Sweet! Since my homeworks are less and less "hit" and more and more "miss", I’ll just go ahead and send what I’ve found there. Homeworks were great while they lasted but people moved on. It is what it is.
In my humble opinion, I think eliminating theta was a mistake. Having all the parameters as a function of the angle gives great insight on what is happening (yes, I'm a physicist), the equations are simpler, and the solutions are beautiful: y=1+(sin 2θ)/(1+sin θ) Optimum when (cos θ)^2=sin θ ymax=1+2/φ (φ-1)^(3/2) xmax=(φ-1)+(φ-1)^(1/2) where φ is the golden ratio.
@@sergiokorochinsky49 I'm thinking set up an equation with y as a function of theta as opposed to x and then plug in different values of theta until you get a maximum. I don't see why that wouldn't work?
@@leif1075 Oh, so you want to plot y(θ) and determine the maximum graphically. That method has a name: "brute force". It always works, it gives you a fairly good approximation with a reasonable amount of work, but you will never get the exact result. Why wouldn't you use derivatives?
@@casc3601 The heighest angle isn't obvious at least. Let us call the heighest point of the square (a,b). If you tilt the square further such that the heighest point is closer to the left square you get a slightly smaller difference of the heights b-1, but also a slightly smaller distance (a-x) which you have to divide by. So it is not exactly clear. If the slope is maximal for x=sqrt(2) (which I would guess to without doing the math), it is the correct solution though. Noteably other comments have pointed out that the x value in the video is negative, hence doesn't solve the problem. Then x=sqrt(2) is actually the solution (that however doesn't mean the slope has to be maximal for that value of x for similar reasons!)
@@freewilly1337 thanks, I went back and what I got for the angle at the bottom of the tilted square was about 49.5 degrees, so slightly off. No idea where the negative x value came from
I tried this without taking it in terms of x, and got the same result. Ended up just reducing the trigonometric equations down, which turns into just finding the max value of y = 1 + sin2θ/(cosθ + 1), not too bad at all.
@@mahdipourahmad3995 Only in the initial choice for theta and the choice to remain in terms of theta for the calculus step rather than convert to being in terms of x.
I would call 7:57 a mistake, or at least further clarification is needed. A plus/minus sign is really needed here, and then a minus/plus sign should be used for cosine. It changes depending on whether 0
I agree. Others also pointed out that resulting value of x is negative, which makes no sense whatsoever given the construction and coordinates being used. So you found the bug, it seems to me.
Another observation: plugging the decimal value of Michaels x (-0.168) into his y formula actually does not yield the the expected value of roughly 1.6 and so his x plugged into y(x) actually does not result in the expression he gave as final solution. It does work though, if you switch 2x² -2 to 2-x² in the y-formula.
@@petersievert6830 Ah, yes. You are right. I thought you meant to check both cases, and my point was just that one of them is relatively easy to reject.
It is interesting to note that the optimal value of x found in the video, x = 1/2 * (-1 + sqrt(5) - sqrt(2sqrt(5) - 2)), is negative! (it is about -0.168) so the optimal situation does not look like the diagram on the board, in that not everything lives in the first quadrant.
If you rotate the left square and trace the upper most point, you will get an arc. The latch is dropping onto this arc. So the touching point will be on the right half of the arc.
@@AndreasHontzia I don't understand the reply. The line passes through the highest vertex of the rotated square. At 8:15, the y-value at that vertex is 'x' -- which is said to be negative. If the y-value at that vertex is negative, how is the y-intercept of the line at a positive y-value? The Desmos page in the post by MathyJaphy shows 'x' at about 1.4.
I used Pythagoras and symmetry instead of trigonometry. Bottom corner = (a,0), Left corner = (0,b), x = a + b, a² + b² = 1. This gives y = 1 + (2ab)/(b+1). Maximum y when b² + b - 1 = 0.
Same here. After deriving it gets to solving s³ + 2s² - 1 = 0 with s = sin (theta) and thus s = (sqrt(5) - 1 )/2. Fun little fact: cos²(theta) = sin(theta).
I got to this equation y = 1 + sin(2 theta) / (1 + sin(theta)) by similarity of the two right triangles above the right square. short leg of smaller one = c + s - 1 short leg of larger one = y - 1 long leg of smaller one = s + 1 long leg of larger one = c + s + 1 with s = sin(theta) and c = cos(theta). Thus (y - 1) = (c + s - 1) (c + s + 1) / (s+1) = [(c + s)² - 1] / (s+1) = 2 c s / (s+1) = sin (2 theta) / (sin(theta) + 1)
(√5-1)/2 is 1/φ, or 2/(1+√5), c is √s. Its useful to know the relationships between φ, φ^2, 1/φ, 1/φ^2, etc. For instance s^2+c^2 is 1/φ^2+1/φ, which is (2-φ) + (φ-1) = 1, quick sanity check on s and c.
The problem when eliminating theta is, that the quadratic equation gives two solutions. This reflects that x is the same for theta as for 90°-theta. Choosing the solution with sin(theta)>1/2, i.e. theta>45°, is IMHO the wrong one, as for that solution the "top" point of the first square is to the left compared to the (1-theta)1. It's probably better to keep theta as free variable.
If someone might be interested in what difference is made to the maximum of y by rotating the square from where the diagonal is vertical, at the vertical, I got y = (4 + √ 2 )/(2 + √ 2 ) = 1.5858 The answer shown, 1 + √ (10√ 5 - 22 ) = 1.6006 The difference is about 0.9 % I guessed it would be more than that. As the square rotates, the lowest point slides away and the length of the whole figure declines. Eventually, I could see that from the treatment in the video.
There IS a fun way to compute this! If you only work with the trigonometry, you get y = 1 + sin (2θ)/(1+ sin θ). The derivative of that, expressed as a polynomial in sin θ = a, has (a^2 + a - 1) as a factor, with the other factor discardable. You get y when you set a as the inverse of the golden ratio. Pretty elegant.
I found it simpler to obtain y = f(theta). I obtained y = 1 + (sin(2 * theta)/(sin(theta) + 1)) and after taking the derivative and solving for sin(theta) = phi - 1 implies cos(theta) = sqrt(phi - 1), where phi is the golden ratio, implies y(max) = 1 + (2 * (phi - 1)^(3/2)/phi). Note: All your coordinates can now be expressed in terms of phi to accurately graph the problem.
That was the long way for sure. I envisioned the rotated square as a cam and went right to the maximumization equation looking at the change in Y with the change of the base length X. X is bounded from 2 to 2+sqrt2. Basic engineering application of this math.
Nice ! But i uesd theta(no elimination) and it simplified everything and gave the equation t^3+2t^2=1 where t= sin(theta) and solving it gives t=-1,(√(5)-1)/2
Let the angle of rotation be θ then noting height and width of rotating square is Sinθ + Cosθ (let S = Sin(θ) etc ) Similar triangle gives us the required height is Y = 1 + 2SC/(S+1). Differentiating and equating to zero gives us the cubic S^3 +2S^2 - 1 = 0 Wolfram alpha gives us S = (√5 - 1)/2 . C is just found using pythagoras. Then carefully substitute back into expression for Y (tedious but straight forward)..
Interesting that x is 1/phi-sqrt(1/phi) if I calculated that right? But that's negative (about -0.168)? Perhaps the wrong root was taken for the quartic equation.
I agree. The overall result may or may not be correct. Michael suffers from pure mathematician's disease - he despises the real world. If he checked his results along the way, he would have noticed this.
Simpler to express in trig form. s,c are sin, cos . ss = sin^2 etc. y-1=2sc/(s+1) differentiating and set to zero gives.scc=(1+s)(cc-ss) Or sss+2ss-1=0 Which factorises as (ss+s-1)(s+1)=0 Or s=1-ss=cc Hence c=sqrt(s) Also s = golden ratio -1= ¥-1 0.608... y = 1 +( 2(¥-1)^1.5)÷¥ =~1.600566 the same as the video. That ¥ should be phi x =~ 1.404 ie close to 1.414 which it would be with the left square at 45°.
The problem becomes easier, when the diagram is mirrored along the y-axis and shifted such that the non-tilted square has its lower left corner at (0,0). If we denote with x the height of the adjacent corner of the tilted square, we can find y(x) = 2*x*sqrt((1-x)/(1+x)) + 1. This has an optimum at x = (sqrt(5) - 1)/2 with gives an optimal value for y of (sqrt(5)-1)*sqrt(sqrt(5)-2) + 1.
