Geometry from an exam for teachers.

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[@rcb3921]

One of the extremely useful (if slightly embarrassing to me) take-away-thoughts from these videos by Penn is this: after defining terms, take a second moment to clearly state your goal *in those terms*. When working out a problem, I have often struggled (and been thwarted), trying to simplify every expression and uniquely identify every term. And it just isn't always necessary.

[@lexyeevee]

in general i enjoy problems where you're trying to solve for an expression rather than a lone variable; it makes the algebra a little different and more interesting from the usual though in this case, if it helps: you could define the ratio as k and write the side lengths as y and ky, then solve for k instead; at the end you'd be able to divide through by y² and have a regular quadratic in terms of k

[@giosanchez2714]

woah did not expect the answer to be such a clean ratio

[@kumoyuki]

me too. especially one with such a clear smell of the golden ratio, either...

[@Utesfan100]

@@kumoyuki Or the silver ratio even

[@inv41id]

I just love how I eyeballed from the thumbnail and thought to myself "that looks like about five" and that just turns out to be the exact solution

[@rohitmadashri7250]

You're a genius just like me bro. I eyed balled fermats theorem and knew the solution doesn't exist.

[@Jack_Callcott_AU]

@@rohitmadashri7250 Wow.. I thought that too.

[@swingardium706]

6:15 I would solve differently from here. If you divide all the terms by x² you get a quadratic equation in the variable y/x. You can use the quadratic formula from there.

[@jayantanayak4981]

Thought the same. 😅

[@ryantamburrino3289]

That's how I did it! A bit cleaner if you ask me.

[@emmanuellaurens2132]

@@jayantanayak4981 I thought the opposite ! I was going to divide by y² :)

[@joshuaromeo9187]

Me (who's terrible at math): yes truly it was the best course of action

[@an4r4k34]

I actually got this one right! Albeit in a completely different way. (Spoilers for solution below) I used the actual equation of a circle and diagonal line to find the larger square. To find the smaller square I set a as half the length of the smaller square and b the length of the smaller square. Then you can solve for b with a system of equations as follows (in this case the radius of the circle is 10) 2a = b x^2 + y^2 = 100 x = a y = 5*2^.5 (half the length of the large square) + b This will find the x and y coordinates of point G (in his diagram). It's kind of confusing without a graph to explain but I'm proud to have figured it out (I'm 15 btw)

[@txikitofandango]

Since we're looking for x/y, if you don't feel like factoring 5y^2 + 4xy - x^2 = 0, you can always divide through by y^2. Then you have a quadratic in x/y.

[@AtariDays80]

I think the (y+x) = 0 part of this is saying you could inscribe a square on the other side of that top chord. But of course, such an inscribed square would just be overlaid on the big square.

[@iabervon]

It's a good check that you got all your numbers right if the negative root gives an invalid solution with a size that's obviously right.

[@ardabaser1349]

I don't understand how such a square could have two of its corners on ABCD? Unless that what you meant. I Might have mixed that up.

[@gwynbleidd3849]

it's not saying this at all. If it were, it'd be (y-x)=0

[@Adderkleet]

@@ardabaser1349 The square would technically have 2 corners on line AB, because the points A and E (and B and F) would be in the exact same point. |Y| = |X|, so AB = EF. It's clearly the "incorrect" root/solution. But it does make sense mathematically.

[@williampommeranz6821]

@@gwynbleidd3849 my thoughts are for the case where y = -x, the inscribing circle is being reduced to a point. a circle of zeto radius. undefined and supurflous. just a thouht.

[@mbarekouazragh9982]

Dear Michael, Please correct me if I’m wrong but it seems that with what was given, there is a step missing ! You should prove that EF is centered inside AB !!! Must prove that the only possibility of choosing two points in segment AB and two points in the arc AB to have a square is to center EF inside AB.

