Everyone is saying how this one is easy, just drag the chord down to the diameter - however, I think it's great that we're getting a proof that applies to every case and not just the simplest one.
I agree having other methods to solve these problems is great. However, it is also helpful to recognize when the problem does not give you details that "lock" in a particular design - meaning you can interpret the question in the way that is most beneficial and still get the correct answer without a lot of computation. Very common with competition problems. Both being able to solve every case as well as being able to recognize the trick are good skills to have!
I mean, dragging the chord down is the type of thing I would do cus I know the people who made the problem aren't crazy, but I would do it with tears in my eyes
@@gabitheancient7664 If your goal is to prove the result, then the methods presented in the video are good. If that's not your goal, if you simply need to supply an answer, noting that this must be true regardless of parameters and finding the simplest solution can be just as valid, and often faster.
I just had this problem in the Intermediate maths challenge. Now I am doing the Hamilton maths challenge. This question took me 20 seconds after watching your previous videos. I dragged the chord down to the diameter
@@faizanahsan7155 The problem is set up such that ANY chord=6 creates the same area. You don't have to prove it, you just have to realize that with so few constraints on the construction it has to be valid for all chord placements. If the chord is the diameter it simplifies the problem greatly.
If you split the shape in half along the line of reflection (the shared diameters of the circles), the resulting half of the shaded area is an arbelos. An arbelos has an interesting property that its area is equal to the area of a circle that has a diameter the length of the chord tangent to the inner semicircles and perpendicular to their shared diameter. This can be proven with the chord power theorem (a subset of Ptolemy's theorem). This means as long as the chord length is constant, the area of the shaded area will be the same regardless of the diameters of the circles. The chord also represents the geometric mean of the diameters of the two semicircles. With that, the area of the shaded area can be written 2*pi*(PQ/4)^2 or 2*pi*(3/2)^2 which gets you 9/2*pi en.wikipedia.org/wiki/Arbelos
I actually took the UKMT IMC 2023 recently in school and I was so stuck on this problem, even after getting a copy of the question paper afterwards and staring for multiple hours at my rough workings trying to find a solution. Thank you so much for explaining this, this really is one of the best communities on UA-cam for explaining things like this in an actual video instead of just saying "here's the mark scheme, have fun". I actually qualified for the Pink Kangaroo (a follow-on round to the IMC you get invited to if you do well enough) and because I'm the only person in my school to qualify, I'm under lots of pressure to do well. Now that I know the chord-chord theorem from this video, I'm a bit more confident than before that I can hopefully can get a good score
hey its max, i submitted this problem and was in imc 2023. my friend qualified for pink kangaroo too. I wish you the best of luck mate, but try not to stress. Have fun :)
I just wanted to let you know that because we’re not taught the chord-chord theorem, you can also use trigonometry to solve the question (you can try if you want), and that’s how I solved it. Unfortunately, I wasn’t allowed to enter because I’m in year 11 and my school said I should be focusing on my revision, but I wish you good luck in the pink kangaroo and with your GCSEs
for a lot of these that are missing info, you can just assume they are equal to each other, or to 0 and it usually works. in this case, if you assume a = b, then blue area is just get a circle of diameter 6 with 2 circles of diam 3 removed. so 3*3*Pi-1.5*1.5*Pi*2 = 9Pi/2
@@dj_laundry_list No, you specifically have to come up with a proof that it doesn't matter which value you choose. Just assuming things is never enough. A solution like this is incomplete.
Diámetro círculo exterior =2r+2R → Área azul =B → B/π =(R+r)²-R²-r² =2Rr →→ Potencia del punto medio de PQ respecto a la circunferencia exterior =3*3=2R*2r → 9=4Rr → 2Rr=9/2 → B=9π/2 =Área azul. Gracias y saludos cordiales
I did mine as a right angle but I named the distance from the center of the middle circle to the chord "d". And big radius "r". Then used area and Pythagorean formulas to get r^2-d^2 and cancelled them out.
Here's another (simplified) way: 1. Regardless of the diameters of each of the smaller circles, their total area within the larger circle will always be the same value, including when they both have the same diameter 2. Radius of large circle is r 3. Assume the two internal circles have same diameter, r/2 4. Shaded area A = (Area of large circle) - 2 x (Area of one smaller circle) 5. Shaded area A = pi*r*r - (pi/2)*r*r 6. Given PQ = 6, in this scenario PQ = 2r = 6 7. A = 9pi - 9(pi/2) = 9(pi)/2
I think I used this approach without even knowing it. I don't really know math that well and what is equal to what in a specific scenario, but just by looking at the image, I assumed the following: 1. a = r, 2. b = 2r, 3. R = 3r - this being the radius of the largest circle. Having that set up, I also used the first approach with the triangle and the result was the same. I just didn't have any proof of why b = 2a, it just seemed like that would be the case looking at the given image. But I know that is not good enough explanation for math lol
@@charleslivingston2256 it always puzzles me how naive people are in thinking only they worked out the obvious solution and need to let us know in the comments
Figured I'd follow this challenge and pause, wrote this up before continuing the video x,y,z denote the radii of the circles where x contains y and z. y is the smaller of the two other circles. centre of PQ is a distance of z-y from the centre of larger circle forms a right angle triangle which means: (z-y)²+3² = x² = (z+y)² therefore z² - 2zy + y² + 9 = z²+ y² + 2zy cancelling out to 4zy = 9 what is the shaded region? area = πx² - πy² - πz² for now we can set aside pi as a common factor, and then substitute y+z in again, so area/π = (y+z)² - y² - z² = y² + z² + 2zy - y² - z² they cancel out to leave us with area = π2zy meaning the area is π(9/2)
I solved it similar to your right triangle method, except I expressed the radius of the two smaller circles in terms of the radius 'r' of the big circle. When I then calculated the area of light blue region as the area of the large circle minus the area of the two interior circles, found all the r's cancelling, leaving me with the correct answer.
I got the answer by adjusting the radius of the outer circle so that I get and equilateral triangle PQR where i created R as the meeting point of the outer circle and the bigger inner circle.
I did it like this The Chord length is gives as 6 Suppose the chord runs through the centre, then its the diameter. That gives 2 similar circles above and below with half the diameter each, which is 6. So the sum of the areas is 2*3.14*3*3/4 = 14.13 square units (Never forget the units)
Your solution only shows that the answer is that when you make the chord go through the centre. You don't have any proof for it being the same in the general case.
@@wombat4191 there is this theory that if you draw a diameter through the centre of a chord, the ratios are same, you can use this. Also it's possible to visualise the circles on either sides add up to the same area when the chord is moved upwards and downward
@@d4django but you didn't use the chord theorem to prove anything. Just a theorem existing is not enough, you have to show how it applies to your case and why the way you calculated it applies to any case. As for the circles adding up to the same area, that's simply not true. The further the chord is from the center, the larger the circles' combined area. As the chord approaches the upper edge of the outer circle (meaning that the radius and area of the upper inner circle approach 0), the lower inner circle area approaches infinity.
Take the distance from the radius point to the tangent point, call it x. The empty area is the sum of the two empty circles. pi((r-x)/2)^2 + pi((r+x)/2)^2 = 0.5 pi (r^2 + x^2) The shaded area is the full circle minus the empty area, so pir^2 - 0.5pi (r^1 + x^2) = 0.5pi ( _r^2 - x^2_ ) Now we use intersecting chord theorem, (r+x)(r-x) = 3*3 -> _r^2 - x^2_ = 9 Substitute 9 into the equation for the shaded area to get 4.5pi
Great problem. I did first find the answer by assuming the chord was a diameter of the big circle. Of course, that doesn't prove the answer is invariant to the diameter of the big circle, so then I used the second method (chord theorem) to solve it.
