@@kindiakmath We haven't studied arc functions so i thought it is the final answer, now i searched a bit and counted: its arctg 1 and it is 45, 45+45=90 wow thats correct
These solutions are great, especially the complex value one. I love it when complex numbers are used to make problems so much simpler than the real-value approach. I used very standard methods (cosine rule and double angle formula) to get the same result, but not as elegant as these three.
Professor Stankova (numberphile youtube channel ) says when she solved this problem at a school level without knowledge of trigonometry and complex numbers
The blue angle (γ) is obviously 45°. For the red and green angle, I used the cotangent sum formula, cot(α + β) = (cot α cot β -1)/(cot β + cot α). cot(α + β) = (3*2 - 1)/(2+3) = 1, so α + β = 45°. Adding this to the blue angle we get α + β + γ = 90° .
I first solved it with trig but I enjoyed the complex solution the most. It's really clever. I instantly knew where you were going when you summed the arguments. Great, now I have to look for other cool ways to apply complex numbers to other seemingly unrelated stuff
I'm a simple person. Knowing that the angles will not change with proportion, I drew it in CAD. Measured the angles with a precision of .001 degrees and summed them. 90 degrees. Took about 2 minutes. CAD is great at solving practical geometric problems, I do it all the time.
@@misterkite How much beauty of maths have you gleaned from this googling exercise? I mean, you didn't even have to solve this problem in the first place at all. Sometimes you need an answer quickly, but this wasn't a practical computation required to build a shed or something, it was a maths problem for the sake of solving it with maths. These 10 seconds have been traded to eternity for nothing in exchange. You could've found the answer in the comments even faster, or drag forward the video xport slider. It would be as satisfying to one's heart and useful for exercising one's mind as googling it up. Which is, fair to say, not at all. :(
@@misterkite Well, let's just agree to disagree about what mathematics is. It's a talk much longer than that which YT comments are suitable for. Sounds fine? :-)
I wonder if anybody here can solve the puzzles I just posted on my channel. One of them is practically unsolvable even by the most powerful computer in the world.
I just used pythagoras and sin law in my head, pulled out a calculator and got an answer very close to 90. I rounded, so my guess is it's probably 90 but i'm too lazy to verify
I thought that this guy probably has at least hundred of thousands of subs already until I take a look and it was only 1k. You are massively underrated. Keep up the great work!
180 deg from inverse tan (1) + inverse tan (2) + inverse tan (3). My calculations were in degrees mode. (I’m typing this before watching video, my answer was based off the thumbnail)
Thank u very much for making this video. As a 9th grader I immediately came up with the trigonometric solution. However i was blown away that it was solvable with pure algebra. I would've just plugged it into a calculator and would've thought that was it. Sadly, I didn't quite have the background knowledge to understand the 3rd explanation (idk what an argument is, I just know that i=squrt(-1)). Btw is my English good
Before you started. 90 degrees. There are always more complicated ways of solving a problem, but in engineering you must find the most simple. Basic trig will do the job.
So to get things out of the way, blue is certainly 45*. Square/2. If I'm allowed to use Sin Cos Tan, then simply apply tan^-1 to each triangle using the squares as algebraic representations of their length ratios. I'm gonna look for more before watching. Second conceived way is maybe you can link it to gradients, which can be thought of in some way as a form of angles, then convert it.
There is an even easier visual proof, for which you'll only need graph paper. First draw a 1 x 1 square with the origin in its lower left corner, then extend its diagonal (doubling it towards the upper right) and draw a (proportionally larger) 2 x 1 rectangle on that extended diagonal, then triple the diagonal of that rectangle and build an even larger 3 x 1 rectangle on that one. The diagonal of that rectangle will coincide with the Y-axis, and the three angles will together make up the 90° angle between the X- and Y-axis,.
1 minute head maths gave me solution 2. Solution 3 with complex numbers was really cool too. Solution 1 while learnable for an 8th grader, becomes less apparent as you learn more maths.
Or simple area reasoning: Given the 3 identical squares. The blue line halves 1 square = area 1/2 square. The green line halves 2 squares, area = 1 square. The red line halves 3 squares, area = 1.5 squares. The total area of the 3 differently formed triangles is 3 squares. The angle that would give you an area of 3 squares is of course 90 degrees, so thus the sum of the angles must be 90 degrees.
The red triangle and the green triangle are similarity triangle therefore the sum of red and green angle -> x+y + 135=180 then x+y would be 45 and obviously blue angle is 45 therefore x+y+z = 90
I would have never derived the pictorial solution and although I am comfortable with complex numbers I did not see that third solution at all. Very, very good. As far as I am concerned this kind of analysis puts the "applied" into mathematics.
For some reason I went: (1 + 2/3 + 1/3) * 45° = 90° I’m not sure why I correlated the areas of the rectangles like that or if it even makes sense but that’s how arrived at 90
45+22.5+15=82.5 degrees is my instinctive reaction. The Blue triangle is an isoceles with a right angle, meaning the remaining 90 degrees are divided by 2 for an even 45 degrees. If this is X x Y, then the green triangle is X x Y/2, making it 22.5 degrees. Likewise, Red is X x Y/3, making it 15 degrees. I'd like to see why I'm wrong though. Had I a calculator, I'd work each out as Tan(^-1)(Opposite/Adjacent)
I thought you'd have 3 different answers from seperate methods, but this shows just how comprehensive we can get with maths! I'm glad to learn something every day, and equally glad I don't have to solve these equations outside of videogames anymore!
My thought process is just arctan(1/3) + arctan (1/2) + arctan(1) which is the m of the functions (y = mx + b) if you use radians then 180/pi * arctan() of all these angles
My first thought was, that it should be propably 90degrees. Inv-tan of 1/3 + inv-tan of 1/2 + 45 gives you 90 degrees! But I had to use my calculator. The complex number solution was impressive, we hadn‘t learned that in school. Very cool!
Great thinking, but i really wanna know how can someone develop a thought on doing the construction in the first steps.. Btw, i did proceed using trigonometry, by assuming a side length x, and then using cot(theta).
That's a great question... one should always ask how people come with certain solutions. Often, proofs obscure the thought process which makes seem certain solutions as magic. But there is always some reasoning behind it... Let's think about it!
Is there a general solution where we have green and blue angles with distance between them, to calculate distance between blue and red if we have red angle?
I really enjoyed the complex numbers proof. There is one caveat though. An argument of a complex number is the angle between the number and the real axis, plus any integer multiplication of 2π / 360°. For example, what is the the angle between i⁴ and the real axis? i⁴ = 1, therefore it's 0. On thr other hand the angle between i and the real axis is π/2 / 90°, and the sum it 4 times is 2π / 360° In other words the sum of the angles can be any number of the form 90° + k*360° We can ignore negative k's, since we are adding three positive angles. Morover, each of the angles is less than 90° angles in a right triangle, which aren't the rightangle itself. 0 = 90° + 360° > 270° Therefore we indeed reach the conclusion that the angle is 90°
0:16 I did this: I draw that. First is the third one, I put everything to "1 centimeter". So first is arcsin(1/sqrt2). The second is only diff. is arcsin(1/sqrt5) and the third arcsin(1/sqrt10). Add together is 45+26,57+18,43 is 90 degrees. :D It was 4 minutes. So now I will continue the video.