[@SgtSupaman]

It is basic symmetry that doesn't really require proving simply due to the definitions of a circle and a square. If a circle touches all four corners of a square, you can't possibly fit a second square outside of the first square and inside the circle (with two corners touching the circle) unless the second square is centered on the side of the first square. In fact, even if the second one were a rectangle, it would still have to be centered on the larger square for two corners to be touching the circle.

[@hydra1668]

I guessed and I got it right. I just looked at a triangle made between the top side point B and Point G and saw that it made a triangle that was roughly twice as long as EF. I have no idea about the the maths there, but I'mma take my win and head out.

[@puzzLEGO]

there's a significantly easier way to do this. I added another square on the bottom of the same size as the smaller one. Then I used the circle diameter rather than the radius. If I use d for diameter instead of r for radius, It comes out to d^2 = 2x^2 and then (2y + x)^2 + y^2 = d^2, which I solved simultaneously with a clean quadratic (same as the one he ended up with but no fractions)

[@collection6062]

I would use a ruler

[@mdravenstein2370]

Yeah the first thing that came to me was like adding another square below.

[@buzzzysin]

I find it weird(ly pleasing) that a "nice" integer number like 5 pops out of what I expected to be at least a rational solution at best.

[@Frenchstud69]

@7:29 "...and that's a good place to stop." perfect outro

[@weirdfrog1196]

The intersecting chords theorem simplifies this greatly

[@Macieks300]

0:05 I don't know why that "region in Russia"'s name wasn't mentioned but it's Adygea and its name was spelled wrong on the blackboard

[@JavierSalcedoC]

Not even a region either, but a Republic

[@petersievert6830]

Maybe he didn't like to mispronounce it...

[@orionspur]

The sequence of side-length ratios for n-gons is interesting: 3, 5, 5+2√5, 13, etc. The last integer in the sequence is 13, at n=6. (read more for spoiler) For n odd, ratio = cot²(π/2n) For n even, ratio = 4 csc²(π/n) - 3 And that's a good place to stop.

[@aceofhearts37]

You can also approximate this very easily by noting that (x/2 + y) is just shy of the radius. Giving you an approximation of y/x = (sqrt(2)-1)/2 which is very close to 1/5. Yes, I'm an engineer.

[@mehdimarashi1736]

Here we go! You can be a very successful civil engineer. Less than 5-10% error and we are golden!

[@hannahbrown2235]

I thought the same thing but didn’t realize at first why that way didn’t get me exactly 5 😅

[@crtwrght]

I found a different triangle for any interested, also making use of a circle theorem. If you mirror EFGH to the bottom of the circle to square E`F`G`H` then the line from G to G` has length 2y+x (In my working I just called x=1 and r=1/sqrt(2) but I'll stick with the x&y here). The triangle HGG` is right-angled and hence by circle theorems HG` is a diameter with length sqrt(2)x. Pythagoras then directly gives us the equation y^2 + (2y+x)^2 = (sqrt(2)x)^2 which simplifies to the same quadratic.

[@insouciantFox]

If you continued this construction ad infinitum, what would the total area of the squares be?

[@bobh6728]

25/24

[@fahrrad67]

That's some amazing new chalkboard technology! I wish my school had it. When I knock on the chalkboard in my classroom, nothing happens and the students just look at me like I'm weird. :-)

[@Qermaq]

I kinda took a shortcut. We've all drawn a circle radius 7 and noted how close the circle comes to {5,5}. For it to really pass through that point, the radius has to actually be sqrt50. A circle with radius sqrt50 hits the lattice at two unique points after rotation and reflection over axes - {5,5} and {1.7}. And it clicked that this yields a 10x10 and a 2x2 square.

[@trueriver1950]

You would do well in a maths olypiad...

[@luminous2585]

That was really interesting. I almost got stuck trying to follow you near the end when you determined that 5y = x because I thought that at least one of the factors had to equal 0 but I didn't know why it couldn't have been (y+x) until I realized that this would mean y=-x and that can't be since both of these values represent positive lengths.

[@OscarCunningham]

When trying to solve the quadratic, one way to do it would be to look at the diagram and see that x = -y corresponds to the 'solution' where the little square is the same as the big square. So x+y must be a factor of the quadratic, and then you can divide to get the other one.