There is a third way. In the problem there are no defined distance between the chord and the center of the large circle. When all distances are valid I can choose any, and the answer should be the same. So I can assume the chord is the diameter. Then the two small circles are equal size. The area we are seaking are large circle - 2 small circles. 3² *pi-2*1,5²*pi=4,5*pi
Just because the problem doesn't give some specified distance, you can't simply assume it doesn't matter and leave it at that. I't okay to assume, but you always need to end up with a proof for that assumption. Without such proof, your solution is incomplete and you have only shown that the area is 4,5*pi when the chord goes through the center.
@@wombat4191 hmm but since there's only one solution we don't have to prove other cases. The large circle can have any diameter, and for that give large circle the smaller circles can have any diameter.. The answer is always 4.5*pi. It is intuitive once you do these kind of problems
@@suhail_69 you still have to prove that there's only one solution. It's not an established fact at any point of the problem until you show it to be a fact. Intuitivity in turn means nothing in maths, you always need a proof. It's also intuitive to think that there are infinite prime numbers, but you still need something like Euclid's theorem to prove it.
@@wombat4191 It didn't say proof.It was a solution and an answer. If the area was depended of the distance, there are no direct answer to the question. Other than a formula or an interval. Then the task lack data. If something is not defined or as follow in the question you can chose the simplest one. If you want a proof for all circles, it should be in the question.
an elegant and general solution would be: let the vertical diameter be MN and O the cut of the chords. now consider the right triangles: MQO, QON and MNQ. now consider the semicircles of diameter each of its sides. We know that the Pythagorean theorem holds with semi circles (for simplicity let's call them by their diameters, for example the semicircle string MQ, we just call it MQ), so: MO+OQ=MQ NO+OQ=NQ MQ+NQ=MN adding member to member the three equations: MO+NO+2OQ=MN multiplying by 2 2(MO+NO)+4OQ=2MN => 4OQ=2MN-2(MO+NO) 9π/2=painted area
There is a very interesting method using Feynman's trick. Since the diameter of the outer circle is not given the answer is independent of it. So choose this diameter to be 6. Then the given chord is nothing but the diameter of the outer circle. Hence both the inner circles are of diameter =6/2=3. Hence the required area = pi(3×3 -1.5×1.5×2)=pi×4.5.
How do you prove that the diameter doesn't matter though? You just assumed that, but you have no proof for it whatsoever, which means that you have only really solved the problem in one specific edge case.
Simplest solution: It doesn't matter where the chord is. With two circles inside a larger circle in this fashion, the position of the chord does not affect the area of the two smaller circles. Meaning that you could assume the chord is the diameter of the larger circle and there are two equal smaller circles within. By sliding the chord up or down and resizing the two smaller circles, their area remains constant. With the cord (length 6) in line with the diameter, this gives is a larger circle with a radius of 3. The two smaller circle will both have a radius of 6/4. Larger circle area minus two smaller circles areas gives the answer: (3^2 * pi) - 2 (1.5^2 * pi)
Aka Method 3: Reductio Absurdum - reduce the problem to a silly level. In this case as stated the problem seems inpossible as the radius of the big circle could be anything. As a solution exists, then it is independent of the radius of the big circle (with the proviso that it must be at least 3 units to be able to contain a chord of length 6 units). The "silly" level is when PQ is a diameter and the result follows.
@@qcom1008 Presh's solution does just that. If you substitute "R" for the radius of the large circle it becomes (R^3*PI)/2 which works for any radius and any chord.
I did the algebra from the first method, but I figured out the "2ab" figure by using geometric means. Half of the chord is 3, which is the geometric mean of the 2 diameters. From there I found out what 2ab had to equal.
Chord-chord power theorem: 4ab = 3.3 = 9. Shaded area = (big circle - small circle - other small circle) = π(a+b)^2 - πaa - πbb = 2πab. We can forget Pythagoras! :))
I did it by a third method. Assume the problem has a solution -- we are told it does! Then we will get the same answer for any outer circle that has a chord of length 6. Pick an outer circle with a diameter of 6. It's radius is three and the radius of each smaller circle is 3/2. Use the area formula and subtract. This problem reminds me of the following curious problem. Take a sphere and drill out a cylindrical hole where the center line of the cylinder goes through the center of the sphere. The length of the hole measured along the inner edge of the hole is 6. What is the remaining volume (the volume of the sphere after removing the hole)? Since we are not told the radius of the sphere nor the radius of the cylinder, it doesn't sound like there is enough information to solve the problem. To show that there really is a solution, you have to introduce variables for the two radii. Then use the formuli for the volume of a sphere, cylinder, and cylindrical cap. Strangely, the radii cancel out and you are left with a specific numerical answer. But there is quick way to solve it if you assume that the question really does have an answer. If it has an answer, you will get the same answer for any spherical radius and any cylindrical radius as long as the hole has length 6. If you pick the most convenient sizes, the answer falls out almost trivially.
Lool at the dots, they are equal in length apart. There are 6 equal lengths, smaller circle covers 2 bigger 1 covers 4. We know the line goes through the 2nd dot out of 6, meaning its 1/3rd length along. Simply what sin(x) gives 1/3? Once we find that we can take the cosine of that x and divide half the line by it. This gives us the radius of the entire circle. Once we have this we can easily solve the area of smaller and bigger circle. V easy.
As an Adidas slogan once said "Impossible is Nothing"! The shaded area is 9π/2 Assume the radius of big circle is R and radiuses of insscribed circles are X and Y. The area shaded in blue is S S = πR² - π(X² + Y²) (1) (6/2)² = 4XY (2) according to crossing chordes theoreme Lets multiply left and right part of (2) on -π/2 and add it to (1) We get S - 9π/2 = πR² - π(X² + 2XY + Y²) Since X + Y = R we get S - 9π/2 = 0 so S = 9π/2
suppose that small circles are equal and R(big) = 6/2 = 3 r(small) = 3/2 we have 2 small circles in big circle, so shaded area equals to S=R-2r S = pi*3^2 - 2*pi*(3/2)^2 = 9/2 = 4.5
There was a third way too. That is by considering that the chord is actually the diameter of the bigger circle. This is because there was no restriction in the question for this so considering that the radius of the smaller circles is 1.5 and the bigger circle is 3. now doing simple Maths we get bigger circle - 2 smaller circle area = 3^2*pi-2*1.5^2*pi=9pi-4.5pi=4.5pi By this was too the answer comes same.
Another way to solve is as follows: Notice if small circle grow and big circle get smaller, if cord PQ is unchanged, at some point PQ will become a diameter. This is exactly when two circles are the same size. then you do pi * (PQ/2)^2 - 2 * pi * (PQ/4)^2 = pi * 3 * 3 - 2 * pi * 1.5 * 1.5 = pi (9 - 4.5) = pi * 4.50 = 9 pi / 2 not so scientific, but still works since it does not violate the problem conditions.
But you don't have any proof that your solution applies to all cases and not just that specific one you chose. You would probably only get like 1 or 2 out of 6 points with that answer in a maths test.
@@nmmm2000 but you always have to show proof for your assumptions at the end. The raw solution is very rarely sufficient in math problems like these, you have to show that the solution works with the given parameters, not just one specific instance of the given parameters. In the maths test this was taken from, you would only get a fraction of full score with answer like this.
Nice! Job well done. The problem strongly tells how important geometry (dimensions of lines, polygons and other shapes), trigonometry (triangle theorems) and algebra (polynomial theorems and algebraic manipulations) are as significant branches of mathematics. Only mathematics have clearer theories and unchanging principles over decades while other sciences remain obscure as days passes by.
Can't remember where I saw it but surely there's already a formula for this... For the required area A... [ A = Pi times (Chord Length 'PQ' squared) divided by 8 ]... as far as i am aware this holds true for any Chord Length even zero...
