe the
statement and we want to prove it well the question is at what level of your mathematical career or mathematical life you are because
this problem has not only one not only two but at least 54 solutions.
54 solutions? No kidding.
Are we gonna do all of them? No. we will do the most brilliant one the
most simple one that a fifth grader may be able to grasp so the idea of our solution at least the attempt when we cut with
scissors can be carried forward mathematically so we may want to move all three angles into the same
geometric location in our picture so we need to find the best geometric
location where to move those three angles well there is no best geometric location
until you know what the solution is so we might as well just try to move
them all the way in this corner next to gamma because we
already have a right angle here this guy we have gamma here and so the other two angles alpha and
beta must somehow fit and here comes the brilliant idea for
the solution as I a said there is something missing in
this picture what’s missing is half of the picture
we will replicate these 3 squares up and then we’ll start
seeing the truth and now i think I’m going to
start labeling things so people can refer to them later on I’m just labelling all of the important points I will choose point H and connect it to two other points D and E and I will say that no more
constructions will be done in this solution we somehow
have to make sense of what this picture is telling us well for
starters we see 3 angles exactly where
we wanted them to be where this ninety degree angle was we
have our original gamma and then 2 more could those two angles be indeed alpha and beta well you can kinda guess that this angle is bigger than the next one by the way it looks so perhaps that will
be the 45 alpha because alpha was bigger
than beta could this be forty five and let’s try as detectives figure out
where it could come from this picture it is involved in this large triangle did we see a similar situation earlier in
the problem certainly right here where we saw our original alpha and where we decided
that alpha is 45 degrees because this was ABE was a right isosceles triangle do we have a right isosceles triangle her we might could EH be equal to HD these are the two extra
segments that we draw on top the answer is yes because these two segments are hypotenuses Hypoteni or hypotenuses? so EH and HD the two segments
which we drew are the hypotenuses of two right triangles two that we also draw on top so I’m going
to mark them in blue so these two sides are hypotenuses of two right triangles E F H and H I D the ones which I drew in blue now
they are right because they have a right angle here coming from top but they are also exactly the
same size and shape this right triangle has
one leg two and the other leg 1 and this one also has a leg of two and leg of 1 which means that their third size, the two hypotenuses must be the same so
indeed EH is equal to HD and we are on our way to show that
this big green triangle in the middle is right isosceles it’s already
isoseles but the question is why is this guy right okay but what else is around this angle there is this guy here this angle in this other one what do they add up to and the answer is also very simple one of thise angles we already know I claim this angle is our angle beta because there is a third triangle which
is precisely the same shape and size as those because it is right here again a right triangle with legs 1 and two and so therefore this
angle beta must be equal to this single beta and guess what it must be equal to this angle beta they
all share exactly the same angle so accidentaly we have shown that we have moved angle beta to where we wanted it gamma is there and if that is really 45 we will be done and again backtracking we wanted to show
that this big triangle was right isosceles we already know that this angle here is
beta what is this guy well this guy this second angle participates in a right triangle where the other acute angle is beta what could this
guy be we know that the angles in a triangle
add up to 180 this guy is ninety therefore these 2 must add up to 90 simple arithmetic shows this angle here must be ninety minus
beta ninety minus beta something and beta must overall give you 180 what is the remaining angle here this curly one if you start adding up
let’s do it beta plus ninety minus beta plus something must be 180 so beta goes away and that’s
something turns out to the our 90 degrees you’ve done it! that’s right so our angle
alpha is indeed moved here along with beta and we can conclude the the three angles add up to 90 degrees and this is an amazing construction which uses only squares isosceles triangles in congruent right triangle with legs 1 and 2 and that is indeed a brilliant approach which mathematicians take to solve
problems they complete it to what’s missing there and then the problem becomes easier

100 thoughts on “The Three Square Geometry Problem – Numberphile

  1. 180 degrees minus 45, then it's a 2/3 and 1/3 solution (carry the decimal as far as you want). The solution (rounded up) is 30 and 15 degrees.
    It's not rocket surgery.