I just did method 2 with an inverse trig calculator. As a designer/engineer, just round to whatever decimal places fits your usage. Out of laziness, I got 89 degrees. Good math shown, but problems like these are solved way more easily in the real world
As a funny side note, since we know geometrically that the three angles add up to pi/2, we can use the tan approach to get the triple angle sum formula for tangent, which looks like this: tan(a+b+c) = (A+B+C - ABC)/(1-AB-BC-AC), where A = tan(a), B = tan(b), C = tan(c). Now we use A = 1, B = 1/2, C = 1/3. Since we know that the angles add up to pi/2, then tan(a+b+c) is infinity, so the denominator (1-AB-BC-AC) must be 0. Plugging in the values of A, B, and C gives us 1 - 1/2 - 1/3 - 1/6 = 0, or 1/2 + 1/3 + 1/6 = 1. So we proved that 1/2 + 1/3 + 1/6 = 1, using geometry and trigonometry.
I don't really get it? I mean I liked the first solution but just tipping arctan(1)+arctan(1/2)+arctan(1/3) in my calculator gives me the 90 degrees solution? I expected it to have rounding errors or something but works fine. Is there some rule about not doing that that wasn't mentioned? Or something which makes doing that invalid?
There is also a 4rth method. Knowing that this is a math problem, the answer will most likely be a common angle like 30,45,90, 180 etc instead of an abscure angle like 105, and just looking at the problem it looks like the angles add up to 90, so i say its 90 degrees. I guess its not a proof, but i was able to get the answer right
90 degrres answer the angles of the there triangles are 45 , 26.565 and 18.435 There likely a more elegant way but using trig Let the side of the square = 1 Hence the angle of the triangle formed by the blue line = 45 degrees since it is the diagonal of the square The side of the triangle formed by the green line are 1, 2 and 2.23607 2.23607 (comes from using pythaogrean theorem) Since there are three sides SSS using the law of cosine yields 18.435 degrees The side of the triangle formed by the organe line are 1,3 and 3.16628 Since there are three sides and using the law of cosine yields 26.565 degrees Hence the sum of the total angles = 45 + 18.435 + 26.565 = 90 degrees (answer)
let one side equal to x. such that the area of the largest triangle is 3/2 x^2 which is the same as 3/2x times root 10x^2 times sin theta of the largest triangle (1/2 ab sin C). then you can get a equation for the first theta in x then you just repeat for the rest. then just input a value for x. it doesn't matter which value you put since all of them equal to 90
im still convinced it is 82.5 degrees even after watching until the end. In my head the blue is obviously 45°. The green angle is half of the blue angle because it is in a 2 to 1 side ratio. And the red angle is a third of the blue angle because it is in a 3 to 1 side ratio. Could you explain why it isnt just 45° + 22.5° + 15°?
You would be right IF these were circles, with those diagonals as radii. Then the angles would add up to be 82.5 deg. But each square diagonal is not simply proportional to each other when you increase the side of the square. The ratios have a sin in them as well.
@@kindiakmath I'm with Beppo here I did the same thing. and I checked it, or at least I thought I did by looking up the tangent angle ratio (opposite over adjacent sides correct?) which gave me 22.5 and 15 as well. So i got 82.5 as well. I'm VERY rusty on my Trig so going from distant memory. I'm not saying I'm right, just that I don't logically understand where I went wrong. Is the tangent of the middle angle NOT 22.5 degrees? Scratching my head here.
@@kindiakmath Ya found a trig table from an old textbook ( not sure what website gave me 22.5) but i found the correct one. 26 and change 18 and change for the smallest one. Add it all up 90. Counterintuitive, but it is what it is. thanks for responding.
My solution is just the complex number solution, but without any computation. Simply draw a circle (with radius equal to the diagonal of the square) around the bottom left point of the box. Extend the first angle to the circle. Construct a line from this intersection to the intersection of the top right corner of the first box and the circle. Draw the second angle from the center of the circle to this new intersection. Finally you can extend the vertical line on the left edge of the box to the circle to see that the sum of the angles is pi/2.
I'm not entirely sure what I did, but I just looked at it, thought about some random stuff for a second, and got the answer, definitely felt a lot simpler than the solutions in the video, but I also couldn't put my method into words so it's not a very good solution xD
I like "picture"'s solution most. I love those proofs without using any algebra. I am loking for a similar proof of the expression for tangens of an angle sum, i.e. tan(α+β)= ..., without using trigonometrics but only its definition. If have tried using the the tangens projection in a unit circle, but failed.
But sir, you can simply take ange of 1 square that is just 45 Then sice it is streced out exactly same kength, 45 is divided into 2 that is 22.5 And again, because of 3rd square, so it is 11.25 All added up should give us 78.75 degree
I propose a mixed solution, Complex + Geometrical: Once you add the first two angles (green and blue) by multiplication of (1+i)(2+i)=1+3i you can see that this intermediate value can be represented by a point depicted by the same initial three squares flipped 90°, i.e. something symmetrical along the 45° blue segment, and the third angle (red) taken from the horizontal base is exactly the same angle taken from the new vertical base. This means the third angle (red) is the complement to the sum of the first two (green and blue), i.e. the sum is 90°. It seems you went first summing the second and third angles just to deceive us... 🙂
I see all these people in the comments going through all these mathematically accurate answers and I’m just sitting here like “Well, they’re 3 parts of the same right angle, they prolly add up to 90”
If you get comfy working with Direction Vectors in place of angles, and learn how to compute the direction sum d1 (+) d2, this can be solved computationally (using Dataflow Geometry). Answer: [ 0, 1 ] (1/4 rev or 90 deg)
@@kindiakmath From a computational perspective, the dot product v1•v2 does not involve trig sin, cos, it's simple arithmetic: v1.x*v2.x + v1.y*v2.y. The "other" definition of the dot product: || v1 || || v2 || cos(ø) depends on a trig function and knowing angle ø. In pure math, though, you see the 2nd definition more often....there's a post-trig theory dependency. The point is, using the first definition, you don't need to obtain any angle values or trig functions to solve the problem.
Or ... (few lines) ... a = red angle; b = green angle; c = blue angle global shape: 3 identical squares (side of each square is 1) • angle a = sin⁻¹(1/√(1 + 3²)) = 18.43° • angle b = sin⁻¹(1/√(1 + 2²)) = 26.56° • angle c = 45° (obviously) a + b + c = 90° 🙂
The use of complex numbers is of course a great way of "thinking outside the box" but one must not forget that the argument properties of complex numbers (in multiplication) are proven through trigonometry. So is it REALLY a different method than the trig approach?