PQ is a Chord Deonote the vertical chord RS which intersects PQ at T, the midpoint of PQ The lengths PT x QT = RT x ST PQ is 6, so PT x QT = 3 x 3 = 9 which means that RT x ST = 9 Let "n" be the diameter of the upper circle, n > 0, which gives the lower circle a diameter on 9/n and the large circle a diameter of (n+9/n) Upper Areas = pi(n/2)^2 Lower Area = pi(9/2n)^2 Larger Circle Area = pi(n/2+9/2n)^2 Let a = n/2 and b = 9/2n Blue Area = pi(a+b)^2 - a^2pi - b^2pi = pi(a^2+2ab+b^2 -a^2-b^2) = 2abpi =2 x n/2 x 9/2n x pi = 9pi/2 for all values where n > 0.
Let r be the top circle radius, R be the bottom circle radius, and 𝑅 be the full circle radius. By Intersecting Chords Theorem, for any two chords that intersect in a circle, the products of the lengths of the segments on opposite sides of the intersection will be the same. Let D be the intersection point and AB be a diameter of the full circle bisecting both inner circles from top to bottom. AD would therefore be 2r, DB would be 2R, and AB would be 2𝑅. PD(DQ) = AD(BD) 3(3) = 2r(2R) 9 = 4rR r = 9/(4R) 2𝑅 = 2r + 2R 𝑅 = r + R Area = π𝑅² - πr² - πR² Area = π(r+R)² - πr² - πR² Area = π(r²+2rR+R²) - πr² - πR² Area = πr² + 2πrR + πR² - πr² - πR² Area = 2πrR = 2π(9/(4R))R Area = 18π/4 = 9π/2
That's a really nice problem. I'm not in touch with geometry anymore and literally took me 15mins so it was a really good brain opener. Some people used Ptolemy's theorem but we can also solve it just by careful observation. Notice that the quadrilateral formed by joining P, Q, the centre of smallest circle and the centre of largest circle is a rhombus. Let O be the centre of smallest circle and O' be the centre of largest circle. So OM=O'M => O'M=2OM => Radius of largest circle = 3(radius of smallest circle). Since centres are concurrent then => Radius of smallest circle + Radius of bigger circle = Radius of largest circle. Simple Pythagorean theorem and done.
So, this "drag down the chord to make it the diameter" hack is not the right way to think if you're serious about math. It might get you the answer in a multiple choice question if that question has all numbers as choices. However, if that question is an Olympiad question where you need to derive the answer, this hack will get you a 0. The reason is that you're assuming without knowing whether this question has an answer. In other words, you're assuming without proving that the answer is invariant to the position of the chord. "If a question has been asked, it must have a determinate answer" is not a logically valid proof. It might be a safe practical assumption depending on the context, but it's not a mathematically rigorous approach. The key mathematical insight here is not the area but the proof that this area is invariant to the position of the chord.
Since a/b is not specified, it's simple to assume a=b, which makes the chord a diameter of the large circle, so a=b=c/4 and r=c/2=3. Also you have the total area of the small circles is 2(pi)a^2, and of course the area of the large circle is (pi)r^2. So then you have (pi)(r^2-2a^2) = (pi)((c/2)^2 - 2(c/4)^2) = (pi)(c^2/8) = (pi)(36/8) = (pi)(9/2).
That way you only end up with an answer to a specific situation where the chord is the diameter, with nothing to show that the same solution works for any diameter.
@@wombat4191 But the problem statement implies that the answer is independent of the actual sizes of the two circles, so we're free to assume any ratio we want, and we might as well assume one where a/b approaches/is 1. Finding special cases to solve in cases of problems like this is often a quick way to get the answer. If you wanted to PROVE that your answer is independent of the ratio of a/b then of course this approach falls short. But that wasn't the task here.
@@jagmarz Problem statement implying something means nothing. What if it was a trick question and the correct answer was to prove that it's unknowable? That of course is not the case, but what I mean is that a problem being supposedly solvable isn't a proof for anything. At best it tells you that a proof exists and you have to find it. As for "that wasn't the task here". yeah it wasn't explicitly. However, when you take a specific case of a problem and solve that, you of course have to prove that your specific solution applies to the general case as well. The solutions Presh showed did not specify the problem any further, so his answers work for all cases by default.
LOL this is a question from the British Intermediate Maths Challenge (IMC), and I took part in it. Btw I got 128 out of 135. The only question I got wrong is Q20, which is a tricky but interesting question. I recommend Presh to make a Shorts video about that question. Q25 is easy
Method 1 fails for all circles as b is assumed to be _greater_ than a. What about the chord where PQ is the diameter of the circle and a is thus _equal to_ b?
Well, it's trivial to prove that b ≥ a covers all possible cases through symmetry, but Presh could have at least mentioned it. And the method works perfectly well for a = b, because pythagorean theorem actually also works when the "triangle" is just a line.
@@wombat4191 I was just pointing out (badly) that method 1 was incomplete as it omitted the possibility that b=a. In this case it made no difference (as the result also held), but in others it can make a difference. There is a classic example of proving that 1=2 that has one step dividing by a-b which is fine _unless_ a=b, at which point you are dividing by 0; needless to say, you end up showing that a=b by dividing a-b which implies you must have divided by 0, thus making the result wrong!
Thank you for these problems I find it super interesting. As a note of feedback, I encourage you to refrain from using contextual "this" because its not always very clear what "this" is. i.e. instead of "this distance" you might use "the distance in purple"
First get the diameters of the circles by using Euclid's Height Theorem: h² = pq (|PQ|/2)² = d ⋅ D 3² = 5d d = 9/5 D = 18/5 A = π/25 ⋅ (27² - 18² - 9²) = 324π/25 ≈ 40,72
I discovered your channel this year and I really like the way you solve the exercises with simplicity. I would like you to help me solve an exercise from the Olympiads of Algeria (high school, for 17 years old). m is A natural number. The number m + 7 is a perfect square and m-34 is a perfect square. Find the value of m. THANK YOU FOR ALL YOUR VIDEOS!!
I mean this is the easiest way. The difference between the two squares is 41 (34+7) and by the logic that (x+1)^2=x^2+2x+1, and we can remove the x^2 as we want the difference, we know that x is the lower number where 2x+1=41, and that leaves x as 20, so we can substitute in to m-34=20 -> m=54. Sorry for the clunky explanation
@@necros4031 m-34 = x^2, not m-34=x. So with x=20, x^2 is 400, and m=434, from which we can also find (x+1)^2 = 441 = 434+7, which meets the requirement. Keep in mind you assumed the two square numbers are sequential, so there may have been other solutions, but there are none in this case.
I made the assumption that it didn't matter how big the circles were so I shrunk it down until the diameter of the big circle was 6. That made the other two circles the same size. I got the answer in a few seconds after that.
You made an assumption, but did you end up with a proof? Your solution is not complete without a proof, now you only solved the problem for one specific case.
Because the problem is claiming to have an answer, we can surmise that the area must be the same no matter how draw the circles, provided only that PQ has a length of 6. As such, it would be *very convenient* if PQ happened to go through the center of the large circle. This means the large circle could have a radius of 3, and the two small circles would be half the width, with a radius of 1.5. So, plugging in some circle area equations, we get area = pi * 3^2 - pi * 1.5^2 - pi*1.5^2 = 4.5 * pi. But why should this area be invariant at all? That's the real question.
Hi Presh ! Please analize this method without Pitagoras or cords theorems, but only using some logic : the problem not define segments a and b, then , we can assume de problem is "resolvable" in any case, and furthermore, that the final value of this solutions are the same. Well, we can then resolve for a=b=1/2(p+q). From there , it is simple aplication o area of big circle (R=3) minus 2 times area of minor one (r=1,5). Regards !
Assume the writer has formulated the problem well. Then, since the ratio of the smaller circles' area is not specified, assume any ratio will give the same answer for the blue area. Hence we can assume that the circles have equal area. Then PQ is the diameter, hence the big circle has area 9*pi, and each of the two circles has are (9/4)*pi, then (9 - (9/4) - (9/4))*pi = (9/2)*pi gives the area.
