Did I make this too hard?

[@javiernovillomartin]

I think the solution should be 1/((2+√2)*√2), when you're trying to simplify the last expression on the board. Great video!

[@nilsastrup8907]

Correct, and that is equal to 1/(2sqrt(2)+2), so he might have just forgotten to wright the sqrt(2)

[@javiernovillomartin]

@@nilsastrup8907 that's it 😄

[@PascoeMichael]

Yes, you are right! s = 1/(2(1+sqrt(2)) = (sqrt(2) - 1)/2; Bye!

[@The_Math_Enthusiast]

Yes

[@xxxx015]

Now approximate that to 1 so it's nicer and easier to work with ,,🤣🤣😂

[@Bodyknock]

Two small errors: 9:25 He accidentally wrote 2r^2 twice when he meant to write 2r the second time 12:41 The denominator is missing a 2 in front of the root 2.

[@leif1075]

Another error: as r goes to infinity, s goes to infinity too because you can just extend the length of the square s until it reaches the other side of the triangle..I don't see why not?..unless the 1 length is fixed and you can't increase the height of the triangle from 1 to anything else..?

[@Bodyknock]

@@leif1075 The 1 was a fixed length, that was part of the assumptions in the set up.

[@Walczyk]

@@leif1075 no

[@Walczyk]

@Leonardo Balestriere its cool that it can be rewritten as sqrt(2) - 1

[@Gianmarcos-ny1qd]

@@leif1075 when r goes to infinity the area of the triangle and the circular sector are equal to r^2/2 so that means that the area of the square (s^2) is 0

[@goodplacetostop2973]

13:06 The 100K subs Q&A was the sacrifice to make to have these top-tier intros

[@jbthepianist]

No homework?

[@goodplacetostop2973]

@@jbthepianist There is the homework, just scroll the comments

[@jbthepianist]

@@goodplacetostop2973 Sorry I missed it lol

[@goodplacetostop2973]

@@jbthepianist Don’t worry it’s fine.

[@HideyukiWatanabe]

10:30 You can also use AMGM; 1/s = 2r+1/r + 2 >= 2sqrt(2r/r)+2 = 2+2√2, and so s

[@gandev]

Neat!

[@allenminch2253]

Absolutely awesome problem combining geometry, algebra, and calculus! Thanks for sharing!

[@failsmichael2542]

Denote four vertices of the square by M, N, P, Q where M, N belongs to AB, P belongs to BC and Q belongs to the arc. Draw QH // AB (H belongs to AC) Easy to see that BN/BA = NP/AC = s/r => BN = s/r Thus QH = AM = 1 - s(1 + 1/r) Applying Pythagoras theorem in triangle QHC yields (r - s)^2 + [1 - (r + 1)/r * s]^2 = r^2 which simplifies to (2r^2 + 2r + 1)s^2 - 2r(r^2 + r + 1)s + r^2 = 0 You can check that there are two roots of this equation, namely s = r and s = r/(2r^2 + 2r + 1). But we must have s < r so the only admissible value for s is s = r/(2r^2 + 2r + 1) Finally, by AM-GM 2r^2 + 1 >= 2sqrt{2} r so we can conclude that s

[@barsercan7203]

This channel is getting better every single day. Great intro and problem!

[@khaloscar]

This is exactly the sort of problems I am looking for. The problems my math course give is dull and pretty linear. This problem was complex and interesting with lots of nooks and crannies. In my mind I call this savvy maths and it is exactly what I need to develop. Thank you.

