a really cool geometry puzzle!

[@bigjazbo9217]

At 6:40, I think length of hypotenuse of smaller triangle should be sqrt(2)*(1-r)-r-x.

[@ThAlEdison]

Yeah, me too

[@flexeos]

same here

[@oloyt6844]

Same it instantly stuck out to me because obviously you just subtract r and x to get the new hyptenuse.

[@thorntontarr2894]

I just saw your comment after I placed my second comment.

[@trumanmohr1153]

Yeah I think the correct answer is supposed to be (56 - 39 sqrt(2) - 32 sqrt(3) + 23 sqrt(6)) * π / 16. The answer that he gives for the total area is less than the area of the first circle of radius r

[@jagmarz]

6:43 - I really don't understand where the sqrt(2)(1-r-x) comes from? Is that supposed to be sqrt(2)(1-r) - r - x instead?

[@filbranden]

11:14 we have 12 - 9 which is 9??

[@normanstevens4924]

Which is true for small values of 9.

[@minwithoutintroduction]

@@normanstevens4924 hhh

[@JBMJaworski]

Filipe you are right! The denominator is 3. Professor Michael Penn made a mistake.

[@GuilhermeBelmont]

He mistook to know who saw until th end

[@HelloWorld-dq5pn]

6:36 Shouldnt it be sqrt2(1-r)-r-x?

[@Dshado]

How would you know that the quarter circle intercects at the tangent point of lines and the biggest circle?

[@JosuaKrause]

yeah, this is kind of crucial to the proof

[@jounik]

Wasn't that part of the construction? Edit: I see what you mean. "Because of the rays" is superfluous, the circle is drawn _so that_ it intersects the rays at the same points it intersects the quarter circle. Edit again, nope, that doesn't work for most other angles. See below.

[@Dshado]

@@jounik well depending on the angle between the lines it wouldn't work, right?

[@jounik]

@@Dshado You are right and I am wrong. It does need to be shown to work out.

[@JosuaKrause]

@@jounik No the construction was 30 degree angles. Nothing there to force the points to align.

[@thichhochoi766]

Did he miscalculate the hypotenuse of the second right triangle? The first one is Sqrt(2)*(1-r) but how the second one = sqrt(2)(1-r-x)? Should it be sqrt(2)(1-r) - r -x ?

[@michaelempeigne3519]

typical penn thinking.

[@RAG981]

I agree. He has done it wrongly.

[@Jack_Callcott_AU]

Small computation error at the very end...should be: (2*√3-3)/3.

[@fonaimartin98]

For those who are wondering, why the unit circle passes through the points where the rays meet circle 1: 1) Because the angle in which the ray 'A' and the side of the square meet just happens to be 120°, the inner bisector of that angle is perpendicular to the other ray 'B'. 2) As this bisector does clearly contain the center of the smaller circle (as the center must be equidistant from those lines whose angle is bisected), the point in which it intersects the other ray 'B' is actually the point where that ray 'B' touches the circle. 3) Moreover, ray 'A', ray 'B' and this bisector are forming a triangle congruent to that triangle formed by the sides of the square and ray 'A'. This means the point of tangency is a distance of 1 away from the bottom left corner. Also have I found a nice shortcut to this problem: there is actually no need to calculate the radii of the smaller circles. 1) We know that there is a given amount of line length for all these circles to fill. 2) We know that they actually fill that line length without any left over, because there are infinitely many of them. If they did not fill the line, we could add another circle, which is a clear contradiction. 3) We can speculate that they form a geometric series due to self-similarity reasons. This means the common ratio can actually be found, and can be immediately applied in the area sum.

[@ВасилийКоровин-г9э]

The ease with which Michael left it without explanation reminded me of an anecdote: Professor in math class: -...so it's obvious, that... - Excuse me, professor, why is it obvious? Professor stared at the blackboard for a while, then ran away off the classroom. He returned 20 minutes later: - I just talked to my colleagues... It's obvious indeed.

[@fonaimartin98]

@@ВасилийКоровин-г9э certainly one of my favourite anecdotes :) What actually bothers me more about this solution is that noticing these 3 objects passing through the same point is like... exceedingly hard; compared to the advantage caused by noticing it, at least. I mean the other way is brute-force calculating the radius of the inscribed circle of a deltoid. Does not seem that fancy.