@@wombat4191 Assumptions needn't be proved. The fact that our answer relies on the assumption that the writer has formulated the problem well, means that either the problem is well-formulated and our answer is correct, or the problem has several possible answers, in which case our answer is one of many such. The limitation of using that assumption is that the solution won't tell us whether the problem is well-formulated. Note: I have used the word "assume" 3 times for comedic effect, but really we are only assuming one thing.
@@ventsiR where did you ever hear that "assumptions needn't be proved" in mathemathics? That's what maths is all about, proving stuff. Otherwise you could just assume anything and treat it as a true fact. Being able to assume that a problem is solvable is not a proof, it only gives you the knowledge that a proof exists. It is still your burden as a solver to find that proof. By specifying some previously undetermined values you essentially solve a different problem, a specific case of the original one. After that you of course must prove that your answer applies generally and not just in that specific case.
What is the software he use to make this transition of photos, like multiplying ( inserting ) pi in both sides of the equation with a smooth animation ?
Easy. I pushed the chord all the way up so that the upper circle had no area and the lower circle was the same as as the big circle. Blue area equals zero.
Another method: The question implies that the result does not depend on the radius r or the outer circle, only on the cord length. By setting r=6, the cord will fall in the middle, forming a diameter of the outer circle, and the two small circles will both have radius 3/2. So the radius of the shaded region becomes: outer - 2*inner = π3^2 - 2*π(3/2)^2 = 9π/2
a=ø of small circle b=ø of medium circle c=(a+b)=ø of large circle PQ =**6** ab= (**6**/2)²= 9 Blue = πc²/4 -πa²/4 -πb²/4 = (π/4)(c² -a² -b²) = (π/4)[(a+b)² -a² -b²] = (π/4)[(a² +b² +2ab -a² -b²) = (π/4)(2ab) =abπ/2 ab=9 Blue = 9π/2
By the intersecting chords theorem we know that r×R=(6/2)×(6/2) And we know that the area A to be proportional to r×R (and we know the coefficient), I feel like the solution has been a bit overcomplicated
@@wombat4191 oh it was specified.../facepalm no idea how i missed that. I could see it now in the thumbnail even. Ok, universe makes sense again. Thanks ^^
In order for that problem to have an answer that isn't expressed as a ratio in terms of a and b, then the shaded area MUST be a constant value regardless of the position on PQ, INCLUDING the case where PQ is the diameter. So, set PQ as the diameter, and the correct answer is simple math. In fact, you missed a chance to show that for ALL cases with similar inscribed circles, and a chord of length N, the shaded area is N^2/8. THAT would have been more valuable.
Question. I am trying get myself ready for college but I graduate from High School 20+ Years ago and I barely remember math. Is there any website where I can learn Math from the scratch in an organize way? I appreciate your help.
Co-geom. configs.: P → (-3,0); M → (0,0); Q → (3,0); r₁ → radius of upper inner circle; r₂ → radius of lower inner circle; C → the enclosing circle Findings: ~ S is at (0,2r₁) & T is at (0,2r₂) ~ radius of C: r₁+r₂ ~ centre of C: (0,2r₁-(r₁+r₂)) ⇔ (0,r₁-r₂) ~ eqn. of C: x² + [y-(r₁-r₂)]² = (r₁+r₂)² ⇔ x² + y² - 2(r₁-r₂)y = 4r₁r₂ ...(*) ~ as Q is on C, by (*), 9 = 4r₁r₂ ⇔ r₁r₂ = 9/4 ~ area req. = (r₁+r₂)²π-r₁²π-r₂²π = 2r₁r₂π = 9π/2
My solution is easier and faster: We notice from the question that the blue area is independent of the exact sizes of the two small circles. Without Loss Of Generality, we let the two smaller circles be the same size. Then the chord of length 6 is a diameter of the larger circle, radius 3. Each smaller circle has radius 1.5. Subtraction of the 2 small circle areas from the large circle area gives the solution of "4.5pi". And That's The Answer.
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@@sylvenara Bro i was born in kuwait. The law here is i can stay in kuwait till 60 years of age.... there are about 15 or more indian schools in kuwait and each school has the CBSE curriculum... i am studying in AMSB indian school ( dawoodi bohra school ) so thats how i an here.... reaching till class 10 its so satisfying hopefully passing grade 10 is also a great year 😀
See your math skills as a tool box. You now have alot of tools. Use them on problems where it is not obvious which tools to use. When you know your tool box and see what tool to chose, then you will succeed. When you have a large problem to solve, try to simplify. As in this case make the chord the diameter. Or if large numbers, try a problem with smaller or fewer numbers. Then use that method for the larger problem. Good luck.
Everyone is saying how this one is easy, just drag the chord down to the diameter - however, I think it's great that we're getting a proof that applies to every case and not just the simplest one.
I agree having other methods to solve these problems is great. However, it is also helpful to recognize when the problem does not give you details that "lock" in a particular design - meaning you can interpret the question in the way that is most beneficial and still get the correct answer without a lot of computation. Very common with competition problems. Both being able to solve every case as well as being able to recognize the trick are good skills to have!
But it does apply to every case though! As the position of the chord doesn't matter, you can just solve this in a simpler way by doing the above
I mean, dragging the chord down is the type of thing I would do cus I know the people who made the problem aren't crazy, but I would do it with tears in my eyes
@@gabitheancient7664 If your goal is to prove the result, then the methods presented in the video are good. If that's not your goal, if you simply need to supply an answer, noting that this must be true regardless of parameters and finding the simplest solution can be just as valid, and often faster.
@@Qermaq I know but, my math heart...
I just had this problem in the Intermediate maths challenge. Now I am doing the Hamilton maths challenge. This question took me 20 seconds after watching your previous videos. I dragged the chord down to the diameter
Can You Please Explain How You Dragged The Chord Down To The Diameter?
@@faizanahsan7155 Just pretend PQ is the diameter then the two circles have radius 1.5
@@faizanahsan7155 The problem is set up such that ANY chord=6 creates the same area. You don't have to prove it, you just have to realize that with so few constraints on the construction it has to be valid for all chord placements. If the chord is the diameter it simplifies the problem greatly.
Thanks
just connect SQ and TQ. You got: [(2a)^2+3^2] + [(2b)^2+3^2] = (2R)^2 rearrange: 4(R^2-a^2-b^2) = 18 then time pi and divide 4 you got your answer.
If you split the shape in half along the line of reflection (the shared diameters of the circles), the resulting half of the shaded area is an arbelos. An arbelos has an interesting property that its area is equal to the area of a circle that has a diameter the length of the chord tangent to the inner semicircles and perpendicular to their shared diameter. This can be proven with the chord power theorem (a subset of Ptolemy's theorem). This means as long as the chord length is constant, the area of the shaded area will be the same regardless of the diameters of the circles. The chord also represents the geometric mean of the diameters of the two semicircles.
With that, the area of the shaded area can be written 2*pi*(PQ/4)^2 or 2*pi*(3/2)^2 which gets you 9/2*pi
en.wikipedia.org/wiki/Arbelos
For Amercians, the IMC is a test designed to be taken by 13-15 year olds. The questons get harder towards the and this is the last question.