[@pilosopher974]

that’s such a cool observation 😎

[@TheBrutalDoomer]

I used the chord product theorem, setting big square side to 2 and small square side to 2x, you get AE * EB = HE * (continuation of HE), so (1+x)(1-x) = 2x(2+2x), which results in 2x=0.4, so ratio is 1 to 5.

[@BraxtonMeyer]

that triangle for the second one came out of left field for me.The way I thought about it may have worked but in retrospect seems more convoluted.

[@MrConverse]

6:02 & 6:07, *x squared over two

[@goodplacetostop2973]

7:28

[@CTJ2619]

Very well done video

[@nikkola807]

I decided to go with the intersection of chords. Let x = AE, y = EF. Build additional chord to the end using HE as the starting line, then construct the equation of intersection for chords: x*(x+y) = y * (2*x+y+y) => x*(x+y) = y*(2*(x+y)) => x = 2y. Then lets construct and look at the desired ratio (2x+y)/y and insert our result (2*2*y + y)/y = 5

[@shjescaresme]

I just wrote the same solution, I had missed your comment. I think it's much nicer this way!

[@trueriver1950]

x+y = 0 does indeed make sense, even tho it is not the desired answer. y = -x describes a square going downwards from the line rather than upwards, of side length x and sharing one (at least one) side with sq ABCD. It's actually identically sq ABCD The algebra is telling you that the square ABCD actually has at least two corners touching the circle, and at least one side that lies along line AB and so fits all the requirements that you captured in your use of Pythagoras. In my view then you choose the solution to the quadratic based on the direction that y takes as positive (up) rather than negative (down), which is the one constraint you didn't include yet.

[@manucitomx]

I love geometry! Thank you, professor

[@bestcreations4703]

I decided to stop the video and work on it myself from scratch, I initially wrote two equations to define the height, y, of the small square by recognising it as the height of the larger square plus two of the width of the smaller, as done in the video. From there I ended up doing the same for x, since after all it is a square, resulting in a similar equation there. From that point I substituted and easily found the same answer. Very interesting how two people can solve the same problem correctly in different ways.

[@learncheung2959]

Let the midpoint of AB be M. Note it is easy to prove the points H, M, C are collinear. ∠HAB=∠HCB (angles in same segment) and so ΔHAE and ΔMCB are similar. We have AE:HE=CB:MB=2:1. (x−y)/2:y=2:1 and so x=5y

[@charlesbromberick4247]

This problem is a natural for the Smith Prerogative, and a smart choice would be making the radius of circle = 1. This done the algebra is simplified and we verify Mike´s result of 5.

[@JavierSalcedoC]

Pretty cool that the answer is 5

[@Ptaku93]

The region, if anyone's interested, is miswritten on the side of the blackboard - it's Adygea

[@MuyGordoGato]

Seems like a complicated way to do it. Just assume the large square is centered with the coordinates 0,0 at the center. Draw a line from the upper right corner of the small square crossing the y axis in the center of the bottom of the small square. The equation for this line is y= 1/2x + 1/2. The radius of the circle is sqrt(2)/2, so it's equation is sqrt(2)/2 = x^2 + y^2. Substitute for y from the line equation into the circle equation and solve for x. You will get two values, one negative and the other is 0.1, which represents half of one side of the small square. So, if the side of the small square is 0.2 and the large one is 1, the ratio is 5.

[@zmaj12321]

I remember solving this problem when I was first getting introduced to comp math. Good times.

[@yapsiauwsoengie6507]

I set the center as (0,0) coordinate. A line with tangent value 2 will across over DG so we can define G as intersection of the line with the circle easily.

[@Mrmoocows99]

I don't know how I found this video, but I listened to this guy draw on a chalk board and do math, for seven and a half minutes and retained none of the information.