I did direct calculation, atan(1) + atan(1/2) + atan(1/3) = 90°. I made the assumption that it was kosher to use a basic calculator that had the three basic trig functions and their inverses, I wasn't about to derive arctangents.
I do like Numberphile's answer, though I would have used the tan method. However I just looked at it and thought 'yup, they all add up to 90'. Also, the 'complex' method is just that, in fact 'too complex'!
I solved it using the reflection. I assumed those rays as light rays and so when they reflect back, angle of incidence = angle of reflection, and all 3 rays will meet after 12 squares at one single point, and so you can compare it with original rays from they emerged and you'd find those 2 smaller angles adds to bigger angle. Or give angles name x, y and z at the point where they meet, say y and z are smaller angles, so solve for y and z from that triangle (basic algebra), and you get y + z = x, And x is 45 degrees (it's obvious from the 1st ray and square) And so x + y + z = x + x = 90 degrees.
@@kindiakmath like the same as if you morphed the plane so that i -> i+j, if you get what i mean, rotate and enlarge thhe 1:2 triangle and place its base on the hypotenuse of the 1:1 triangle, if you get it you'll realize it's far, far more elegant than your solutions imo
Let the angles be a, b, c, their respective points A, B, C and the point of connection P. Let AX be the diagonal of the first square (from left to right). It can be shown that the triangle AXP is similar to the triangle BCP - common 135° angle and all sides are proportional with ratio sqrt(2). Therefore, we know the angle XÂP is equal to b. Because AX is a diagonal, a + b = 45°. So, a + b + c = 90°. Nice one! What about extending the squares to the left, can we show that the limit sum of the angles is divergent?
@@kindiakmathCool!! How did they do it?? I thought of comparing it term by term to the harmonic series. Starting with arctan(1/n) >= 1/n, and then going to x >= tan(x) for (0,1]. Which would be a cute proof, but i dont remember inequalities that well to make it formal :/Cool!! How did they do it?? I thought of comparing it term by term to the harmonic series. Starting with arctan(1/n) >= 1/n, and then going to x >= tan(x) for (0,1]. Which would be a cute proof, but i dont remember inequalities that well to make it formal :/
@@alicwz5515 i think we can try arctan(x) > (pi/4) * x and then sum over x = 1, 1/2, 1/3, ... on the RHS, which diverges; by the comparison test, LHS diverges too
Can ya help me with this, im struggling with calc? I know the standard limit lim h->0 ((x^h-1)/h) = ln(x) But why is, lim h->0 (((x+h)^h-1)/h) = ln(x) Do we use partial limit, or is there any similar way to prove this, as before? Also, when is partial limit valid?
One approach i have is that denoting f(t) = x^t, the given limit is equivalent to f'(0) = ln(x). In a similar fashion, defining g(t) = (x+t)^t, the limit in question is g'(0). Using differentiation techniques should get us to g'(0) = ln(x).
i'll be fairly honest and would love to ask you what's your definition of natural logarithm... its pretty much impossible to explain since everything has tons of equivalent formations...
@@pragyanpranay3681 i would define logarithm as the inverse of exponential! i've defined the exponential function in a different video on my channel here: th-cam.com/video/p_v3K0OQg08/w-d-xo.html there's a linked document for a more technical discussion too.
This is a geometry problem, hence no trig allowed. Note that this was solved in the 1970’s by a Russian high schooler using a nice construction - see Martin Gardner’s math column in Scientific American around 1970.
The 1rst solution is understandable by 12 years children. As teacher, it's my favourite The 3rd solution is the most elegant. As mathematician, it's my favourite
This is trivially soluble with the tangent of sum of three angles formula setting tan A -> a, tan B -> 1/2, tan C -> 1/3 and taking the limit from below as a -> 1
Tana=1/2 Tanb=1/3 Tan(a+b)=(1/2+1/3)/(1-1/2x1/3) which is equal to 1 and tan45 is 1 so angle a and b sums to 45. Third one is 45 from tanc=1 so 45+45=90
The solution came up as 90° with just mind math. This was my thought process: 45° + tan⁻¹ (½) + tan⁻¹ (⅓) Let tan⁻¹ (½) be 'tan 𝑎 = ½' and tan⁻¹ (⅓) be 'tan 𝑏 = ⅓'. And let β = tan⁻¹ (½) + tan⁻¹ (⅓) Using a trig identity, β = tan⁻¹[(½+⅓)/(1−½×⅓)] ⇒ β = tan⁻¹[1] ⇒ β = 45° So, the sum of all angles, θ = 45° + β = 45° + 45° = 90° An Indian highschool student can solve this easily.
I'm not sure what is hard about solving this at all. Since all sides are the same length and since Tan is a ratio, you can use 1 (or any whole number) for that length. ArcTan of 1/1 is 45 (blue angle). ArcTan of 1/2 (green angle) is 26.565. ArcTan of 1/3 (red angle) is 18.435. So they add up to 90.
The way I attempted was similar to the elegant complex number solution, only instead of complex numbers, I used vectors and cosθ to find out the angles 😄
I looked at the image for about 5 seconds and said the answer was 90° (no math or anything) I just saw the answer in my head. My teachers hated me in HS for the reason I dint show my work. I told them, I dont know how and I dont know why, what I know is thats the answer and the work hurts my head... I also have a minor case of Dyscalculia
@@kindiakmath I edited my comment because I sound kinda smug. I draw diagonals on the first and second square, from bottom to top then top to bottom. The two diagonals and the third square's diagonal you will get something like an N. You obviously have the 45 degree angle in the first square with the diagonal line but you also get the 2x1 right triangle that is formed by the diagonal in the first square and half of the diagonal of the second square. All the angles are in that single corner like in the video and so the sum is 90 degrees.
@@kindiakmath I use corresponding angle property.where the blue region was 3x , green was 2x and red was x. where x equals 15 degree. 6*15 = 90 Degree.
** I am wrong** I solved this in about 10 seconds, Saying "Square" as opposed to rectangles indicates 4 sides of equal lengths, thus the blue angle is 45*, then red was 1/3 that angle (15*, so green must be 30*) total sum of 90*. It helps that I deal with angles every day by trade as a carpenter, so you kind of have to be able to extrapolate these things naturally on the fly! great question though! Edit: quick aside I wrote this comment before watching the solutions, why is it necessary to draw an additional 3 squares above the first set to prove these angles? you only need two facts to prove this. Let's call the angles as such and the Bases of these angles alike and same. Blue will be C, Green is B, Red is A. All three "triangles" share the same Height but have different Base lengths which are proportional to angle/base C (Blue). "Triangle B(Green)" has a base that is 2x the length of the blue base. "triangle A(Red)" has a base that is 3 times the length of the blue triangle. Finding the red angle being proportional to the blue angle is just 1/3 of 45, so 15*, Green angle being 2/3 of 45* and proportion to the blue angle makes it 30*, 15+30+45. I don't believe I brought any incorrect assumptions to the table with this line of thought, but please tell me if I am wrong!
