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There's a very simple way of solving problems of this type, although you do have to assume that a unique solution exists. Since the inner radius y is not specified, its value cannot have a bearing on the red area (or else there would not be a unique solution). So let's set y to zero. y=0. Then the 20 m stick becomes a diameter of the red circle, whose radius is 10, and area 100 pi. You can easily do this sort of thing in your head.
Yes, you can imagine the inner radius shrinking, and the tangent being pulled with it, but in order for the tangent to touch the outer circle, the outer circle must also shrink. So you can imagine that there are infinitely-many such setups with different inner and outer radii, and because the question implicitly tells us there's a definite answer, that means all these setups have the same answer. So take the limit for the inner radius as it goes to 0, where it will be like you said, and the limit for the area will be the same area as we started with (because it's constant).
@@lawrencetorrance7051That condition is implicit when the problem doesn't give any of the other quantities. Because either the answer is constant (this you can safely proceed with that presumption), it it's not (in which csee the problem is faulty and any answer is going to be wrong anyway so using the presumption doesn't make the situation any worse).
@@ddichny I was responding to the claim in the original message that you have to assume the solution is unique. In this case, the solution is not unique. There is an infinite number of X, Y pairs that satisfy the conditions.
Solved it in my head from the thumbnail by making an assumption. Since we're not given either diameter, I assume that it doesn't matter just how big either circle is, as long as 20m line fits. In that case, let's reduce the size of the inner circle until it's infinitely small and vanishes. Now we have a circle with a 20m line making a diameter. Using the pi r squared formula, its area must be 100π
I've watched lots of math videos, and when I saw the thumbnail for this, I thought, "No flippin' WAY!!" My immediate objection was that the size of the inner circle could be anything, compared to the outer circle. But apparently, that doesn't matter! The red area would be the same, no matter how big the inner circle is, as long as the line remains 20 meters in all cases. That blows my mind. WOW! What a great problem!! Thanks. You made my day!
I must have seen a similar problem before, because I went straight for Pythagoras, and got 100pi as the answer I would expect. However, I didn't do it in detail, to verify that the elements canceling out was what I expected. It just felt like, "I know this problem, they will cancel out in some way" Also, I did the twist where I pretend the small circle was zero, and got 100pi. If there where any solution, to cancel out: it would have to work with small circle as zero.
@@thorbjrnhellehaven5766 "I did the twist where I pretend the small circle was zero..." Yes, I did that, too. That's when I verified that it doesn't matter what size the inner circle is, as long as the line is 20m. The farther away from center you move the line, of course, the larger the inner circle becomes. The outer circle becomes larger, too. The inner circle "grows" faster than the outer circle. Nothing surprising about any of that. But the fact that the _difference_ of the two circles' areas remains constant is what blew my mind. My first thought was that the relative sizes of the two circles _has_ to matter. But amazingly, it doesn't!
This was exactly my initial thought; There are an infinite amount of inner and outer circle combinations (to be exact any and all circles with a radius greater than 10 meters (20 meters / 2) that you can create that fit this description. So my initial intuition was that we would need an additional datapoint to solve this problem. But apparently for all those infinite combinations, the difference between outer and inner circle will always be 100 pi m2, even as the radius of the outer circle (and with it the radius of the inner circle) would be approaching infinity! This is UTTERLY MIND BLOWING. Math is so awesome! I do think it would have been appropriate though for Math Queen to point out and acknowledge this fact/effect. I think this realization (not the actual solution) is the most amazing part of this problem!
Very cool indeed. Similar first thoughts, but I was expecting there to be some kind of trick to find one of the other two angles and go from there. Kinda wish I'd thought about it a bit longer. I may have to code a visualization of the simultaneous growth.
Love seeing a specialist doing their craft! The problem got me a headache when I saw it but as soon as you drew x touching with the tangent, it was an "oh" moment. I think I might brush-up my math skills as a grown adult now that it's no longer something others are forcing on me. Thank you!
Great solution. It seems simpler when you get to 4:34 to just substitute in what you know x squared equals (y squared + 100) and then you just have an equation with y as the only unknown. They cancel out and you’re left with A= 100pi
I solved the general form of this problem about 25 years ago. I loved how elegant the solution is. If the length of the stick is 2L, the area is pi L^2.
It needs to be, if this problem has a unique solution. You can find it by looking at the extreme case that the inner circle vanishes. (You still have to prove that, but the way this problem is stated, you can kicd of assume it.)
Yes! Using the units when inserting the values can also help with discovering if there was a mistake -- always check that the units come through correctly! (I wrote notes about that so many times when grading PHYS 100 homework exercises ...)
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@@PH-md8xp It kinda does. If the problem doesn't give you the radius of either circle, then it's reasonable to conclude that the inner radius is arbitrary; otherwise, the problem wouldn't be solvable.
@@truthwalker2545No, if the inner circle has a radius larger than 0, then the stick is no longer the diameter of the outer circle. You have to make the outer circle larger to keep the stick at 10m.
@@stephenandrusyszyn3444 not true. as she shows in the video the radius of the inner circle can be anything as long as the radius of the outer circle is follows the ratio 100 = x^ - y^2 where x is outter and y is inner. so if you pick y to be 8, then x has to be 10, to fit this equation. In other words there are an infinit number of circles where the stick's midpoint touches the inner circle, and the edges touch the outer circle The only part that OP's solution doesn't prove is that the area of the donut is the same for any value of the inner circle. It's taken on faith that the problem well formulated, and therefore if the area of the donut is the same regardless of the radiuses, then the area must be piR^2, where R is half 20
I do so Like the fact that every step is laid out. I've often been stuck in maths at one step, that I just can't see, therefor cannot follow to the conclusion. Thanks
Others have mentioned this in roundabout ways but when a chord touches a point on the inner ring , the area of the annulus will always be pi(r)(sqr) with r represented by the chord. Always, no matter where that chord crosses.
Because the inner circle doesn't have any conditions for its size, we can think of it as an area of 0. If we do that, the 20m will go from one edge to another through the center, and that will give us that the ring has a diameter of 20m. A = pi × r² A = pi × (20m ÷ 2)² A = pi × 100m²
A maths professor gave me the napkin ring problem as a joke, understanding the joke is a whole undergraduate course in mathematics - this is a sub problem of that.
@@vinaybhade7046 yep, it is Pythagoras - the simplest proof is topological. the cord is constant. The area of the ring is a difference of two disks pi(R^2-r^2) = pi([cord/2]^2) Always a right angle triangle sides R r and cord/2 R^2 = [cord/2]^2 + r^2
This is one of the types of problems that would have driven me nuts when I was younger. None of my teachers ever told me that this was an option. All the problems they showed in class were ones where you found the value for each variable, but then of course for homework and quizzes there were problems like this, and I could just never wrap my head around it.
