A tricky geometry problem!
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- Опубликовано: 8 сен 2024
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Nice problem, and clever solution! By the way, at minute 14:36, shouldn't it be: 2(-5+4√2)-4 = -14 + 8√2 ???
Yes, you're right. 👍
Yes!
@@megauser8512 oui il y a une erreur de calcul.
Yeah I was confused how he got 20sqrt(2) instead of 8sqrt(2) there as well.
The 20*sqrt(2) part needs to be 8*sqrt(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
14:36 it should indeed be +8sqrt2
Will you be posting a riddle?
Did the -2 disappear when he brought everything up to the board at 7:10 ?
@@skylardeslypere9909 And at that point you can see that r=1 is a solution to the equation and dividing by (r-1) gives the answer in the neater format of (9-4sqrt2)/49 nice and simply.
@@jimbolambo103 haven't tried it for myself so I'll believe your word for it!
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.
Since you get (1-y)^2 + r^2 = (sqrt(2)-1+r)^2 when writing the distance from (x,y) to (1,1) in 2 different ways, I don't think that it is much easier.
@ Boba Fett @ Mega User -- For clearer readability, it would be better to use a blank space around your subtraction and addition signs everywhere:
x = 1 - r; x^2 + y^2 = (1 + r)^2 --> y^2 = (1 + r)^2 - (1 - r)^2 = 4*r
(1 - y)^2 + r^2 = (sqrt(2) - 1 + r)^2
Your diagrams look so well-defined and vibrant. I don't know how you do it.
He inspire me to be maths teacher in future 😏😎
@@thedarkknight1865 wtf
Yeah it looks like he’s got some new colors, they look great
i think he also turned the chalk on its side to make the lines thicker
One of my Math Professors at WVU had the most beautiful diagrams and graphs. He also called everyone "Professor". RIP Dr. Karwowski
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.
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
I looked for this method but didn't find it until i'd written out mine, yours is shorter but very similar (k=2t)
18:20 Here is a little mistake. -140-(-40)*sqrt(2)=-100*sqrt(2), so here shouldn't be the plus.
Great video :)
Theres another mistake before that
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
Thx
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'
diagonal which is the Rgreen and Rorange
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.
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
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.
If you follow the algebra, you can draw a diagram which is isomorphic to the algebraic reasoning. However, I assure you it would likely look chaotic.
At least all of the operations used are technically feasible using geometric constructions.
@@geekoutnerd7882 Well, that's kind of my point. There's no interesting geometric insight to be extracted from that process. It's more akin to brute force.
@@geekoutnerd7882 Right, and technically all math proofs can be broken down into many pages of dense set theory axioms. But it's the higher level ideas and creative abstractions on top of the axioms that are interesting.
If a circle is touching another circle and a line, then its center is equidistant from both of these. Then, if you shift the line and replace the fixed circle with its centerpoint, your are again equidistant from these. It's the definition of a parabola. So the original problem has a more geometric solution by defining two parabolas and finding their intersection. This will also require Cartesian coordinates and a quadratic equation, but anyway. Also, if one notices that there is another circle fulfilling all conditions and its coordinates are integer, it even allows bypassing the quadratic equation.
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
Lol yeah JEE type which I really really dislike
@@agamanbanerjee9048 Brother jee problems are lot more than calculations,it needs a vision to see it
@@arnabchatterjee4847 ~orokom monehoy~
@@agamanbanerjee9048 What's JEE?
@@synaestheziac JEE (Joint Entrance Examination) is an all India common engineering entrance examination conducted for seeking admission to various prestigious engineering colleges including IITs (Indian Institute of Technology), NITs (National Institute of Technology), and many others all over the country. It is regarded internationally as one of the most challenging engineering admission tests.
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.
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...
Yeah lol
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.
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!)
The smaller circle has radius 2^(1/2) because the diagonal of the unit square is 2^(1/2).
you mean 2^(1/2) - 1
I would say that we know by the Pythagorean Theorem that the diagonal of a square of sidelength a has a length of a*sqrt(2). Here, we have a=1, so the diagonal of our unit square is sqrt(2). Then, from this, we subtract the radius of the unit circle, which is 1 by definition. This gives us, for the orange circle, a radius of sqrt(2)-1.
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.
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.
hmm i vaugely rember something called the inversion circle, I think michael covered it once. Anyway, I think people who know this theorem should notice this comment to confirm your calculations and then maybe michael can pin this comment since it appears more elegant.
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.
The relationship between k and r is that
k=2*sqrt(r)
Its the diagonal of the unit square minus the radius of the largest quarter circle
Really enjoyed your use of color to stress steps and important ideas. Might borrow/use this in my upcoming first year as a teacher.
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).)
Isn't this one of those problems which can be easily solved using circle inversion?
Any chance you could do some videos on that?