[@vineetkaddu1214]

The way I did was to inscribe a circle in a trapezium formed in this particular triangle. I then calculated the ratio of areas of said circle and trapezium. Then I calculated the area of the triangle with height that terminates at the outer tangent of last second circle. Then I multiplied the area of triangle by the ratio of area of circle to trapezium, since that triangle is basically the sum of all trapeziums. I then added the area of the last circle. IDK if I made any mistake in calculating the final answer but i am too lazy to type my value in the calculator 😅

[@Jack_Callcott_AU]

Regarding that first fact that you mentioned I wondered about that too, and I think Michael should have made a remark about it, because it's not obvious to me that the point of tangency where the circle meets the rays should intersect with the unit circle quadrant at the same points. Is this just an accident of the right angle = 90° being trisected. ...The way I solved the problem in hand was to find the ratio of the radius of 2nd circle to the radius of the 1st circle, which is x/ r which I call the "scaling factor" = SF ( in this case SF = √3-1) Because the circles are all similar the ratio of the area of any given circle to the area of the one before it is C_n/C_(n-1) = SF^2 = (x/r)^2 = 4-2*√3. so C_n = (4-2*√3)^n*C_0 where C_0 is the area of the first circle = (7-4*√3)*π. Thus we have a formula for the total area as a geometric series which can be easily calculated. 📽⛑

[@gerryiles3925]

As several others have pointed out, the sqrt(2)*(1-r-x) is incorrect. Using the correct value gives a considerably more complex expression for the common ratio: x = r * (sqrt(6) + sqrt(3) - sqrt(2) -2) / (sqrt(6) - sqrt(3) - sqrt(2) + 2) This ratio is approx. 0.58879 instead of 0.73205 giving a very different final answer. This video really needs a re-record and re-post...

[@ivandebiasi6657]

When you calc the second ratio (third radius over second radius) how can you prove it is equal to the first one? I obtain that y/x = [sqrt(2)(1-r)-r-2x] / [sqrt(2)(1-r)-r] while the first is x/r = [sqrt(2)(1-r)-r-x] / [sqrt(2)(1-r)-r]... i made a mistake or there is no common ratio?

[@gerryiles3925]

@@ivandebiasi6657 The common scaling ratio should be "obvious" by the construction. For y/x it looks like you've forgotten to subtract a y from the numerator and an x from the denominator, you need to divide OCy by OCx, where O is the "origin" and Cx and Cy are the centers of the 2nd and 3rd circles, check these lengths carefully...

[@ivandebiasi6657]

@@gerryiles3925 You are right but it can be reduced to the form I wrote by multiplying the denominators, eliminating the common factors and dividing again

[@ivandebiasi6657]

This is a general fact: if (a/b) = (s-a)/(p-b) then a/b = s/p

[@gerryiles3925]

Duh, yes, that's true, I was only looking at the details of the expression and saw the lengths weren't right, didn't even consider you'd already somewhat simplified it. If you manipulate your expressions for the ratios to be of the form "smaller radius" = "scale factor" * "larger radius", so, x = ar and y = bx where a and b are the scale factors completely reduced to not include any of the "variables" then you should get the same common ratio, e.g. above I rearranged the "x/r =" expression to solve for x in terms of r, then I factored out one r and substituted in the rest to get x = r * scale factor. If you do a similar (though considerably more complex) process on your y/x equation to get y = x * scale factor then you should get the same ratio.

[@awildscrub]

2 questions: 1. 3:15 how would you know that the circle, quarter circle and straight line all intersect at the same point? 2. 6:29 how did you arrive at the conclusion that the length is √2(1-r-x)

[@oloyt6844]

yeah we want a proof, maybe try prove it yourself :)

[@thorntontarr2894]

And @ 6:40 by inspection the smaller hypotenuse is sqrt2 * (1-r) - (r+x).