You might observe that it’s actually impossible to compute either circle’s radius. There are an infinite number of radii that satisfy the initial given problem, but the area is always the same for all these radii.
I think this realization (that with this problem description the difference of the area between the outer and inner circles will always be 100 pi m2 no matter how large the outer circle gets) is by far the most interesting part of this problem. Much more interesting than how the solution is calculated. I think it was a missed opportunity of Math Queen to not discuss this fact. Especially because this channel is about showcasing the awesomeness of math. This is a mind blowing fact/realization!
Since you can draw a line of length 20 in any large enough circle (R>=10) and because we have no other given values, the outcome must be independent of the radius R. Choose a convenient value for R: R=10. Then the inner circle becomes a point. Area A becomes Pi*R^2 = 100 * Pi.
But maybe the answer to the question is “not enough information”. Merely calculating the answer for the case of y=0 doesn’t prove the general solution.
Where I come from, you're generally required to find all solutions. Finding just some doesn't count. You can't assume that there is only one solution and that all necessary constrains were provided. The correct answer could be a function. E.g., what's the volume of a cuboid with one edge being 20. If we designate the edges as a, b and c and assume that a = 20, then the answer is 20bc. You can't say it's 100 even though that would be one solution (b=5, c=1).
I came to the same result by reasoning in the following manner: The only value given in the diagram is the length of the straight line. Therefore, the value of the area should always be constant regardless of where the line is placed, or the size of the circle(s), provided that its ends touch the outer circle, and it makes a tangent with the inner white circle, _regardless of the size of the white circle_. Therefore, this should also be the case when the area of the white circle is zero, so that the line in fact forms a diameter of the outer circle. The area of such a circle with a diameter of 20m is 100*pi metres.
This is the correct method IMO. In a question like this where only one variable is specified, a more direct solution can often be found by pondering what happens if other variables are set to zero/infinity wherever the question allows.
This problem and solution utterly blew my mind! There are an infinite amount of inner and outer circle combinations (to be exact any and all circles with a radius greater than 10 meters (20 meters / 2) that you can create that fit this description. So my initial intuition was that we would need an additional datapoint to solve this problem. But apparently for all those infinite combinations, the difference between outer and inner circle will always be 100 pi m2, even as the radius of the outer circle (and with it the radius of the inner circle) would be approaching infinity! This is UTTERLY MIND BLOWING. I do think it would have been appropriate though for Math Queen to point out and acknowledge this fact/effect. I think this realization (not the actual solution) is the most amazing part of this problem! Math is so awesome!
@josip.harasic you're wrong sir. There are an infinite amount of circles that fit this description. For instance, you could take a circle with a radius of 15m (diameter of 30m) and then you could place the 20m stick to touch the edges of the circle to form the white inner circle. You could also take a circle with radius of 50m (diameter of 100m) and place the same 20 m stick to touch the edges. That would form a bigger inner white circle than in the first example. In both cases the solution would be the same 100 pi m2, because no matter the radius of the outer circle (as long as its diameter is larger than the 20m stick) , the difference of the outer and inner circle will always be 100 pi m2 You can do this for an outer circle that has a diameter of 1 million meters and put the same 20m stick to touch the edges. You would have a relatively very large inner circle, just a bit smaller than the outer circle. Yet the difference will be the same 100 pi m2. Even if the radius of the outer circle were approaching infinity, you can still make that 20m stick to touch the edges to form a very very large inner circle that is relatively ALMOST as big as the outer circle. Yet the difference between the outer and inner circle will remain that same 100 pi m2 even as the area of both the inner and outer circles both approach infinity! This is a mind blowing fact!
You could also set up an integral using 4 * Σ (0 to r) * √(r^2 - x^2) - 4 * Σ (0 to r₁) * √(r₁^2- x₁^2). Σ being the sum of all values between 0 and the radius of each subsequent circle. A bit more complicated, I think. But interesting none the less. Many blessings Math Queen. Thank you for your videos, they are always fun and informative.
What a wonderful problem! When I saw this I had no clue how to solve it and wondered if it was even possible (although I assumed it was). Genius level stuff. Thank you!
I wish you were my math teacher Math has always been my absolute worst subject and I’ve failed nearly every math class I’ve ever taken (except one now lol) But your explanations make so much sense to me, and you actually somehow make math fun! Like a puzzle to solve, not something to just get through. Idk it’s both intriguing and engaging and I appreciate it, thank you!
decided to try this one on my own before watching the solution. i got so close! i set up the triangle and the pythagorean theorem, but i got stuck trying to square root the 100 + y^2 instead of moving the constant to its own side. cool problem!
Very insightful that you were able to solve it without ever solving for the radius of either circle. In the end, makes sense, since the answer is the same for any two circles the 20-meter stick is touching in the way the problem describes. That means not only do you not have to solve for the radius of either circle, you actually cannot solve for either radius with the information given. Good one!
Since no dimensions other than the chord length are given, there is nothing preventing you from assuming the trivial case where the radius of the inner circle = 0. Thus the 20m chord becomes the diameter of the red circle, leading to a radius of 10m. pi * (10m)^2 = 100pi meters^2
I once saw this problem in the Asimov/Greenberg Math puzzles book, where a line drawn like that describes a diameter of an equivalent sized circle (without the white inner circle). With that, you know diameter, radius, and pi R Squared. Works for any such similar circles and sizes.
Interesting that the area is independent of the relationship between x and y. So, the limit of the area as y approaches zero is still 100*pi. (As another pointed out, that would simply be a circle with radius = 10.) But the limit as y approaches x is the same. You would end up with a circle that is infinitely thick but with infinite circumference, because the chord would still have to span a space with non-zero area.
That was brilliant. I used to be able to figure this stuff out in high school. It's like a layer cake with 2 stacks, it boils down to 2 key tricks. You gotta love the way it looks impossible but apply the formula for area that you know, plus the pythagorean theorem, then voila, the 2 systems of equations just solve each other. I used to love those math genius tricks. Yes, Travis, it's 2 things at once... 😅
You forgot to mention the theorem that allows us to assume that the side length is 10. The theorem: If a radius of a circle is perpendicular to a chord of the same circle, then the radius bisects the chord. I still remember having to be that rigorous in high school. It's entirely possible you don't know how to state the theorem in English, but you still speak very well. After all, I don't know how to state it in German.
I bisected the 20m with the unknown diameter of the larger circle, defined the smaller and larger parts as x and y, and used the chord theorem for x*y=10*10 = 100. Looks like not enough to solve, but then you can also define the small radius r as (y-x)/2. And also y would be 2r + x. If you then compute the red area, and manipulate the equations you can actually find that the red area is pi * x * y making it 100pi. Surprising!