I actually took the UKMT IMC 2023 recently in school and I was so stuck on this problem, even after getting a copy of the question paper afterwards and staring for multiple hours at my rough workings trying to find a solution. Thank you so much for explaining this, this really is one of the best communities on UA-cam for explaining things like this in an actual video instead of just saying "here's the mark scheme, have fun". I actually qualified for the Pink Kangaroo (a follow-on round to the IMC you get invited to if you do well enough) and because I'm the only person in my school to qualify, I'm under lots of pressure to do well. Now that I know the chord-chord theorem from this video, I'm a bit more confident than before that I can hopefully can get a good score
hey its max, i submitted this problem and was in imc 2023. my friend qualified for pink kangaroo too. I wish you the best of luck mate, but try not to stress. Have fun :)
@@c6u thanks for the extra bit of motivation and I wish your friend good luck as well
Same bro it was confusing, I got into grey kangaroo (I’m in yr9)
I just wanted to let you know that because we’re not taught the chord-chord theorem, you can also use trigonometry to solve the question (you can try if you want), and that’s how I solved it. Unfortunately, I wasn’t allowed to enter because I’m in year 11 and my school said I should be focusing on my revision, but I wish you good luck in the pink kangaroo and with your GCSEs
for a lot of these that are missing info, you can just assume they are equal to each other, or to 0 and it usually works. in this case, if you assume a = b, then blue area is just get a circle of diameter 6 with 2 circles of diam 3 removed. so 3*3*Pi-1.5*1.5*Pi*2 = 9Pi/2
yea thats what i did. this is how u should solve these type of problems in competitions at least
Exactly, it's not like we need to write a proof on an exam question. For arbitrary values, we can just pick the most convenient one
@@dj_laundry_list No, you specifically have to come up with a proof that it doesn't matter which value you choose. Just assuming things is never enough. A solution like this is incomplete.
Diámetro círculo exterior =2r+2R → Área azul =B → B/π =(R+r)²-R²-r² =2Rr →→ Potencia del punto medio de PQ respecto a la circunferencia exterior =3*3=2R*2r → 9=4Rr → 2Rr=9/2 → B=9π/2 =Área azul.
Gracias y saludos cordiales
I did mine as a right angle but I named the distance from the center of the middle circle to the chord "d". And big radius "r". Then used area and Pythagorean formulas to get r^2-d^2 and cancelled them out.
Here's another (simplified) way:
1. Regardless of the diameters of each of the smaller circles, their total area within the larger circle will always be the same value, including when they both have the same diameter
2. Radius of large circle is r
3. Assume the two internal circles have same diameter, r/2
4. Shaded area A = (Area of large circle) - 2 x (Area of one smaller circle)
5. Shaded area A = pi*r*r - (pi/2)*r*r
6. Given PQ = 6, in this scenario PQ = 2r = 6
7. A = 9pi - 9(pi/2) = 9(pi)/2
I think I used this approach without even knowing it. I don't really know math that well and what is equal to what in a specific scenario, but just by looking at the image, I assumed the following:
1. a = r,
2. b = 2r,
3. R = 3r - this being the radius of the largest circle.
Having that set up, I also used the first approach with the triangle and the result was the same. I just didn't have any proof of why b = 2a, it just seemed like that would be the case looking at the given image. But I know that is not good enough explanation for math lol
For getting 4ab=9, you could have also used the intersecting chords theorem which directly gives (2a)*(2b)=3*3
That was the second method he showed.
@@charleslivingston2256 it always puzzles me how naive people are in thinking only they worked out the obvious solution and need to let us know in the comments
@@jacoboribilik3253 why tf do u care
Figured I'd follow this challenge and pause, wrote this up before continuing the video
x,y,z denote the radii of the circles where x contains y and z. y is the smaller of the two other circles.
centre of PQ is a distance of z-y from the centre of larger circle
forms a right angle triangle which means: (z-y)²+3² = x² = (z+y)²
therefore z² - 2zy + y² + 9 = z²+ y² + 2zy
cancelling out to 4zy = 9
what is the shaded region? area = πx² - πy² - πz²
for now we can set aside pi as a common factor, and then substitute y+z in again, so
area/π = (y+z)² - y² - z² = y² + z² + 2zy - y² - z²
they cancel out to leave us with area = π2zy
meaning the area is π(9/2)
I solved it similar to your right triangle method, except I expressed the radius of the two smaller circles in terms of the radius 'r' of the big circle. When I then calculated the area of light blue region as the area of the large circle minus the area of the two interior circles, found all the r's cancelling, leaving me with the correct answer.
I got the answer by adjusting the radius of the outer circle so that I get and equilateral triangle PQR where i created R as the meeting point of the outer circle and the bigger inner circle.
I did it like this
The Chord length is gives as 6
Suppose the chord runs through the centre, then its the diameter. That gives 2 similar circles above and below with half the diameter each, which is 6.
So the sum of the areas is 2*3.14*3*3/4 = 14.13 square units (Never forget the units)
Your solution only shows that the answer is that when you make the chord go through the centre. You don't have any proof for it being the same in the general case.
@@wombat4191 there is this theory that if you draw a diameter through the centre of a chord, the ratios are same, you can use this. Also it's possible to visualise the circles on either sides add up to the same area when the chord is moved upwards and downward
@@d4django but you didn't use the chord theorem to prove anything. Just a theorem existing is not enough, you have to show how it applies to your case and why the way you calculated it applies to any case.
As for the circles adding up to the same area, that's simply not true. The further the chord is from the center, the larger the circles' combined area. As the chord approaches the upper edge of the outer circle (meaning that the radius and area of the upper inner circle approach 0), the lower inner circle area approaches infinity.
@@wombat4191 yes. You are right. I did not cover all scenarios.
I would only approach it this way if it's only a multiple choice question
Take the distance from the radius point to the tangent point, call it x.
The empty area is the sum of the two empty circles. pi((r-x)/2)^2 + pi((r+x)/2)^2 = 0.5 pi (r^2 + x^2)
The shaded area is the full circle minus the empty area, so pir^2 - 0.5pi (r^1 + x^2) = 0.5pi ( _r^2 - x^2_ )
Now we use intersecting chord theorem, (r+x)(r-x) = 3*3 -> _r^2 - x^2_ = 9
Substitute 9 into the equation for the shaded area to get 4.5pi
wonderful question - wonderful explanation!! thank you Presh.
Great problem. I did first find the answer by assuming the chord was a diameter of the big circle. Of course, that doesn't prove the answer is invariant to the diameter of the big circle, so then I used the second method (chord theorem) to solve it.
There is a third way.
In the problem there are no defined distance between the chord and the center of the large circle. When all distances are valid I can choose any, and the answer should be the same. So I can assume the chord is the diameter. Then the two small circles are equal size. The area we are seaking are large circle - 2 small circles. 3² *pi-2*1,5²*pi=4,5*pi
Great idea 💡
Just because the problem doesn't give some specified distance, you can't simply assume it doesn't matter and leave it at that. I't okay to assume, but you always need to end up with a proof for that assumption. Without such proof, your solution is incomplete and you have only shown that the area is 4,5*pi when the chord goes through the center.
@@wombat4191 hmm but since there's only one solution we don't have to prove other cases. The large circle can have any diameter, and for that give large circle the smaller circles can have any diameter.. The answer is always 4.5*pi. It is intuitive once you do these kind of problems
@@suhail_69 you still have to prove that there's only one solution. It's not an established fact at any point of the problem until you show it to be a fact. Intuitivity in turn means nothing in maths, you always need a proof. It's also intuitive to think that there are infinite prime numbers, but you still need something like Euclid's theorem to prove it.
@@wombat4191 It didn't say proof.It was a solution and an answer. If the area was depended of the distance, there are no direct answer to the question. Other than a formula or an interval. Then the task lack data. If something is not defined or as follow in the question you can chose the simplest one. If you want a proof for all circles, it should be in the question.
an elegant and general solution would be:
let the vertical diameter be MN and O the cut of the chords.
now consider the right triangles:
MQO, QON and MNQ.
now consider the semicircles of diameter each of its sides.
We know that the Pythagorean theorem holds with semi circles (for simplicity let's call them by their diameters, for example the semicircle string MQ, we just call it MQ), so:
MO+OQ=MQ
NO+OQ=NQ
MQ+NQ=MN
adding member to member the three equations:
MO+NO+2OQ=MN
multiplying by 2
2(MO+NO)+4OQ=2MN =>
4OQ=2MN-2(MO+NO)
9π/2=painted area
I used the 2nd method. Was alot easier than the question seemed at 1st.
4th
I used a bit of logic and used the chord to be a diameter... and solved for the area so i got the answer in an easier manner
But you got no proof that your answer is valid in all cases.
