Nice problem, and clever solution! By the way, at minute 14:36, shouldn't it be: 2(-5+4√2)-4 = -14 + 8√2 ???

4:40 Homework 7:10 It should be 2h+2k-2 on LHS but it’s fixed at 8:53 14:36 +20√2? Where does the 20 comes from? 18:37 Another homework (solution at 18:56) 19:49 Good Place To Stop

Your diagrams look so well-defined and vibrant. I don't know how you do it.

At 6:17, the simplification is easier if we substitute eqt(3) into eqt(1). Then we can get k^2=4r and this is eqt(4). Then substitute eqt(3) into eqt(2), we can get k in terms of r and this is eqt(5). Then we substitute eqt(5) into eqt(4) and finally we can get the same quadratic equation of r. Finally, we can get r=1 (rejected) or r= (9-4√2)/49.

thanks for taking the time and effort to make all these videos. Brings back memories of the math classes I took in the 1960's. (But I wasn't clever enough to come up with your solutions!)

Really enjoyed your use of color to stress steps and important ideas. Might borrow/use this in my upcoming first year as a teacher.

Great video Michael

I can suggest one thing to simplify the solution: from these two equations for x and y being center of a red circle: x = 1-r ; x^2 + y^2 = (1+r)^2 -> y^2 = (1+r)^2 - (1-r)^2 = 4*r you can get that y = 2*sqrt(r), and from this point it's much easier to solve it for y , and then get the solution from it.

No one has suggested a simpler solution besides using Descartes' Theorem yet. After getting -1+√2 for the radius of the smaller quarter circle by looking at the diagonal, let r be the radius of the small circle. The right side of the square will have length 1 = √((1+r)^2-(1-r)^2) + √((√2-1+r)^2 - r^2) = √(2r(-1+√2) + 3-2√2) + 2√r. Move 2√r to the other side, square, rearrange, and let t = √r to get (3-√2)t^2 - 2t + (√2 - 1) = 0. Notice that t=1 is an extraneous solution (we have r

18:20 Here is a little mistake. -140-(-40)*sqrt(2)=-100*sqrt(2), so here shouldn't be the plus. Great video :)

Diagonal of the square is sqrt(2), and is made up of radius of big quarter circle and of small quarter circle. Radius of big quarter circle is 1. Thus, radius of small quarter circle is sqrt(2) - 1.

Really enjoyed solving this one. I normally post something about my solution that was different to yours. But in this case, our methods were pretty much the same.

Would have loved to see a more geometric argument instead of just bashing with coordinate geometry. In my opinion most geometry problems can be solved like this but it turns into just pure algebra without interesting geometric insights.

for the homework part, calculate the diagonal of the square, then minus the radius of the quarter circle from it to get 'sqrt(2)-1'

Love it!

Thanks for sharing. It could perhaps simplify calculations a bit if we use two right triangles to obtain the system of two equations with two unknowns that can be then solved for r. The vertices of the first triangle are (0,0), centre of the small circle, and its orthogonal projection onto the x-axis. The vertices of the second triangle are (1,1), centre of the small triangle, and its orthogonal projection onto the y-axis. The side lengths of the first triangle are then 1+r (hypotenuse), 1-r, and x, and the side lengths of another triangle are sqrt(2) - 1 + r (hypotenuse), r, and 1 - x. Just a thought. Thanks again. Great content as always.

If anyone wants to know the answer since he wrote down 20 instead of 8, before denesting it simplifies down to (7-4sqrt(2)+-sqrt(24-16sqrt(2)))/(11-6sqrt(2)). The denested final answer in the video is the one associated with this correct answer when you take the negative sign. If you take the positive sign, it just simplifies down to 1. I also want to note that the answer of r=1 corresponds to the larger quarter circle reflected across the top side of the square.

Draw diagonal (0,0) (1,1). Because it's a square, it measures √2. Observe that this diagonal has the length of the green radius + yellow radius, then, yellow radius = √2 - green radius = √2 - 1

14:35 It should be +8sqrt(2) and not +20sqrt(2) classic michael... 18:27 it shoul be -100sqrt(2) and not +100sqrt(2). Again classic michael...

Circle inversion wrt the circle-circle contact point also works like charm.