[@carstenmeyer7786]

6:42 As a lot of others have mentioned, √2(1 - r - x) is not correct. Let's do it right, using notation for recursive relations: *radius: x -> r_k, r -> r_{k-1} with r_1 = r = 2 - √3* *diagonal piece: d_k, d_{k-1} with d_1 = √2 * (1 - r)* By similarity, we have *r_k / d_k = r_1 / d_1 for all k ≥ 1 (1)* The diagonal pieces also satisfy the recursion *d_{k+1} = d_k - r_k - r_{k+1} for all k ≥ 1 (2)* Combine (1) and (2) into a recursion for *r_k:* *r_k / d_k = r_{k+1} / d_{k+1} = r_{k+1} / ( d_k - r_k - r_{k+1} )* Solve for *r_{k+1}* and expand the resulting fraction by *1/d_k* to get a common ratio *q:* *r_{k+1} = r_k * ( 1 - r_k/d_k ) / ( 1 + r_k/d_k ) // use (1)* *= r_k * ( 1 - r_1/d_1 ) / ( 1 + r_1/d_1 ) =: r_k * q* Finally simplify *r_1/d_1* and then the common ratio *q:* *r_1 / d_1 = ( 2 - √3 ) / ( √6 - √2 ) = 1 / [ ( √6 - √2 ) ( 2 + √3 ) ] = 1 / ( √6 + √2 )* *=> q = ( √6 + √2 - 1 ) / ( √6 + √2 + 1 )*

[@byronwatkins2565]

At 6:35, performed correctly I think the point of tangency is easier. The distance between these is (x+r)cos(15) or the difference in radii is r-x=(r+x)sin(15)

[@alcodark]

enlighten me pls wiht the sqr2(1-r-x) ...coz i only see sqr2(1-r) -r-x

[@jounik]

Ditto. That'd be at 6:36 .

[@AnkhArcRod]

Spot on! Was looking for this comment.

[@natepolidoro4565]

True

[@adogonasidecar1262]

Agreed. Was struggling with that and went to check comments and bingo

[@aljuvialle]

@@jounik I still don't get the arithmetics, how from `sqrt(2) * (1 - r) -x -r` (shown on the big picture) Michael got `sqrt(2)* (1 - x - r)` on a small picture

[@kilianklaiber6367]

Minute 6:43. Is there a mistake? The hypothenuse of the smaller triangle with a side length equal to x should be the length of the hypothenuse of the larger triangle [= sqrt(2)*(1-r)] minus (r+x). That's not equal to sqrt(2)*(1-r-x), which I see on the black board?!

[@ZotyLisu]

I really liked this channel as it stood out for me because of the rigorousness which is rare for math on KZbin (everything was always proven, every possible case examined, etc). But lately these videos seem to have more and more errors, also many things go unproven, and cases go unconsidered. I'm not writing this in bad faith, just please consider if more quality over quantity wouldn't be overall better for the channel and education

[@mathcanbeeasy]

At 6:41, √2*(1-r-x) is completely wrong.

[@stewartcopeland4950]

After correction, the total area of the circles is about 0.345 or: pi*(2 - 3^0.5)^2/(1 - (6^0.5 + 3^0.5 - 2^0.5 - 2)^4)

[@humanat]

This is what I got as well

[@koenth2359]

I too calculated an exact solution (see my post) that expresses it as (28-39√2-32√3+23√6) π/16 which evaluates to about 0.345243225544...

[@alijabbar6278]

x/r = (4 + sqrt(2) - sqrt(6))/(4 - sqrt(2) + sqrt(6)) ≈ 0.588709677 after proceed same path the sum of the circles area = pi* (56 - 39 sqrt(2) - 32 sqrt(3) + 23 sqrt(6))/16

[@digxx]

Thanks, that what I got, too.

[@egillandersson1780]

Hello Michael ! I'm a long time follower of your channel and I like usually all of your videos. You taught me a lot. But these one, ... ???? First, at 0:51 : it seems correct but not obvious for me (and for some others as I can read) Then at 6:41 : why √2 (1-r-x) instead of √2 (1-r)-r-x ? And at 11:14 : 12-9=9 !!!! For the ratio x/r, I find (√6 - √2 + √3 - 2)/(√6 - √2 - √3 + 2) ≃ 0.588 which is quite different from √3 - 1 ≃ 0.732 Of course, it looks graphically obvious that the circles are geometric progression but not with the ratio you found. But I could be wrong too. Say me, please !