[@Christian_Martel]

1:08 This is good place to give a try…. Set x = half-side of small square and c•x = half-side of large square With x>0, c>0 then coordinates of B and G are: B ( cx, cx ) and G ( x, cx + 2x ) Both B & G are at radius r from centre, so (cx)^2 + (cx)^2 = x^2 + (cx+2x)^2 … algebra stuff …. (x^2 )•(c^2 -4c -5)=0 Soln-> x=0, c=5, c=-1 Only valid soln c=5 … right, professor?

[@redShiftish]

I did this with circle trigonometry instead - solved for the angle whose terminal side intersects the circle at point G. Wasn't very messy and arrived at the same answer.

[@sillymel]

I think I have a more elegant solution (simpler algebra): Let J be the midpoint of both segment AB and segment EF, x = JB, and y = JF. Note that triangles JGB and JFG are similar (see geometrical underpinnings below) and AB/EF = 2x/2y = x/y. Since triangles JGB and JFG are similar, JB/JG = JG/JF, that is x/JG = JG/y, so x = (JG)^2 / y. By the Pythagorean Theorem, (JG)^2 = (JF)^2 + (FG)^2 = y^2 + (2y)^2 = y^2 + 4y^2 = 5y^2, so x = 5y^2 / y = 5y, or x/y = 5. Details: Let point J be the midpoint of segment AB and point K be the midpoint of segment GH. First, I want to show that J is also the midpoint of segment EF. (You just assumed this in the video.) Let point O be the center of the circle, line 1 be the perpendicular bisector of segment GH, and line 2 be the perpendicular bisector of segment AB. Since segments AB and GH are both chords of circle O, point O is on both of their perpendicular bisectors, namely lines 1 and 2, so lines 1 and 2 intersect at point O. Since quadrilateral EFGH is a square, it is also a rectangle, and thus a parallelogram, so line EF is parallel to line GH by the definition of parallelogram. (It's fairly easy to show that every rectangle is a parallelogram. Just note that opposite sides are both perpendicular to either of the other two sides, and thus opposite sides are parallel.) Now, note that by the definition of perpendicular bisector, line 1 is perpendicular to line GH, and line 2 is perpendicular to line AB. Since line 1 is perpendicular to line GH and lines GH and EF are parallel, line 1 is also perpendicular to line EF as a line being perpendicular to one perpendicular line implies it is perpendicular to the other. Now, since points A, B, E, and F are collinear, line AB and line EF are the same line, so lines 1 and 2 are both perpendicular to line AB. Now, assume lines 1 and 2 are different lines. Then lines 1 and 2 are parallel since lines that are perpendicular to the same line are parallel. However, lines 1 and 2 can't be parallel since they intersect at point O. Thus, the assumption that lines 1 and 2 are different lines is false, i.e lines 1 and 2 are the same line. Now, since line 1 is the perpendicular bisector of segment GH, line 1 bisects segment GH, so line 1 intersects line GH at the midpoint of segment GH, namely point K. Similarly, line 2 intersects line AB at at point J, which means that point J is on line 1 since line 1 and 2 are the same line. Now, lines EH and FG are both perpendicular to line GH since quadrilateral EFGH is a rectangle. Since line 1 is also perpendicular to line GH, lines 1, EH, and FG are all parallel as lines that are perpendicular to the same line are parallel. Recall that point K is the midpoint of segment GH. This means that segments GK and KH are congruent and point K is between points G and H by the definition of midpoint. Also, note that lines GH and EF are transversals of lines 1, EH, and FG. Since segments GK and KH are congruent, point K is between points G and H, points K and J are in line 1, and lines 1, EH, and FG are parallel, segments EJ and JF are congruent as parallel lines that intercept congruent segments on one transversal intercept congruent segments on any transversal. Since point K is between points G and H, points J and K are on line 1, point J is on line EF, and lines 1, EH, and FG are parallel, point J must be between points E and F. (If it wasn't, line 1 would have to intersect line EH or line FG.) Finally, since point J is between points E and F and segments EJ and JF are congruent, point J is the midpoint of segment EF. Next, I want to show that point J is between points D and G. Recall that J is the midpoint of both segments AB and EF. By algebra and the