That was my initial thought as well - as the intersection point between a hyperbola and a parabola as in the appended image. However there are four intersections so I suspect a nasty quartic and MP's solution is probably simpler.
imgur.com/a/fkcqCzD
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
Using Michael's equation:
0 = (1-r)² - (1+r)² +((3-√2)r+(√2-1))²
0 = (1-2r+r²) -(1+2r+r²) + r²(11-6√2) + 2r(-5+4√2) + (3-2√2)
0 = r²(11-6√2) + r(-14+8√2) + (3-2√2)
0 = (11-6√2)r² -2(7-4√2)r + (3-2√2)
QF: r = ( 2(7-4√2) - √(4(7-4√2)² - 4(11-6√2)(3-2√2)) ) / (2(11-6√2))
r = ((7-4√2) - √((7-4√2)² - (11-6√2)(3-2√2)) ) / (11-6√2)
r = ((7-4√2) - √(49+32-56√2) - (33+24-22√2-18√2)) ) / (11-6√2)
r = ((7-4√2) - √(24-16√2)) / (11-6√2)
Note: √(24-16√2) = √((4-2√2)²) = 4-2√2
r = (3-2√2) / (11-6√2)
Rationalize the denominator by multiplying by (11+6√2)/(11+6√2):
r = (3-2√2)(11+6√2) / (121-72)
r = (33-24-22√2+18√2) / 49
r = (9-4√2) / 49
There. That's it.
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).
root 2 -1 is obvious from inspection, give that the long diagonal is part of a right triangle with both sides equal 1.
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.
I think the end solution is still correct (don't quote me on that, but I got the same thing and it matched up with a quick sketch that I drew and measured, so fingers crossed?), but yeah, the expression he gets to before that is mucked up. I assume he worked through the problem before the video and already knew the answer he was trying to get to, so just skips his way to the end
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.
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.
Circle inversion wrt the circle-circle contact point also works like charm.
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 ?
Great video Michael
at 7:15 shouldnt it be 2h+2K-2 instead of +2?
At least it got fixed after the next cut.
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.
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 👿:)
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
“This monstrosity” lol
How would you resolve the plus -minus ambiguity?
I did it using own method and got numerical value of radius as roughly ~ 0.044
CAD?
its yellow and green quarter circles
Could this have been done more simply with circle inversions?
Daily dose of Michael Penn is increasing my brain cells one video at a time
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!
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.
You are correct. At 14:35 There was a factor 2*4, not 5*4.
7:31 genius, lol
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
i really can't see the yellow in the quarter circle.... full green to me... (les gouts et les couleurs...)
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.
final answer is correct: r = 0,06822...
Right! The complicated solution is off because r would be 4.421... or negative.
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?
Let's take a minute to appreciate the work to draw the problem and that's a good place to stop.
Hi,
For fun:
3:40 : "ok, great".
Love it!
brilliant!
الكتابة غير واضحة خصوصا الطباشير الأزرق.
واصل تحياتي لك شكرا جزيلا .
مجهود جبار
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.
From far that problem looks unsolvable but your explanation made it look like 5th grade math
This is nice problem that requires some calculations but it is not tricky
two Pythagoras
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.
why don't we take a ruler to measure it, it much easier
Because you would only end up with an approximate value for r.
(9 - 4√2)/49 is an exact answer.
Can be done using the descartes theorem. A theorem you’ve actually proved on this channel before
Can you send me the link to that video?
Funny how the trivial solution (h,k,r)=(0,2,1) ends up being the extraneous one.
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!
Maybe you want the other root provided by the quadratic formula. Your answer might be for a circle that obeys the stated tangency requirements but does so in some unexpected configuration, perhaps enclosing some of the other arcs or extending them to full circles to reach the tangent points.
@@kevinmartin7760 the other solution was negative so I discounted it. I may have made an error somewhere. It is a nicer solution.
What a "mess"... why not just draw it to scale and measure it?
Crazy Messy.
No.1 Fan
You were wromg in some paths... i think this is a very long solution, maybe we can find something synthesize
It isn't yellow, it's green
Dude he says yellow-orange and i see green-yellow
*Perhaps one of the hardest problems you're doing so far* 🤕
I'd say it's not hard, albeit lengthy and boring.
@@agamanbanerjee9048 Yeah! So true 😅😅😅
Like number 1000
no need to call the point P(h,k) it is simply (h,k)=P
Simple Math errors
There is a compact and countable set in same time if there's give me an exemple
I assume you mean subset of the real numbers. Any finite set is countable and compact; a slightly less trivial example is the closure of {1/n : n is a natural number}
@@franciscomorilla9559 this conjucture is true?
Any countable compact set is finie.
@@franciscomorilla9559
I have an exercise i need preuve of this conjucture if is true
@@franciscomorilla9559 l work just in real line
@@noumanegaou3227 I just told you an example of a countable compact set which is not finite, the closure of
{1/n:n in N}
Where N are the natural numbers. It is compact since it is bounded (bounded below by 0; above by 1) and closed (it is the closure of a set); by the Heine-borel theorem closed and bounded subsets of the real numbers are compact. It is obviously not finite since it is in bijection with the natural numbers
PLEASE SOLVE THEORETICAL PHYSICS IN A EASY WAY OR ALL CHAPTERS OF MATHEMATICAL PHYSICS
ALL CHAPTERS?!
only the interesting one
Ok all you math whizzes. Put up your own video with the same problem
I believe the answer is 113/49
Can anyone pls confirm?
No, r must be smaller than 1.
How is the radius of the small circle going to be over twice the side length of the square?