[@nathanielsilver6752]

You never explain why the 30-degree angles of the rays forces the top circle to be tangent to the rays at the points where the rays intersect with the (big) circle.

[@allykid4720]

r/1 = r = tan15 = (1 - cos30)/sin30 = 2 - sqrt3; (r - x)/(r + x) = sin15 = sqrt[(1 - cos30)/2]; x = r (1 - sin15)/(1 + sin15); x2 = r (1 - sin15)^2/(1 + sin15)^2; x3 = r (1 - sin15)^3/(1 + sin15)^3....

[@eytanmann6208]

unless I missed something - but at 6:40 from start - you note that the hypotenus is sqrt(2)*(1-r-x) - this is a mistake - the correct formula is sqrt(2)(1-r)-r-x -> the x should have been OUT

[@zacharyzug]

Hrmmmmm - seems to me that the 30-degree angles never really came into play - any symmetrical angles would work for this approach (you'd still get a 45-degree angle for the large triangle)... In fact, I'm wondering if it's possible to guarantee that those angles could be 30, or if there's another bit of fun calculating what they actually are?

[@tubebrocoli]

Yep, instead he assumed that tangents of the bigger inscribed circle at the point of intersection with the radius-1 quarter circle go through the corner of the square at the bottom-left.

[@notananimenerd1333]

We can use the 30 deg angle thing to find the value of bigger circle’s radius in one step… draw perpendicular from the centre of bigger circle to the point of contact and connect the centre of the circle to the vertice of square from where the lines are drawn.. then in that triangle using tan15 we get tan15=r/1 as base will be radius of quarter circle which directly gives r=2-sqrt3 as tan15=2-sqrt3

[@diego_sabbagh]

fascinating problem, thank you. just one thing (at 11:00) should be reviewed

[@cosmicvoidtree]

I’m curious as to what happens when you vary the angle of the rays in the square.

[@eejuN_1973]

The best-looking answer I could ever get is this pi * v2 /16 * (v3 - 1) * [ (v2 - 1)*(v2 + v3) ] ^2 "v" is for sqrt Btw adding the value of the 30-degree angle is redundant in the problem statement, as only at this angle is the given figure possible. Ofc the uniqueness of this angle needs to be proven and its value to be found, which only adds to the problem's appeal

[@Anonymous-zp4hb]

Get a final answer for the area of: ( pi / 16) ( 7 - 4sqrt(3) ) ( sqrt(6) - sqrt(2) ) ( sqrt(6) + sqrt(2) + 1 )^2 Was discouraged with the way this expression looks! Told myself that area would be about 1/3 Plugged it into the calculator (starts with 0.34524...) The relief :)

[@tobiaskyrion6019]

I arrived at the ratio for the radii in the form r_k+1 = r_k*(2 - sqrt(2 - sqrt(3)))/(2 + sqrt(2 - sqrt(3))). The final value for the sum of circle areas is (pi/8)*((23/2)*sqrt(6) - 16*sqrt(3) - (39/2)*sqrt(2) + 28).

[@TimMaddux]

I would be curious to see what you get for the area from angles (set to 30deg for the video) ranging between 0deg and 45deg.

[@bobzarnke1706]

More generally, if angle between the square's diagonal and one of the lines tangent to the circles is θ, then the length of the diagonal of the square (√2) is also r csc θ + r√2, making r = (√2 sin θ)/(1+ √2 sin θ). Above, θ = π/12 and sin(π/12) = (√3-1)/(2√2), giving r = 2-√3. Adding lengths along the diagonal up to the center of the first circle gives: x csc θ + x + r = r csc θ; so, x = kr, where k = (1-sin θ)/(1+sin θ) and the circles have radii r, kr, k²r, ... So, the sum of the areas of all the circles is πr²/(1-k²). If a tangent line, the large circle and the quarter circle all intersect at a point, then r = tan θ as well as r = (√2 sin θ)/(1+ √2 sin θ) above. Setting these equal implies that √2 (cos θ - sin θ) = 1, or 2 cos(θ+π/4) = 1, making θ = π/3 - π/4 = π/12, which is the only angle at which the tangent and two circles intersect.