[@udic01]

12:39 the denominator should have 2sqrt(2). And after you simplify you get (sqrt(2)-1)/2

[@davidgould9431]

There's a nice geometric consequence of this (see below). Solving it: I found it a *tiny* bit easier to flip the diagram left-right and put the origin at C, giving equations y = x/r and x² + y² = r². I then substituted the (x,y) coordinates into the circle equation as Michael did. The quadratic was, by then, just as gruesome. Another (slight) improvement was setting s = 1/( 2r + 2 + (1/r)) = (2r + 2 + (1/r))⁻¹ and finding s' is just a tad easier (for me - your mileage may vary). Geometry. Once you've solved and got r = 1/√2 = √2/2 and s = 1/√2 - 1/2, you find the distance of the top of the square from B, s/r = 1 - r which suggests a nice construction: Extend CA to CD where |CD| = 1 and complete the square, passing through B, of course. The unit square has diagonal √2 so our circle passes through the unit square's centre because its radius is ½√2. It then turns out that the little square's diagonal is on the diagonal of the unit square, with the corner on the circle at the unit square's centre. That's easy to see if you consider the circle centred at A with radius r. So we have three squares with colinear diagonals: the unit square; a square side length r and the square side length s at the top corner of the latter. Which I thought was a lovely, neat end to the problem.

[@bmenrigh]

First off, GREAT video. The drone intro and chalkboard on the tree is an amazing setting. Second, the math was great. Third, I always try to guess at answers to you geometry maximization problems an this case, I said "It's probably one but maybe it's sqrt(2) or 1/sqrt(2)" and I'm pleased that one of my intuitive guesses was correct.

[@maharanirani54]

When I saw the trees I thought it was an ad lol😅. I always like the videos, thank you.

[@AmritGrewal31]

Some chimp on a nearby tree: "interesting..." 🦧

[@ChrisConnett]

I solved this in a very different way by introducing a radius from C to the lower right of the square and then working with everything out in terms of θ, the angle between AC and the auxiliary radius. My derivative was a bit uglier, but got to the same answer. s = (cosθ - cos²θ) / (1 - cosθ + sinθ); s has a maximum at θ=π/4, which I think is a beautiful result and shows off a lovely symmetry.

[@JNCressey]

I would have done the start a little differently. Working a bit more with the geometry, instead of working with coordinates and the functions that plot the lines. Here's how: # Labeling Let D, E, F, G be vertices of the square, such that: • D and F are on the line AB with the order going AFDB, • E is on the hypotenuse BC, and • G is on the circle (centre=C,radius=r). Let H be the point on the line AC such that GH is perpendicular to AC. Let x be the length HC, and Let y be the length GH. (edit: originally I wrote x and y as r*cos(θ) and r*sin(θ). But only ended up using the Pythagoras theorem, so they're simpler as x and y.) # Pythagoras theorem observation Observe GHC is a right triangle with its right angle at H. Therefore x^2+y^2=r^2. # Similar triangle observation Observe DBE is a similar triangle to ABC. DE=s and AC=r gives us the scale s/r. AB=1, therefore DB=s/r. # Form equations from the lengths We have • AC=r, • AH=s [because its parallel to DE], and • HC=x. Therefore we can make the equation x=r-s. We have • AB=1, • AF=y [because it's parallel to GH], • FD=s, and • DB=s/r. Therefore we can make the equation y=1-s-s/r. # Combine into one equation We can substitute the two equations we just got for x and y into the Pythagoras theorem x^2+y^2=r^2 to get the following equation: (r-s)^2+(1-s-s/r)^2=r^2. Simplifying and collecting the orders of s, we get this equation as in 9:30 in the video: (2r^2+2r+1)*s^2 -(2r^3+2r^2+2r)*s + r^2 = 0. Then I would have continued the same.

[@mrphlip]

Very nice, and well explained, this is the approach I took too.

[@robertingliskennedy]

top notch Michael, great presentation

[@hiddeneagle1408]

Really dig you trying out these new settings! You're an inspiration to us, Mr. Penn!

[@patrickpablo217]

fun bonus fact: when the square is maximized this way, the point where the square touches the circle is simply at y = 1/2 and x is 1/2 *back* from the center of the circle. This also means that if you draw the diagonal line from the top left corner of the square down through the lower right corner where the square touches the circle, and you then continue that line down to the x-axis, that diagonal goes through the center of the circle. ...unless, of course, I messed up the algebra somewhere :)

[@GeorgeFoot]

Nice problem, beautiful how simple the answer becomes

[@drkankenstein6943]

Wow, great set up 💥💥

[@The_Math_Enthusiast]

I would say: Please don't change this intro. I would have paid for this!!