Thank you for your video. I have one suggestion to make the solution a bit clearer for students. You mentioned that the tangent cord is perpendicular to the radius of the circle, but didn't specifically mention that the radius is a perpendicular bisector of the tangent cord. When you diagrammed the right triangle you mentioned that you only needed half of cord's length for the side of the triangle, but didn't provide much of an explanation as to why it was exactly half.
If you draw the radius called x to both ends of the stick you receive isosceles triangle. The height of such triangle is bisecting the base which is obvious to show by triangle identity.
Hi, method correct, (a similar problem was on a gcse question set i gave students recently) although most exams would expect the correct units to be added to your answer.
This was a puzzle given about 15 years ago on the Saturday morning NPR "Car Talk" program with Click and Clack, the Tap-it Brothers. I recall that the area of the ring is exactly the area of a circle whose diameter is equal to the length of the given line. And it doesn't matter the outer diameter or the inner diameter, as long as the given line is tangent to the smaller circle and its end points are on the outer circle. The area of the "washer" is equal to the area of circle whose diameter is equal to the length of the given line. Furthermore, there is a three dimensional version of this problem too. Take a solid metal sphere and drill a hole down through the middle of the sphere. Make the diameter of the hole somewhere between about 5% of the diameter of the sphere and 98% of its diameter. If the diameter of the sphere is only slightly larger than one of your fingers and the hole through the center is about 95% of the diameter of the sphere, then the sphere will look like a ring that one wears on his or her finger. Set the sphere with the hole in it on a table on the flat created by the hole through the sphere. Measure the height of the piece of metal above the table top, which is the same as measuring the length of the hole in the piece. Now here is where the similarity to the two-dimensional case comes in. All spheres with holes that stand at the same height will have exactly the same volume of material, regardless of the original diameter of the sphere or the size of the hole through them, as long as they stand at the same height.
… wow, that was easy, I didn’t see that coming. Thanks! It just goes to show again, that if you can write down as many true things about the problem that you can think of, then the solution stares you right in the face. Thanks!
Super-cute problem. First I was like, "wait, have you given enough information, or did you forget to mention something?" But then I realized by Pythagoras, the solution is independent of the ratio of the radii. :-)
This is a great example where you don't even need to solve for X or Y individually to get your answer. This also indicates that the discs can be drawn to different proportions of their radii, meaning you could have the two have almost the same radius just as long as you can lie a stick across the outer one and tangentially touch the inner disc. It will always be ~ 314.16. Just imagine the inner disc reduced to a radius of 0.
"Hello my lovely" 😊 I really enjoy your maths channel and selection of puzzles. I hope you keep this channel going on for a long time. My maths level is good so many of the puzzles I can solve perhaps a bit easily, ... however I love watching, plus I love that you choose problems that are possible for many people to try. ... Many maths channels focus on difficult problems that most people can't relate to. You are doing a "bang-up job" as they say 🏆😊👍
I drew a diameter perpendicular to the 20m line, then used that rule about the arms of intersecting chords to get 10.10 = ( R - r ) ( R + r ) where R is the outer radius and r the inner. Then difference of squares.
We had a similar license in Denmark 🇩🇰. It wa split so that you had a full license for colour tvs. A light for radio reception, and a cheaper tv license for bw tvs. With the proliferation of Internet based broadcasts it was reformed in the naughties so there is only one license, but being able to receive a signal is sufficient to have to pay the license. Therefore having an Internet connection or a device capable of receiving something via the Internet (or a car with a built in radio receiver) you have to pay.
There is a very good 3d analog. If you take a sphere and drill a hole through the center of the sphere, and the length of the hole is a specific number, such as 1 meter, what is the volume of the ring of material left after the hole is drilled? Since the radius of the shpere is not mentioned, if you trust the question to be complete this means the radius of the sphere (and the diameter of the hole) are irrelvant. So you can set the diameter to zero and the result is a sphere of radius is 1/2 meter, and solve for that.
Let us name the bigger radius as R & smaller (only white) as r. As per one theorem (that I can't recall the "name" now, but learnt) the 20m line is bisected by the bigger diameter (=2R), at distance r from the (common) center. Let the 20 m line b PQ & mid-point, or point of intersection by 2R be O. The two parts, (being segmented at O) are (R+r) & (R -r) Then PO = OQ = 20/2 = 10, being the bisected parts (halves). The theorem says, PO.OQ = (R+r)(R -r) PO.OQ = R^2 -r^2 10 . 10 = R^2 -r^2 R^2 -r^2 =100 The coloured (pink) area is the difference of areas of both circles = πR^2 - πr^2 = π(R^2 - r^2). [substituting the value for (R^2 - r^2)] Hence, coloured (pink) area = π(100) sq.m.
Without looking at the video or comments: I can visualize that the red circle would shrink, keeping that chord at 20m, if the white circle was made smaller. As there is no other information I'll assume (but not prove) that the red area must therefore stay the same. So shrink the white circle to a dot, gives a red circle of diameter 20m (radius 10m). Thus area is PI r^2 or about 314 m^2.
This math problem is some glimpse into some deeper truth in Nature. It's related to Newton's theorem that for most purposes, you can represent a gravitationally interacting body by a geometrical point of the same mass at its gravitational center. It's also related to the shell theorem. If you have a heavy sphere, and you drill straight holes through it, then letting a test body fall through the holes will take the same time for each of them to reach the other end. Imagine Earth and straight tunnels through Earth like from New York to Shanghai, or from London to Los Angeles, or from the North Pole to the South Pole. If you let a ball roll through any of the holes, and if you ignore friction and air resistance, it will always take about 45 mins for the ball to roll through one of the tunnels, independent of their length.
Very elegant! The thing that's not intuitive (to me!) is that the area of the ring is _independent_ of the diameter of the inner circle. It comes out cleanly in the algebra, but not at all obvious in the geometry. But the formation of the question says that the area of the ring _must be_ independent of the diameter of the circle. There's a degenerate case that's easy to solve: set the diameter of the inner circle to 0 !!!! Then the "ring" becomes a circle with diameter 20, radius 10, area 100*pi. No fancy algebra needed -- but you must believe that the problem is correctly stated, to come up with it. Thank you!
I think that it is important to point out (for any math students who are watching this) that this solution will be true for any values of the smaller circle radius and larger circle radius which can have that particular 20m line relationship.