A solution which is as simple or clearer than the first solution uses Pythagoras theorem only, to find sum of the two smaller angles, to which the blue angle is added to complete the answer, as follows: Instead of drawing three extra squares above the given ones, add three extra squares BELOW the given squares. Join the end of the red line to the right-base point of the second square you added. Join this same point to the top-most point connecting all the three colored lines, making four lines at that point (as a result). Notice that the two lines you just created, together with the red line (given), form an isosceles triangle having respective proportional lengths of; sqrt(5): sqrt(5): sqrt(10). And corresponding proportional angles of; 45: 45: 90 or ratio 1:1:2 per Pythagoras theorem. Clearly, one of the angles is the sum of the red and green angles we seek their sum. They are corresponding angles. Hence the RED and the GREEN angles sum to 45 degrees (and this is the main clincher of this method). Since the BLUE angle is trivially half of 90 degrees or its bisection, it readily follows that: angle( RED+GREEN) + angle(BLUE) = 45 + 45 = 90 degrees. Nothing else was used but Pythagoras theorem. Note: Diagram could not be drawn here, hence my wordy description. But it's as simple or simpler than the first solution when sketched as described.
I think I did this intuitively. I solved it very quickly by eye by applying my knowledge of that theorem. The 45 degrees is obvious. The base lengths give me the rest of the info that I need to estimate the other angles. I solved it, but I didn't prove it. I would hope that anyone who did grade six math would get this without much trouble.
im not vey good at explanations or typing things out but i can tell you there is a much more easy way to find out the answer. if only one side of the right triangle grows by factor of one there is amuch more simplistic wway to solve. not sure how to write it out with a keyboard but its very similar to the second answer you gave
my first thought was "45 + arctg 1/2 + arctg 1/3"
and then what is arctg 1/2 + arctg 1/3? :)
@@kindiakmath We haven't studied arc functions so i thought it is the final answer, now i searched a bit and counted: its arctg 1 and it is 45, 45+45=90 wow thats correct
@@futo8343 awesome!!
@@kindiakmath solved this using complex no my ans was argument of (I+i)(2+i)(3+i} which is one
@@futo8343arctan(x) + arctan(y) = arctan ( (x+y)/(1-xy) )
Derived from tangent angle addition formula
These solutions are great, especially the complex value one. I love it when complex numbers are used to make problems so much simpler than the real-value approach. I used very standard methods (cosine rule and double angle formula) to get the same result, but not as elegant as these three.
Yes!! It's actually crazy how complex numbers can be used to solve so many problems.
Professor Stankova (numberphile youtube channel ) says when she solved this problem at a school level without knowledge of trigonometry and complex numbers
The blue angle (γ) is obviously 45°. For the red and green angle, I used the cotangent sum formula,
cot(α + β) = (cot α cot β -1)/(cot β + cot α).
cot(α + β) = (3*2 - 1)/(2+3) = 1, so α + β = 45°. Adding this to the blue angle we get
α + β + γ = 90° .
6:50 arg(5*(1+i)^2) =2*arg(1+i) avoids an unnecessary multiplication step.
that is actually genius
I first solved it with trig but I enjoyed the complex solution the most. It's really clever. I instantly knew where you were going when you summed the arguments. Great, now I have to look for other cool ways to apply complex numbers to other seemingly unrelated stuff
the complex one was my favorite strat!
Oh that complex solution is really nice! Super straight forward and generalizable.
I just used complex numbers for this, they make angle additions really easy
thats the beauty!
FAST solution:
Draw three triangels with the coordinates:
the blue angle : 0,0 2,0 2,2
the green angle: 0,0 2,2 2,3
the red angle : 0,0 2,3 0,3
done
I'm a simple person. Knowing that the angles will not change with proportion, I drew it in CAD. Measured the angles with a precision of .001 degrees and summed them. 90 degrees. Took about 2 minutes. CAD is great at solving practical geometric problems, I do it all the time.
"Let no one ignorant of geometry enter."
or just google "atan(1) + atan(1/2) + atan(1/3) in degrees" took 10 seconds
@@misterkite How much beauty of maths have you gleaned from this googling exercise? I mean, you didn't even have to solve this problem in the first place at all. Sometimes you need an answer quickly, but this wasn't a practical computation required to build a shed or something, it was a maths problem for the sake of solving it with maths. These 10 seconds have been traded to eternity for nothing in exchange. You could've found the answer in the comments even faster, or drag forward the video xport slider. It would be as satisfying to one's heart and useful for exercising one's mind as googling it up. Which is, fair to say, not at all. :(
@@cykkm you had to know to use atan and that the result would be in radians. My way required just as much math.
@@misterkite Well, let's just agree to disagree about what mathematics is. It's a talk much longer than that which YT comments are suitable for. Sounds fine? :-)
Never would have thought of trying for a solution using complex numbers. Fascinating. My mind always goes to trig functions first.
My favorite solution out of the three! And trig is the most intuitive approach
I see all these fancy solutions and im over like "stack da angles on top of each other, oh look its 90°!"
It's always illuminating to see a problem being solved in different ways!
I think it’s Polya who talked about how it’s more important to solve one problem on ten ways rather than solve ten different problems
@@kindiakmathInteresting thought and approach to a problem
@@TranquilSeaOfMath thanks!
I wonder if anybody here can solve the puzzles I just posted on my channel. One of them is practically unsolvable even by the most powerful computer in the world.
I just used pythagoras and sin law in my head, pulled out a calculator and got an answer very close to 90. I rounded, so my guess is it's probably 90 but i'm too lazy to verify
I thought that this guy probably has at least hundred of thousands of subs already until I take a look and it was only 1k. You are massively underrated. Keep up the great work!
thanks mate!
Arctan(1)+arctan(1÷2)+arctan(1÷3)
I went straight for the complex numbers - a fantastically simple way to solve this tricky problem :-)
180 deg from inverse tan (1) + inverse tan (2) + inverse tan (3). My calculations were in degrees mode. (I’m typing this before watching video, my answer was based off the thumbnail)
My answer was based on the fact that y=tan(D)x+b is the formula for a linear equation with slope D degrees (graphing calculator in degrees mode)
The complex solution is a two-liner. Brillant.
I drew a square with 2x longer side that the normal square and using similar traingles you could see the anlges go into the right angle of the square
Thank u very much for making this video. As a 9th grader I immediately came up with the trigonometric solution. However i was blown away that it was solvable with pure algebra. I would've just plugged it into a calculator and would've thought that was it. Sadly, I didn't quite have the background knowledge to understand the 3rd explanation (idk what an argument is, I just know that i=squrt(-1)). Btw is my English good
Good enough for me:)
@@kindiakmaththx 👍
i feel like by knowing two legs of a triangle there is only one possibibity for the angles = you can figure them out with a simple formula
Before you started. 90 degrees.
There are always more complicated ways of solving a problem, but in engineering you must find the most simple.
Basic trig will do the job.