I found it simpler to set the origin at the upper right point, where the diagonal line hits the flat square (with x increasing to left for simplicity). This give 3 collinear points: (0,0), (1+sina,sina+cosa-1), (sina+cosa+1,y-1); where 0≦a≦π/4. Working out the slope, y=1+(sina+cosa-1)(sina+cosa+1)/(sina+1). I know that can be simplified, but at this point I did a numerical optimization for y, yielding y≅1.600566212, to the limit of my spreadsheet's precision. This matches the Michael's solution, again to the limit of my spreadsheet's precision.
Hi Michael. Here is a nice Diophantine Equation: (Japan 2020, Junior finals question 3) Find all tuples of positive integers (a,b,c) such that Lcm(a,b,c)=(ab+bc+ca)/4. Thanks.
Thanks Michael for this beautiful problem and all contributor's posts. I focused on the fact that someones found a rational value (8/5) while others found a different answer related with the golden ratio. Both answers are beautiful and I know I'm speaking about a difference less about 1e-3, but it seems that both didn't compute the same thing. The former maximize the max of the slope of the upper line. Unfortunately, the max slope of the upper line doesn't give exactly the max value for Y. The slope can be for instance a bit larger, and Y could be a bit smaller if the right square is closer. So we first have the max of the slope for Y = 8/5 (theta about 36,87° and tan(upper line vs horizon)=1/4) and then the max of Y = 1,600566... (theta about 38,17° and sin(theta)=(sqrt(5)-1) / 2).
y = 1 + (2sinθcosθ)/(cosθ + 1)= 1 + (sin2θ)/(cosθ + 1). considering dy/dθ = 0 leads to a trigonometric equation which boils down to t^4 + t² - 1 = 0 where t = tan θ. Optimal t = sqrt((sqrt(5)-1)/2). drawing a right-angled triangle and plugging in values for sinθ and cosθ [try not to use Mathematica], gives the result y = 1+ sqrt(10sqrt(5)-22)
managed this one without trig. Let A be the x-coordinate of the bottom corner of the left square, and let B be the y-coordinate of the left corner. Then A^2 + B^2 = 1. The line passes through (A, A+B) and (1+A+B, 1). The y-coordinate of the line between these two points simplifies to y = 1 + A - (B-1)^2/A. Note that dB/dA = -A/B when you use the chain rule. Setting dy/dA = 0 and simplifying a bit, you get (B-1)^2/A^2 + 2(B-1)/B + 1 = 0. It is easier to get everything in terms of B here, so you end up with the cubic B^3 - 2B + 1 = 0, which factors as (B-1)(B^2 + B - 1) = 0. Golden ratio time! B = (sqrt5 - 1)/2, and since B^2 = 1 - B, you get A^2 = B, so A = sqrt(B). From there, y = 1 + A - (B - 1)^2/A = 1 - (B^4 - B)/A = 1 - A(B^3 - 1) = sqrt((sqrt5 - 1)/2)(3 - sqrt5) + 1, which is equivalent to Michael's answer
Using theta you'll find y -1 = 2(cos θ)(sin θ)/(sin θ + 1). Max for sin θ = φ (the golden ratio! Wow!) so cos θ = φ^(1/2). Wolfram Alpha is not necessary.
For anyone who is confused by your answer and differs from Michael's, one simple way to check if you're on the right track is this: what is the y-coordinate of the top most corner of the tilted square? If your answer is not dependent on the tilt angle or some other variable you chose to use, and is instead just sqrt(2), you are not on the right track.
Oh, pulling out the ancient calculus memories at 12:00. The derivative of the square root of a function is the derivative of the function divided by 2 times the square root of the function. That is, d sqrt(f)/dx = (df/dx) / (2 sqrt(f)) Thus, d(sqrt(2 - x^2))/dx has f = 2 - x^2 Therefore, df/dx = -2x (derivative of 2 is 0, derivative of -x^2 is -2x) Thus, (df/dx) / (2 sqrt(f)) = -2x / 2 sqrt(2 - x^2) = -x / sqrt(2 - x^2) = -x(2 - x^2)^-0.5
Actually, results in your way gives 1.60, while computating the intersection with the y axis of the straight passing trough A(√2/2 √2) and B(1+√2,1) gives 3-√2 (≈1.59). Since the maximum y is defined for the highest y that A can assume, the process should be correct
I get the same end result for y (≈1.601) with a slightly different approach, however, your x (≈-0.168) must be some wrong root since plugged into y the result is ≈0.399.
I did not read all the comments but your x solving from the derivative is negative, and that is impossible. The problem is that you made a mistake when you used the plus sign for sin(theta) in set of minus sign. Following with this change of sign, making the derivative and solving for x, the value is aprox. 1.4041865 and then the value of y is 1.6005
If we stick to parametrizing by theta, we get the top point as (cos(θ), cos(θ) + sin(θ)) and the upper right point as (cos(θ)+sin(θ)+1,1). Then, the slope of the line segment is (cos(θ)+sin(θ)-1)/(-sin(θ)-1), so y = (cos(θ)+sin(θ)-1)/(sin(θ)+1) * (cos(θ) + sin(θ) + 1) + 1 from the upper right point and the point (0,y). This simplifies as y = 2sin(θ)cos(θ)/(sin(θ)+1) + 1, and its derivative is 2[ ( (cos^2(θ) - sin^2(θ))(sin(θ) + 1) - sin(θ)cos(θ)cos(θ) )/(sin(θ)+1)^2 ]. y is maximized only when this derivative is 0, so we need that the numerator of this fraction is 0. Using cos^2(θ) = 1 - sin^2(θ) and letting s = sin(θ) for brevity, we get (1 - 2s^2)(s+1) - s(1 - s^2) = 0, or s - 2s^3 + 1 - 2s^2 - s + s^3 = 0, or -s^3 + 1 - 2s^2 = 0, or s^3 + 2s^2 - 1 = 0. This factors as (s+1)(s^2 + s - 1) = 0; since s = sin(θ) is between 0 and 1 as θ is between 0 and π/2, the only solution is s = (sqrt(5) - 1)/2. Curiously, for this angle, cos^2(θ) = (sqrt(5) - 1)/2 and 1/(sin(θ) + 1) = (sqrt(5) - 1)/2, so we get that y = (3 - sqrt(5)) sqrt( (sqrt(5) - 1)/2 ) + 1. sqrt( (sqrt(5) - 1)/2 ) doesn't seem to have a nice form, so that's our answer. This is equivalent to the answer Dr. Penn gets, so that's good. tl;dr: θ = sin^(-1)(1/ϕ)
I find this answer in term of theta more apropriate, as we would naturally measured the rotation of the first square instead of the measure of x (length which doesn't have a real physical signification for the square)
As other suggested - this can be solved without calculus: for max Y of left-most line, we want the top-most line to have the (acute) angle as steep as possible (while still going thru the top-most vertex of the tilted square) AND to stretch to the left as far as possible the left-most line (while still having the left-most line go thru the left-most vertex of the tilted square). Fortunately these two requirements do not contradict and they both are achieved when the tilted square has been rotated 45° - i.e. when one of its diagonals is orthogonal to the lower-most line. Then the (x,y) values of the upper-most vertex of the tilted square can be found - which enables finding the expression of the upper-most line. Inserting in this expression the x-value of the left-most line (it is also found from the achieved "optimally tilted" square) gives the desired Max Y.
I defined x as the x-coordinate of the bottom vertex of the tilted square and used only the pythagorean theorem to determine the coordinates of the important points, no angles involved. That way I got y = ((4x^2 - 4x^3) / (1 + x))^1/2 + 1. To determine the maximum point I knew the +1 and the ^1/2 dont change the x-value of the maximum so I simplified y to (4x^2 - 4x^3)/(1 + x) which is easy to differentiate. The maximum is at x = (-1 + sqrt(5))/4 and I was too lazy to calculate y. The fun part was that I constructed the whole situation in geogebra and also displayed the target function in the same graph which allowed simple error control.