That's the most intelligent solution.
I did Pythagorean theorem using triangles MSP and PMT. Gives 9/2= r2-a2-b2, multiply both sides by pi and voilà
There is a very interesting method using Feynman's trick. Since the diameter of the outer circle is not given the answer is independent of it. So choose this diameter to be 6. Then the given chord is nothing but the diameter of the outer circle. Hence both the inner circles are of diameter =6/2=3. Hence the required area = pi(3×3 -1.5×1.5×2)=pi×4.5.
How do you prove that the diameter doesn't matter though? You just assumed that, but you have no proof for it whatsoever, which means that you have only really solved the problem in one specific edge case.
Simplest solution: It doesn't matter where the chord is. With two circles inside a larger circle in this fashion, the position of the chord does not affect the area of the two smaller circles. Meaning that you could assume the chord is the diameter of the larger circle and there are two equal smaller circles within. By sliding the chord up or down and resizing the two smaller circles, their area remains constant. With the cord (length 6) in line with the diameter, this gives is a larger circle with a radius of 3. The two smaller circle will both have a radius of 6/4. Larger circle area minus two smaller circles areas gives the answer: (3^2 * pi) - 2 (1.5^2 * pi)
Exactly how I did it
it seems at first intuitive, but how do you prove that moving the chord doesnt affect the shaded area
Aka Method 3: Reductio Absurdum - reduce the problem to a silly level.
In this case as stated the problem seems inpossible as the radius of the big circle could be anything. As a solution exists, then it is independent of the radius of the big circle (with the proviso that it must be at least 3 units to be able to contain a chord of length 6 units). The "silly" level is when PQ is a diameter and the result follows.
what if the smaller circle gets smaller and smaller and the bigger one gets bigger.
at some point, the area in blue will be reduced to 0
@@qcom1008 Presh's solution does just that.
If you substitute "R" for the radius of the large circle it becomes (R^3*PI)/2 which works for any radius and any chord.
I did the algebra from the first method, but I figured out the "2ab" figure by using geometric means. Half of the chord is 3, which is the geometric mean of the 2 diameters. From there I found out what 2ab had to equal.
I saw this question on UA-cam before IMC 2023
Leaky😁
@@MissFiren I don't think so.
search "Find the area. 【FunMath, 9th grade - 53】"
@@mein7422 I know. That was a joke.😁
Thank you very much. For sure, not a leak, I swear.
Chord-chord power theorem: 4ab = 3.3 = 9. Shaded area = (big circle - small circle - other small circle) = π(a+b)^2 - πaa - πbb = 2πab.
We can forget Pythagoras! :))
I did it by a third method. Assume the problem has a solution -- we are told it does! Then we will get the same answer for any outer circle that has a chord of length 6. Pick an outer circle with a diameter of 6. It's radius is three and the radius of each smaller circle is 3/2. Use the area formula and subtract.
This problem reminds me of the following curious problem. Take a sphere and drill out a cylindrical hole where the center line of the cylinder goes through the center of the sphere. The length of the hole measured along the inner edge of the hole is 6. What is the remaining volume (the volume of the sphere after removing the hole)? Since we are not told the radius of the sphere nor the radius of the cylinder, it doesn't sound like there is enough information to solve the problem. To show that there really is a solution, you have to introduce variables for the two radii. Then use the formuli for the volume of a sphere, cylinder, and cylindrical cap. Strangely, the radii cancel out and you are left with a specific numerical answer. But there is quick way to solve it if you assume that the question really does have an answer. If it has an answer, you will get the same answer for any spherical radius and any cylindrical radius as long as the hole has length 6. If you pick the most convenient sizes, the answer falls out almost trivially.
Assumption isn't enough, you need a proof as well.
Lool at the dots, they are equal in length apart. There are 6 equal lengths, smaller circle covers 2 bigger 1 covers 4. We know the line goes through the 2nd dot out of 6, meaning its 1/3rd length along. Simply what sin(x) gives 1/3? Once we find that we can take the cosine of that x and divide half the line by it. This gives us the radius of the entire circle. Once we have this we can easily solve the area of smaller and bigger circle. V easy.
Didn't know Intersecting chords theorem actually, once gotten to know that, I took out the answer vigorously
Thanku, learnt a new thing
super easy. barely an inconvenience. 3 squared is equal to the multiplication of the diamenters. (rR)=9/4. and then π(r+R)^2-πr^2-πR^2=π9/2
As an Adidas slogan once said "Impossible is Nothing"!
The shaded area is 9π/2
Assume the radius of big circle is R and radiuses of insscribed circles are X and Y. The area shaded in blue is S
S = πR² - π(X² + Y²) (1)
(6/2)² = 4XY (2) according to crossing chordes theoreme
Lets multiply left and right part of (2) on -π/2 and add it to (1)
We get
S - 9π/2 = πR² - π(X² + 2XY + Y²)
Since X + Y = R we get
S - 9π/2 = 0 so
S = 9π/2
suppose that small circles are equal and
R(big) = 6/2 = 3
r(small) = 3/2
we have 2 small circles in big circle, so shaded area equals to S=R-2r
S = pi*3^2 - 2*pi*(3/2)^2 = 9/2 = 4.5
There was a third way too. That is by considering that the chord is actually the diameter of the bigger circle. This is because there was no restriction in the question for this so considering that the radius of the smaller circles is 1.5 and the bigger circle is 3. now doing simple Maths we get bigger circle - 2 smaller circle area = 3^2*pi-2*1.5^2*pi=9pi-4.5pi=4.5pi
By this was too the answer comes same.
Another way to solve is as follows:
Notice if small circle grow and big circle get smaller, if cord PQ is unchanged,
at some point PQ will become a diameter.
This is exactly when two circles are the same size.
then you do
pi * (PQ/2)^2 - 2 * pi * (PQ/4)^2 = pi * 3 * 3 - 2 * pi * 1.5 * 1.5 = pi (9 - 4.5) = pi * 4.50 = 9 pi / 2
not so scientific, but still works since it does not violate the problem conditions.
But you don't have any proof that your solution applies to all cases and not just that specific one you chose. You would probably only get like 1 or 2 out of 6 points with that answer in a maths test.
@@wombat4191 This is true. but assuming the solution does not contradict the problem conditions, you can proceed with the solution.
@@nmmm2000 but you always have to show proof for your assumptions at the end. The raw solution is very rarely sufficient in math problems like these, you have to show that the solution works with the given parameters, not just one specific instance of the given parameters. In the maths test this was taken from, you would only get a fraction of full score with answer like this.
You are using the fact that all mathematics is a transformation. Congratulations! This retired math instructor would give you full credit.
Since the radii of the two circles are not specified, we can simply have the chord be equal to the diameter.
But that way you get no proof that the solution is the same in any situation.
That's the most intelligent solution.
@@stephenlesliebrown5959 no, it's an incomplete solution.
thanks for the chord-chord theorem.
You can also use geometric mean theorem
I get the answer very quickly assuming that a=b=3/2, but it's not a proove. our two mthods are great ! Thank you
Nice! Job well done. The problem strongly tells how important geometry (dimensions of lines, polygons and other shapes), trigonometry (triangle theorems) and algebra (polynomial theorems and algebraic manipulations) are as significant branches of mathematics. Only mathematics have clearer theories and unchanging principles over decades while other sciences remain obscure as days passes by.
Can't remember where I saw it but surely there's already a formula for this... For the required area A... [ A = Pi times (Chord Length 'PQ' squared) divided by 8 ]... as far as i am aware this holds true for any Chord Length even zero...
Let c denote the length of the chord, then the shaded area is A=πc²/8. Grind c=6.