  2. Pfft., did you notice that after you mapped those secondary triangles there was a bit left over?
    That proves to Earth is s a spheroid.
    Thake that Flat Earthers…!!!.?//:~|)

  3. I haven't done maths in over 30 years but I could have worked that out with my Casio calculator as a kid in seconds. Tan is O/A. What's a fifth grade in old skool English years?

  4. Ehm, It's so simple, i dont know why you would go for taht
    1 angle has 90 degrees in it, we know for sure that alpha is 45 degrees
    L + B + Y = 90 degrees
    90 degrees = 45 degreres + B + Y
    45 degrees = B + Y
    we can find them using the tool
    pretty sure it's 25 /20, or any combination that goes to 45
    ofc this is not wthout any proof, but cmon 😀
    5th graders wouldn't have open answers
    they would have
    a/b/c/d answers

  5. Where did you go to the fifth grade? Here in USA one would start 5th grade at age 10. I never saw such constructions with rule and protractor until I was in high school, as a junior age 16. And I was on an AP track. We never made empirical measurements on a geometric construction in order to check conjectures. If only!

  6. without calculator:
    you can figure out that "A" is 45 degrees because is a right isosceles triangle. Then, Tan(B)=1/2; Tan(Y)=1/3. Trigonometric identity: Tan(B+Y)= (Tan(B)+Tan(Y))/(1-Tan(B)Tan()Y).
    You'll come to that Tan(B+Y)=1, that means B+Y=45 because Tan(A) was also 1, so A+B+Y=45+45=90.

  7. OR…. Assuming 90 degrees for each segment 90 – A (45)=45. 90 – B (26)=64. 90 – C (18)=72. Therein lies the difference of 45 + 64 + 72 = 181. Add 181 to 89 gives you 270, or 3x 90 = 270. Works for me.

  8. My favorite from grammar school was an algebra one wherin I prove that 1.999 repeating= 2 which it certainly does. It shows the discord between math and reality because if you filled a jug 9/10 of the way full then 9/10 of what was left, then 9/10 of what was left, etc, etc, then you'd never actually fill the jug but mathematically it does.

  9. Dear Lady! I love your style of presentation!!!!! What a charming experience to see it all come together in such an inviting way as you guide us around in such an expert way!!!! You are like a breath of fresh air for sure for sure!!! Thank you for this great little exercise. David Kernberger

  10. Use 3 unit squares;
    let A=alpha,B=beta and C=gamma.

    doing the algebra;
    from (1),
    sin(A+B+C)=1,hence A+B+C= sin^-1(1)=90 degrees(half a pie)

  11. This is why most people give up on math…….such a long, boring and ignorant proof…..just go to your table of tangents and get it over with.

  12. tan(a+b) = (tan(a) + tan(b))/(1-tan(a)tan(b)), plug in 1/2 and 1/3 you got 1, so the sum of angle beta and gamma is 45, and 45+45 = 90.
    But this would be a 10th grader solution :/

  13. I have a question. What is the diameter of the circle that the motorcycle makes ???? if we turn the steering wheel alpha degree? This "phenomenon" seems to me very complicated. Thank you

  14. gonna guess it bade around phi just for bants, and not use arctan at all… == 45 + 27.8115294938 + 17.1884705063 == 90
    A = 45
    B = 45 * 0.61803398875
    G = 45 * 0.61803398875 * 0.61803398875

  15. i literally gasped when i saw the beautiful geometry picture, who came up with the idea is obviously genius !

  16. A much easier solution based on simalar triangles, using only the 3 given squares (no other construction needed), that a 5th grade student should be able to find :

    1) Angle CÊJ = Beta (alternate interiors).

    2) Triangles BCE and BDE are similar (BE/BC = BD/BE = sqrt(2) and angle B is common to both triangles (if necessary : B=135°). Hence : angle BÊC = gamma.

    3) Then : alpha + beta + gamma = 90°

    Elegant, isn't it ? 🙂

  17. 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.

  18. 4:30 Uh… put the papers on top of the square lol. It's funny how someone that is smart can be lacking in common sense sometimes.