definition of midpoint, this means that EF = 2(JF) and AB = 2(AJ). Since quadrilaterals ABCD and EFGH are squares, segments AB and AD are congruent, and segments EF and FG are congruent, so by more algebra and the definition of congruent segments, FG = 2(JF) and AD = 2(AJ). By even more algebra, FG/JF = 2 = AD/AJ. Since quadrilaterals ABCD and EFGH are rectangles, angles DAB and GFE are both right angles, so angles DAB and GFE are congruent as all right angles are congruent. Now, since point J is on rays AB and FE (look at the diagram in the video), angles DAB and DAJ are the same angle, and angles GFE and GFJ are the same angle, so angles DAJ and GFJ are congruent by substitution. Since FG/JF = AD/AJ and angles DAJ and GFJ are congruent, triangles DAJ and GFJ are similar by the SAS Similarity Theorem, so angles AJD and FJG are congruent by the definition of similar triangles. Recall that angles DAB and GFE are right angles. This means that lines AB and AD are perpendicular, and lines EF and FG are perpendicular. Since lines AB and EF are the same line, lines AD and FG are perpendicular to the same line, which means that lines AD and FG are parallel. Note that lines JG and FG are not parallel (they intersect at point G), so lines AD and JG are not parallel, either, since lines AD and FG are parallel. Since lines AD and JG are not parallel, they must intersect. Call the point of intersection point L. Note that point J must be between points L and G, otherwise lines AD and FG would be forced to intersect, which is impossible. By the diagram in the video, point J is between points A and F. Now, since point J is between points L and G, and point J is between points A and F, angles FJG and AJL are vertical angles, so they are congruent by the Vertical Angle Theorem. Since angles AJD and FJG are also congruent, angles AJD and AJL are congruent by the Transitive Property of Congruent Angles. Also, segment AJ is congruent to itself by the Reflexive Property of Congruent Segments, and angle JAD is congruent to angle JAD by the Reflexive Property of Congruent Angles. Now, note that point L must be on ray AD (otherwise line GJ would have to cross line AB twice, which would make lines AB and JG the same line, which is impossible since point G is not on line AB), so angles JAD and JAL are the same angle. Thus, angle JAD is congruent to angle JAL by substitution. Now, since angles JAD and JAL are congruent, segement AJ is congruent to itself, and angles AJD and AJL are congruent, triangles AJD and AJL are congruent by ASA, so angles ADJ and ALJ are congruent by the definition of congruent triangles. Now, by way of contradiction, assume that points D and L are distinct. Then since angles ADJ and ALJ are congruent, lines DJ and LJ are parallel by the CAP Theorem. However, lines DJ and LJ can't be parallel as they intersect at point J. Thus, the assumption that points D and L are distinct is false, and they are in fact the same point. Recall that point J is between points L and G. By substitution, this means that point J is between points D and G, which is what I wanted. Finally, I want to show that triangles JGB and JFG are similar. Note that by the Inscribed Angle Theorem, the measure of angle DOB is twice the measure of angle DAB, as well as twice the measure of angle DGB, so the measure of angle DAB equals the measure of angle DGB by algebra. Since angle DAB is a right angle, this means angle DGB is also a right angle by more algebra and the definition of right angle. Now, since point J is between points D and G, angles DGB and JGB are the same angle, so angle JGB is a right angle by substitution. This means that triangle JGB is a right triangle. SInce point J is between points A and B, lines AB and JB are the same line. Recall that lines AB and EF are the same line, which means that lines JB and EF are the same line. Since lines EF and FG are perpendicular, this means that lines JB and EF are perpendicular, so segment FG is an altitude of triangle JGB, specifically the altitude to the hypotenuse. Finally, as the altitude to the hypotenuse of a right triangle divides the triangle into similar triangles, triangles JGB and JFG are similar, and I'm done.

[@MurkySchlaffler]

i wish my teachers in school had a rewind button like here, lol. Good video.