So to get things out of the way, blue is certainly 45*. Square/2.
If I'm allowed to use Sin Cos Tan, then simply apply tan^-1 to each triangle using the squares as algebraic representations of their length ratios.
I'm gonna look for more before watching.
Second conceived way is maybe you can link it to gradients, which can be thought of in some way as a form of angles, then convert it.
gradients from this POV is directly linked to arctangents! love the attempt~
Nice. I did it on paper using only Pythagoras and the knowledge about triangles with sides of same ratios.
4:25 Apparently this is only true if rs
good catch!!
There is an even easier visual proof, for which you'll only need graph paper. First draw a 1 x 1 square with the origin in its lower left corner, then extend its diagonal (doubling it towards the upper right) and draw a (proportionally larger) 2 x 1 rectangle on that extended diagonal, then triple the diagonal of that rectangle and build an even larger 3 x 1 rectangle on that one. The diagonal of that rectangle will coincide with the Y-axis, and the three angles will together make up the 90° angle between the X- and Y-axis,.
I used the Pythagorean theorem and solved for the angles for each triangle with the law of cosines
1 minute head maths gave me solution 2. Solution 3 with complex numbers was really cool too. Solution 1 while learnable for an 8th grader, becomes less apparent as you learn more maths.
totally agree with your analysis
The complex number solution was spectacular. Definitely my favorite. Using the sum or arctangent was my first thought though
i've the same feelings as you!
Or simple area reasoning: Given the 3 identical squares. The blue line halves 1 square = area 1/2 square. The green line halves 2 squares, area = 1 square. The red line halves 3 squares, area = 1.5 squares. The total area of the 3 differently formed triangles is 3 squares. The angle that would give you an area of 3 squares is of course 90 degrees, so thus the sum of the angles must be 90 degrees.
the complex one was a very elegant solution
The red triangle and the green triangle are similarity triangle therefore the sum of red and green angle -> x+y + 135=180 then x+y would be 45 and obviously blue angle is 45 therefore x+y+z = 90
Bruh, arctg function rules.
I would have never derived the pictorial solution and although I am comfortable with complex numbers I did not see that third solution at all. Very, very good. As far as I am concerned this kind of analysis puts the "applied" into mathematics.
yes! i'm a fan of the appliedness too
For some reason I went: (1 + 2/3 + 1/3) * 45° = 90°
I’m not sure why I correlated the areas of the rectangles like that or if it even makes sense but that’s how arrived at 90
45+22.5+15=82.5 degrees is my instinctive reaction.
The Blue triangle is an isoceles with a right angle, meaning the remaining 90 degrees are divided by 2 for an even 45 degrees.
If this is X x Y, then the green triangle is X x Y/2, making it 22.5 degrees.
Likewise, Red is X x Y/3, making it 15 degrees.
I'd like to see why I'm wrong though.
Had I a calculator, I'd work each out as Tan(^-1)(Opposite/Adjacent)
I thought you'd have 3 different answers from seperate methods, but this shows just how comprehensive we can get with maths! I'm glad to learn something every day, and equally glad I don't have to solve these equations outside of videogames anymore!
Amazingly arctangents have some kind of nice properties even without calculators :)
My thought process is just arctan(1/3) + arctan (1/2) + arctan(1) which is the m of the functions (y = mx + b)
if you use radians then 180/pi * arctan() of all these angles
My first thought was, that it should be propably 90degrees. Inv-tan of 1/3 + inv-tan of 1/2 + 45 gives you 90 degrees!
But I had to use my calculator. The complex number solution was impressive, we hadn‘t learned that in school. Very cool!
Complex numbers have crazy uses
The 3rd solution is awesome!
Great thinking, but i really wanna know how can someone develop a thought on doing the construction in the first steps..
Btw, i did proceed using trigonometry, by assuming a side length x, and then using cot(theta).
Whoever came up with the first solution originally is a true genius. :) Perhaps the nice properties of squares could motivate such a strategy.
That's a great question... one should always ask how people come with certain solutions. Often, proofs obscure the thought process which makes seem certain solutions as magic. But there is always some reasoning behind it... Let's think about it!
@@academyofuselessideas i think my first attempt at this was the complex numbers one; but it felt too high-schooley for elementary school kids
Is there a general solution where we have green and blue angles with distance between them, to calculate distance between blue and red if we have red angle?
Probably some trigonometry
I really enjoyed the complex numbers proof. There is one caveat though. An argument of a complex number is the angle between the number and the real axis, plus any integer multiplication of 2π / 360°.
For example, what is the the angle between i⁴ and the real axis? i⁴ = 1, therefore it's 0. On thr other hand the angle between i and the real axis is π/2 / 90°, and the sum it 4 times is 2π / 360°
In other words the sum of the angles can be any number of the form
90° + k*360°
We can ignore negative k's, since we are adding three positive angles. Morover, each of the angles is less than 90° angles in a right triangle, which aren't the rightangle itself.
0 = 90° + 360° > 270°
Therefore we indeed reach the conclusion that the angle is 90°
Yes!! Really appreciate the contextualisation of the argument, pun intended.
@@kindiakmath ❤️❤️
Also, ty for the video!
0:16 I did this: I draw that. First is the third one, I put everything to "1 centimeter". So first is arcsin(1/sqrt2). The second is only diff. is arcsin(1/sqrt5) and the third arcsin(1/sqrt10). Add together is 45+26,57+18,43 is 90 degrees. :D It was 4 minutes. So now I will continue the video.
0:26 Ok. Three solutions within 8 minutes than I failed. XD
arc sin (1/(sqrt(x))) --> First x number: diagonal lenght. (1^2+1^2)=>sqrt2, second: (1^2+2^2)=>sqrt5, third (1^2+3^2)=>sqrt10.
You'll find the sum of arcsin's useful! To rigorously justify that the angle sum is 90 degrees
I just did method 2 with an inverse trig calculator. As a designer/engineer, just round to whatever decimal places fits your usage. Out of laziness, I got 89 degrees. Good math shown, but problems like these are solved way more easily in the real world
this is really simple - it is simple common sense and does not require any degrees.
I really like the complex one, it’s much simpler and from a computing perspective would require the least number of instructions.
Yes!!
I calculated the ratio of the 3 rectangles and found their arc-tangents to get 3 angles. I then added those and got 90°
I would like to see a general formula for n number of squares…not sure if that’s possible?
That’s a really good question; my friend investigated this question on his own time too
It's Σ arctan(1/n) with n going from 1 to the number of squares. The sigma is how you write down sums in maths.
@@TheMrDisastrous thanks! I did have an inkling that it would be that
As a funny side note, since we know geometrically that the three angles add up to pi/2, we can use the tan approach to get the triple angle sum formula for tangent, which looks like this: tan(a+b+c) = (A+B+C - ABC)/(1-AB-BC-AC), where A = tan(a), B = tan(b), C = tan(c).
Now we use A = 1, B = 1/2, C = 1/3.