Let the radius of the inner circle be r and the radius of the outer circle R. The area of the ring is the area of the outer circle minus the area of the inner. A = πR² - πr² = π(R²-r²) --- [1] Let M and N be the ends of the line segment, O the center of the circles, and T the point of tangency between MN and the inner circle. As OT is a radius of the inner circle, then its length is r, and as it is also the point of tangency between the inner circle and MN, then ∠OTM = ∠NTO = 90°. As any line through the center of a circle that intersects a chord of that circle perpendicularly is also a bisector of that chord, then T is the midpoint of MN, and MT = TN = 20/2 = 10. As OM and ON are radii of the outer circle, then OM = ON = R. As OM = ON, MT = TN, and OT is common, then ∆OTM and ∆NTO are congruent triangles by SSS congruency. Triangle ∆OTM: OT² + TM² = OM² r² + 10² = R² R² - r² = 100 A = π(R²-r²)
The curious thing about this problem is that the area of the ring isn't depenendent on the radius of either the inner or outter circle (as long as the outter circle has a radius of at least 10 to fit the stick).
With the single given dimension it's natural to say the problem is ill-formed because there is no unique solution, but the saving grace is that it is not asking for the radius of anything. It's asking for AREA, which has a unique solution. No assumptions are required.
If you make the assumption that the problem is solvable irrespective of the relative sizes of the inner and outer circle, you can simply reduce the second circle to the limit where it becomes a point in the middle, and the chord becomes the diameter of the circle. Then it's trivial to see that the area is 100.pi
There is a much more simple solution. The construction of the question dictates a relationship of the circles both infinitely large and infinitely small. If the inner circle is infinitely small it becomes zero. The diameter of the outer circle becomes 20m. Therefore radius is 10m. A=pi*r² or 314.16
This was interesting, because at first glance, the problem looked under defined. But then I realized that given any chord in a circle, there's always an implied inner circle dictated by a line from the center intersecting the chord. If the circle was made smaller, then the chord would have to be longer. But there's some interesting posts here about an inner circle with an area of 0. In which case there is no inner circle.
Correct. But then the outer circle has a radius of 10m and the inner (non existing) circle a radius from 0m and still the difference of the areas is 100pi m^2 😊
@@hobbyist6181 It's always nice to see it work out nicely like that :) That's the kind of universal logic I like to see in the programs I write, but then there's all these pesky edge cases that usually ruin my elegant ideas.
By symmetry, the 20m stick can form an equilateral triangle with two other tangents, and then we can use area of that triangle to solve for inner and outer radii which yields R = 2r After that, it's just pi (R² - r²) Good problem nevertheless
Its a fun one, cause x could be any size greater than 10, so we can't determine x or y, only the relationship between them, which suffices to find the area.
The weird thing is that the ratio of radii is irrelevant. If we knew that, we could imagine a vanishingly small inner circle and we'd get a ring area equal to the area of the larger circle. But is there a way to see that the ratio of radii is irrelevant? Very nice puzzle!
Assume a diameter of the outer circle to be any number larger than 20. Then determine the diameter of the inner circle by using the graphical equation of a circle. I assumed a big diameter of 30m and calculated the inter circle diameter to be 22.36…m and you get the same solution
Without attempting any proof or calculation, I’d say that the if area the ring does not depend on the area of the two circles, then it must also be the area when “inner circle” has “area” zero. That is, when the “ring” is a circle with diameter 20. But now I have to show my work.
I can see now that others have reached the same conclusion (and I hope to avoid similar vitriolic discussions). As we’ve seen I made a correct guess, but it still requires a follow up with proof. And that’s fairly easy with a careful choice of a right triangle.
Let's calculate the area of the red ring ▶ Ared= πr₂² - πr₁² By drawing the radius r₁ from the center O, to the middle of the stick, and the radius r₂ where the stick touches the circumference of the large circle, we obtain a right triangle 💡 The right triangle ΔAOB: [AO]= r₁ [OB]= r₂ [BA]= 20/2 [BA]= 10 ⇒ By applying the Pythagorean theorem, we get: [AO]²+[BA]²= [OB]² r₁² + 10²= r₂² ⇒ r₂² - r₁² = 100 Ared= πr₂² - πr₁² Ared= π(r₂² - r₁²) Ared= 100 π Ared≈ 314,16 square units
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There's a very simple way of solving problems of this type, although you do have to assume that a unique solution exists.
Since the inner radius y is not specified, its value cannot have a bearing on the red area (or else there would not be a unique solution).
So let's set y to zero. y=0.
Then the 20 m stick becomes a diameter of the red circle, whose radius is 10, and area 100 pi.
You can easily do this sort of thing in your head.
Yes, you can imagine the inner radius shrinking, and the tangent being pulled with it, but in order for the tangent to touch the outer circle, the outer circle must also shrink.
So you can imagine that there are infinitely-many such setups with different inner and outer radii, and because the question implicitly tells us there's a definite answer, that means all these setups have the same answer.
So take the limit for the inner radius as it goes to 0, where it will be like you said, and the limit for the area will be the same area as we started with (because it's constant).
I did the same, it's much faster.
You mean, you have to assume solutions are not unique.
@@lawrencetorrance7051That condition is implicit when the problem doesn't give any of the other quantities. Because either the answer is constant (this you can safely proceed with that presumption), it it's not (in which csee the problem is faulty and any answer is going to be wrong anyway so using the presumption doesn't make the situation any worse).
@@ddichny I was responding to the claim in the original message that you have to assume the solution is unique. In this case, the solution is not unique. There is an infinite number of X, Y pairs that satisfy the conditions.
Solved it in my head from the thumbnail by making an assumption. Since we're not given either diameter, I assume that it doesn't matter just how big either circle is, as long as 20m line fits. In that case, let's reduce the size of the inner circle until it's infinitely small and vanishes. Now we have a circle with a 20m line making a diameter. Using the pi r squared formula, its area must be 100π
This is brilliant logic, thank you.
I've watched lots of math videos, and when I saw the thumbnail for this, I thought, "No flippin' WAY!!" My immediate objection was that the size of the inner circle could be anything, compared to the outer circle. But apparently, that doesn't matter! The red area would be the same, no matter how big the inner circle is, as long as the line remains 20 meters in all cases. That blows my mind.
WOW! What a great problem!!
Thanks. You made my day!
I must have seen a similar problem before, because I went straight for Pythagoras, and got 100pi as the answer I would expect.
However, I didn't do it in detail, to verify that the elements canceling out was what I expected.
It just felt like, "I know this problem, they will cancel out in some way"
Also, I did the twist where I pretend the small circle was zero, and got 100pi. If there where any solution, to cancel out: it would have to work with small circle as zero.
@@thorbjrnhellehaven5766 "I did the twist where I pretend the small circle was zero..."