[@LeAstronaut]

At 6:36, I think it is supposed to be the square root of 5, because you used the property (x^2-y^2) = (x-y)(x+y)

[@anon6514]

Circle: y = sqrt(rr-xx) Line AB: y = r/root2 Take the difference: y = sqrt(rr-xx) - r/root2 The inscribed quadrilateral is a square under this condition: y = 2x In terms of x: 2x + r/root2 = sqrt(rr-xx) Square it: [ 5 ] xx + [ 2r root2 ] x + [ -rr/2 ] = 0 Solve quadratic: x = ( -2 +|- 3 ) r / 5root2 The negative root is the x-value at pointA ( a rectangle with no height? * ) So the solution is in the positive root: x = r / 5root2 The side-length of the mini-square is double: 2x = r root2 / 5 Length of line AB is r root2 So the ratio is 1/5 * Edit: It doesn't have no height. I was thinking x is negative but y is zero? But actually, it's the y value at point D So, really the negative root is just the larger square... Neat.

[@BenHermanGaming]

I just graphed sin(θ)-sqrt(2)/2=cos(θ)*2 well solving HG and GF when r = 1 then used cos(θ)*2 to get the side length

[@bilalabbad7954]

Geometry is interesting but in the same time it is hard to deal with

[@stmisbehavin662]

Taking this a tiny bit further, it's easy to show from this ratio that the square root of two has to be just a little bit larger than 1.4. (Extra credit: √2 - 1.4 is just a bit less than 1/70. Use the third square in this sequence to show why!)

[@MuharremGorkem]

Bonus Question: Draw all 4 such small squares (Gen-1) on all sides. In the second round draw also 4 more even smaller Gen-2 squares sitting in between of Gen-1 squares and the circle. Continue in this fashion. Now sum up areas of all squares including the Gen-0 starting square. What is the ratio of this sum and the area of the circle?

[@Wiebejamin]

I did this a much more complicated way. Used some trig formulae from that second triangle then used sin^2+cos^2=1. Wound up getting 2/5, I think I made an arithmetic error due to the hasty scrap paper I'm using, must've thrown an extra 2 in somewhere.

[@datalaforge]

Great video! @7:00 I don’t remember how to reverse foil like that.

[@kenhaley4]

It would be a little simpler to start (w.l.o.g.) by assuming that y = 1, and solve the equation for x.

[@qveciana]

Well...I worked a little more but I obtained the result. Put A(0.0), draw a generic centered rectangle EFGH there and resolve when the rectangle is a square. It's just awesome to get the ratio 5 !!! Amazing mathematics !!

[@killianobrien2007]

Not 5!!!, just 5

[@qveciana]

@@killianobrien2007 Yes, OH my God !!! Why 5 ? For the Jackson five ? 'Cause we have 5 fingers in each hand !!!🙌

[@MarCamus]

I constructed another small square so that we have 2 intersecting chords and then I used the intersecting chords theorem to get an equation with X and Y and then I solved for x/y

[@stephenlesliebrown5959]

Thought "ratio of the two squares" was asking for the blue area to the green area so got 25, otherwise ok.

[@ingewaldjohnsson7683]

Yes ,i thought so to!

[@oesoy]

First time watching his vids and i'm impressed with how abrupt he ended his video

[@Nikioko]

Side ratio: r - r/√2 : 2r/√2. OK, roughly :D

[@donaldasayers]

Did it faster than the video. Started with a big square of side 2 and a little one of side 2y to go straight to the same quadratic.

[@scriptguru4669]

did it by eye

[@heygooooooooo]

Very smooth edits in this video!

[@infiniteseries6210]

The following question came to my mind: Put another square on top of the little one inscribed in the smaller circle segment and so on. Does the sum of the areas of the infinitely many squares converge or not? To what value if it does? After the third step or so you get messier and messier nested expressions with square roots. Mmmhh. Is there a way to figure it out?