The relationship between k and r is that k=2*sqrt(r)

Since the orange and yellow circles are tangent, one of the couples of the radii are colinear, and it so happens it's one of the square's diagonals. The measure of the diagonals of a square of side a is a×sqrt(2) (via the pythagorean theorem). Sincethe side of the square is 1, the diagonals are sqrt(2). And since the radius of the yellow circle is 1: radius of orange circle=diagonal of square-radius of yellow circle=sqrt(2)-1 QED

brilliant!

Thank you for the solution. And could you please tell me why you set three unknown variables and seem pretty sure you can find three equations to solve them out. Thank you!

This solution was concerned more with the calculation and equations part with very little use of geometry but then too great problem to give it in a test and waste the time of students

An interesting related problem is to find the curve of the midpoints of all circles that are tangential to two tangential circles in terms of their radii r1 and r2.

Could this have been done more simply with circle inversions?

How would you resolve the plus -minus ambiguity?

root 2 -1 is obvious from inspection, give that the long diagonal is part of a right triangle with both sides equal 1.

Isn't this one of those problems which can be easily solved using circle inversion? Any chance you could do some videos on that?

Its the diagonal of the unit square minus the radius of the largest quarter circle

Radius of orange circle is equal to √2-1 because diagonal of unit sqaure is equal to √2 and it's also equal to sum of radius of yellow and orange circle. Substracting radius of the unit circle gives us the result that we've got. And that's a good place to stop.

Daily dose of Michael Penn is increasing my brain cells one video at a time

The smaller circle has radius 2^(1/2) because the diagonal of the unit square is 2^(1/2).

I solved this by a very different method, inversion of circles. Draw the inversion circle so that its center is at the tangent point of the two arcs and its radius so that it touches the side (1,0) - (1,1) of the square. The two arcs then invert to a pair of parallel straight lines parallel to the (1,0) (0,1) diagonal. The right edge of the square inverts to a circle that tangent to the right side of the square and goes through the arcs tangent point. It is then trivial to find the image of the orange circle. It is tangent to both straight lines and its diameter is equal to the distance between the lines. It is also tangent to the circle that image of the right side. The diameter of the orange circle can then be computed by inverting its image. The radius is (2-√2)/(10-√2). Descarte's theorem does not apply here, the conditions are wrong.

“This monstrosity” lol

Its a great video but there was a mistake in calculating your middle part of quadratic solution at 14:36 which ultimately resulted in the wrong answer in the end, another comment also pointed it out. Just letting you know Mike and thanks for the vid.

at 7:15 shouldnt it be 2h+2K-2 instead of +2?

r is solution of : (11-6 sqrt 2) r*r - 2 (7-4 sqrt 2) r + 3-2 sqrt 2) = 0 (video is not correct as dar as the second coefficient is concerned). So r=1 is also solution of the problem (with h=0 and k=2). The searched circle is then "above" the two others. And the other solution is r=(9-4 sqrt 2)/49 (as said in the video).

Draw a line connecting (0,0) and center of the smallest circle and likewise connect that center to (1,1) First line is (1+r); second is (√2-1+r). Now use these lines as hypotenuses, making right-angled triangles with x-axis(y=0) and x=1, respectively. The base of the first RAT is (1-r) while the base of the second RAT is simply r. Now just using Pythagorus's Theorem, find the heights of each RAT, which must sum to one. First is √((1+r)²-(1-r)²) = 2√r. Second is √((√2-1+r)²-r²) = √(3-2√2+(2√2-2)r). So, we have 1 = 2√r + √(3-2√2+(2√2-2)r). Hence (1-2√r)² = (√(3-2√2+(2√2-2)r))² Hence 1-4√r+4r = 3-2√2+(2√2-2)r Hence 0 = 2(3-√2)r - 4√r + 2(√2-1) This is quadratic in √r. Quad. Formula gives √r = (4 - √(16 - 16(3-√2)(√2-1)) / (4(3-√2)) [Note: Subtract the √(b²-4ac) since addition will give a result obviously too large.] Hence √r = (1 - √(1 - (3-√2)(√2-1)) / (3-√2) Hence √r = (1 - √(6-4√2) ) / (3-√2) Now, √(6-4√2) = √(4-4√2+2) = (√(2-√2))² = 2-√2 Hence √r = (√2-1) / (3-√2) Rationalize the denominator by multiplying top and bottom by (3+√2) Hence √r = (√2-1)(3+√2) / 7 Hence √r = (2√2-1)/7 Squaring then gives: r = (9-4√2) / 49

If one calls radius of upper 1/4 circle a, and projects OP and segment P (1,1) onto the (1,0)(1,1) segment, the resulting equation is: 2sqrt(r) + sqrt(a(a+2r)) = 1, or sqrt(a(a+2r)) = 1 - 2sqrt(r). Squaring both sides gets you a simpler equation in d = sqrt(r): ((3-sqrt(2))d^2 - 2d + sqrt(2) - 1 = 0. Thus, d = (2sqrt(2) - 1)/7 and r = (9-4sqrt(2))/49.