Yes, I did that, too. That's when I verified that it doesn't matter what size the inner circle is, as long as the line is 20m.
The farther away from center you move the line, of course, the larger the inner circle becomes. The outer circle becomes larger, too. The inner circle "grows" faster than the outer circle. Nothing surprising about any of that. But the fact that the _difference_ of the two circles' areas remains constant is what blew my mind. My first thought was that the relative sizes of the two circles _has_ to matter. But amazingly, it doesn't!
This was exactly my initial thought; There are an infinite amount of inner and outer circle combinations (to be exact any and all circles with a radius greater than 10 meters (20 meters / 2) that you can create that fit this description. So my initial intuition was that we would need an additional datapoint to solve this problem. But apparently for all those infinite combinations, the difference between outer and inner circle will always be 100 pi m2, even as the radius of the outer circle (and with it the radius of the inner circle) would be approaching infinity! This is UTTERLY MIND BLOWING. Math is so awesome!
I do think it would have been appropriate though for Math Queen to point out and acknowledge this fact/effect. I think this realization (not the actual solution) is the most amazing part of this problem!
Very cool indeed. Similar first thoughts, but I was expecting there to be some kind of trick to find one of the other two angles and go from there. Kinda wish I'd thought about it a bit longer. I may have to code a visualization of the simultaneous growth.
Love seeing a specialist doing their craft! The problem got me a headache when I saw it but as soon as you drew x touching with the tangent, it was an "oh" moment. I think I might brush-up my math skills as a grown adult now that it's no longer something others are forcing on me. Thank you!
Great solution. It seems simpler when you get to 4:34 to just substitute in what you know x squared equals (y squared + 100) and then you just have an equation with y as the only unknown. They cancel out and you’re left with A= 100pi
I solved the general form of this problem about 25 years ago. I loved how elegant the solution is. If the length of the stick is 2L, the area is pi L^2.
It needs to be, if this problem has a unique solution. You can find it by looking at the extreme case that the inner circle vanishes.
(You still have to prove that, but the way this problem is stated, you can kicd of assume it.)
I really like seeing problems like this. Thanks for the video, Susanne!
Love these problems and your very clear explanations!
As an engineer, my brain says "don't forget the units "square meters" in the final solution! 😄
Yes! Using the units when inserting the values can also help with discovering if there was a mistake -- always check that the units come through correctly! (I wrote notes about that so many times when grading PHYS 100 homework exercises ...)
Teachers teach their subject, but above all they teach who they are to their pupils. May many pupils benefit from you and your lessons, they will become good pupils and good people!
let the inner circle have an area of 0. then the R of the outer circle is 10.
Exactly!
@@lusoverse8710
Doesn’t solve the problem though, does it?
@@PH-md8xp It kinda does. If the problem doesn't give you the radius of either circle, then it's reasonable to conclude that the inner radius is arbitrary; otherwise, the problem wouldn't be solvable.
@@truthwalker2545No, if the inner circle has a radius larger than 0, then the stick is no longer the diameter of the outer circle. You have to make the outer circle larger to keep the stick at 10m.
@@stephenandrusyszyn3444
not true. as she shows in the video the radius of the inner circle can be anything as long as the radius of the outer circle is follows the ratio 100 = x^ - y^2
where x is outter and y is inner. so if you pick y to be 8, then x has to be 10, to fit this equation.
In other words there are an infinit number of circles where the stick's midpoint touches the inner circle, and the edges touch the outer circle
The only part that OP's solution doesn't prove is that the area of the donut is the same for any value of the inner circle. It's taken on faith that the problem well formulated, and therefore if the area of the donut is the same regardless of the radiuses, then the area must be piR^2, where R is half 20
A great lesson in logical thinking and creatively looking at a problem. Thank you for your great teaching ability.
These videos are great. just the right level of difficulty to make them approachable
I do so Like the fact that every step is laid out. I've often been stuck in maths at one step, that I just can't see, therefor cannot follow to the conclusion. Thanks
Where were you in my youth! You explain it so much better, than my school books...
This is great knowledge if you ever need two discs of the same weight but one needs a hole of a certain size in it.
Others have mentioned this in roundabout ways but when a chord touches a point on the inner ring , the area of the annulus will always be pi(r)(sqr) with r represented by the chord. Always, no matter where that chord crosses.
1:56 - rearrange A = pi*x^2 - pi*x^2:
1) A = pi(x^2 -y^2)
4:09 - rearrange y^2 + 10^2 = x^2:
2) 100 = x^2 - y^2
substituting 2) into 1:
A = pi(100)
A = 100*pi
Because the inner circle doesn't have any conditions for its size, we can think of it as an area of 0.
If we do that, the 20m will go from one edge to another through the center, and that will give us that the ring has a diameter of 20m.
A = pi × r²
A = pi × (20m ÷ 2)²
A = pi × 100m²
A maths professor gave me the napkin ring problem as a joke, understanding the joke is a whole undergraduate course in mathematics - this is a sub problem of that.
Even I understood it this way. But is there a proof that the areas will always be the same if the length of the line is constant.
@@vinaybhade7046 yep, it is Pythagoras - the simplest proof is topological.
the cord is constant. The area of the ring is a difference of two disks pi(R^2-r^2) = pi([cord/2]^2)
Always a right angle triangle sides R r and cord/2
R^2 = [cord/2]^2 + r^2
@@carly09et ok. Got it
This is the only channel where I get to be somebody's "lovely". I feel so wanted.
This is one of the types of problems that would have driven me nuts when I was younger. None of my teachers ever told me that this was an option. All the problems they showed in class were ones where you found the value for each variable, but then of course for homework and quizzes there were problems like this, and I could just never wrap my head around it.
You might observe that it’s actually impossible to compute either circle’s radius. There are an infinite number of radii that satisfy the initial given problem, but the area is always the same for all these radii.
Good comment - this was throwing me
Nice use of the plural there, sir! 👍🏻
Mensa Swede? That's kinda cringe, bro. 😊
I think this realization (that with this problem description the difference of the area between the outer and inner circles will always be 100 pi m2 no matter how large the outer circle gets) is by far the most interesting part of this problem. Much more interesting than how the solution is calculated.
I think it was a missed opportunity of Math Queen to not discuss this fact. Especially because this channel is about showcasing the awesomeness of math. This is a mind blowing fact/realization!
The question is about showing a very interesting phenomenon..
And that geometric simplication is a thing!
Since you can draw a line of length 20 in any large enough circle (R>=10) and because we have no other given values, the outcome must be independent of the radius R.
Choose a convenient value for R: R=10. Then the inner circle becomes a point. Area A becomes Pi*R^2 = 100 * Pi.