I dont understand why, as others suggest, that you cannot use x=sqrt2 due to the maximum of y being related to the maximum of x, since the largest x can be is the diagonal of the rotated square of sides 1, also sin(theta) then equals cos(theta) and theta =45 with those being known values. Unless the objective is solely to solve using first term diff calculus or without the sides of the square being known
In the expression for x, isn't the sign wrong in front of √(2√5 - 2)? I get an evaluation for x of -0.168 with that expression; but a more reasonable value of 1.404 with the sign changed to give: x = ½ (√5 - 1 + √(2√5 - 2) ) That also is then in agreement with the results below of a maximum when cos² θ = sin θ so that sin θ = φ - 1. Also I believe the identity sin θ + cos θ = √2 sin (θ + ¼π) (well known to the electrical engineers at least) would come in quite handy in simplifying the calculations.
Why didn't you just say that the diagonal of the sqaure is root(2) an then solvetheproblem via similar triangles. First triangle from xE(root(2)/2,1) and the other xE(0,1)?
It is not correct, because, you cannot be sure that diagonals of square are paralllel or pependicular to the axes. Your way to solve this problem is only a special case. I know this pain. Long time ago I was preparing to the IMO. Almost every task was about including every case of solution. Skipping any case was painful for your score xd
I thought this as well, but the diagram is misleading in that the height is actually greatest when you tilt the square slightly to the left. If you look at how he did it, and then plug in sint+cost=x (where t is theta) you will find that the value of Y is actually greatest when theta is about 38 degrees, which would mean that you can not just assume that the height is sqrt2 because the square is tilted at a different angle than expected.
I did this a bit differently, using the angle of tilt (alpha) of the almost-vertical diagonal from the vertical as the parameter. Then if y = 1 + z, and c = cos(alpha), s = sin(alpha), I got: z = (2c^2 - 1)/(1 + sqrt(2)(c - s)/2) Unwilling to solve dz/dalpha = 0, I just computed z for a range of alpha and found the maximum at alpha = 0.119 to be z = 0.600566, which is sqrt(10sqrt(5) - 22). I notice that my equation for z looks a lot like yours if I put 2c^2 = x^2, s^2 = 1 - c^2 = 1 - x^2/2, s = (1 - x^2/2)^(1/2) This didn't quite give your equation, then I realised I needed to use s = - (1 - x^2/2)^(1/2). Now I get exactly your equation. :) By the way: 2c^2 - 1 = cos(2.alpha) = sin (pi/2 - 2.alpha), therefore if I put pi/2 - 2.alpha = 2.theta then alpha = pi/4 - theta. If I substitute for alpha in my equation for z I get Sergio's equation. :))
Taking the other quadratic solution essentially exchanges the values for sin(theta) and cos(theta) and represents the square tilted in the other direction (with its "vertical" diagonal tilted to the left instead of to the right) but (I think) ultimately yields the same relationship between x and y. Note that both these tilted squares yield the same value for x, but the two theta values are complementary. But this probably should have been explained.
This equation is not so difficult to solve by hand After squaring both sides you can check for rational roots or get rig of repeated root with differentiation Remaing quartic can be easily solved by rewriting it as difference of two squares and then as a product of two quadratics After solving polynomial equation check the solution because squaring both sides may produce extraneous solutions
Two points on a line (root 2 /2 , root 2) and (1 + root 2 , 1), so solve for y intercept. Calculus is only needed to demonstrate this is the max via perturbation.
Sir i think you have calculated minimum value of y instead of maximum you should check it by second derivative test And also value of "y" you have found i wrong if we plug in
I believe you are correct! If you plot y as a function of x, there's a minimum at the value found above, but that's the only critical point. This function actually has a max at one of the end points of its natural domain which seem to be x = 1 and x = sqrt(2). The latter looks to give you the max.
Solved it just using linear equations and pythagoras. y is the y-axis intercept of the linear function running through the two points where the square touches the slope. The resulting function is quite ugly, so I also turned to wolfram alpha to help with the derivative and setting it equal to zero. Getting the same solution felt satisfactory nonetheless.
The negative result would make the y-value for every point on the rotated square less than 1. The straight line would then cross the y-axis at a value less than 1 -- which is lower than when theta = 0, so there is no maximum y-value to be found when theta is between 0 and -pi.
shouldn't the left square hit the line at ( sqrt2/2, sqrt 2) and the right square hit the line at (1+sqrt2, 1)? If so, we have two points to make a line with, and we then just solve for when x=0.
So I did it without using angles, by specifying the x value of the corner of the rotating square, then plugging in values to find the maximum y value to be around 1.6005236512374004 while Michael gives a value of about 1.600566212001556. So yes, you can get pretty close. The advantage of using calculus is you can PROVE the answer the maximum, and show it is an absolute maximum and not a local maximum.
Not really, without calculus is not possible to prove that the max value you find is the actual biggest one ( mainly because you would have to check literaly every single angle, but there are infinitly many, so......)
@@joaopedrobmenezes2977 no you can maximize y value by construction the two squares are at 45 degree to each other - some logic can demonstrate it is a local max.
I love your videos in general but this solution is a bad one, simply because I haven't heard of a math competition where in the solution you can plug in wolfram alpha. I would recommend a little simpler solution which is based on labeling the same angle theta. Label the big trapezoid (whose base y we need to calculate) ABCD by starting from the bottom left corner. Label the topmost point of the tilted square as P. Then in terms of theta you can easily calculate the distances from P to the lines AB and BC. Let F be a point on the line segment AD such that FC is perpendicular to AD. Again you can easily calculate FC and angle MCD in terms of theta. In the end you will find that y = AD = 1 + sin (2 * theta) / (sin (theta) + 1). Calculating the first derivative and its roots will bring you to a cubic equation sin^3 (theta) + 2 * sin^2 (theta) - 1 = 0 which is very easy to solve because it transforms to t^3 + 2*t^2 - 1 = 0 which has one integer root and only one positive root. From here you can find sin (theta), cos (theta), sin (2 * theta) and therefore AD.
Consider that after some angle < π/4 any further increase in the angle (and thus in x) decreases the slope more than it is lifted meanwhile, which leads to a net decrease in y.
Let rotated squire touches axes at (a,0) and (0,b), a^2+b^2 = 1 Points (0,y) (b,a+b), (a+b+1,1) are on the same line => (y-1)/(a+b+1) = (a+b-1)/(a+1) y = 1 + ((a+b)^2-1)/(a+1) = 1 + 2ab/(a+1) = 1 + 2a sqrt((1-a)/(1+a)) y' = 0 => a^2+a-1 = 0 => a = (sqrt(5)-1)/2 - golden ratio cojugate ymax = 1 + sqrt(10sqrt(5)-22)
My intuition was to go for y1 being the full height of the diagonal (sqrt(2)) but that was not the correct solution. It is much simpler to solve though. :-)
Is the Y maximum value when the diagonal of the square is perpendicular to the slope? I'm pretty sure it is, but please enlighten me if it's not. Then you can crunch the numbers from there.
I get the same result after 2 hrs of trying, using phytgorean theorem (initially i tried to use trigonometry but i get lost) .... i even had to recall derivatives of products. Thanks for the exercise! Edit: you get new subscriber
Can you just make the intersection of the first square and the line (sqrt2/2, sqrt2) to find the slope of the line then use that to find out what it equals at x=0? The tallest that intersection of the first square and the line can be will be the diagonal of the square from two corners right?
I think you mean we put the left hand square so that its diagonal is perpendicular. That's the answer, right? No... You can twist the left hand square clockwise a little bit and it will lift the slope of the line we are trying to find. It's the slope we need to find - not the height of the left hand square.
@@samstep4279 No it won't shift he slope. If you rotate the left hand square clockwise, the bottom corner will go clockwise to, thus lowering the slope, if the left hand square is rotated that the diagonal is vertical, that point where it touches the slope is the most upper point it can get. It is a square that you rotate and not just a vertical line.
@@lennyganado3975 First of all I am not an flat earth believer, far from it ! Second, I am an engineer and thus I looked at this in a realistic way of problem solving (that what is often times too big of a problem for mathematicians being able to do) ! Third, the designed problem used in this video is in a closed frame ! All the outlines off thegeometrical figures are all of the same width and colour, thus making them all geometrical objects laying against each other and in one large geometrical object, making this a realistic engineering problem than anything else where the law of physics rule above anything else (a concept that is apparently unknown to mathematicians) ! So use lines of different width, colour, etc if you want to indicate what is fixed or adjustable in any design or drawing (something that apparently is also unknown to mathematicians) ! Fourth, if you start with insults, you must really be a compulsive and impulsive pric* I bid you a good day sir.
