Cheat method to solve: when a problem is phrased like this, it suggests that you will get the same answer no matter what values are used for A and B. So you can pick convenient values. You could set A = 0, and then B fills up exactly half the of the circle. Or you can set A = B, and it's a very convenient shape to look at (both semi-circles fit into a square in the middle of the circle).
First I realised that the area must be constant, then I adjusted my view and realized when one semicircle was infinitesimal the others base must be the diameter of the circle. Then, for some reason I didn't just say "oh that means it's half" but instead realised if I went the other way and made both semicircles equal in size they would fit the square that fits in the circle. Then I messed up the maths and ended up with a nonsensical answer.
Yes, but it would be lacking proper mathematical problem development, therefore you would only be answering for two particular cases, not solving as a whole (although here it's indifferent). Many teachers would not accept that as a correct answer.
Same here. Intuition is not proof though; although the question heavily implies it, initially we do not know that the area is the same regardless of the radii of the semi-circles.
There's the assumption that as no lengths are given for a & b and it's solvable that it holds for all a & b >= 0, which means you can use a limit condition of a (or b) = 0.
@@cigmorfil4101 no, you cannot prove it that way. You can "guess" the result and then try to prove it regardless of the radii; but the question could be wrong in what it implies and makes you assume wrong conclusions. For example, "given an ellipse with a and b as its long and short axes, what is its perimeter and area?" You would reason that since a circle's area is pi*r*r and its perimeter is pi*(r+r), and given that a a and b have the "same role" as the circle's radius, the logical answers would be pi*a*b (true) and pi*(a+b) (false). It does works for particular cases : a point, where a=b=0, a circle, where a=b=r, and a parabola, where a= infinity... but it is not true for the general case.
There is a beautiful alternative trigonometry solution. You express both radii a and b of the semicircles as r*cos(alpha) and r*cos(beta), where r is the radius of the big circle. With symmetry you can show that cos(beta)=sin(alpha). With the formula sin(alpha)^2+cos(alpha)^2=1 you get Area = 1/2pi*a^2 + 1/2pi*b^2 = 1/2pi*r^2 * (sin(alpha)^2 + cos(alpha)^2) =1/2
My solution, using the inscribed angle-central angle theorem: ibb DOT co SLASH 9YqGHTC (please replace DOT and SLASH by . and / respectively; it seems that RUclips doesn't like links in the comments.) As the central angle is 90º, the two shaded triangles must be congruent (they have same hypothenuse R and same angles). So, the result follows quickly. Thanks for sharing!
Hi, please find below another way for solving. If we consider A1A2 and B1B2 the parallel diameters of the two semicircles, X the tangent point and O the center of the circle of radius R that contains them, we will have: -the angle A1B2B1 equals 45 degrees (from the isosceles triangle XB2B) -this angle with the tip on the circle subscribes the same chords as the center angle A1OB1 so this will be 90 degrees. -The right triangles A1XB1 and A1OB1 have the common side A1B1 (hypotenuse). Thus (A1B1) exp2 = 2 (a) exp2 + 2 (b) exp2 = 2 (R) exp2. Result: (a) exp2 + (b) exp2 = (R) exp2. So the area of the two semicircles is half the area of the circle that contains them
My strategy was the same, but by moving in this direction we did an assumption: "area doesn't depend on the form of inscribed semicircles". This assumption comes implicitly from the task description. I think those assumptions can be made ONLY for educational tasks, but NOT FOR research/real-world tasks
I did that too but I was unhappy with that solution. not because I thought it was wrong but because it assumes that there is a static solution to be found. imo if the problem doesn't explicitly instruct the solver that there is a unique solution regardless of (in this case) semicircle size then that should not be assumed. the solution the video presents does not assume that there is a solution and is a way more satisfying solution imo
@@robinlindgren6429 I used this method also, but then also thought about the case where both semicircles are the same size. This then creates a square in the middle of the large circle where two of the sides of the square are the diameter of the semicircles. The diagonal of that square is the diameter of the large circle. That famous theorem that Presh likes to use shows that the sides of the square are 1/sqrt(2) the size of the diameter of the circle. Meaning that in this case also the area of the semicircles is half the large circle. (1/sqrt(2))^2 =1/2) The question implies a static answer or an answer dependent on the size of one of the semicircles. both edge cases show the same answer, making a answer dependant on the size of the semicircles unlikely. Now I will watch the rest of the video.
@@robinlindgren6429 Yeah, this problem would actually be pretty cool, if it said "radius of top semicircle is A and radius of second one is B, go solve". Only for us later to find out it was all but misdirection :D
Connect the two bases of the semicircles with two diagonal chords and two outer chords (length called p). Inscribed angle of an outer chord is 45 deg, so central angle of an outer chord is 90 deg. So p^2 = 2.r^2 = (a+b)^2+(a-b)^2; r^2 = a^2 + b^2; 0.5.pi.(a^2+b^2) = 0.s.(pi.r^2); area of semicircles is half of area of full circle...
The distance between centers of small circles is "a + b". Using Pythagoras theorem, we can write this distance also as sqrt(r^2 - a^2) + sqrt(r^2 - b^2) . Now rearrange this equation with terms containing "a" and "b" on same side and square both sides and simplify. You will have your answer.