But maybe the answer to the question is “not enough information”. Merely calculating the answer for the case of y=0 doesn’t prove the general solution.
Where I come from, you're generally required to find all solutions. Finding just some doesn't count. You can't assume that there is only one solution and that all necessary constrains were provided. The correct answer could be a function. E.g., what's the volume of a cuboid with one edge being 20. If we designate the edges as a, b and c and assume that a = 20, then the answer is 20bc. You can't say it's 100 even though that would be one solution (b=5, c=1).
You could assume that.. or prove it.
It's magic!! LOL
Very well explained and at a relaxed speed! Love your presentations!!
Tangent and secant theorem:
R-r/10=10/R+r
R^2-r^2=100
Pi*R^2-Pi*r^2=100*Pi
I can't do any of these problems but it's fascinating to see how it's figured out. Make's no sense to be but enjoyable to watch
The tangent with 90°: Triangle in the circle: R² = r² + 10² AND A= π.(R² - r²) -> A = π.10² = 314,159
Love to see your new international channel - I wish you every possible success!
I came to the same result by reasoning in the following manner: The only value given in the diagram is the length of the straight line. Therefore, the value of the area should always be constant regardless of where the line is placed, or the size of the circle(s), provided that its ends touch the outer circle, and it makes a tangent with the inner white circle, _regardless of the size of the white circle_.
Therefore, this should also be the case when the area of the white circle is zero, so that the line in fact forms a diameter of the outer circle. The area of such a circle with a diameter of 20m is 100*pi metres.
This is the correct method IMO. In a question like this where only one variable is specified, a more direct solution can often be found by pondering what happens if other variables are set to zero/infinity wherever the question allows.
This problem and solution utterly blew my mind! There are an infinite amount of inner and outer circle combinations (to be exact any and all circles with a radius greater than 10 meters (20 meters / 2) that you can create that fit this description. So my initial intuition was that we would need an additional datapoint to solve this problem. But apparently for all those infinite combinations, the difference between outer and inner circle will always be 100 pi m2, even as the radius of the outer circle (and with it the radius of the inner circle) would be approaching infinity! This is UTTERLY MIND BLOWING.
I do think it would have been appropriate though for Math Queen to point out and acknowledge this fact/effect. I think this realization (not the actual solution) is the most amazing part of this problem! Math is so awesome!
No, there is only one solution for circles radius, because we have the given string length "20" and his position to the two circles.
@josip.harasic you're wrong sir. There are an infinite amount of circles that fit this description.
For instance, you could take a circle with a radius of 15m (diameter of 30m) and then you could place the 20m stick to touch the edges of the circle to form the white inner circle.
You could also take a circle with radius of 50m (diameter of 100m) and place the same 20 m stick to touch the edges. That would form a bigger inner white circle than in the first example.
In both cases the solution would be the same 100 pi m2, because no matter the radius of the outer circle (as long as its diameter is larger than the 20m stick) , the difference of the outer and inner circle will always be 100 pi m2
You can do this for an outer circle that has a diameter of 1 million meters and put the same 20m stick to touch the edges. You would have a relatively very large inner circle, just a bit smaller than the outer circle. Yet the difference will be the same 100 pi m2. Even if the radius of the outer circle were approaching infinity, you can still make that 20m stick to touch the edges to form a very very large inner circle that is relatively ALMOST as big as the outer circle. Yet the difference between the outer and inner circle will remain that same 100 pi m2 even as the area of both the inner and outer circles both approach infinity! This is a mind blowing fact!
You could also set up an integral using 4 * Σ (0 to r) * √(r^2 - x^2) - 4 * Σ (0 to r₁) * √(r₁^2- x₁^2). Σ being the sum of all values between 0 and the radius of each subsequent circle. A bit more complicated, I think. But interesting none the less. Many blessings Math Queen. Thank you for your videos, they are always fun and informative.
What a wonderful problem! When I saw this I had no clue how to solve it and wondered if it was even possible (although I assumed it was). Genius level stuff. Thank you!
I found it!! It is between the white background and the white disc.
Great and usefull video. Love the way you explain the approach used. Thank you Suzanne.
I wish you were my math teacher
Math has always been my absolute worst subject and I’ve failed nearly every math class I’ve ever taken (except one now lol)
But your explanations make so much sense to me, and you actually somehow make math fun! Like a puzzle to solve, not something to just get through. Idk it’s both intriguing and engaging and I appreciate it, thank you!
As math admirer I really love your videos my perfect comfort entertainment thank you so much keep shinning with that wholesome smile of yours ! :)
decided to try this one on my own before watching the solution. i got so close! i set up the triangle and the pythagorean theorem, but i got stuck trying to square root the 100 + y^2 instead of moving the constant to its own side. cool problem!
That was slick. Very well explained. But I must say, I enjoyed the way you write your letter A. Very elegant
That was a beautifully clean solution, mine was not so clear-cut, but yielded the correct answer.
Very insightful that you were able to solve it without ever solving for the radius of either circle. In the end, makes sense, since the answer is the same for any two circles the 20-meter stick is touching in the way the problem describes. That means not only do you not have to solve for the radius of either circle, you actually cannot solve for either radius with the information given.
Good one!
Great presentation!
Since no dimensions other than the chord length are given, there is nothing preventing you from assuming the trivial case where the radius of the inner circle = 0. Thus the 20m chord becomes the diameter of the red circle, leading to a radius of 10m. pi * (10m)^2 = 100pi meters^2
I once saw this problem in the Asimov/Greenberg Math puzzles book, where a line drawn like that describes a diameter of an equivalent sized circle (without the white inner circle). With that, you know diameter, radius, and pi R Squared. Works for any such similar circles and sizes.
I used sin and cos and tan method. It also works as well, thank you so much for the brain teasers. You're way is so much easier.
The solution is beautifully cheeky.
Interesting that the area is independent of the relationship between x and y. So, the limit of the area as y approaches zero is still 100*pi. (As another pointed out, that would simply be a circle with radius = 10.) But the limit as y approaches x is the same. You would end up with a circle that is infinitely thick but with infinite circumference, because the chord would still have to span a space with non-zero area.
That was brilliant. I used to be able to figure this stuff out in high school. It's like a layer cake with 2 stacks, it boils down to 2 key tricks. You gotta love the way it looks impossible but apply the formula for area that you know, plus the pythagorean theorem, then voila, the 2 systems of equations just solve each other. I used to love those math genius tricks. Yes, Travis, it's 2 things at once... 😅
You forgot to mention the theorem that allows us to assume that the side length is 10. The theorem: If a radius of a circle is perpendicular to a chord of the same circle, then the radius bisects the chord. I still remember having to be that rigorous in high school. It's entirely possible you don't know how to state the theorem in English, but you still speak very well. After all, I don't know how to state it in German.