@@seriously1184 It's a pity that an engineer such as yourself missed the point that as you rotate the first square a little clockwise from its highest point, not only does the bottom corner move left, but so does the rightmost corner and so _the second square moves left in its entirety._ That movement left of the "hinge" of the slope increases the slope initially, and that is why the value of y will in fact increase a little as the first square rotates clockwise from its highest point.
For those of you that did the calculus in theta- what's the optimal value for theta? My intuition says 45 degrees but I would expect the answer to be nicer if it were the case
Hi, Michael! I love your videos because you demonstrate clear solutions to challenging problems, skip the tedious too-obvious steps, and admit your mistakes. BUT I don't understand how at 7:30 you got sin theta = 0.5(x+sqrt(2-x^2)). According to the quadratic formula, shouldn't it be 0.5(2x +- sqrt(4x^2 -4•2(x^2 -1))) = x +- sqrt(2-x^2) ? I see that my result CAN'T be right because the sqrt must be tiny and sin theta must be close to 0.5x, but I don't understand what I'm doing wrong. Help!
i just assumed the square was rotated by 45 degrees and then i made two points and wrote the straight line. This could be solved more simply with euler's calculus of variations imo c
Excuse me if I'm being dumb, but if the question was asking what the maximum of y was, wouldn't you just set the theta equal to 45 degrees as that would maximize the vertical height of the tilted square, which would then maximize the height of y? Like, there IS a way to solve this numerically, unless I'm being dumb. Edit: Forgot to mention that since it gives side lengths, you can solve it. Not sure if I need to mention that, but still. Edit 2: Every response to this keeps on getting more confusing. Feel free to keep commenting, I think eventually someone will figure it out, because clearly I can't.
Because that is not the actual solution. The 45 degree angle would be the maximum solution for x, but it is not for y. (even if somewhat close to it of course) By tilting the square just a little bit clockwise, x gets just a little tiny bit smaller, but y gets higher, because the top of the square is moving to the right more than downwards. Of couse you can do this numerically as well, but that's a whole other story. So basically, you could do a search with the angle looking for a value of theta between 0 and 45 degrees and do a binomial search until you reached whatever your degree of error is. But numerical solutions are in general no exact solutions, which is what Michael is looking for here.
@@petersievert6830 but y is the height, and if y is on the line from the top right corner of the regular square to the top of the tilted square, then to maximize the angle of that line to the horizontal, you would just tilt the tilted square to 45 degrees... Also, x would be maximized at angles larger than 45 and approaching 90, as the most the distance can be is that gap plus 1.
@@nateexists6292 Be awere that the X-coordinate of the top of the square is not fix. Moving the top to the right while keeping almost the same height, enlarges y.
@@BrianStewart126 I am not sure, what you are referring to exactly. The numerical solution? If there are multiple local maxima between x=1 and x=sqrt(2) (which corresponds to my theta ranging from 0° to 45° , then it could obviously fail, sure. But it seems intuitive for me, that when you tilt the square clockwise from x=sqrt(2) to x=1, then y would rise for a certain range and fall afterwards. (If that assumption is wrong, I am in troubel, sure.) btw I was imagining some kind of search, that makes do without the derivative, rather the engineer's way suggesting some kind of binomial search and possibly not even very effcient and not the analytical way, bec if you got that far to get the derivative dy/dx resp dy/dtheta , then you surely are able to proceed without the need for a numerical solution) Apart from that, I only wanted to give some intution for the idea, why x=sqrt(2), thus the bottom angle=45° is not the solution. That was not meant as a recipe to find the actual solution at all.
your expression for y should be (sin(2\theta))/(1+sin\theta)+1. Differentiate, then when setting the numerator to zero, convert it to an equation in sin\theta only. You should get a factorizable cubic equation in sin\theta.
Three friends set up a meeting in front of the Nilton Santos stadium to watch a a soccer match. They agreed that each one should arrive in a moment chosen between 3 p.m. and 4 p.m. and that none of them will wait more than 30 minutes for too much, within the stipulated time. What is the probability that the three friends will meet between 3 p.m. and 4 p.m.? Answer : 1/2
I did this problem completely differently and ended up with 1+ [Sqrt(2)(6-2Sqrt(5))(Sqrt(Sqrt(5) -1))]/4 which calculates to the exact same number that Dr. Penn got. Anyone know how to prove that those two answers are equal??
Yeah, the x value you get for this is negative = - 0.168117.... So not sure if this is a valid answer or an extraneous solution from the quartic equation.
I don't get the question. The maximum? What maximum? You say later that you turn the tilted square, and it makes it pretty clear, but I really think it should have been included in the question explicitly.
The question asks for the maximum value of y. That means over all values of anything that can vary, and the diagram has only one thing that can vary, the angle of the first square. No ambiguity to my mind.
Hi everyone! I made a google form to suggest problems. It will be linked in every video in the description from now on.
Love your content...
❤️❤️❤️
Sweet! Since my homeworks are less and less "hit" and more and more "miss", I’ll just go ahead and send what I’ve found there.
Homeworks were great while they lasted but people moved on. It is what it is.
Can we submit anything, even on topics like geometry, graph theory or combinatorics?
Anything is fair game!
@@goodplacetostop2973 I miss your homeworks (Miss as I want to see them again)
In my humble opinion, I think eliminating theta was a mistake.
Having all the parameters as a function of the angle gives great insight on what is happening (yes, I'm a physicist), the equations are simpler, and the solutions are beautiful:
y=1+(sin 2θ)/(1+sin θ)
Optimum when (cos θ)^2=sin θ
ymax=1+2/φ (φ-1)^(3/2)
xmax=(φ-1)+(φ-1)^(1/2)
where φ is the golden ratio.
You can solve this without calculus right?
@@leif1075, I doubt it. But I'm all ears... What do you propose?
@@sergiokorochinsky49 I'm thinking set up an equation with y as a function of theta as opposed to x and then plug in different values of theta until you get a maximum. I don't see why that wouldn't work?
@@leif1075 Oh, so you want to plot y(θ) and determine the maximum graphically. That method has a name: "brute force". It always works, it gives you a fairly good approximation with a reasonable amount of work, but you will never get the exact result. Why wouldn't you use derivatives?
@@sergiokorochinsky49 you could get the exact result sometimes. In this case i think you would.
1:06 My body is telling me x=√2 but my mind is telling me it’s not that easy. Let’s see how wrong I am...
14:54 Good Place To Stop
Yeah, intuition can be misleading when it comes to rotating squares.
Yep. Calculated the case for x=sqrt2 and got to y=3-sqrt2 but that's not correct lol
Why isn't that correct? X=√2 would give you the largest x+1 and the largest angle for the top line, no?
@@casc3601 The heighest angle isn't obvious at least. Let us call the heighest point of the square (a,b). If you tilt the square further such that the heighest point is closer to the left square you get a slightly smaller difference of the heights b-1, but also a slightly smaller distance (a-x) which you have to divide by. So it is not exactly clear. If the slope is maximal for x=sqrt(2) (which I would guess to without doing the math), it is the correct solution though.
Noteably other comments have pointed out that the x value in the video is negative, hence doesn't solve the problem. Then x=sqrt(2) is actually the solution (that however doesn't mean the slope has to be maximal for that value of x for similar reasons!)
@@freewilly1337 thanks, I went back and what I got for the angle at the bottom of the tilted square was about 49.5 degrees, so slightly off. No idea where the negative x value came from
Fun to see the golden ratio arising in an optimization problem (the cosecant of the tilt angle theta is the golden ratio here).
I tried this without taking it in terms of x, and got the same result. Ended up just reducing the trigonometric equations down, which turns into just finding the max value of y = 1 + sin2θ/(cosθ + 1), not too bad at all.
I think in the denominator it should be sinx+1
@@imacup5047 Depends what you take θ to be, of course. I was taking it as the angle Michael marked as π/2 - θ, which of course would change things.
@@TheProloe hmm
where exactly did ur solution divert from Michael's?
@@mahdipourahmad3995 Only in the initial choice for theta and the choice to remain in terms of theta for the calculus step rather than convert to being in terms of x.
I would call 7:57 a mistake, or at least further clarification is needed. A plus/minus sign is really needed here, and then a minus/plus sign should be used for cosine. It changes depending on whether 0
I agree. Others also pointed out that resulting value of x is negative, which makes no sense whatsoever given the construction and coordinates being used. So you found the bug, it seems to me.