I did this question in the UKMT :) glad to see I got it right
same
i sat this question and because of you it was really easy Thanks presh
PQ is a Chord
Deonote the vertical chord RS which intersects PQ at T, the midpoint of PQ
The lengths PT x QT = RT x ST
PQ is 6, so PT x QT = 3 x 3 = 9
which means that RT x ST = 9
Let "n" be the diameter of the upper circle, n > 0, which gives the lower circle a diameter on 9/n and the large circle a diameter of (n+9/n)
Upper Areas = pi(n/2)^2
Lower Area = pi(9/2n)^2
Larger Circle Area = pi(n/2+9/2n)^2
Let a = n/2 and b = 9/2n
Blue Area = pi(a+b)^2 - a^2pi - b^2pi = pi(a^2+2ab+b^2 -a^2-b^2)
= 2abpi
=2 x n/2 x 9/2n x pi
= 9pi/2 for all values where n > 0.
Let r be the top circle radius, R be the bottom circle radius, and 𝑅 be the full circle radius. By Intersecting Chords Theorem, for any two chords that intersect in a circle, the products of the lengths of the segments on opposite sides of the intersection will be the same. Let D be the intersection point and AB be a diameter of the full circle bisecting both inner circles from top to bottom. AD would therefore be 2r, DB would be 2R, and AB would be 2𝑅.
PD(DQ) = AD(BD)
3(3) = 2r(2R)
9 = 4rR
r = 9/(4R)
2𝑅 = 2r + 2R
𝑅 = r + R
Area = π𝑅² - πr² - πR²
Area = π(r+R)² - πr² - πR²
Area = π(r²+2rR+R²) - πr² - πR²
Area = πr² + 2πrR + πR² - πr² - πR²
Area = 2πrR = 2π(9/(4R))R
Area = 18π/4 = 9π/2
Please could you do videos where you solve hard problems live so we can see your thought process
That's a really nice problem. I'm not in touch with geometry anymore and literally took me 15mins so it was a really good brain opener.
Some people used Ptolemy's theorem but we can also solve it just by careful observation.
Notice that the quadrilateral formed by joining P, Q, the centre of smallest circle and the centre of largest circle is a rhombus.
Let O be the centre of smallest circle and O' be the centre of largest circle.
So OM=O'M => O'M=2OM => Radius of largest circle = 3(radius of smallest circle).
Since centres are concurrent then => Radius of smallest circle + Radius of bigger circle = Radius of largest circle.
Simple Pythagorean theorem and done.
I wonder, if somebody recognized Archimed's arbelos in this problem?
Yes,I did. Archimedes called half of the blue area "arbelos", shoe makers knife, and calculated its area.
I solved it a bit different method but similar to first one. Never heard of chord chord.
I sat the imc this year (from which this question is from), almost no questions need algebra, in this one you just assume the chord is the diameter
please show me how do you resonned when you're in front of you this type of of exercise
I also defined "a" the same way and got 9pi/2, but why define both a and b? Medium circle has radius 2a based on the dots.
So, this "drag down the chord to make it the diameter" hack is not the right way to think if you're serious about math. It might get you the answer in a multiple choice question if that question has all numbers as choices. However, if that question is an Olympiad question where you need to derive the answer, this hack will get you a 0. The reason is that you're assuming without knowing whether this question has an answer. In other words, you're assuming without proving that the answer is invariant to the position of the chord. "If a question has been asked, it must have a determinate answer" is not a logically valid proof. It might be a safe practical assumption depending on the context, but it's not a mathematically rigorous approach. The key mathematical insight here is not the area but the proof that this area is invariant to the position of the chord.
I'm amazed to discover that the shaded area is always the same size regardless of how big the outer circle is. That feels significant, somehow.
Since a/b is not specified, it's simple to assume a=b, which makes the chord a diameter of the large circle, so a=b=c/4 and r=c/2=3. Also you have the total area of the small circles is 2(pi)a^2, and of course the area of the large circle is (pi)r^2. So then you have (pi)(r^2-2a^2) = (pi)((c/2)^2 - 2(c/4)^2) = (pi)(c^2/8) = (pi)(36/8) = (pi)(9/2).
That way you only end up with an answer to a specific situation where the chord is the diameter, with nothing to show that the same solution works for any diameter.
@@wombat4191 But the problem statement implies that the answer is independent of the actual sizes of the two circles, so we're free to assume any ratio we want, and we might as well assume one where a/b approaches/is 1. Finding special cases to solve in cases of problems like this is often a quick way to get the answer. If you wanted to PROVE that your answer is independent of the ratio of a/b then of course this approach falls short. But that wasn't the task here.
@@jagmarz Problem statement implying something means nothing. What if it was a trick question and the correct answer was to prove that it's unknowable? That of course is not the case, but what I mean is that a problem being supposedly solvable isn't a proof for anything. At best it tells you that a proof exists and you have to find it.
As for "that wasn't the task here". yeah it wasn't explicitly. However, when you take a specific case of a problem and solve that, you of course have to prove that your specific solution applies to the general case as well. The solutions Presh showed did not specify the problem any further, so his answers work for all cases by default.
LOL this is a question from the British Intermediate Maths Challenge (IMC), and I took part in it. Btw I got 128 out of 135. The only question I got wrong is Q20, which is a tricky but interesting question. I recommend Presh to make a Shorts video about that question.
Q25 is easy
How did you find the follow-on round(s)?
Gostei dessa, muito bom. O raio do círculo de baixo fica em função do raio do círculo de cima, ou vice-versa, e a área azul é invariante, fantástico!
Method 1 fails for all circles as b is assumed to be _greater_ than a. What about the chord where PQ is the diameter of the circle and a is thus _equal to_ b?
It's a degenerate case but the math works just fine for it
Well, it's trivial to prove that b ≥ a covers all possible cases through symmetry, but Presh could have at least mentioned it. And the method works perfectly well for a = b, because pythagorean theorem actually also works when the "triangle" is just a line.
@@wombat4191 I was just pointing out (badly) that method 1 was incomplete as it omitted the possibility that b=a.
In this case it made no difference (as the result also held), but in others it can make a difference.
There is a classic example of proving that 1=2 that has one step dividing by a-b which is fine _unless_ a=b, at which point you are dividing by 0; needless to say, you end up showing that a=b by dividing a-b which implies you must have divided by 0, thus making the result wrong!
Thank you for these problems I find it super interesting. As a note of feedback, I encourage you to refrain from using contextual "this" because its not always very clear what "this" is. i.e. instead of "this distance" you might use "the distance in purple"
First get the diameters of the circles by using Euclid's Height Theorem: h² = pq
(|PQ|/2)² = d ⋅ D
3² = 5d
d = 9/5
D = 18/5
A = π/25 ⋅ (27² - 18² - 9²)
= 324π/25
≈ 40,72
Why are the three centres collinear?
Could you please make a video about the basic of algebra 1 2
Love it!!!!!!!!!!
Fantastic!
I discovered your channel this year and I really like the way you solve the exercises with simplicity. I would like you to help me solve an exercise from the Olympiads of Algeria (high school, for 17 years old).
m is A natural number. The number m + 7 is a perfect square and m-34 is a perfect square. Find the value of m.
THANK YOU FOR ALL YOUR VIDEOS!!
I mean this is the easiest way. The difference between the two squares is 41 (34+7) and by the logic that (x+1)^2=x^2+2x+1, and we can remove the x^2 as we want the difference, we know that x is the lower number where 2x+1=41, and that leaves x as 20, so we can substitute in to m-34=20 -> m=54. Sorry for the clunky explanation
@@necros4031 m-34 = x^2, not m-34=x.
So with x=20, x^2 is 400, and m=434, from which we can also find (x+1)^2 = 441 = 434+7, which meets the requirement.
Keep in mind you assumed the two square numbers are sequential, so there may have been other solutions, but there are none in this case.
x² = m+7
y²= m-34
x²-y²=41
(x+y)(x-y)=41
But 41 is prime and x and y are natural numbers
Then: x+y=41 and x-y=1
That gives us x=21
21²-7 = m, m = 434
I made the assumption that it didn't matter how big the circles were so I shrunk it down until the diameter of the big circle was 6. That made the other two circles the same size. I got the answer in a few seconds after that.