Ok, I have it. As Michael said, we're looking for center P(h,k) and radius r, using 3 equations. The last one gives h = 1 - r. The second (and first) one yield k = \sqrt{2} - 1 + (3 - \sqrt{2})r. Now we plug both these expressions in first equation h² + k² = (1 + r)². We get : (11 - 6\sqrt{2})r² + 2(-7 + 4\sqrt{2})r + (3 - 2\sqrt{2}) = 0 There are two ways to solve this equation. The first one is to remark that r = 1 is solution (!). Then using the formula about the sum or the product of the roots, we can find the second root given by Michael, which is r' = (9 - 4\sqrt{2}) / 49. Clearly the second root is the one shown in Michael's figure, but the first one is also relevant : radius r = 1 yields center P(0,2), and a circle which is tangent to both initial circles and vertical straight line. The second way is to calculate the discriminant. We get : \Delta = 4(24 - 16\sqrt(2)). At this point, I had a lot of problems to find the solution given by Michael, because the formula for the roots of the equation use \sqrt{\Delta}. In fact, there exist two integers a, b such that \Delta = (a + b\sqrt{2})². Indeed, we must have a² + 2b² = 4*24 and 2ab = 4*(-16). The latter allows to get b = -32/a, that we plug in the former : a² + 64/a² = 96. This yields equation a^4 - 96a² + 64 = 0 ... discriminant ... blabla ... We find that \Delta = (8 - 4\sqrt{2})², so \sqrt{\Delta} is cool, and we do retrieve roots r = 1 and r' = (9 - 4\sqrt{2}) / 49. My point is that the set of numbers of the form x + y\sqrt{2} with rational numbers x, y is intriguing. Sometimes, the square root of such a number has same nature. Probably that this set of numbers is well known. Any information about this ? Did Michael published a video about this ?

The radio from the orange circle is (√2)-1 because the distance between (0) and (1,1) is √2 that say than √2=r1+r2

Dropping a perpendicular from P to the lower side of the square, meeting it at Q, gives a more geometric solution using Pythagoras' theorem, equivalent to solutions given by several other commenters. Using coordinates in the diagram to identify unlabelled vertices, in Δ (0,0)PQ, (1+r)² = PQ²+(1-r)², giving PQ = 2√r. In Δ (1,1)P(1,k), (√2-1+r)² = r² + (1-PQ)², giving (3-√2)r-2√r+(√2-1) = 0, which is a quadratic in √r. Ignoring the solution √r=1, gives √r = (√2-1)/(3-√2) = (2√2-1)/7 and r = (9-4√2)/7. (√r=1 is a valid solution in the sense that it satisfies the same constraints as the small circle, ie, being tangent to the two quarter circles and the side of the square. It's easy to see by reflecting the square about the vertical line (1,0)(1,1).)

3:35 how do we know that the radious of the green circle is 1 did I miss somthing can you please reply to this comment with the time he mentioned its one. thanks.