Thanks - came looking for that and appreciate you mentioning the theorem.
Thanks - came to commentary to check that presumption !
I bisected the 20m with the unknown diameter of the larger circle, defined the smaller and larger parts as x and y, and used the chord theorem for x*y=10*10 = 100. Looks like not enough to solve, but then you can also define the small radius r as (y-x)/2. And also y would be 2r + x. If you then compute the red area, and manipulate the equations you can actually find that the red area is pi * x * y making it 100pi. Surprising!
Another fantastic video. Thanks.
Thank you for your video. I have one suggestion to make the solution a bit clearer for students. You mentioned that the tangent cord is perpendicular to the radius of the circle, but didn't specifically mention that the radius is a perpendicular bisector of the tangent cord. When you diagrammed the right triangle you mentioned that you only needed half of cord's length for the side of the triangle, but didn't provide much of an explanation as to why it was exactly half.
If you draw the radius called x to both ends of the stick you receive isosceles triangle. The height of such triangle is bisecting the base which is obvious to show by triangle identity.
That is fantastic! I want to go back to school to study Chemistry. I have to get my Math back on track.
Hi, method correct, (a similar problem was on a gcse question set i gave students recently) although most exams would expect the correct units to be added to your answer.
Wow! What a lovely solution!
This was a puzzle given about 15 years ago on the Saturday morning NPR "Car Talk" program with Click and Clack, the Tap-it Brothers. I recall that the area of the ring is exactly the area of a circle whose diameter is equal to the length of the given line. And it doesn't matter the outer diameter or the inner diameter, as long as the given line is tangent to the smaller circle and its end points are on the outer circle. The area of the "washer" is equal to the area of circle whose diameter is equal to the length of the given line.
Furthermore, there is a three dimensional version of this problem too. Take a solid metal sphere and drill a hole down through the middle of the sphere. Make the diameter of the hole somewhere between about 5% of the diameter of the sphere and 98% of its diameter. If the diameter of the sphere is only slightly larger than one of your fingers and the hole through the center is about 95% of the diameter of the sphere, then the sphere will look like a ring that one wears on his or her finger. Set the sphere with the hole in it on a table on the flat created by the hole through the sphere. Measure the height of the piece of metal above the table top, which is the same as measuring the length of the hole in the piece. Now here is where the similarity to the two-dimensional case comes in. All spheres with holes that stand at the same height will have exactly the same volume of material, regardless of the original diameter of the sphere or the size of the hole through them, as long as they stand at the same height.
"Car Talk" was a fun to listen to. I learned and enjoyed the comedy. Miss them.
… wow, that was easy, I didn’t see that coming. Thanks! It just goes to show again, that if you can write down as many true things about the problem that you can think of, then the solution stares you right in the face. Thanks!
Much easy than what I was about to try to do
Super-cute problem. First I was like, "wait, have you given enough information, or did you forget to mention something?" But then I realized by Pythagoras, the solution is independent of the ratio of the radii. :-)
This is a great example where you don't even need to solve for X or Y individually to get your answer.
This also indicates that the discs can be drawn to different proportions of their radii, meaning you could have the two have almost the same radius just as long as you can lie a stick across the outer one and tangentially touch the inner disc. It will always be ~ 314.16. Just imagine the inner disc reduced to a radius of 0.
"Hello my lovely" 😊 I really enjoy your maths channel and selection of puzzles. I hope you keep this channel going on for a long time. My maths level is good so many of the puzzles I can solve perhaps a bit easily, ... however I love watching, plus I love that you choose problems that are possible for many people to try. ... Many maths channels focus on difficult problems that most people can't relate to. You are doing a "bang-up job" as they say 🏆😊👍
I drew a diameter perpendicular to the 20m line, then used that rule about the arms of intersecting chords to get 10.10 = ( R - r ) ( R + r ) where R is the outer radius and r the inner. Then difference of squares.
We had a similar license in Denmark 🇩🇰. It wa split so that you had a full license for colour tvs. A light for radio reception, and a cheaper tv license for bw tvs. With the proliferation of Internet based broadcasts it was reformed in the naughties so there is only one license, but being able to receive a signal is sufficient to have to pay the license. Therefore having an Internet connection or a device capable of receiving something via the Internet (or a car with a built in radio receiver) you have to pay.
Smart problem and tricky solution.🎉
Easy peasy. Thank you.
There is a very good 3d analog. If you take a sphere and drill a hole through the center of the sphere, and the length of the hole is a specific number, such as 1 meter, what is the volume of the ring of material left after the hole is drilled?
Since the radius of the shpere is not mentioned, if you trust the question to be complete this means the radius of the sphere (and the diameter of the hole) are irrelvant. So you can set the diameter to zero and the result is a sphere of radius is 1/2 meter, and solve for that.
Let us name the bigger radius as R & smaller (only white) as r.
As per one theorem (that I can't recall the "name" now, but learnt) the 20m line is bisected by the bigger diameter (=2R), at distance r from the (common) center.
Let the 20 m line b PQ & mid-point, or point of intersection by 2R be O.
The two parts, (being segmented at O) are (R+r) & (R -r)
Then PO = OQ = 20/2 = 10, being the bisected parts (halves).
The theorem says,
PO.OQ = (R+r)(R -r)
PO.OQ = R^2 -r^2
10 . 10 = R^2 -r^2
R^2 -r^2 =100
The coloured (pink) area is the difference of areas of both circles = πR^2 - πr^2 = π(R^2 - r^2). [substituting the value for (R^2 - r^2)]
Hence, coloured (pink) area = π(100) sq.m.
I'll know when I'm good at math when I start seeing triangles everywhere, thanks for the video!
Without looking at the video or comments: I can visualize that the red circle would shrink, keeping that chord at 20m, if the white circle was made smaller. As there is no other information I'll assume (but not prove) that the red area must therefore stay the same. So shrink the white circle to a dot, gives a red circle of diameter 20m (radius 10m). Thus area is PI r^2 or about 314 m^2.
very enjoyed this one
Simply! the hats off!
Well done! This is a good one. Thanks!
Glad you like it :) Thanks for watching!
This math problem is some glimpse into some deeper truth in Nature. It's related to Newton's theorem that for most purposes, you can represent a gravitationally interacting body by a geometrical point of the same mass at its gravitational center. It's also related to the shell theorem. If you have a heavy sphere, and you drill straight holes through it, then letting a test body fall through the holes will take the same time for each of them to reach the other end. Imagine Earth and straight tunnels through Earth like from New York to Shanghai, or from London to Los Angeles, or from the North Pole to the South Pole. If you let a ball roll through any of the holes, and if you ignore friction and air resistance, it will always take about 45 mins for the ball to roll through one of the tunnels, independent of their length.