Since we know that the angles add up to pi/2, then tan(a+b+c) is infinity, so the denominator (1-AB-BC-AC) must be 0. Plugging in the values of A, B, and C gives us 1 - 1/2 - 1/3 - 1/6 = 0, or 1/2 + 1/3 + 1/6 = 1.
So we proved that 1/2 + 1/3 + 1/6 = 1, using geometry and trigonometry.
Love this
I don't really get it? I mean I liked the first solution but just tipping arctan(1)+arctan(1/2)+arctan(1/3) in my calculator gives me the 90 degrees solution? I expected it to have rounding errors or something but works fine. Is there some rule about not doing that that wasn't mentioned? Or something which makes doing that invalid?
the complex method is absolutely amazing
same!!
There is also a 4rth method. Knowing that this is a math problem, the answer will most likely be a common angle like 30,45,90, 180 etc instead of an abscure angle like 105, and just looking at the problem it looks like the angles add up to 90, so i say its 90 degrees. I guess its not a proof, but i was able to get the answer right
thats a handy heuristic nonetheless!
90 degrres answer
the angles of the there triangles are 45 , 26.565 and 18.435
There likely a more elegant way but using trig
Let the side of the square = 1
Hence the angle of the triangle
formed by the blue line = 45 degrees since it is the diagonal of the square
The side of the triangle formed by the green line are 1, 2 and 2.23607
2.23607 (comes from using pythaogrean theorem)
Since there are three sides SSS using the law of cosine yields 18.435 degrees
The side of the triangle formed by the organe line are 1,3 and 3.16628
Since there are three sides and using the law of cosine yields 26.565 degrees
Hence the sum of the total angles = 45 + 18.435 + 26.565 = 90 degrees (answer)
the challenge here is to compute without calculators :)
let one side equal to x. such that the area of the largest triangle is 3/2 x^2 which is the same as 3/2x times root 10x^2 times sin theta of the largest triangle (1/2 ab sin C). then you can get a equation for the first theta in x then you just repeat for the rest. then just input a value for x. it doesn't matter which value you put since all of them equal to 90
if we assume they are squares, and that each side is equal to 'x' then wouldn't it just be tan(1/1)+tan(1/2)+tan(1/3)? (all tans are inverse)
i was literally thinking about this irl yesterday while looking at the tiles in my floor
im still convinced it is 82.5 degrees even after watching until the end.
In my head the blue is obviously 45°.
The green angle is half of the blue angle because it is in a 2 to 1 side ratio.
And the red angle is a third of the blue angle because it is in a 3 to 1 side ratio.
Could you explain why it isnt just 45° + 22.5° + 15°?
The challenge is computing the angle to be 22.5; trig functions are a little more complicated than just halving the angle
You would be right IF these were circles, with those diagonals as radii. Then the angles would add up to be 82.5 deg. But each square diagonal is not simply proportional to each other when you increase the side of the square. The ratios have a sin in them as well.
@@kindiakmath I'm with Beppo here I did the same thing. and I checked it, or at least I thought I did by looking up the tangent angle ratio (opposite over adjacent sides correct?) which gave me 22.5 and 15 as well. So i got 82.5 as well. I'm VERY rusty on my Trig so going from distant memory. I'm not saying I'm right, just that I don't logically understand where I went wrong. Is the tangent of the middle angle NOT 22.5 degrees? Scratching my head here.
@@jamesedward9306 definitely not 22.5 haha
@@kindiakmath Ya found a trig table from an old textbook ( not sure what website gave me 22.5) but i found the correct one. 26 and change 18 and change for the smallest one. Add it all up 90. Counterintuitive, but it is what it is. thanks for responding.
My solution is just the complex number solution, but without any computation. Simply draw a circle (with radius equal to the diagonal of the square) around the bottom left point of the box. Extend the first angle to the circle. Construct a line from this intersection to the intersection of the top right corner of the first box and the circle. Draw the second angle from the center of the circle to this new intersection. Finally you can extend the vertical line on the left edge of the box to the circle to see that the sum of the angles is pi/2.
I'm not entirely sure what I did, but I just looked at it, thought about some random stuff for a second, and got the answer, definitely felt a lot simpler than the solutions in the video, but I also couldn't put my method into words so it's not a very good solution xD
Same
I think it is better to evaluate:
A = tan(x+y+z) = ( tan(x+y) + tan(z)) / (1-tan(x+y)*tan(z) with tan(x) = 1, tan(y) = 1/2 and tan(z)=1/3 →
tan(x+y) = ( tan(x) + tan(y)) / (1-tan(x)*tan(y) = ( 1 + 1/2) / ( 1- 1*1/2) = 3 →
A = (3 + 1/3)/( 1-3*1/3) = (10/3) / 0 → A tend towards infinity → *x+y+z = π/2*
this works too! i like it since it uses fewer arctangents
I like "picture"'s solution most. I love those proofs without using any algebra.
I am loking for a similar proof of the expression for tangens of an angle sum, i.e. tan(α+β)= ..., without using trigonometrics but only its definition. If have tried using the the tangens projection in a unit circle, but failed.
But sir, you can simply take ange of 1 square that is just 45
Then sice it is streced out exactly same kength, 45 is divided into 2 that is 22.5
And again, because of 3rd square, so it is 11.25
All added up should give us 78.75 degree
That assumes a clean cut relationship between the sides and the angles (ie trig) which just isn’t as clear as we would wish it to be :/
I propose a mixed solution, Complex + Geometrical: Once you add the first two angles (green and blue) by multiplication of (1+i)(2+i)=1+3i you can see that this intermediate value can be represented by a point depicted by the same initial three squares flipped 90°, i.e. something symmetrical along the 45° blue segment, and the third angle (red) taken from the horizontal base is exactly the same angle taken from the new vertical base. This means the third angle (red) is the complement to the sum of the first two (green and blue), i.e. the sum is 90°.
It seems you went first summing the second and third angles just to deceive us... 🙂
I see all these people in the comments going through all these mathematically accurate answers and I’m just sitting here like
“Well, they’re 3 parts of the same right angle, they prolly add up to 90”
If you get comfy working with Direction Vectors in place of angles, and learn how to compute the direction sum d1 (+) d2, this can be solved computationally (using Dataflow Geometry). Answer: [ 0, 1 ] (1/4 rev or 90 deg)
thats interesting! i'm contemplating a dot product approach but that ultimately boils down to taking cosines, still.
@@kindiakmath From a computational perspective, the dot product v1•v2 does not involve trig sin, cos, it's simple arithmetic: v1.x*v2.x + v1.y*v2.y. The "other" definition of the dot product: || v1 || || v2 || cos(ø) depends on a trig function and knowing angle ø. In pure math, though, you see the 2nd definition more often....there's a post-trig theory dependency. The point is, using the first definition, you don't need to obtain any angle values or trig functions to solve the problem.
I wonder if Wildberger's rational trigonometry can also be used to solve this problem. In other words, use spread instead of angle.
ngl i would LOVE a rat trig solution to this
I was watching at 1:57. I look away for a bit and i see 4:15.