But it's easy to see that theta
@@TJStellmach That's the point Tim. Michael choose the sign in a way, that theta is indeed not
Another observation: plugging the decimal value of Michaels x (-0.168) into his y formula actually does not yield the the expected value of roughly 1.6 and so his x plugged into y(x) actually does not result in the expression he gave as final solution. It does work though, if you switch 2x² -2 to 2-x² in the y-formula.
@@petersievert6830 Ah, yes. You are right. I thought you meant to check both cases, and my point was just that one of them is relatively easy to reject.
It is interesting to note that the optimal value of x found in the video, x = 1/2 * (-1 + sqrt(5) - sqrt(2sqrt(5) - 2)), is negative! (it is about -0.168) so the optimal situation does not look like the diagram on the board, in that not everything lives in the first quadrant.
Who cares
@@TheMartian11 If x is negative, then how can the left vertex of the rotated square be on the y-axis?
i get x ~ +1.4
If you rotate the left square and trace the upper most point, you will get an arc. The latch is dropping onto this arc. So the touching point will be on the right half of the arc.
@@AndreasHontzia I don't understand the reply. The line passes through the highest vertex of the rotated square. At 8:15, the y-value at that vertex is 'x' -- which is said to be negative. If the y-value at that vertex is negative, how is the y-intercept of the line at a positive y-value?
The Desmos page in the post by MathyJaphy
shows 'x' at about 1.4.
I used Pythagoras and symmetry instead of trigonometry. Bottom corner = (a,0), Left corner = (0,b), x = a + b, a² + b² = 1. This gives y = 1 + (2ab)/(b+1). Maximum y when b² + b - 1 = 0.
y can be expressed as 1+sin(2 theta)/(1+siin(theta)), solve: sin(theta)=(root(5)-1)/2
Correct! This gives optimal value for y = 1 + sqrt(-22 + 10 sqrt(5))
Same here. After deriving it gets to solving s³ + 2s² - 1 = 0 with s = sin (theta) and thus s = (sqrt(5) - 1 )/2.
Fun little fact: cos²(theta) = sin(theta).
I got to this equation y = 1 + sin(2 theta) / (1 + sin(theta)) by similarity of the two right triangles above the right square.
short leg of smaller one = c + s - 1
short leg of larger one = y - 1
long leg of smaller one = s + 1
long leg of larger one = c + s + 1
with s = sin(theta) and c = cos(theta).
Thus (y - 1) = (c + s - 1) (c + s + 1) / (s+1) = [(c + s)² - 1] / (s+1) = 2 c s / (s+1)
= sin (2 theta) / (sin(theta) + 1)
(√5-1)/2 is 1/φ, or 2/(1+√5), c is √s. Its useful to know the relationships between φ, φ^2, 1/φ, 1/φ^2, etc. For instance s^2+c^2 is 1/φ^2+1/φ, which is (2-φ) + (φ-1) = 1, quick sanity check on s and c.
The problem when eliminating theta is, that the quadratic equation gives two solutions. This reflects that x is the same for theta as for 90°-theta. Choosing the solution with sin(theta)>1/2, i.e. theta>45°, is IMHO the wrong one, as for that solution the "top" point of the first square is to the left compared to the (1-theta)1.
It's probably better to keep theta as free variable.
If someone might be interested in what difference is made to the maximum of y
by rotating the square from where the diagonal is vertical,
at the vertical, I got y = (4 + √ 2 )/(2 + √ 2 ) = 1.5858
The answer shown, 1 + √ (10√ 5 - 22 ) = 1.6006
The difference is about 0.9 %
I guessed it would be more than that. As the square rotates, the lowest point slides away and the length of the whole figure declines. Eventually, I could see that from the treatment in the video.
There IS a fun way to compute this!
If you only work with the trigonometry, you get y = 1 + sin (2θ)/(1+ sin θ).
The derivative of that, expressed as a polynomial in sin θ = a, has (a^2 + a - 1) as a factor, with the other factor discardable.
You get y when you set a as the inverse of the golden ratio. Pretty elegant.
I did it this way as well.
I found it simpler to obtain y = f(theta).
I obtained y = 1 + (sin(2 * theta)/(sin(theta) + 1)) and after taking the derivative and solving for sin(theta) = phi - 1 implies cos(theta) = sqrt(phi - 1), where phi is the golden ratio, implies y(max) = 1 + (2 * (phi - 1)^(3/2)/phi).
Note: All your coordinates can now be expressed in terms of phi to accurately graph the problem.
That was the long way for sure. I envisioned the rotated square as a cam and went right to the maximumization equation looking at the change in Y with the change of the base length X. X is bounded from 2 to 2+sqrt2. Basic engineering application of this math.
Should have kept using the angle theta instead of eliminating theta. Then you get y = 1 + 2/phi^(5/2). Much more simple.
Nice ! But i uesd theta(no elimination) and it simplified everything and gave the equation t^3+2t^2=1 where t= sin(theta) and solving it gives t=-1,(√(5)-1)/2
Correct!
So your theta is 38.2deg, that sounds plausible. Theta should be a little less than 45deg.
Let the angle of rotation be θ then noting height and width of rotating square is Sinθ + Cosθ (let S = Sin(θ) etc )
Similar triangle gives us the required height is Y = 1 + 2SC/(S+1).
Differentiating and equating to zero gives us the cubic S^3 +2S^2 - 1 = 0
Wolfram alpha gives us S = (√5 - 1)/2 . C is just found using pythagoras.
Then carefully substitute back into expression for Y (tedious but straight forward)..
Interesting that x is 1/phi-sqrt(1/phi) if I calculated that right? But that's negative (about -0.168)? Perhaps the wrong root was taken for the quartic equation.
I agree. The overall result may or may not be correct. Michael suffers from pure mathematician's disease - he despises the real world. If he checked his results along the way, he would have noticed this.
Simpler to express in trig form.
s,c are sin, cos . ss = sin^2 etc.
y-1=2sc/(s+1)
differentiating and set to zero gives.scc=(1+s)(cc-ss)
Or sss+2ss-1=0
Which factorises as
(ss+s-1)(s+1)=0
Or s=1-ss=cc
Hence c=sqrt(s)
Also s = golden ratio -1= ¥-1 0.608...
y = 1 +( 2(¥-1)^1.5)÷¥ =~1.600566 the same as the video.
That ¥ should be phi
x =~ 1.404 ie close to 1.414 which it would be with the left square at 45°.
"Y" is this in my reccomended?
Wouldn't the X need to be on the range [1,sqrt(2)] to be a valid solution given the way the diagram is presented?
The problem becomes easier, when the diagram is mirrored along the y-axis and shifted such that the non-tilted square has its lower left corner at (0,0). If we denote with x the height of the adjacent corner of the tilted square, we can find y(x) = 2*x*sqrt((1-x)/(1+x)) + 1. This has an optimum at x = (sqrt(5) - 1)/2 with gives an optimal value for y of (sqrt(5)-1)*sqrt(sqrt(5)-2) + 1.
I found it simpler to set the origin at the upper right point, where the diagonal line hits the flat square (with x increasing to left for simplicity). This give 3 collinear points: (0,0), (1+sina,sina+cosa-1), (sina+cosa+1,y-1); where 0≦a≦π/4. Working out the slope, y=1+(sina+cosa-1)(sina+cosa+1)/(sina+1).
I know that can be simplified, but at this point I did a numerical optimization for y, yielding y≅1.600566212, to the limit of my spreadsheet's precision. This matches the Michael's solution, again to the limit of my spreadsheet's precision.
Hi Michael. Here is a nice Diophantine Equation: (Japan 2020, Junior finals question 3) Find all tuples of positive integers (a,b,c) such that Lcm(a,b,c)=(ab+bc+ca)/4. Thanks.
Thanks Michael for this beautiful problem and all contributor's posts. I focused on the fact that someones found a rational value (8/5) while others found a different answer related with the golden ratio. Both answers are beautiful and I know I'm speaking about a difference less about 1e-3, but it seems that both didn't compute the same thing. The former maximize the max of the slope of the upper line. Unfortunately, the max slope of the upper line doesn't give exactly the max value for Y. The slope can be for instance a bit larger, and Y could be a bit smaller if the right square is closer. So we first have the max of the slope for Y = 8/5 (theta about 36,87° and tan(upper line vs horizon)=1/4) and then the max of Y = 1,600566... (theta about 38,17° and sin(theta)=(sqrt(5)-1) / 2).