[@subhashk49]

How are you assuming that the point of intersection of the two lines with the top circle is also the point of intersection of the line with the quarter circle ? Although the radius r1 of top circle calculated by me by a different method is the same as calculated by you by making the assumption ab initio. After that I calculated the radius of next circle and ratio r1/r2 by noting that (r1-r2)/(r1+r2) is equal to to sin 15 degree. Then using the square of r1/r2 as the common ratio of infinite geometric series to calculate the total area. I landed up with not simple expression as you.

[@mohammadw.alomari1322]

@Michael Penn Q1: The real interesting problem is to find the total areas of the remaining curved area between the infinite circles and the tangent lines? Another pretty Q2: what is the limit of total areas of the circles in the lower sector having the same radius r / the total area of the circles with 1 eighth the radius (r) (=1/8 r) each; lying above the sector and the side of the above square, as r goes to 0. Q3: If the angle between the two lines is pi/2. how many circles with the same radius r could be? Q4: In the area between the horizontal line (the above side of the square) and the curve (above arc) (a portion of the upper square. What is the ratio of filling circles with the same radius r? What can you conclude? These questions are much harder than the one you proposed, with respect.

[@Okkk517]

The angle between the tangent lines should not be given. In other words , it can be deduced from the setup of the problem. Moreover your final answer to the problem is too small (.055 pi) compared the expected area just by looking at the figure).

[@AndreasHontzia]

Observation: If you segment the yellow angle (that encloses the circles) by lines perpendicular to the 45° angle of the unit square, you will get isoceles trapezoids, each cointaining one circle. These trapezoids and their circles are similar to the other trapezoids and their circles. So the ratio of areas of one circle to its trapezoid is constant. This ratio can be used to calculate the area consumed by all the circles of the whole yellow triangle constructed by all of the trapezoids. Is this a valid solution idea?

[@kevinstewart2572]

Thanks for the fun problem! I wonder if errors are tossed in now and then just to ensure we're actually paying attention. :-) Michael, this problem inspired a question you may want to address in a separate video, as I explain: Starting from the formula for the total area of the circles, A: A = πr^2 / (1 - (x/r)^2), this can be written here as A = π(7 - 4√3) / (480√3 - 588√2 + 340√6 - 832), but (after consulting Wolfram alpha) a simplified version is A = (π/16)(56 - 39√2 - 32√3 + 23√6). The question naturally arises here: Can you demonstrate a generalized technique, and rationale therefor, to use in rationalizing fractions with surds, e.g., here of the type 1 / (p + q√a + r√b + s√(ab)), with the usual constraints for these values over integers, etc.? Thanks for considering.

[@aksiiska9470]

si on veut eviter le sin cos tan, on doit tounjours appliquer la loi du ancien grecq thales et la symmetrie pour trouver la intersection cercle-ligne droite et pour trouver le centre du cercle

[@aljuvialle]

let's take n'th circle with radius l with hypotenuse L, and let's find radius of next circle x. Length of hypotenuse of the smaller triangle is (L - l - x) by construction. Caclulating proportions we'll get x = l(L-l)/(L+l). using that L = l/sin(15), we'll get x = l * c, where c = (1 - sin(15))/(1 + sin(15)). next can sum up total sum from 0 to inf of (r * c^n). r is const, 0 < c < 1. so result will be r / (1 - c) = r * (2* sin(15))/(1 + sin(15)). Then plug in values of the sin(15) and the biggest r values and get the result. sin(15) = (sqrt(3) - 1)/2*sqrt(2) and r = 2 - sqrt(3).

[@koenth2359]

Using the double angle formula, I showed that sin(15°) = (√6 - √2)/4 Also, sin(15°)=(1-α)/(1+α) where α is the ratio of radii of adjacent circles. This gave me α = 15+10√2-8√3-6√6 (which is about 0.58879) The diagonal is √2 = r (√2+1+2(α+α^2+α^3+...)) = r(√2+1+2α/(1-α)) This gave me r=2-√3 which is about (0.26795) Now A = πr^2(1+α^2+α^4+...) = πr^2/(1-α^2) 1/(1-α^2)= (8+3√2+5√6)/16 (which is about 1.53063) Area of all circles = πr^2/(1-α^2) = (56-39√2-32√3+23√6) π/16 which is about 0.345243225544...