[@PlutoTheSecond]

I think he should do one of those cheesy lifestyle show intros like Michael Stevens did on D!NG.

[@petersievert6830]

I arrived at 9:47 on my own, saw that convoluted quadratic formula and then tought: well, maybe not the right approach :D

[@robertgerbicz]

Without derivation: r/(2*r^2+2*r+1)

[@stevenwilson5556]

It hurts me that your amazing videos have such small view counts… your channel is making me fall in love with math even more than I did before I started watching. You make relatively challenging problems seem easy.

[@emersonschmidt7882]

Great video. I, like others, have used geometry and arrived at the second degree equation in your board...

[@thibaud5764]

Love your new editing

[@bobzarnke1706]

A more geometric first few steps. Label the 2 right side vertices of the square D and E and extend DE to meet AC at F; then FC = (r-s). Let EF = x. ΔDFC is similar to ΔBAC; so (s+x)/(r-s) = 1/r. ΔEFC is right angled; so x^2 + (r-s)^2 = r^2. The rest of the proof is the same.

[@txikitofandango]

I did it the same way as you. Got an incredibly messy quadratic equation. And, with a little encouragement from Wolfram Alpha, it all worked out. Amazing.

[@manucitomx]

The peacefulness of the intro set us up for the barrage of square roots. Well done! Thank you, professor.

[@mryip06]

I prefer the intercept form x/r + y/1 = 1 for eqt. of BC

[@anshumanagrawal346]

Yes

[@tonyhaddad1394]

Michael amazing problem 🙏🙏

[@kujmous]

I had this happen the other day. I had done all the hard steps to find the critical value, but I slipped up in evaluating the answer. If this were a problem worth twenty points, I'd say nineteen were earned. Excellent explanation of your thought process, and your intro was SO BEAUTIFUL. A chalkboard in a forest is a good place to stop.

[@Chrisuan]

Really cool approach with the coordinate system and line/circle equations. I'd have used geometry/trig functions and probably made it a lot harder :D

[@vinc17fr]

One can do simpler in the following way. Let y denote the distance between A and the square. Due to the similar triangles, one has (1 − y − s)/s = 1/r, which gives y = 1 − s − s/r. And the point on the circle, one has (r − s)² + y² = r², i.e. (r − s)² + (1 − s − s/r)² = r², which gives (1 + (1 + 1/r)²)·s² − 2(r + 1 + 1/r)·s + 1 = 0. Solving for s gives s = 1 / (2r + 2 + 1/r), which is maximum when 2r + 2 + 1/r is minimum. Derivative (much simpler than in the video!): 2 − 1/r² = 0, giving r = 1/√2, and s = (√2 − 1) / 2.

[@SyranoTV]

This outside setup is a really great place for mathematics !

[@wjx8439]

here's a neat pattern I found after solving the quadratic equation for s. [note that there is a mistake when terms are collected, below is the correct one] (2r²+2r+1)s²-(2r³+2r²+2r)s+r²=0 we can rearrange the terms after finding out that the coefficients of s² and s are awfully similar. (2r²+2r+1)s²-(2r³+2r²+r)s-rs+r²=0 (2r²+2r+1)(s²-rs)-rs+r²=0 (2r²+2r+1)(s)(s-r)-(r)(s-r)=0 as s=r will lead to s=0, we discard the solution, in fact you can also substitute s=r into the original equation [s=1-s/r-√(2rs-s²)] and find out that s=r is actually an extraneous root produced after the equation squaring both sides. and finally, (2r²+2r+1)(s)-r=0 or just s=r/(2r²+2r+1). bonus: since 1/s=2r+2+1/r≥2√2+2 by AM-GM inequality, s≤1/(2√2+2)=(√2-1)/2, with equality reached iff 2r=1/r, or r=1/√2.