We didn't take limitations throughout this process for x. One of them is 0≤x≤√2. An amazing problem thought!
y = 1 + (2sinθcosθ)/(cosθ + 1)= 1 + (sin2θ)/(cosθ + 1). considering dy/dθ = 0 leads to a trigonometric equation which boils down to
t^4 + t² - 1 = 0 where t = tan θ. Optimal t = sqrt((sqrt(5)-1)/2). drawing a right-angled triangle and plugging in values for sinθ and cosθ [try not to use Mathematica], gives the result y = 1+ sqrt(10sqrt(5)-22)
A value for theta would be nice to see how far from 45 degrees it is.
I agree.
It’s around 38.5° or ~2/3 (radians). Then y should be slightly greater than 1.60 for unit squares.
It's 38.17270763...°
x
managed this one without trig. Let A be the x-coordinate of the bottom corner of the left square, and let B be the y-coordinate of the left corner. Then A^2 + B^2 = 1.
The line passes through (A, A+B) and (1+A+B, 1). The y-coordinate of the line between these two points simplifies to y = 1 + A - (B-1)^2/A.
Note that dB/dA = -A/B when you use the chain rule.
Setting dy/dA = 0 and simplifying a bit, you get (B-1)^2/A^2 + 2(B-1)/B + 1 = 0.
It is easier to get everything in terms of B here, so you end up with the cubic B^3 - 2B + 1 = 0, which factors as (B-1)(B^2 + B - 1) = 0. Golden ratio time!
B = (sqrt5 - 1)/2, and since B^2 = 1 - B, you get A^2 = B, so A = sqrt(B).
From there, y = 1 + A - (B - 1)^2/A = 1 - (B^4 - B)/A = 1 - A(B^3 - 1) = sqrt((sqrt5 - 1)/2)(3 - sqrt5) + 1, which is equivalent to Michael's answer
The final maximum value of y is just a tad over 1.6. (More specifically, my calculator gives it as 1.600566212...)
Great. Thanks
Using theta you'll find y -1 = 2(cos θ)(sin θ)/(sin θ + 1). Max for sin θ = φ (the golden ratio! Wow!) so cos θ = φ^(1/2). Wolfram Alpha is not necessary.
The top point of the tilted square is at 1/2*(x PLUS sqrt[2 - x^2]). Then you get x = [sqrt(5)-1]/2 PLUS sqrt[[sqrt(5)-1]/2].
For anyone who is confused by your answer and differs from Michael's, one simple way to check if you're on the right track is this: what is the y-coordinate of the top most corner of the tilted square? If your answer is not dependent on the tilt angle or some other variable you chose to use, and is instead just sqrt(2), you are not on the right track.
So what is the correct y value of the top coordinate?
@@casc3601 1.600566212
Oh, pulling out the ancient calculus memories at 12:00.
The derivative of the square root of a function is the derivative of the function divided by 2 times the square root of the function. That is, d sqrt(f)/dx = (df/dx) / (2 sqrt(f))
Thus, d(sqrt(2 - x^2))/dx has f = 2 - x^2
Therefore, df/dx = -2x (derivative of 2 is 0, derivative of -x^2 is -2x)
Thus, (df/dx) / (2 sqrt(f)) = -2x / 2 sqrt(2 - x^2) = -x / sqrt(2 - x^2) = -x(2 - x^2)^-0.5
Actually, results in your way gives 1.60, while computating the intersection with the y axis of the straight passing trough A(√2/2 √2) and B(1+√2,1) gives 3-√2 (≈1.59). Since the maximum y is defined for the highest y that A can assume, the process should be correct
I get the same end result for y (≈1.601) with a slightly different approach, however, your x (≈-0.168) must be some wrong root since plugged into y the result is ≈0.399.
Y max=3-sqrt(2)
I would place the origin of coordinates at an invariant (fixed point) , let say the point (X, 0) in this case would be my (0,0).
I did not read all the comments but your x solving from the derivative is negative, and that is impossible. The problem is that you made a mistake when you used the plus sign for sin(theta) in set of minus sign. Following with this change of sign, making the derivative and solving for x, the value is aprox. 1.4041865 and then the value of y is 1.6005
If we stick to parametrizing by theta, we get the top point as (cos(θ), cos(θ) + sin(θ)) and the upper right point as (cos(θ)+sin(θ)+1,1). Then, the slope of the line segment is (cos(θ)+sin(θ)-1)/(-sin(θ)-1), so y = (cos(θ)+sin(θ)-1)/(sin(θ)+1) * (cos(θ) + sin(θ) + 1) + 1 from the upper right point and the point (0,y). This simplifies as y = 2sin(θ)cos(θ)/(sin(θ)+1) + 1, and its derivative is 2[ ( (cos^2(θ) - sin^2(θ))(sin(θ) + 1) - sin(θ)cos(θ)cos(θ) )/(sin(θ)+1)^2 ]. y is maximized only when this derivative is 0, so we need that the numerator of this fraction is 0. Using cos^2(θ) = 1 - sin^2(θ) and letting s = sin(θ) for brevity, we get (1 - 2s^2)(s+1) - s(1 - s^2) = 0, or s - 2s^3 + 1 - 2s^2 - s + s^3 = 0, or -s^3 + 1 - 2s^2 = 0, or s^3 + 2s^2 - 1 = 0. This factors as (s+1)(s^2 + s - 1) = 0; since s = sin(θ) is between 0 and 1 as θ is between 0 and π/2, the only solution is s = (sqrt(5) - 1)/2. Curiously, for this angle, cos^2(θ) = (sqrt(5) - 1)/2 and 1/(sin(θ) + 1) = (sqrt(5) - 1)/2, so we get that y = (3 - sqrt(5)) sqrt( (sqrt(5) - 1)/2 ) + 1. sqrt( (sqrt(5) - 1)/2 ) doesn't seem to have a nice form, so that's our answer. This is equivalent to the answer Dr. Penn gets, so that's good.
tl;dr: θ = sin^(-1)(1/ϕ)
I find this answer in term of theta more apropriate, as we would naturally measured the rotation of the first square instead of the measure of x (length which doesn't have a real physical signification for the square)
As other suggested - this can be solved without calculus: for max Y of left-most line, we want the top-most line to have the (acute) angle as steep as possible (while still going thru the top-most vertex of the tilted square) AND to stretch to the left as far as possible the left-most line (while still having the left-most line go thru the left-most vertex of the tilted square). Fortunately these two requirements do not contradict and they both are achieved when the tilted square has been rotated 45° - i.e. when one of its diagonals is orthogonal to the lower-most line. Then the (x,y) values of the upper-most vertex of the tilted square can be found - which enables finding the expression of the upper-most line. Inserting in this expression the x-value of the left-most line (it is also found from the achieved "optimally tilted" square) gives the desired Max Y.
Using the value of Y that he got when he did the question, the square is tilted 38 degrees, not 45 degrees as you would intuitively think.
selection of problems is getting really good. thanks for the awesome content man!
A graph y=f(theta) would be the cherry on the cake. And also a good sanity check.
I defined x as the x-coordinate of the bottom vertex of the tilted square and used only the pythagorean theorem to determine the coordinates of the important points, no angles involved.
That way I got y = ((4x^2 - 4x^3) / (1 + x))^1/2 + 1.
To determine the maximum point I knew the +1 and the ^1/2 dont change the x-value of the maximum so I simplified y to (4x^2 - 4x^3)/(1 + x) which is easy to differentiate. The maximum is at x = (-1 + sqrt(5))/4 and I was too lazy to calculate y.
The fun part was that I constructed the whole situation in geogebra and also displayed the target function in the same graph which allowed simple error control.
I dont understand why, as others suggest, that you cannot use x=sqrt2 due to the maximum of y being related to the maximum of x, since the largest x can be is the diagonal of the rotated square of sides 1, also sin(theta) then equals cos(theta) and theta =45 with those being known values. Unless the objective is solely to solve using first term diff calculus or without the sides of the square being known
In the expression for x, isn't the sign wrong in front of
√(2√5 - 2)?
I get an evaluation for x of -0.168 with that expression;
but a more reasonable value of 1.404 with the sign changed to give:
x = ½ (√5 - 1 + √(2√5 - 2) )
That also is then in agreement with the results below of a maximum when
cos² θ = sin θ
so that
sin θ = φ - 1.
Also I believe the identity
sin θ + cos θ = √2 sin (θ + ¼π)
(well known to the electrical engineers at least)
would come in quite handy in simplifying the calculations.