[@digxx]

You sure about the first term 28? Shouldn't it be 56?

[@koenth2359]

@@digxx Thanks! Yes you are right, final answer should be A = (56 - 39√2 -32√3 +23√6)π/16, corrected it in the main post.

[@surrealistidealist]

So exhilarating!!! Thank you!!!

[@jardozouille1677]

Approximatively correct but clear :)

[@JBMJaworski]

Denominator at the end is 3.

[@mrflibble5717]

Tks Michael, I always enjoy your vids. Pls inform when the item is fixed and reposted. The error just increases the complexity, but is a really interesting problem! Rod

[@minwithoutintroduction]

11:14 (-3+2sqrt(3))(3+2sqrt(3))=12-9=3

[@dmitriturevski5620]

In the beginning I was hoping that it would produce a tricky series so that pi would cancel out in the end.

[@dmitriturevski5620]

@@JohnPretty1 That's the point. There's the Leibniz formula for pi as a sum of the infinite series for example, so we could have something similar here with the sum of radiuses squared converging to some nice constant / pi making pi cancel out when calculating the total area.

[@donaldasayers]

Nice puzzle.

[@tubebrocoli]

You assumed that the tangents of the bigger inscribed circle at the point of intersection with the radius-1 quarter circle go through the bottom-left corner of the square. I assume that this can be proven from the angles you added at the beginning?

[@willyh.r.1216]

Love it, thx.

[@jagmarz]

The problem statement feels over-constrained to me. Knowing the rays are at 30 degrees was unnecessary to finding r, so that's one extra bit, but it's also not clear to me (at least as of 4:43) why it's necessary that the rays tangential to the largest circle where it intersects the quarter-circle must be radii of it, although maybe that's just by construction? Is this a complete problem statement? "Given a unit square with corner C and inscribed quarter circle Q of unit radius with center C, construct circle O tangential to two adjacent sides of the square opposite corner C such that the tangent lines (A and B) of O at the intersections with Q are radii of C. Construct infinite circles o(n) where each successive circle o(n+1) is tangent to A, B and o(n), where o(0) = 0. Find the some of the area of all circles o(n) (n=0 .. infinity)."

[@Alidaher2468]

Here is a fun problem for a new video Find all natural powers of 2 that differ by a perfect square .

[@carstenmeyer7786]

We need to find all *m, n ∈ ℤ* such that *2^m - 2^n = q^2, q ∈ ℕ_0 (*)* The RHS is non-negative, so we have *m ≥ n.* Consider the case *n < 0:* *n < 0 and m > n:* LHS cannot be an integer -- contradiction! *n < 0 and m = n: 2^m - 2^n = 0^2 =>* solution! Now consider *n ≥ 0* and define *d := m - n ∈ ℕ_0:* *q^2 = 2^n * (2^d - 1) =: 2^n * o, o ∈ ℕ odd* All prime powers of a perfect square are even, so *n = 2N* has to be even to satisfy the equation. Then we can rewrite *q = r * 2^N, r ∈ ℕ_0 => 2^d = r^2 + 1* Consider the equation above "mod 4". For *d ≥ 2,* there cannot be a solution, since - the LHS is divisible by 4, since *"2^d ≡ 0 mod 4"* - the RHS is not divisble by 4, since *"r^2 + 1 ∈ {1; 2} mod 4"* Checking *d ∈ {0; 1}* manually, both lead to solutions. The entire solution set is given by *D = { (n, n) | n ∈ ℤ } ∪ { (2n + 1, 2n) | n ∈ ℕ_0 }*

[@Alidaher2468]

@@carstenmeyer7786 good but you considered that m and n belong to integers while i said natural powers of 2 , so m and n are positive integers

[@carstenmeyer7786]

@@Alidaher2468 I'm sorry, must have read "powers of 2" instead of "natural powers of 2"... That slightly shortens the proof: Omit the case "n < 0" at the beginning and "n = 0" later. The other steps remain exactly the same.

[@Alidaher2468]

@@carstenmeyer7786 yes great solution 👌

[@mab9316]

MAGIC

[@adipy8912]

Finally a new geometric proplem

[@Horinius]

@Michael As others have pointed out, your calculations are all wrong from @6:40 onwards. Please remake a new video with correct answer.