You made an assumption, but did you end up with a proof? Your solution is not complete without a proof, now you only solved the problem for one specific case.
That's the most intelligent solution.
I actually participated in it and got this question right. It was actually decent.
I think we can solve easily assuming PQ as diameter
That way you solve only one specific case, with no proof that it works universally.
That's the most intelligent solution. To make it clearer to non-mathematicians just change "assuming" to "choosing".
there is more optimal solutions
assume the chord is dia of large circle then r=6/2=3,a=b=3/2=1.5.
the solve for it will get same answer
Thanks for a video. That was really nice
Because the problem is claiming to have an answer, we can surmise that the area must be the same no matter how draw the circles, provided only that PQ has a length of 6. As such, it would be *very convenient* if PQ happened to go through the center of the large circle. This means the large circle could have a radius of 3, and the two small circles would be half the width, with a radius of 1.5. So, plugging in some circle area equations, we get area = pi * 3^2 - pi * 1.5^2 - pi*1.5^2 = 4.5 * pi.
But why should this area be invariant at all? That's the real question.
Make more videos like this 👍🏻.......
Which will be beneficial in our competitive exams.
Hi Presh ! Please analize this method without Pitagoras or cords theorems, but only using some logic : the problem not define segments a and b, then , we can assume de problem is "resolvable" in any case, and furthermore, that the final value of this solutions are the same. Well, we can then resolve for a=b=1/2(p+q). From there , it is simple aplication o area of big circle (R=3) minus 2 times area of minor one (r=1,5). Regards !
Assuming is not enough, you have to prove that it works for any case. The two methods shown in the video are the easiest ones that give the proof.
Assume the writer has formulated the problem well. Then, since the ratio of the smaller circles' area is not specified, assume any ratio will give the same answer for the blue area. Hence we can assume that the circles have equal area. Then PQ is the diameter, hence the big circle has area 9*pi, and each of the two circles has are (9/4)*pi, then (9 - (9/4) - (9/4))*pi = (9/2)*pi gives the area.
You have no proof for your assumptions, and therefore your solution is incomplete.
@@wombat4191 Assumptions needn't be proved.
The fact that our answer relies on the assumption that the writer has formulated the problem well, means that either the problem is well-formulated and our answer is correct, or the problem has several possible answers, in which case our answer is one of many such.
The limitation of using that assumption is that the solution won't tell us whether the problem is well-formulated.
Note: I have used the word "assume" 3 times for comedic effect, but really we are only assuming one thing.
@@ventsiR where did you ever hear that "assumptions needn't be proved" in mathemathics? That's what maths is all about, proving stuff. Otherwise you could just assume anything and treat it as a true fact.
Being able to assume that a problem is solvable is not a proof, it only gives you the knowledge that a proof exists. It is still your burden as a solver to find that proof.
By specifying some previously undetermined values you essentially solve a different problem, a specific case of the original one. After that you of course must prove that your answer applies generally and not just in that specific case.
What is the software he use to make this transition of photos, like multiplying ( inserting ) pi in both sides of the equation with a smooth animation ?
How do you know that the cord length is 6? Where did they number come from???
Are you serious?
Easy. I pushed the chord all the way up so that the upper circle had no area and the lower circle was the same as as the big circle. Blue area equals zero.
Another method:
The question implies that the result does not depend on the radius r or the outer circle, only on the cord length.
By setting r=6, the cord will fall in the middle, forming a diameter of the outer circle, and the two small circles will both have radius 3/2.
So the radius of the shaded region becomes:
outer - 2*inner = π3^2 - 2*π(3/2)^2 = 9π/2
Nice one
I just seen a similar problem on yt where the shaded area is 6
a=ø of small circle
b=ø of medium circle
c=(a+b)=ø of large circle
PQ =**6**
ab= (**6**/2)²= 9
Blue = πc²/4 -πa²/4 -πb²/4
= (π/4)(c² -a² -b²)
= (π/4)[(a+b)² -a² -b²]
= (π/4)[(a² +b² +2ab -a² -b²)
= (π/4)(2ab)
=abπ/2
ab=9
Blue = 9π/2
is it impossible to find a and b values?
Yes, not without more information.
By the intersecting chords theorem we know that r×R=(6/2)×(6/2)
And we know that the area A to be proportional to r×R (and we know the coefficient), I feel like the solution has been a bit overcomplicated
I haven't seen geometry in years.. but where did the 3 came from exactly?
It's half of the chord length, which was specified to be 6.
@@wombat4191 oh it was specified.../facepalm no idea how i missed that. I could see it now in the thumbnail even. Ok, universe makes sense again. Thanks ^^
great question
In order for that problem to have an answer that isn't expressed as a ratio in terms of a and b, then the shaded area MUST be a constant value regardless of the position on PQ, INCLUDING the case where PQ is the diameter. So, set PQ as the diameter, and the correct answer is simple math. In fact, you missed a chance to show that for ALL cases with similar inscribed circles, and a chord of length N, the shaded area is N^2/8. THAT would have been more valuable.
Se vc olha direto vc buga, mas se pensar bem consegue
Solved this in 1 minute, it's super easy
I noticed that Method 1 kinda derived Method 2.
How this possible chord is equal to diameter and why....... 🤨🤨🧐🧐🧐🧐
Question. I am trying get myself ready for college but I graduate from High School 20+ Years ago and I barely remember math. Is there any website where I can learn Math from the scratch in an organize way? I appreciate your help.
Khan Academy is probably your best bet. Also "Brilliant"
to clarify, "Brilliant" is another website where you can learn math.
Watch Professor leonard videos
Or Watch the math sorcerer video for self study guide
Go to adult school and retake the course
Anybody have these types of questions pdf or link pls share
Co-geom. configs.:
P → (-3,0); M → (0,0); Q → (3,0);
r₁ → radius of upper inner circle;
r₂ → radius of lower inner circle;
C → the enclosing circle
Findings:
~ S is at (0,2r₁) & T is at (0,2r₂)
~ radius of C: r₁+r₂
~ centre of C: (0,2r₁-(r₁+r₂)) ⇔ (0,r₁-r₂)
~ eqn. of C: x² + [y-(r₁-r₂)]² = (r₁+r₂)²
⇔ x² + y² - 2(r₁-r₂)y = 4r₁r₂ ...(*)
~ as Q is on C, by (*), 9 = 4r₁r₂ ⇔ r₁r₂ = 9/4
~ area req. = (r₁+r₂)²π-r₁²π-r₂²π = 2r₁r₂π = 9π/2
brilliant
My solution is easier and faster:
We notice from the question that the blue area is independent of the exact sizes of the two small circles. Without Loss Of Generality, we let the two smaller circles be the same size. Then the chord of length 6 is a diameter of the larger circle, radius 3. Each smaller circle has radius 1.5. Subtraction of the 2 small circle areas from the large circle area gives the solution of "4.5pi". And That's The Answer.
Nice
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I got this one by the long way
Hello Presh
I'm a 12th grade student
I'm currently preparing for jee
Could you please provide me any tips to improve at maths
I am starting 10th next week... CBSE Student studying in Kuwait 😀
Good luck.
@@huzefa6421 Kuwait? How
@@sylvenara Bro i was born in kuwait. The law here is i can stay in kuwait till 60 years of age.... there are about 15 or more indian schools in kuwait and each school has the CBSE curriculum... i am studying in AMSB indian school ( dawoodi bohra school ) so thats how i an here.... reaching till class 10 its so satisfying hopefully passing grade 10 is also a great year 😀
See your math skills as a tool box. You now have alot of tools. Use them on problems where it is not obvious which tools to use. When you know your tool box and see what tool to chose, then you will succeed. When you have a large problem to solve, try to simplify. As in this case make the chord the diameter. Or if large numbers, try a problem with smaller or fewer numbers. Then use that method for the larger problem.
Good luck.