Very elegant! The thing that's not intuitive (to me!) is that the area of the ring is _independent_ of the diameter of the inner circle. It comes out cleanly in the algebra, but not at all obvious in the geometry. But the formation of the question says that the area of the ring _must be_ independent of the diameter of the circle. There's a degenerate case that's easy to solve: set the diameter of the inner circle to 0 !!!! Then the "ring" becomes a circle with diameter 20, radius 10, area 100*pi. No fancy algebra needed -- but you must believe that the problem is correctly stated, to come up with it. Thank you!
I see I'm not the first person to have this insight.
I think that it is important to point out (for any math students who are watching this) that this solution will be true for any values of the smaller circle radius and larger circle radius which can have that particular 20m line relationship.
Let the radius of the inner circle be r and the radius of the outer circle R. The area of the ring is the area of the outer circle minus the area of the inner.
A = πR² - πr² = π(R²-r²) --- [1]
Let M and N be the ends of the line segment, O the center of the circles, and T the point of tangency between MN and the inner circle.
As OT is a radius of the inner circle, then its length is r, and as it is also the point of tangency between the inner circle and MN, then ∠OTM = ∠NTO = 90°. As any line through the center of a circle that intersects a chord of that circle perpendicularly is also a bisector of that chord, then T is the midpoint of MN, and MT = TN = 20/2 = 10.
As OM and ON are radii of the outer circle, then OM = ON = R.
As OM = ON, MT = TN, and OT is common, then ∆OTM and ∆NTO are congruent triangles by SSS congruency.
Triangle ∆OTM:
OT² + TM² = OM²
r² + 10² = R²
R² - r² = 100
A = π(R²-r²)
Dooooope tusind tak!
Got the answer intuitively, doesn't matter the angles in the chord measurement as it is always the same area
The curious thing about this problem is that the area of the ring isn't depenendent on the radius of either the inner or outter circle (as long as the outter circle has a radius of at least 10 to fit the stick).
This one was briliant... speaking of briliant....
With the single given dimension it's natural to say the problem is ill-formed because there is no unique solution, but the saving grace is that it is not asking for the radius of anything. It's asking for AREA, which has a unique solution. No assumptions are required.
If you make the assumption that the problem is solvable irrespective of the relative sizes of the inner and outer circle, you can simply reduce the second circle to the limit where it becomes a point in the middle, and the chord becomes the diameter of the circle. Then it's trivial to see that the area is 100.pi
Es geht auch mit dem Höhensatz bzw. Sehnensatz:
(20 m/2)² = (R − r) ⋅ (R + r)
100 m² = R² − r²
A(ring) = π ⋅ R² − π ⋅ r²
= π ⋅ (R² − r²)
= 100 ⋅ π m²
≈ 314,16 m²
Good stuff.
There is a much more simple solution. The construction of the question dictates a relationship of the circles both infinitely large and infinitely small. If the inner circle is infinitely small it becomes zero. The diameter of the outer circle becomes 20m. Therefore radius is 10m. A=pi*r² or 314.16
The method that jumped out to me was -
δA≈10²δθ/2,
∴A=50∫dθ, between the limits zero and 2π
∴ A=100π
What's interesting is in the limit of infinite radius the area of the thin shell that is formed is also the same
The difference in area would be the same regardless of the value of y, including the special case where y=0, and the 20m stick becomes the diameter.
I first heard this problem on the radio show Car Talk, in their “puzzler” segment.
This was interesting, because at first glance, the problem looked under defined. But then I realized that given any chord in a circle, there's always an implied inner circle dictated by a line from the center intersecting the chord. If the circle was made smaller, then the chord would have to be longer. But there's some interesting posts here about an inner circle with an area of 0. In which case there is no inner circle.
Correct. But then the outer circle has a radius of 10m and the inner (non existing) circle a radius from 0m and still the difference of the areas is 100pi m^2 😊
@@hobbyist6181 It's always nice to see it work out nicely like that :) That's the kind of universal logic I like to see in the programs I write, but then there's all these pesky edge cases that usually ruin my elegant ideas.
By symmetry, the 20m stick can form an equilateral triangle with two other tangents, and then we can use area of that triangle to solve for inner and outer radii which yields R = 2r
After that, it's just pi (R² - r²)
Good problem nevertheless
Its a fun one, cause x could be any size greater than 10, so we can't determine x or y, only the relationship between them, which suffices to find the area.
The weird thing is that the ratio of radii is irrelevant. If we knew that, we could imagine a vanishingly small inner circle and we'd get a ring area equal to the area of the larger circle. But is there a way to see that the ratio of radii is irrelevant? Very nice puzzle!
Well done Tiger 🐅....👍👍💯💯
Do we need to know if both circles are at the same center point?
Math queen I love you soooo much. How about a derivative or integral problem from real math, Calculus ? Waiting for the hard stuff.
Thank you for your kind feedback! I will upload different levels of math problems over time, but there is a derivative video coming soon. 😊
Coffee, 'Hello my lovelies' and a math problem are a great way to start my day!
Aaaaw, this sounds like a wonderful morning! I’m proud of you starting your day like this!
Assume a diameter of the outer circle to be any number larger than 20. Then determine the diameter of the inner circle by using the graphical equation of a circle. I assumed a big diameter of 30m and calculated the inter circle diameter to be 22.36…m and you get the same solution
A,= ¼πc² = ¼π20² = 100π cm² ( Solved √)
Fast and easy !!!
Without attempting any proof or calculation, I’d say that the if area the ring does not depend on the area of the two circles, then it must also be the area when “inner circle” has “area” zero. That is, when the “ring” is a circle with diameter 20. But now I have to show my work.
I can see now that others have reached the same conclusion (and I hope to avoid similar vitriolic discussions). As we’ve seen I made a correct guess, but it still requires a follow up with proof. And that’s fairly easy with a careful choice of a right triangle.
Let's calculate the area of the red ring ▶
Ared= πr₂² - πr₁²
By drawing the radius r₁ from the center O, to the middle of the stick, and the radius r₂ where the stick touches the circumference of the large circle, we obtain a right triangle 💡
The right triangle ΔAOB:
[AO]= r₁
[OB]= r₂
[BA]= 20/2
[BA]= 10
⇒
By applying the Pythagorean theorem, we get:
[AO]²+[BA]²= [OB]²
r₁² + 10²= r₂²
⇒
r₂² - r₁² = 100
Ared= πr₂² - πr₁²
Ared= π(r₂² - r₁²)
Ared= 100 π
Ared≈ 314,16 square units