[@mbarekouazragh9982]

A area of the initial square, the infinite sum is S = A x (1+1/5+1/5^2+1/5^3+…) geometric series 1/(1-1/5)=5/4 So S converges to 5A/4

[@chandlergloyd4230]

the answer is whatever you want because youre a teacher so youre always right

[@rob876]

At 6:21, you have 5y^2/4 + xy - x^2/4 why did you not just divide this by y^2 to give 5/4 + x/y - 1/4(x/y)^2 = 0? then multiply by 4 and set x/y = u to give 5 + 4u - u^2 = 0 => (1 + u)(5 - u) = 0 => u = 5, which is the ratio we're after

[@quietx6651]

I used the eye test and ruler to find x/y= 5

[@ABaumstumpf]

I am curious how exactly the problem was phrased. If it just says that 2 points of EFGH are on AB and it has to touch the circle then the biggest square would be the easy solution of just - it is identical.

[@fedorkochemasov4533]

I don't get why we can assume that the smaller side of the second triangle he drew is half of the small square. Yes it looks like that in the picture, but what if the picture doesn't represent the scale or it's actually 0.5001 of the square but we just don't see that? Can we have a proof for that at all?

[@dollface9448]

What if you kept putting squares in the space left at the top? What would the side length of the nth square be?

[@edwardlulofs444]

Fun. Thanks.

[@MrDoctorDen]

Surprisingly enough no pi appeared in final answer, despite circles beeing involved

[@abrahammekonnen]

Nice problem. I've been playing with this since yesterday.

[@priyanshugoel3030]

What if we make a formula for a max size of a square in an arc of given radius and arclength or angle.

[@Turalcar]

Region name looks like Adygea except one of the bottom loops is missing

[@1oo1540]

If the goal is to find the ratios of side lengths, couldn’t you scale r down to 1 for the problem?

[@petersievert6830]

Suggesting to set w.l.o.g. x=1 (Esp teachers should strive to make it as easy as possible to grasp, so as teacher I'd go for that :-p )

[@chessandmathguy]

Beautiful.

[@xshortguy]

I'm curious to know what Michael thinks of Norman J Wildberger's rational trigonometry and if we can use those methods to do this same problem.

[@lucachiesura5191]

other way: it is a problem of optimization of the area not included in the squares under the first square...

[@colefinney4986]

Lol my physics teacher has been working on this for like 2 weeks. It’s always in the background of the lesson

[@dominicdelprincipe2583]

top of the head answer: 6:1! (proceeds to watch video) Haha! Proof that I suck at math! The initial guess was supposed to be 6, so I changed it.

[@swierkupitok7264]

let someone remind me to check just [r-(x/2))/(x/2)] ratio as it seems to be easy way to count it

[@ecstasyofgold888]

Love me some geometry

[@AmbachtAle]

Your drawing was so good I looked at it and thought, the answer is going to be near 5. I like to do this as a check to see if the answer is reasonable. Yes indeed, your answer is reasonable.

[@jero37]

My initial approximation is a little less than 1:(sqrt(2)-1)

[@Louigi36]

Geometry was always my weakest subfield of math, so I just used the equation x² + y² = 1 and solved for x. I'm saying "just", but it took quite a bit longer and I kept having to deal with square roots and accidentally created a second negative solution for x where I still don't quite understand what that value would even correspond to in the drawing, but nevertheless, I got to 5. It was fun.

[@irrelevant_noob]

... but x² + y² isn't necessarily 1, it's only that for the single case of x = 5/√26. o.O

[@Louigi36]

@@irrelevant_noob The goal was to calculate a ratio, and ratios are unaffected by scale (otherwise the question would be unsolvable). Or to put it the other way round: If you have to calculate the ratio, you might as well do it for any one particular case, since the answer will be the same for every case. You can obviously do the same steps with x² + y² = r², but all you're doing is needlessly add another parameter to the equation.

[@irrelevant_noob]

@@Louigi36 well, not with the notations shown in the video you can't, x² + y² = r² is incorrect. But sure, whatever makes you feel good about your approach... ¯\_(ツ)_/¯

[@Louigi36]

@@irrelevant_noob The notations were part of his solution, not part of the question. So anyone who solves it before watching this particular solution will have a different notation. x² + y² = r² is the common notation for the circle equation, maybe that's where your confusion stems from. x and y refers to coordinates, so I was essentially calculating the coordinates of the small square to determine its size.