HELP
Or ... (few lines) ...
a = red angle; b = green angle; c = blue angle
global shape: 3 identical squares (side of each square is 1)
• angle a = sin⁻¹(1/√(1 + 3²)) = 18.43°
• angle b = sin⁻¹(1/√(1 + 2²)) = 26.56°
• angle c = 45° (obviously)
a + b + c = 90°
🙂
possible to find exact expressions for arcsin? so that your proof is airtight haha
The use of complex numbers is of course a great way of "thinking outside the box" but one must not forget that the argument properties of complex numbers (in multiplication) are proven through trigonometry. So is it REALLY a different method than the trig approach?
Haha implementation vs execution I guess
I did direct calculation, atan(1) + atan(1/2) + atan(1/3) = 90°. I made the assumption that it was kosher to use a basic calculator that had the three basic trig functions and their inverses, I wasn't about to derive arctangents.
oh ye young whipper snappers and fancy gizmo gadgets (says the one who used python to make the video in the first place)...
@@kindiakmath
Yes!
I do like Numberphile's answer, though I would have used the tan method. However I just looked at it and thought 'yup, they all add up to 90'. Also, the 'complex' method is just that, in fact 'too complex'!
My engineer ass looking at the video thumbnail like: pi/4 + atan(1/2) + atan(1/3)
I tried cos(a+b) and it ends up with 1/sqrt(2), implying me that the angle sum of the red and green angle is 45 degrees. It accidently work
That’s actually pretty ingenious imo, meaning that a dot product approach is possible too
My first thought, blue= 45°, Green looks like ⅔ of blue and Red looks like ⅓ of blue, so R+G+B=90°, idk if what I did makes sense but I was correct :)
The sort of stuff I was doing at O level in 1968 - too easy as the Aussies would say :)
i just used a calculator with trigonometry... i realy liked the first solution, pure geometrical problem solving.
I solved it using the reflection. I assumed those rays as light rays and so when they reflect back, angle of incidence = angle of reflection, and all 3 rays will meet after 12 squares at one single point, and so you can compare it with original rays from they emerged and you'd find those 2 smaller angles adds to bigger angle. Or give angles name x, y and z at the point where they meet, say y and z are smaller angles, so solve for y and z from that triangle (basic algebra), and you get y + z = x,
And x is 45 degrees (it's obvious from the 1st ray and square)
And so x + y + z = x + x = 90 degrees.
reflection is the key
From Philosophical , Physical and Mathematical Perspective.@@ПавелАртемьев-п2и
i did a construction where multiplying the vectors i+j and 2i+j resulted in i+3j and so the angle of the 2 biggest = 90 - the smallest
Interesting! What kind of multiplication are you referring to?
@@kindiakmath like the same as if you morphed the plane so that i -> i+j, if you get what i mean, rotate and enlarge thhe 1:2 triangle and place its base on the hypotenuse of the 1:1 triangle, if you get it you'll realize it's far, far more elegant than your solutions imo
@@ytkerfuffles6429 wow got it; never thought of it that way though! a rotational approach haha
Let the angles be a, b, c, their respective points A, B, C and the point of connection P. Let AX be the diagonal of the first square (from left to right). It can be shown that the triangle AXP is similar to the triangle BCP - common 135° angle and all sides are proportional with ratio sqrt(2). Therefore, we know the angle XÂP is equal to b. Because AX is a diagonal, a + b = 45°. So, a + b + c = 90°.
Nice one!
What about extending the squares to the left, can we show that the limit sum of the angles is divergent?
haha my friend actually proved that it diverges!! crazy insight you got there
@@kindiakmathCool!! How did they do it?? I thought of comparing it term by term to the harmonic series. Starting with arctan(1/n) >= 1/n, and then going to x >= tan(x) for (0,1]. Which would be a cute proof, but i dont remember inequalities that well to make it formal :/Cool!! How did they do it?? I thought of comparing it term by term to the harmonic series. Starting with arctan(1/n) >= 1/n, and then going to x >= tan(x) for (0,1]. Which would be a cute proof, but i dont remember inequalities that well to make it formal :/
@@alicwz5515 i think we can try arctan(x) > (pi/4) * x and then sum over x = 1, 1/2, 1/3, ... on the RHS, which diverges; by the comparison test, LHS diverges too
Can ya help me with this, im struggling with calc?
I know the standard limit
lim h->0 ((x^h-1)/h) = ln(x)
But why is,
lim h->0 (((x+h)^h-1)/h) = ln(x)
Do we use partial limit, or is there any similar way to prove this, as before? Also, when is partial limit valid?
One approach i have is that denoting f(t) = x^t, the given limit is equivalent to f'(0) = ln(x). In a similar fashion, defining g(t) = (x+t)^t, the limit in question is g'(0). Using differentiation techniques should get us to g'(0) = ln(x).
i'll be fairly honest and would love to ask you what's your definition of natural logarithm... its pretty much impossible to explain since everything has tons of equivalent formations...
@@pragyanpranay3681 i would define logarithm as the inverse of exponential! i've defined the exponential function in a different video on my channel here: th-cam.com/video/p_v3K0OQg08/w-d-xo.html there's a linked document for a more technical discussion too.
@@kindiakmath alright i will see to it once!
This is a geometry problem, hence no trig allowed. Note that this was solved in the 1970’s by a Russian high schooler using a nice construction - see Martin Gardner’s math column in Scientific American around 1970.
neat context
I dunno about favorite, but this has just reaffirmed my seething hatred for trig.
The 1rst solution is understandable by 12 years children. As teacher, it's my favourite
The 3rd solution is the most elegant. As mathematician, it's my favourite
100% agree w you! And the 2nd solution is the straightforward but tedious approach
This is trivially soluble with the tangent of sum of three angles formula setting tan A -> a, tan B -> 1/2, tan C -> 1/3 and taking the limit from below as a -> 1
left as an exercise :))
Tana=1/2 Tanb=1/3
Tan(a+b)=(1/2+1/3)/(1-1/2x1/3) which is equal to 1 and tan45 is 1 so angle a and b sums to 45. Third one is 45 from tanc=1 so 45+45=90
Why does it stump everyone?
You could see that it was 90 degrees just by looking at it.
Moving the triangles in your mind.
The solution came up as 90° with just mind math.
This was my thought process:
45° + tan⁻¹ (½) + tan⁻¹ (⅓)
Let tan⁻¹ (½) be 'tan 𝑎 = ½' and tan⁻¹ (⅓) be 'tan 𝑏 = ⅓'.
And let β = tan⁻¹ (½) + tan⁻¹ (⅓)
Using a trig identity,
β = tan⁻¹[(½+⅓)/(1−½×⅓)]
⇒ β = tan⁻¹[1]
⇒ β = 45°
So, the sum of all angles,
θ = 45° + β
= 45° + 45°
= 90°
An Indian highschool student can solve this easily.