[@V1DE0DR0ME]

This is amazing. I was stuck at the quadratic equation. If we fail to see this pattern, is there a way to solve the quadratic equation manually? I ended up with messy r^6, r^5, r^4, r^3 and r^2 terms within the square root...

[@wjx8439]

@@V1DE0DR0ME b^2-4ac =(-2r^3-2r^2-2r)^2-4(2r^2+2r+1)(r^2) =(4r^2)(r^2+r+1)^2-(4r^2)(2r^2+2r+1) =(4r^2)[(r^2+r+1)^2-(2r^2+2r+1)] =(4r^2)[(r^2+r+1)^2-2(r^2+r+1)+1] =(4r^2)(r^2+r+1-1)^2 =(2r)^2*(r^2+r)^2 =(2r^3+2r^2)^2 hence s=[(2r^3+2r^2+2r)-(2r^3+2r^2)]/(4r^2+4r+2) =r/(2r^2+2r+1) [+ version leads to s=r so i won't write it here]

[@V1DE0DR0ME]

@@wjx8439 Oh wow. That +1 forcing out the squares was brilliant. Thank you.

[@wjx8439]

@@V1DE0DR0ME No worries, hope it helps.

[@ignacioelia759]

The presentation was TOO COOL FOR SCHOOL

[@vinayteja3276]

Bro thanks for teaching daily one new concept

[@CrystalblueMage]

13:07 I think a squared snuck in there, but it wasn't used in the rest so...

[@bastian1833]

Yet another solution: Let D be the point where the square touches the circle. It's coordinates are (s,1-s-t) where t is the length above the square on AB. For similarity, it is t/s = 1/r or t = s/r and therefore D = (s, 1-s-s/r). Plugging this into the equation for the circle yields (r-s)^2 + (1-s-s/r)^2 = r^2 from which it follows that (r-s)^2 = r^2 - (1-s-s/r)^2 and by the third binomial formula (r-s)^2 = (r-1+s+s/r)(r+1-s-s/r). Observe that r+1-s-s/r = r-s+(r-s)/r = (r-s)(1+1/r) and cancelling r-s on both sides of the former equation yields r-s = (r-1+s+s/r)(1+1/r) = r-1+s+s/r + 1-1/r+s/r+s/r^2 = r+s+2s/r-1/r+s/r^2, or rearranged 2s+(2s-1)/r+s/r^2 = 0. This are the zeros of a parabola in 1/r dependent on the parameter s, which has the zeros 1/r = (1-2s +/- sqrt((1-2s)^2-8s^2))/(2s). For most values of s, there are two solutions for r that yield that square with sides s, however, if s is too large, there is no solution at all and the sqrt will be imaginary. If the sqrt is 0 there is only one solution and that's where the square is largest. Therefore, for the optimal solution it is 1/r = (1-2s)/(2s) where for s it holds that (1-2s)^2 = 8s^2 or 1-2s = +/-2s*sqrt(2) and therefore, s = 1/(2+/-2*sqrt(2)). Since s has to be positive, it is s = 1/(2+2*sqrt(2)) and 1/r = (1-1/(1+sqrt(2)))/(1/(1+sqrt(2))) = 1+sqrt(2)-1 = sqrt(2), or r = 1/sqrt(2).

[@harriehausenman8623]

The formatting of this comment is questionable :-)

[@harriehausenman8623]

But the reasoning seems legit ;-)

[@waikeanng145]

I just solve using 30s.....just make a line from diagonal of square to C, to form a 45-90-45 right triangle...since it is diagonal, both have same length... hence, using pythagorean theorem... we get sqrt2 r=sqrt 2 s +r s=(1-1/sqrt2) r ds/dr=1-1/sqrt2(max) done!

[@bastian1833]

You don't know that C is located on the extended diagonal of the square in the first place. Also ds/dr isn't appropriate in this case since when you vary the size of the r this way, you also vary the length AB which should remain fixed to 1. But maybe one can somehow argue by symmetry that C should be on the extended diagonal of the square and than use your calculation as part of the solution.

[@matejcataric2259]

Very nice problem!