So… well before watching the video, I decided to attempt solving this; did so in remarkably the similar way. Similar, but rather attractively different (and less algebra)... № 0.1: Let (𝒔) be the side of the square. '𝒔 = 1' № 0.2: Let (𝒕) be radius of quarter circle. '𝒕 = √(2) - 1' № 0.3: Let (𝒓) be the desired lil' circle radius; № 0.4: Let (𝒌) be height from lower-right to lil circle radius's intercept. Solving first for 𝒌… № 1.1: (𝒔 - 𝒓)² + 𝒌² = (𝒔 + 𝒓)² expand № 1.2: 𝒌² = (𝒔² + 2𝒔𝒓 + 𝒓²) - (𝒔² - 2𝒔𝒓 + 𝒓²) № 1.3: 𝒌² = 4𝒔𝒓 № 1.4: 𝒌 = 2√𝒔𝒓 Dunno if [1.4] will be used, but good to say, anyway. Now, working with upper-corner quarter circle № 2.1: (𝒔 - 𝒌)² + 𝒓² = (𝒕 + 𝒓)² … rearrange № 2.2: (𝒔 - 2√𝒔𝒓)² + 𝒓² = ((√2 - 1) + 𝒓)² … with (𝒔 = 1), so № 2.3: (1 - 2√𝒓)² + 𝒓² = ((√2 - 1) + 𝒓)² № 2.4: 1 - 4√𝒓 + 4𝒓 + 𝒓² = (√2 - 1)² + 2(√2 - 1)𝒓 + 𝒓² Cancel the 𝒓² both sides, rearrange into a pseudo-quadratic № 2.5: 0 = (2 - 2√2) + (2√2 - 6)𝒓 + 4√𝒓; and let № 3.1: 𝒒 = √𝒓 substitute into [2.5] № 3.2: 0 = (2 - 2√2) + (2√2 - 6)𝒒² + 4𝒒 … quadratic, so solving for 𝒒; № 3.3: 𝒒 = [ 0.26120, 1.00000 ], and reversing [3.1] to find 𝒓 № 3.4: 𝒓 = [ 0.06823 ];

At 7:15 there's a +2 that should be -2, but later on you seem to drop the term altogether and I don't know why.

i really can't see the yellow in the quarter circle.... full green to me... (les gouts et les couleurs...)

I did it using own method and got numerical value of radius as roughly ~ 0.044

Thought I was on the wrong track as the equations were getting so gnarly. Turns out I was ok after all. What an evil question 👿:)

الكتابة غير واضحة خصوصا الطباشير الأزرق. واصل تحياتي لك شكرا جزيلا . مجهود جبار

7:31 genius, lol

I used different set of equations and ended up with r=(7−4×√(2)−2×√(6−4×√(2))) / (11−6 ×√(2)) that was the "good place to stop" for me. Interesting how complicated is to simply terms with square roots.

its yellow and green quarter circles

Let's take a minute to appreciate the work to draw the problem and that's a good place to stop.

This is nice problem that requires some calculations but it is not tricky

The ans of the radious is negative . Why ? When I put those numbers in my calculator , it's showing negative numbers for both + and - . Isn't the ans would be 1.8859355..... ? I don't know , maybe I'm wrong

Hi, For fun: 3:40 : "ok, great".

No.1 Fan

how did you know that P, (0,0), and the intersection of big circle and littlest circle would lie on a straight line for 3:55? Surely you cannot make the claim that the distance is 1 + r if you do not know they are on a straight line?

two Pythagoras

From far that problem looks unsolvable but your explanation made it look like 5th grade math

final answer is correct: r = 0,06822...

Funny how the trivial solution (h,k,r)=(0,2,1) ends up being the extraneous one.

I suppose every problem is "tricky" until you see the "trick", but this one doesn´t seem all that "tricky" to me - I like your description "nice little geometry problem" better. I´d say it´s worth a 6 or 7 minute solution.

For me, decimal numbers are sensible and all others are irrational. If Mike would have gone to sensible numbers early-on in the problem, he might not have confused himself so much and his solution might have been errorfree. But then, if even Mike can screw up once in a while, I guess we all deserve a little latitude.

Can be done using the descartes theorem. A theorem you’ve actually proved on this channel before

You were wromg in some paths... i think this is a very long solution, maybe we can find something synthesize

If you follow through the correction of -20 to -8 the answer comes out as r=12(11+6sqrt2)/49~4.77190564267067 Something still not right!

Crazy Messy.

Dude he says yellow-orange and i see green-yellow

What a "mess"... why not just draw it to scale and measure it?

why don't we take a ruler to measure it, it much easier

It isn't yellow, it's green

*Perhaps one of the hardest problems you're doing so far* 🤕

Like number 1000

There is a compact and countable set in same time if there's give me an exemple

Simple Math errors

no need to call the point P(h,k) it is simply (h,k)=P

PLEASE SOLVE THEORETICAL PHYSICS IN A EASY WAY OR ALL CHAPTERS OF MATHEMATICAL PHYSICS

I believe the answer is 113/49 Can anyone pls confirm?

Ok all you math whizzes. Put up your own video with the same problem