[@irrelevant_noob]

@@Louigi36 pretty sure it's not me who's confused... And btw a square doesn't have (2-dimensional) coordinates. -.-

[@kudui21]

This is simple puzzle but I'd fail with selecting red triangle.

[@Degenerecy]

If I showed this to my teacher, he would of failed me for not showing my work via flowchart.

[@ReapersRed]

Was it ever explicitly stated that the square in between the circle and main square was in the center of the side ab?

[@kenhaley4]

No, but it can be easily demonstrated by observing the symmetry. If you move the little sqare to the left or right of center (keeping EF on AB), it will be impossibile for both of the two upper corners (G and H) to lie on the circle.

[@ReapersRed]

@@kenhaley4 ah I forgot that it was tangent on both corners

[@DavidRutten]

An integer!? No way!

[@markfischer5044]

I went from the thumbnail alone so I wound up solving for the ratio of the areas...essentially the same problem.

[@surajthakur340]

At 4:06 onwards, Is y/2 taken just by observation or can it be found out?

[@new-knowledge8040]

Okay, at one point I was into into the English Gematria, where A=6, B=12, C=18, D=24,...Z=156. I was told, by those whom I will not mention, that the 2nd coming of that Jesus Christ fellow somehow meant coming full circle and thus it was related to pi. Say WHAT?? Anyhow, the JC character was known as the messiah, the son and the lord. I knew that based upon the English Gematria, that the MESSIAH=444, and SON=288, and that the LORD=294. And by no coincidence, SQRT=444 as does JESUS=444. So if you take the messiah/son/lord divided by the second coming of Jesus, which is represented by SQRT(2), then that should mean something related to pi. 4.44288294 / sqrt(2) = 3.14159265..... Wow because pi = 3.14159265...... LOL

[@FlatEarthMath]

Excellent! And elegant. :-) I solved it a tiny bit differently, starting with radius = 1, then using the same narrow red triangle on the chalkboard. This saved me from solving a two-variable problem with one equation, but it gave me a whole lot of √2 terms. With x being the unknown small square's side, I ended up with: 5/4 x^2 + x√2 - 1/2 = 0. Plugging that into the quadratic formula gave me x = √2/5 which I compared to the large square's side of √2. So my strategy of avoiding the second variable gave me a lot more work!

[@visiblydisturbed1688]

Is it even possible to have this situation and have the answer not be five?

[@minidialga64]

Idk why youtube recommend me this video when i can barely do calculations with fractions

[@shjescaresme]

Cool problem! I'll share another solution. Let x = AE and y = EF (the side of the small square). Extend GF until it intersects the circumference at G'. Then the power of point F is AF · FB, which is equal to GF · FG'. Then, (x + y)x = y(2x+2y). Dividing by (x+y) gives x = 2y. The side of the large square is 2x+y, so the ratio is (2x+y)/y = (2·2y+y)/y = 5.

[@lyrimetacurl0]

I wonder what the size of the even smaller square above that one would be? Could be worked out in the same way. Could be called the "cat in the hat" problem 😁

[@lyrimetacurl0]

I worked it out to be around 1/141 point something of the biggest, not a clean integer 😣

[@twelfthdoc]

One of the rare instances where I think of possible solutions before we dive in, and my brain intuited that the ratio was probably an unusual number for working with squares and circles, but a relatively simple one, and so immediately came up with the ratio 5:1. It was gratifying that my hypothesis just by eyeballing the problem ended up actually being the solution!

[@KatzeARL]

I saw this as a thumbnail at the end of a video and read "ratio of who cares" lol I care! I care a lot! :D

[@Etothe2iPi]

I can't understand why you are using Pythagoras to determine the ratio between the side and diagonal of a square. everybody should know since 5th grade what it is.

[@vgstep]

Very interesting problem