[@TheBrutalDoomer]

I solved it kinda the same way, just flipped the triangle to have nicer equations of line (y=x/r) and circle (x²+y²=r²)

[@luismex5575]

Great video would you do one more general, instead of a tríangle could be a polygon ? Thanks

[@mienzillaz]

9:23 ups..

[@az0rs]

these calc 1 problems are mind openingz

[@AllanKobelansky]

Very enjoyable video.

[@LightPhoenix7000]

My question is how does this relate to the angle at C? I initially assumed that angle would be 45 since S went to 0 in either direction but is that correct?

[@GinLottus]

If that angle is labeled as C, then cot(C) = r; you can then develope the problem with this equality and solve the side of the square s through C.

[@iabervon]

No, while most maximization problems you see have symmetric shapes like that, the fact that the part you're trying to maximize is bounded by different things on each side means it's not likely to be the case here.

[@michaelfisher9053]

Springtime near Lynchburg, VA, gorgeous (until the mosquitos find you haha).

[@tahasami597]

Thank you

[@qubikstheman]

A nice problem and a nice solution

[@lisandro73]

I don’t know, but I think you miss a root 2 at the end, and the denominator is 2root(2)+2

[@victorhuertatur9049]

What a beautifull resolution

[@ro5197]

When you do all the tough work and then end up writing 1/2+2v2 as 1/v2

[@MarieAnne.]

At the end, denominator should be 2+2√2 → (1+√2+1)√2 = (√2+2)√2 = 2+2√2 And if we rationalize denominator, we get 1/(2+2√2) = (√2−1)/2

[@muddyPassenger]

Don't ya forget to multiply the denominator by root 2? In the very end. Seems it should be 2√2 +2. But I might mistake.

[@mienzillaz]

12:38 ups again..

[@pietergeerkens6324]

With the corrected denominator at 12:48, somehow sin ¼π - sin ⅓π is a much more satisfying answer. For one, it suggests some deeper geometric significance to the solution.

[@GinLottus]

I was looking for this answer... I'm trying to construct this problem in geogebra, but get stucked in the square part, because all of their vertices are mobile. Maybe need to develope more the angle part, instead of the side length.

[@7rgrov198]

Always thought you looked like you did bouldering or rock climbing of some sort. Now im 100% sure

[@davidemasi__]

He actually does rock climbing

[@trueriver1950]

Yeah, he has spoken about his rock climbing in previous videos. He is keen enough to travel internationally to go climbing

[@tgx3529]

I had similar idea, tg alfa=r/1=r. & tg alfa=s/x So s/x=r x+s+y=1 y^2+(r-s)^ 2=r^2, So y=sqrt(2rs-s^2) x=1-s-y x=1-s-sqrt(2rs-s^2) r=s/x r=s/(1-s-sqrt(2rs-s^2))

[@vladimirkobarov7154]

From 9:37 to 9:55, how to make that change? I mean how to get s = r/(2r^2+2r+1)?

[@khoozu7802]

Use formula s= - b+-sqrt(b^2-4ac)/2a

[@moonlightcocktail]

Damn that's his backyard?

[@herlandarmantotampubolon8135]

Natural Environment... Gosh!!.. You are amazing, sir

[@aashitAgrawal]

What a intro 👏👏🔥

[@williamchow4136]

My next math lecture should have intros like this

[@jbtechcon7434]

You really could skip the musical intro. I for one have always liked that you get straight to the problem then stop when the problem is done.

[@MrKrabs-xf2tr]

Is there any other way to do this without Calculus? Maybe utilizing AM-GM or Cauchy of some sort?

[@Blabla0124]

Small error at the end when calculating s for r = 1/sqrt(2): s = 1/(2+2sqrt(2))

[@RAG981]

You did pretty well until the very last value for s which should be one over 2 + 2rt2, which is half of rt2-1. You were trying to finish too quickly! By the way that was quite cheeky to solve the quadratic with so little comment!

[@arielfuxman8868]

maximum value of s should be (sqrt2-1)/2

[@mienzillaz]

Good i had headphones to fully experience that intro..:)

[@stewartcopeland4950]

Correct answer : Smax = (2^0.5 - 1)/2 = 0.207...