  19. في استعادة تستحق ٥ نجوم وحرصها على أن يفهم الجميع بأسلوب دقيق وجريئ

  20. Simply turn the entire first three squares and three lines upside down superimpose on original diagram . (you now have 3 squares and 5 (five) diagonal lines ) IN THE right hand square you now have a diagonal at 45° and below it two lines that duplicate the original two lines (one is in same location as first and second goes up one square and over two squares, just as original ) and the sum of those angles (plus the 45 ) equals the square corner or 90 ° . Truly the most simple 'magic ' solution . The video explanation it the LONG way around .

  21. when you use trigonometry,most of hthe problem is already done for you by other mathematicians. if you derive the ratios yoursoelf then you have solved it

  22. Actually, I have already faced this problem when I was 12. And we only need to draw one more square to solve the problem. It's just kinda like this one

  23. slopes of the 3 lines are tan(a)=1=45°, tan(b)=1/2, tan(c)=1/3.
    Let x = a+b+c. Then x-45° = b+c. Applyint tangent both sides:
    tan(x-45°) = tan(b+c)
    tan(x-45°) = [tan(b) + tan(c)] / [1-tan(b)tan(c)]
    tan(x-45°) = [1/2 + 1/3] / [1-(1/2)(1/3)]
    tan(x-45°) = 1
    x-45° = 45°
    x = 90°. ggwp

  24. Use trigonometric functions or complex multiplication to prove better! Welcome to subscribe to "JiuJiMedia"

  25. Everyone is saying simply arctan(1) + arctan(1/2) + arctan(1/3), but here's a solution that doesn't require a calculator. I'll denote alpha beta and gamma as a b and c, as that's simpler. We know a = 45º, that is simply due to basic geometry. We want to find b+c. We know tan(b) = 1/2 and tan(c) = 1/3, so perhaps we can use this. If we can find tan(b+c), we can find b+c for sure. We look for a way to do this by starting with tan(b) + tan(c) = 5/6, since it's simply 1/2 + 1/3. We seek to use the fact that tan(b+c) = (tan(b) + tan(c))/(1 – tan(b) tan(c)). Since we know the values of tan(b) and tan(c), we substitute it into this equation. Solving, we find that tan(b+c) = 1. This implies that a+b = 45º, so the total angle measure is 90º.

    In my opinion a more interesting solution that simply plugging arctan values into a calculator.

  26. Interesting problem, here's the way I approched the problem.
    tan(Beta) = x/2x =1/2
    tan(gamma) = x/3x = 1/3
    if Beta+Gamma = 45° (pi/4), then tan(Beta+Gamma) = 1
    but tan(Beta+Gamma) = (tan(beta)+tan(gamma)/(1-tan(beta)tan(gamma))
    = (1/2+1/3)/(1-1/6) = 1 !

  27. if you use 45-45-90 triangles and 30-60-90 triangles, you can easily tell that alpha is 45, and beta is 30. then when all the constructions are done, you can see that alpha, beta, and gamma add up to 90. furthermore, we can determine that
    gamma = 90 – (45 + 30) = 90 – 75 = 15. really, in my opinion, the most simple way to solve this

  28. Well, if you would have studied inverse trigonometry you would have taken only a minute or two to solve it rather than those boring 12 minutes

  29. This is the key to nice formula for approximating π using arctg series: arctg ½ + arctg ⅓ = arctg 1 = π/4

  30. I feel like this problem is way simpler than the solution that was presented in the video makes it seem. Looking at 6:19, I'd say there's no need to draw points F, G, H, and I, or make the other three squares…

    α = ∠JEB = 45°
    β = ∠JEC
    γ = ∠JED

    ∠BED = 45° – γ
    ∠BEC = 45° – β

    ∠CED = β – γ

    Once we figure that ∠JED = ∠BEC, then 45° – γ = β (and also, obviously, 45° – β = γ).

    From there, remembering α = 45°
    , it's a simple step to reproduce 45° + 45° = 90°.

    α + β + γ = 90° ∴

  31. there is a much easier and faster way to get the proof done and you don't need the top squares… BUT, this one is very educational. Perfect for a high schooler. Its a fairly difficult SAT question; maybe 5th hardest out of 58.

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