Why didn't you just say that the diagonal of the sqaure is root(2) an then solvetheproblem via similar triangles. First triangle from xE(root(2)/2,1) and the other xE(0,1)?
It is not correct, because, you cannot be sure that diagonals of square are paralllel or pependicular to the axes. Your way to solve this problem is only a special case.
I know this pain. Long time ago I was preparing to the IMO. Almost every task was about including every case of solution. Skipping any case was painful for your score xd
I thought this as well, but the diagram is misleading in that the height is actually greatest when you tilt the square slightly to the left. If you look at how he did it, and then plug in sint+cost=x (where t is theta) you will find that the value of Y is actually greatest when theta is about 38 degrees, which would mean that you can not just assume that the height is sqrt2 because the square is tilted at a different angle than expected.
I did this a bit differently, using the angle of tilt (alpha) of the almost-vertical diagonal from the vertical as the parameter. Then if y = 1 + z, and c = cos(alpha), s = sin(alpha), I got:
z = (2c^2 - 1)/(1 + sqrt(2)(c - s)/2)
Unwilling to solve dz/dalpha = 0, I just computed z for a range of alpha and found the maximum at alpha = 0.119 to be z = 0.600566, which is sqrt(10sqrt(5) - 22).
I notice that my equation for z looks a lot like yours if I put 2c^2 = x^2, s^2 = 1 - c^2 = 1 - x^2/2, s = (1 - x^2/2)^(1/2)
This didn't quite give your equation, then I realised I needed to use s = - (1 - x^2/2)^(1/2). Now I get exactly your equation. :)
By the way: 2c^2 - 1 = cos(2.alpha) = sin (pi/2 - 2.alpha), therefore if I put pi/2 - 2.alpha = 2.theta then alpha = pi/4 - theta. If I substitute for alpha in my equation for z I get Sergio's equation. :))
I miss why dr. Penn didn’t use the other value in the quadratic solution of sin(tetha)
Taking the other quadratic solution essentially exchanges the values for sin(theta) and cos(theta) and represents the square tilted in the other direction (with its "vertical" diagonal tilted to the left instead of to the right) but (I think) ultimately yields the same relationship between x and y. Note that both these tilted squares yield the same value for x, but the two theta values are complementary.
But this probably should have been explained.
@@kevinmartin7760 Thanks Kevin, I appreciate it. I still think he should have explained it
This equation is not so difficult to solve by hand
After squaring both sides you can check for rational roots or get rig of repeated root with differentiation
Remaing quartic can be easily solved by rewriting it as difference of two squares and then as a product of two quadratics
After solving polynomial equation check the solution because squaring both sides may produce extraneous solutions
I just realized, x is wrong. It should be g+sqrt(g)=1.40418... g is the golden ratio.
Btw it's much easier using the angle as the variable.
Two points on a line (root 2 /2 , root 2) and (1 + root 2 , 1), so solve for y intercept. Calculus is only needed to demonstrate this is the max via perturbation.
That's what I thought. When I do that though, I get 1.58ish, though
Sir i think you have calculated minimum value of y instead of maximum you should check it by second derivative test
And also value of "y" you have found i wrong if we plug in
I believe you are correct! If you plot y as a function of x, there's a minimum at the value found above, but that's the only critical point. This function actually has a max at one of the end points of its natural domain which seem to be x = 1 and x = sqrt(2). The latter looks to give you the max.
Solution is exactly y=8/5; the method used is approximate and abstruse, you can solve it in easier way considering other similar triangles
Alas you fell into the trap. y depends on theta in a more intricate manner.
I also found the maximum area of the trapezoid formed.
Solved it just using linear equations and pythagoras. y is the y-axis intercept of the linear function running through the two points where the square touches the slope. The resulting function is quite ugly, so I also turned to wolfram alpha to help with the derivative and setting it equal to zero. Getting the same solution felt satisfactory nonetheless.
Why is sin(theta) = 1/2 (x+sqrt(2-x^2))?
Why not ±?
Brcause theta is
The negative result would make the y-value for every point on the rotated square less than 1. The straight line would then cross the y-axis at a value less than 1 -- which is lower than when theta = 0, so there is no maximum y-value to be found when theta is between 0 and -pi.
Put x=1 to the equation
I like this video , I like the good solution , I try to do and try to learn also.
shouldn't the left square hit the line at ( sqrt2/2, sqrt 2) and the right square hit the line at (1+sqrt2, 1)? If so, we have two points to make a line with, and we then just solve for when x=0.
Wow! Who would've thought a quartic polynomial would show up in a problem like this!
You can solve this without calculus though, can't you, by plugging in values of the angle and seeing which theta gives the maximum y value?
So I did it without using angles, by specifying the x value of the corner of the rotating square, then plugging in values to find the maximum y value to be around 1.6005236512374004 while Michael gives a value of about 1.600566212001556. So yes, you can get pretty close. The advantage of using calculus is you can PROVE the answer the maximum, and show it is an absolute maximum and not a local maximum.
Smells like an upcoming engineer in here.
Not really, without calculus is not possible to prove that the max value you find is the actual biggest one ( mainly because you would have to check literaly every single angle, but there are infinitly many, so......)
@@joaopedrobmenezes2977 no you can maximize y value by construction the two squares are at 45 degree to each other - some logic can demonstrate it is a local max.
I hoped to see also what the theta value was and numeric values of x & y... great.
I love your videos in general but this solution is a bad one, simply because I haven't heard of a math competition where in the solution you can plug in wolfram alpha. I would recommend a little simpler solution which is based on labeling the same angle theta.
Label the big trapezoid (whose base y we need to calculate) ABCD by starting from the bottom left corner.
Label the topmost point of the tilted square as P. Then in terms of theta you can easily calculate the distances from P to the lines AB and BC.
Let F be a point on the line segment AD such that FC is perpendicular to AD. Again you can easily calculate FC and angle MCD in terms of theta.
In the end you will find that y = AD = 1 + sin (2 * theta) / (sin (theta) + 1). Calculating the first derivative and its roots will bring you to a cubic equation sin^3 (theta) + 2 * sin^2 (theta) - 1 = 0 which is very easy to solve because it transforms to t^3 + 2*t^2 - 1 = 0 which has one integer root and only one positive root. From here you can find sin (theta), cos (theta), sin (2 * theta) and therefore AD.
Symmetry seems to suggest the maximum is achieved at theta = pi/2
Consider that after some angle < π/4 any further increase in the angle (and thus in x) decreases the slope more than it is lifted meanwhile, which leads to a net decrease in y.
Do you mean pi/4?
Really close to that 👌
Let rotated squire touches axes at (a,0) and (0,b), a^2+b^2 = 1
Points (0,y) (b,a+b), (a+b+1,1) are on the same line => (y-1)/(a+b+1) = (a+b-1)/(a+1)
y = 1 + ((a+b)^2-1)/(a+1) = 1 + 2ab/(a+1) = 1 + 2a sqrt((1-a)/(1+a))
y' = 0 => a^2+a-1 = 0 => a = (sqrt(5)-1)/2 - golden ratio cojugate
ymax = 1 + sqrt(10sqrt(5)-22)
My intuition was to go for y1 being the full height of the diagonal (sqrt(2)) but that was not the correct solution. It is much simpler to solve though. :-)
7:23 I think I found a mistake. That quadratic equation should be - 2sin^2θ + 2xsinθ - x^2 + 1 = 0
13:00 "change that minus sign to an equals sign" ;-)
Is the Y maximum value when the diagonal of the square is perpendicular to the slope? I'm pretty sure it is, but please enlighten me if it's not. Then you can crunch the numbers from there.
I get the same result after 2 hrs of trying, using phytgorean theorem (initially i tried to use trigonometry but i get lost) .... i even had to recall derivatives of products. Thanks for the exercise!
Edit: you get new subscriber
Can you just make the intersection of the first square and the line (sqrt2/2, sqrt2) to find the slope of the line then use that to find out what it equals at x=0?
The tallest that intersection of the first square and the line can be will be the diagonal of the square from two corners right?
I think you mean we put the left hand square so that its diagonal is perpendicular. That's the answer, right? No... You can twist the left hand square clockwise a little bit and it will lift the slope of the line we are trying to find. It's the slope we need to find - not the height of the left hand square.
@@samstep4279
No it won't shift he slope.
If you rotate the left hand square clockwise, the bottom corner will go clockwise to, thus lowering the slope,
if the left hand square is rotated that the diagonal is vertical, that point where it touches the slope is the most upper point it can get.