[@damianbla4469]

I did this in another way, using triangle geometry and calculus. But I haven't received r=1/sqrt(2) as the special value :( Please help me and tell me where I have made an error. Here is my method: Step 1 The little right triangle (which is above the square) is similar to the big right triangle (which contains the circular sector and the square). The sides of the big right triangle are: 1, r, c (where c=sqrt(r^2 + 1)) The corresponding sides of the little right triangle are: y, s, L (where L = c - r) From the SSS we have: s/L = r/c s = r * L / c s = r * (c - r) / c s = r * (c/c - r/c) s = r * (1 - r/c) s = r - (r^2/c) s = r - [r^2 / sqrt(r^2 + 1)] So "s" is a function of "r": s(r) = r - [r^2 / sqrt(r^2 + 1)] Step 2 ds/dr = 1 - { [2r * sqrt(r^2 + 1) - r^2 * (1 / 2sqrt(r^2 + 1)) * 2r] / [sqrt(r^2 + 1)]^2 } ds/dr = 1 - { [2r * sqrt(r^2 + 1) - r^2 * (1 / sqrt(r^2 + 1)) * r] / (r^2 + 1) } ds/dr = 1 - { [2r * sqrt(r^2 + 1) - r^3 / sqrt(r^2 + 1)] / (r^2 + 1) } Step 3 ds/dr = 0 1 - { [2r * sqrt(r^2 + 1) - r^3 / sqrt(r^2 + 1)] / (r^2 + 1) } = 0 { [2r * sqrt(r^2 + 1) - r^3 / sqrt(r^2 + 1)] / (r^2 + 1) } = 1 [2r * sqrt(r^2 + 1) - r^3 / sqrt(r^2 + 1)] = (r^2 + 1) Multiply both sides by "sqrt(r^2 + 1)" and we get: [2r * (r^2 + 1) - r^3 ] = (r^2 + 1) * sqrt(r^2 + 1) [2r^3 + 2r - r^3 ] = (r^2 + 1) * sqrt(r^2 + 1) [r^3 + 2r] = (r^2 + 1) * sqrt(r^2 + 1) Square both sides and we get: [r^3 + 2r]^2 = (r^2 + 1)^2 * (r^2 + 1) r^6 + 4r^4 + 4r^2 = (r^4 + 2r^2 + 1) * (r^2 + 1) r^6 + 4r^4 + 4r^2 = (r^6 + 2r^4 + r^2) + (r^4 + 2r^2 + 1) r^6 + 4r^4 + 4r^2 = (r^6 + 3r^4 + 3r^2 + 1) r^6 - r^6 + 4r^4 - 3r^4 + 4r^2 - 3r^2 - 1 = 0 r^4 + r^2 - 1 = 0 From the Descartes' Rule of Signs en.wikipedia.org/wiki/Descartes%27_rule_of_signs we know that there is for sure one POSITIVE value of "r" which is solution to the above equation. If f(r) = r^4 + r^2 - 1 then f(0) = (-1) 0 which means that the solution must be between 0 and 1. f(1/2) = (1/16) + (1/4) - 1 = (5/16) - 1 = (-11/16)

[@bastian1833]

I only read the first lines but I noted that the L is not c-r, might the error? the cirle touches the lower right corner of the square but L=c-r would only hold if it would touch its upper right corner.

[@damianbla4469]

@@bastian1833 Yes, that is the error. Thank you very much :)

[@wasp89898989]

Hello. Can someone explain to me how s tends to 0 as r tends to infinity? That part didn't seem trivial to me. Thank you!

[@tomatrix7525]

That intro is sick

[@tonyhaddad1394]

12:48 another intentionally mistake from michael 😭😭😭😭

[@jimskea224]

I got this down to 4 simultaneous quadratic equations and then fed them to Maple. Maple got it right.