I'm not sure what is hard about solving this at all. Since all sides are the same length and since Tan is a ratio, you can use 1 (or any whole number) for that length. ArcTan of 1/1 is 45 (blue angle). ArcTan of 1/2 (green angle) is 26.565. ArcTan of 1/3 (red angle) is 18.435. So they add up to 90.
Any formulaes to combine the arctanfents without approximations?
You can just use the tangent, right? You can just say that one square has a length and width of 1, so its tan(1/3)+tan(1/2)+tan(1)
i think you mean arctangent
90 degrees seems about right to me. Of course I'm a genius and good at solving math problems and puzzles.
The way I attempted was similar to the elegant complex number solution, only instead of complex numbers, I used vectors and cosθ to find out the angles 😄
i love the vectors approach too!
Bluee is 45⁰ but how many degrees are green and red?
Very nice solutions !
thank you!
I looked at the image for about 5 seconds and said the answer was 90° (no math or anything) I just saw the answer in my head. My teachers hated me in HS for the reason I dint show my work. I told them, I dont know how and I dont know why, what I know is thats the answer and the work hurts my head... I also have a minor case of Dyscalculia
The sum is 90 degrees? I used a geometric solution that is kinda similar to what is in the video.
what was your approach? i'm curious to know
@@kindiakmath I edited my comment because I sound kinda smug.
I draw diagonals on the first and second square, from bottom to top then top to bottom. The two diagonals and the third square's diagonal you will get something like an N. You obviously have the 45 degree angle in the first square with the diagonal line but you also get the 2x1 right triangle that is formed by the diagonal in the first square and half of the diagonal of the second square. All the angles are in that single corner like in the video and so the sum is 90 degrees.
@@rosverlegaspo6752 woah cool solution mate! the "N" idea is a really clever one!
@@kindiakmath I use corresponding angle property.where the blue region was 3x , green was 2x and red was x. where x equals 15 degree. 6*15 = 90 Degree.
** I am wrong**
I solved this in about 10 seconds, Saying "Square" as opposed to rectangles indicates 4 sides of equal lengths, thus the blue angle is 45*, then red was 1/3 that angle (15*, so green must be 30*) total sum of 90*. It helps that I deal with angles every day by trade as a carpenter, so you kind of have to be able to extrapolate these things naturally on the fly! great question though!
Edit:
quick aside I wrote this comment before watching the solutions, why is it necessary to draw an additional 3 squares above the first set to prove these angles? you only need two facts to prove this.
Let's call the angles as such and the Bases of these angles alike and same. Blue will be C, Green is B, Red is A. All three "triangles" share the same Height but have different Base lengths which are proportional to angle/base C (Blue). "Triangle B(Green)" has a base that is 2x the length of the blue base. "triangle A(Red)" has a base that is 3 times the length of the blue triangle. Finding the red angle being proportional to the blue angle is just 1/3 of 45, so 15*, Green angle being 2/3 of 45* and proportion to the blue angle makes it 30*, 15+30+45. I don't believe I brought any incorrect assumptions to the table with this line of thought, but please tell me if I am wrong!
In fact the green angle unfortunately isn’t even 30 degrees
@@kindiakmath I see where I went wrong in my line of thought lol! The answer for N squares would be interesting.
A solution which is as simple or clearer than the first solution uses Pythagoras theorem only, to find sum of the two smaller angles, to which the blue angle is added to complete the answer, as follows:
Instead of drawing three extra squares above the given ones, add three extra squares BELOW the given squares. Join the end of the red line to the right-base point of the second square you added. Join this same point to the top-most point connecting all the three colored lines, making four lines at that point (as a result).
Notice that the two lines you just created, together with the red line (given), form an isosceles triangle having respective proportional lengths of; sqrt(5): sqrt(5): sqrt(10). And corresponding proportional angles of; 45: 45: 90 or ratio 1:1:2 per Pythagoras theorem. Clearly, one of the angles is the sum of the red and green angles we seek their sum. They are corresponding angles.
Hence the RED and the GREEN angles sum to 45 degrees (and this is the main clincher of this method). Since the BLUE angle is trivially half of 90 degrees or its bisection, it readily follows that:
angle( RED+GREEN) + angle(BLUE) = 45 + 45 = 90 degrees. Nothing else was used but Pythagoras theorem.
Note: Diagram could not be drawn here, hence my wordy description. But it's as simple or simpler than the first solution when sketched as described.
I think I did this intuitively. I solved it very quickly by eye by applying my knowledge of that theorem. The 45 degrees is obvious. The base lengths give me the rest of the info that I need to estimate the other angles. I solved it, but I didn't prove it. I would hope that anyone who did grade six math would get this without much trouble.
im not vey good at explanations or typing things out but i can tell you there is a much more easy way to find out the answer. if only one side of the right triangle grows by factor of one there is amuch more simplistic wway to solve. not sure how to write it out with a keyboard but its very similar to the second answer you gave
There are some interesting identities involving tan() and atan() that are rarely taught in class. Here are some of my favorites:
tan(x) = -i + 2*i / ( 1 + exp(2*i*x))
tan(x) = 1/tan(x) - 2/tan(2*x)
tan(x/2) = 1/sin(x) - 1/tan(x)
tan(x/2 + pi/4) = tan(x) + 1/cos(x)
tan(x/2)^2 = -1 + 2 / (1 + cos(x))
tan(x) * tan(pi/2 - x) = 1
tan(2*x) = 2 / (1/tan(x) - tan(x))
tan(3*x) = tan(x) * tan(pi/3 - x) * tan(pi/3 + x)
atan(x) = 2 * atan( x / (1 + sqrt(1 + x^2)) )
atan(x) = (1/2) * atan( 2*x / (1 - x^2) ) -1 < x < 1
atan(x) = (1/3) * atan(x*(x^2 - 3) / (3*x^2 - 1)) -1/sqrt(3) < x < 1/sqrt(3)
atan(x) = i/2 * ln( (i + x) / (i - x) )
atan(x) + atan(y) = atan( (x + y) / (1 - x*y) ) mod pi
atan(x) - atan(y) = atan( (x - y) / (1 + x*y) ) mod pi
atan(1/2) = atan(1/3) + atan(1/7)
pi/4 = atan(1/2) + atan(1/3)
pi/4 = 2*atan(1/3) + atan(1/7)
pi = atan(1) + atan(2) + atan(3)
tan(10°) = tan(20°) * tan(30°) * tan(40°)
sqrt(3) = tan(20°) * tan(40°) * tan(80°)
tan(1°/2) = 1 / ( sin(1°) + sin(2°) + ... + sin(179°) )
tan(1°) = 1 / ( sin(2°) + sin(4°) + ... + sin(178°) )
There are several ways to solve the problem using various identities from this list.
woah i'm tempted to give students these identities as exercises in trig...
The arg (argument ) ??
Never heard of this.
now that i look at Olympiad questions i realize the importance of trignometary
It has its uses :)