[@antonryzhov]

It is too complex solution. The area equals to the Pi*([square diagonal] - [small piece in the upper right angle, which is easy to find from r])

[@notmymain2256]

No, the square diagonal - the small piece is the 2×sum of the radius, but sum of areas = pi×sum of *squared* radius

[@notmymain2256]

If it asked for the sum of the circumferences then it's easily 2pi*sum of radius which can be easily derived

[@RAG981]

The above goes completely wrong just before 7 mins because Michaek makes up a silly expression for the hypot in the smaller triangle. I did not like his method anyway. I started by using trig which proved simpler. e.g. r is just tan15. Then I found the next radius (done correctly) was (1-sin15)/(1+sin15) times the first. This continued to be the scale factor for the next, and so for all the rest. Summing the GP for the areas gave an answer of Pi*tan15*(1+tan15)^2/4sin15 or 0.1099Pi or about 0.345.

[@khoozu7802]

That should be pi*(tan15) ^2*(1+sin15)^2/4sin15

[@thorntontarr2894]

Is it not misleading to state @0:46 "this inscribed circle will intersect with the rays at the same place that it intersects this quarter circle"? Your drawing and your statement are really misleading. The point I am making is that the inscribed circle is "tangent" with the rays at the same place that it intersects this quarter circle. Later @ 3:47 you state this tangent property.

[@marcosjuniorquijadagonzale9778]

Si a great solution but the answer is (2*sqrt(3)-3)*pi/3 you are the best 🤘🏻

[@General12th]

Hi Dr. Penn!

[@kmlhll2656]

Wow

[@СергейВасиленко-е2ш]

Задача решена неправильно!?

[@takyc7883]

Nice !

[@ahmedgg8867]

Michael plz Find all n naturel numbers such that 2^n divise (5^n) -1

[@carstenmeyer7786]

We use the "LTE-Lemma" (lifting the exponent). If *2^n* divides *(5^n - 1),* then we must have *n ≤ 𝜈_2(5^n - 1) = 2 + 𝜈_2(n), n odd (1)* *= 2 + 𝜈_2(n) + 𝜈_2(6) - 1 = 2 + 𝜈_2(n), n even* Notice both cases simplify to the same result. Splitting the even part from *n,* we obtain: *n = 2^m * o, m, o ∈ ℕ_0, o odd with m = 𝜈_2(n)* We solve the above for *m* to get an upper estimate: *m = log_2(n / o) ≤ log_2(n) (2)* Combining (1) and (2) we notice *n* has to satisfy the following inequality: *n ≤ 2 + 𝜈_2(n) = 2 + m ≤ 2 + log_2(n) => 2^n ≤ 4 * n* By induction, the inequality above is not satisfied for any *n > 4.* Checking the remaining cases *n ∈ {1; ...; 4}* manually, the only solutions are *n ∈ {1; 2; 4}.*

[@camrouxbg]

Just looking at the diagram in the title slide, it looks like a descent problem to use inversive geometry with. But I might be wrong. It's been a minute and I'd need to sit down and work through it.

[@johnloony68]

The bit he didn’t tell us at the end is that the answer is c. 0.162002

[@rkPixie]

wow too many cluebats on this comments.

[@dinuwarabinudithdesilva5464]

Hey Michael, there is a mistake that you've made here 4:17 which has to be r^2-2r+1=0?

[@emiltonklinga3035]

No, his calculations there are correct.

[@dinuwarabinudithdesilva5464]

@@emiltonklinga3035 When factored r^2-2r+1=0 would only give you (r-1)^2=0

[@angelinageorge2278]

Nope he used x=-b-_/b²-4ac/2a

[@Megumin_Random]

At 4:17, he isn’t dividing by 2. He’s subtracting r^2 and 1 from the other side

[@emiltonklinga3035]

@@dinuwarabinudithdesilva5464 The equation is r^2 + 1 = 2(1-r)^2 which is equivalent to r^2 - 4r + 1 = 0.

[@arjunsinha4015]

Truly one of the Maths problems of all time

[@jorex6816]

Well it is a maths problem haha

[@Scumlord]

Funny joke