It is a square that you rotate and not just a vertical line.
@@seriously1184 and the earth is flat, right? why are you arguing with the literal established correct answer right now lol
@@lennyganado3975
First of all I am not an flat earth believer, far from it !
Second, I am an engineer and thus I looked at this in a realistic way of problem solving (that what is often times too big of a problem for mathematicians being able to do) !
Third, the designed problem used in this video is in a closed frame !
All the outlines off thegeometrical figures are all of the same width and colour, thus making them all geometrical objects laying against each other and in one large geometrical object, making this a realistic engineering problem than anything else where the law of physics rule above anything else (a concept that is apparently unknown to mathematicians) !
So use lines of different width, colour, etc if you want to indicate what is fixed or adjustable in any design or drawing (something that apparently is also unknown to mathematicians) !
Fourth, if you start with insults, you must really be a compulsive and impulsive pric*
I bid you a good day sir.
@@seriously1184 It's a pity that an engineer such as yourself missed the point that as you rotate the first square a little clockwise from its highest point, not only does the bottom corner move left, but so does the rightmost corner and so _the second square moves left in its entirety._ That movement left of the "hinge" of the slope increases the slope initially, and that is why the value of y will in fact increase a little as the first square rotates clockwise from its highest point.
You could've found θ immediately after writing x=cosθ+sinθ...
Let me explain:
x=cosθ+sinθ x²=1+sin(2θ)
sin(2θ)=x²-1
2θ=arcsin(x²-1)
θ=(arcsin(x²-1))/2
Before resolve y' = 0, shouldn't you verify the interval where denominator of y' is not zero?
(x+2) >= 2
I got the answer 3-√2, as the y will be maximum if the tilted square's diagonal is perpendicular to the base line
Same. Haven't watched the video but it was fun to attempt.
For those of you that did the calculus in theta- what's the optimal value for theta? My intuition says 45 degrees but I would expect the answer to be nicer if it were the case
θ~38.17° max [sinθ*cot(θ/2)]
Hi, Michael! I love your videos because you demonstrate clear solutions to challenging problems, skip the tedious too-obvious steps, and admit your mistakes. BUT I don't understand how at 7:30 you got sin theta = 0.5(x+sqrt(2-x^2)). According to the quadratic formula, shouldn't it be
0.5(2x +- sqrt(4x^2 -4•2(x^2 -1)))
= x +- sqrt(2-x^2) ?
I see that my result CAN'T be right because the sqrt must be tiny and sin theta must be close to 0.5x, but I don't understand what I'm doing wrong. Help!
I stand corrected.Slightly nudging the left hand square to the right increases the height of y by 0.0147 units.Good job Michael.
It seems to me that it is intuitively obvious that y will be largest if the second square is rotated 45 degrees giving the width to be 1+2^1/2.
this will get you within 1% of the correct answer, but still incorrect. It is not 45 degrees. it is 45 plus or minus 4 degrees to either side.
Something is wrong:
2 * sqrt(5) - 2 = 2.47
sqrt(2.47...) = 1.57
sqrt(5) = 2.24
Thus:
x = 1/2 * (-1 + 2.24 - 1.57) = -0.17
Therefore:
y = 1 + (2x^2 - 2)/(x + 2 + (2 - x^2)^0.5) = 0.4
However:
1 + sqrt(10 sqrt(5) - 22) = 1.6
i just assumed the square was rotated by 45 degrees and then i made two points and wrote the straight line. This could be solved more simply with euler's calculus of variations imo
c
kind of gnarly!
Excuse me if I'm being dumb, but if the question was asking what the maximum of y was, wouldn't you just set the theta equal to 45 degrees as that would maximize the vertical height of the tilted square, which would then maximize the height of y? Like, there IS a way to solve this numerically, unless I'm being dumb.
Edit: Forgot to mention that since it gives side lengths, you can solve it. Not sure if I need to mention that, but still.
Edit 2: Every response to this keeps on getting more confusing. Feel free to keep commenting, I think eventually someone will figure it out, because clearly I can't.
Because that is not the actual solution. The 45 degree angle would be the maximum solution for x, but it is not for y. (even if somewhat close to it of course) By tilting the square just a little bit clockwise, x gets just a little tiny bit smaller, but y gets higher, because the top of the square is moving to the right more than downwards.
Of couse you can do this numerically as well, but that's a whole other story. So basically, you could do a search with the angle looking for a value of theta between 0 and 45 degrees and do a binomial search until you reached whatever your degree of error is. But numerical solutions are in general no exact solutions, which is what Michael is looking for here.
@@petersievert6830 but y is the height, and if y is on the line from the top right corner of the regular square to the top of the tilted square, then to maximize the angle of that line to the horizontal, you would just tilt the tilted square to 45 degrees... Also, x would be maximized at angles larger than 45 and approaching 90, as the most the distance can be is that gap plus 1.
@@nateexists6292 i don’t think you got what peter said, the max of Y is not achieved at theta=pi/4 because y max =/= x max.
Its pretty close tho.
@@nateexists6292 Be awere that the X-coordinate of the top of the square is not fix. Moving the top to the right while keeping almost the same height, enlarges y.
@@BrianStewart126 I am not sure, what you are referring to exactly. The numerical solution?
If there are multiple local maxima between x=1 and x=sqrt(2) (which corresponds to my theta ranging from 0° to 45° , then it could obviously fail, sure. But it seems intuitive for me, that when you tilt the square clockwise from x=sqrt(2) to x=1, then y would rise for a certain range and fall afterwards. (If that assumption is wrong, I am in troubel, sure.)
btw I was imagining some kind of search, that makes do without the derivative, rather the engineer's way suggesting some kind of binomial search and possibly not even very effcient and not the analytical way, bec if you got that far to get the derivative dy/dx resp dy/dtheta , then you surely are able to proceed without the need for a numerical solution)
Apart from that, I only wanted to give some intution for the idea, why x=sqrt(2), thus the bottom angle=45° is not the solution. That was not meant as a recipe to find the actual solution at all.
Why no animation to show all the possibilities? With parametrization according to the distance between the bottoms of the squares?
Why not set up Penn's (x,0) AS the origin?
Would appreciate a more detailed explanation of how the quadratic equation was done. Which value was a, b and c.
a = 2, b = -2x, and c = x^2 - 1.
The quadratic formula is letting us solve for sin(theta).
I tried the angle method. I was also stuck after differentiating for finding critical points.
your expression for y should be (sin(2\theta))/(1+sin\theta)+1. Differentiate, then when setting the numerator to zero, convert it to an equation in sin\theta only. You should get a factorizable cubic equation in sin\theta.
So What is the angle Alpha, another interesting data point.
can one quickly eliminate that theta=45 is not the answer? or it is the answer
I guess the eqn for y is wrong because it should be a straight line but when i plot it on desmos i get some curve.
Please explain..
Three friends set up a meeting in front of the Nilton Santos stadium to watch a
a soccer match. They agreed that each one should arrive in a moment
chosen between 3 p.m. and 4 p.m. and that none of them will wait more than 30 minutes for
too much, within the stipulated time. What is the probability that the three friends will
meet between 3 p.m. and 4 p.m.?
Answer : 1/2
Very impressive but I seriously didn’t think it would be so complicated.
I enjoyed it until 5:53
For me, that was a good place to stop, lol
I did this problem completely differently and ended up with 1+ [Sqrt(2)(6-2Sqrt(5))(Sqrt(Sqrt(5) -1))]/4 which calculates to the exact same number that Dr. Penn got. Anyone know how to prove that those two answers are equal??
Yeah, the x value you get for this is negative = - 0.168117....
So not sure if this is a valid answer or an extraneous solution from the quartic equation.
The point (X- a*Sin(teta)-b*Sin(teta), 1) is equal (X -2*Sin(teta), 1)...
Finding Y maximum is interesting but finding the Maximum Area would be even more interesting.
great pace yet great clarity - a great place to stop
Why is the value of x negative?
I don't get the question. The maximum? What maximum? You say later that you turn the tilted square, and it makes it pretty clear, but I really think it should have been included in the question explicitly.
The question asks for the maximum value of y. That means over all values of anything that can vary, and the diagram has only one thing that can vary, the angle of the first square. No ambiguity to my mind.
Why do you assume sin(theta) > cos(theta)? It looks like you should assume the opposite.
The math in this problem might be first year level calculus but if this were on my final exam I would have gotten a 0.