[@flwi]

When I watched my first video of your channel yesterday I thought you might be a climber (judging by your physique). Now I saw your chalkboard setup with the ropes and the quickdraws and conclude that you are indeed a climber. And that's a good place to stop ;-)

[@keksauraisks]

Imagine stumbling across this weirdo talking to himself in front of a blackboard in the middle of nowhere

[@LimLux]

What a G intro!

[@CTJ2619]

it appears as though there is an error in your final computation. (1+ SRTQ(2) +1) * SQRT(2) does not equal 2 + SQRT(2).. the denominator should be 2+2*SQRT(2) (i.e.( 1+1+ SQRT(2))* SQRT(2)

[@goodplacetostop2973]

HOMEWORK : ABCD is a square such that AB lies on the line y = x + 4 and points C and D lie on the graph of parabola y^2 = x. Compute the sum of all possible areas of ABCD. SOURCE : 2013 SMT

[@goodplacetostop2973]

SOLUTION *68* Let C = (c²,c) and D = (d²,d), and assume without loss of generality that the points are positioned such that c < d. Viewing this in the complex plane, we have B − C = (D − C)i, so B = (c² + c - d, d² - c² + c). Plugging this into y = x + 4 gives us d² - 2c² + y - 4 = 0. Since AB ‖ DC, the slope of DC is 1, so (c - d)/(c² - d²) = 1 ⇒ c + d = 1. Solving this system of equations gives us two pairs of solutions for (c, d), namely (−1,2) and (−2,3). These give √18 and √50 for CD, respectively, so the sum of all possible areas is 18 + 50 = 68.

[@stewartcopeland4950]

@@goodplacetostop2973 Graphically, we see that the answer is rather 136 (square of 100 + square of 36), a factor 2 compared to your solution

[@goodplacetostop2973]

@@stewartcopeland4950 Hmm that’s weird. Do you have the coordinates of these points?

[@stewartcopeland4950]

@@goodplacetostop2973 By solving analytically, I get the abscissa x = 4 and x = 9 for the points on the positive branch of the parabola: this confirms your result of two squares of 18 and 50 respectively: the graphic approach was misleading

[@MinhTran-fl7qg]

There's an error at 9:25, where you said 2r but you wrote 2r^2.

[@profamitgupta]

You missed a 2^(1/2) in the denominator.

[@NotBroihon]

The hardest part was by far solving that qudratic equation.

[@Lirim_K]

I really wonder what a NOT good place to stop would be?

[@شبلالإسلام-ظ6ض]

thats a nice place wow Where is it ?

[@swapnamoy6134]

You have backyard that big😳

[@ghabhitho]

Wouldn't the answer be = 1/(2*(2)^0.5+2)? Very good video

[@tonyhaddad1394]

4:23 michael starts to complicate the problem :p😂

[@davidgould9431]

Well, I think he started at around 2:30 to complicate it: if you flip the diagram left-right and make C the origin, the equations are y = x/r and x² + y² = r² which were easier to handle. Disclaimer: I solved it and got a nice solution, but haven't watched further than about 3:00 yet.

[@jesusalej1]

Hi guy, you missed up at the end, it happens when you work. But excellent video.

[@scialomy]

The Left/Right sound effet in the opening is really disturbing.

[@raffaelevalente7811]

The solution is $\frac{\sqrt{2}-1}{2} \approx 0.207$

[@macurvello]

Is it somehow relevant that that radius of 1/sqrt(2) is equal to sin(pi/4) ?

[@trueriver1950]

I wondered that, but in fact the 1/sqrt2 comes as a tan or cotan in that triangle, not as a sin. I finally decided after some messing about that it was a red herring, but am willing to be enlightened if anyone can show me the connection...

[@CrossMax122]

Are the colors of the thumbnail randomly chosen?

[@davidemasi__]

The king of the forest

[@jimbrown5583]

These new intros are snazzy

[@BillSmithPerson]

Hipster music at start of math lecture

[@trueriver1950]

The start is very Tibees :)

[@realevan]

I gotta know what psychos woke up early on a Saturday to give this a thumbs-down lol

[@pedrov8868]

Haters