Hey, one I got right! Though I did it more through deduction than proof. Namely, that because there was no numbers given for the two semicircles, it must be that the sum of their sizes is conserved for the problem to be solvable. Then simply imagine how one semicircle would grow as the other vanishes (essentially the animation at the end of the video, played in my head). Because the two diameters are parallel and the semicircles are tangent, the only possible result for the unvanished semicircle is that it must cover exactly the half of the larger outer circle.
There is a simpler way to the solution. Let's x be the distance of the full circle center to the diameter of bottom semicircle. Then x^2 + b^2 = R^2. Also, (a+b-x)^2 + a^2 = R^2. Combining these two we first derive that x=a. Therefore a^2 + b^2 = R^2
2:12 "my _favorite_ right-triangle theorem" -- *YES!!* As to the problem, I intuited from the title that it wouldn't matter what the relative sizes of the semicircles would be, so I let one go to zero ... in which case the other one filled half the circle.
Amazing!! No matter what are the radiuses of the two semicircles, they will always have togheder an area of half of the big circle!! The GRAPHIC illustration at the end of the video is great!!
I did it the other way around, with no math at all...just from the thumbnail. I started by thinking, "If the problem's solvable, then the size ratio of the two semicircles doesn't matter - so in the version where the upper semicircle's area is zero then the lower semicircle will have the full circle's diameter and its area will be 1/2...so the total area is always 1/2." Done. Great channel, Presh! I've been enjoying it for months while quarantined!
This was a fun one to solve and the first one i was able to prove in a different way than than Mr. Talwalkar. I could tell it was 1/2 pretty quickly by shrinking the length of a to 0 but it took a bit of time to find a proof. It might be easier to follow along with my solution if you draw it out. Lets call the left,midpoint, and right points of the diameter of the upper semicircle A, B and c, and we'll call the left, midpoint, and right points of the diameter of the lower semicircle D, E, and F. We'll also label the the point that the semicircles touch as T and label the point at the center of the circle as P. I inscribed a right isosceles triangle ATC inside the upper semicircle. Angle BCT is 45° and doing the same thing on the bottom semicircle means that angle TDE is also 45°. The similar angles and the fact that AC is parallel to DF means that the points D T and C are collinear. So knowing that points C D and F lie on the large circle and that the angle TDE is 45°, we can use the inscribed angle theorem to know that an angle CPF with the vertex at the center of the circle P is 90°. Now consider the angles EPF, FPC, and CPB. They must sum up to be 180° degrees because EPB forms a straight line. If you also consider the angles EPF, FEP, and PFE, they must sum up to 180° because they form a triangle. We have shown that angle FPC is 90°, and we know that angle FEP is perpendicular to the parallel lines AC and DF so it must be 90° as well. This means that angles CPB and PFE are congruent. Now it gets interesting. Knowing that PF and PC have the same length because they are both radii of the large circle, we can use angle-angle-side rule to show that CBP and PEF are congruent right triangles. Line segment BC, which is given a length 'a' in the video, has the same length as line segment PE. So now we have a right triangle PEF which has leg EF = b from the video, leg PE = a, and hypotenuse PF which is equal to the radius of the large circle. Now using the right angle theorem we know that a²+b²= r² and we can finish the solution the same way Mr. Talwalkar did. I liked the hidden bit of trivia that this solution showed along the way that the distance of the bottom line segment to the center of the circle is always equal to the radius of the top circle, and vice versa. Also its nice that it holds true without needing to constrain the puzzle such that b>= a. All in all a very fun problem. *EDIT* After looking at you website, I see that used the same method used by the person on cut-the-knot website.
Very cool problem! I solved it in a totally different fashion, using only Pythagorean Theorem. I drew lines from the center of the big circle to one corner of each semi-circle, as well as a vertical line connecting the centers of the two horizontal lines and passing through the center of the big circle. This made two right triangles both with hypotenuse equal to the big circle's radius. I applied Pythagorean Theorem to both triangles and combined the two equations. I plugged the results into the area formula for the 2 semi-circles and got 1/2.
Here's another sneaky way to do it: (1) Replace the two half circles with isosceles triangles of width 2a and height a (and width 2b and height b) The base angle of both these triangles is 45%. (2) Swing around one of these triangles maintaining its bases contacts with the circle until it kisses the other triangle. (3) We now have an angle of 45+45= 90 between the two chords (bases of the triangles). Draw a line between the other ends of the chords which will give us a hypothenuse which is also a diameter of the circle. (4) Bring back the two semicircles but this time place them on the outside of the chords (just for neatness sake). We already have a semicircle on the hypothenuse which is half the original circle. (5) Finally we have our proof using the area of the semicircle on the hypothenuse is the sum of the areas of the semicircles on the other two sides - yes Pythagoras doesn't just work with squares- honest.
Here's a less algebraic way to see that the combined area doesn't depend on the sizes of the semicircles: Consider the triangle formed by the right-hand vertex of each semicircle together with their point of tangency. This is right-angled because the two edges inside the semicircles are both at 45° to the parallel diameters. The length of the hypotenuse doesn't depend on the sizes of the semicircles, since it's always subtended by a 45° angle from the left-hand vertices of the semicircles. So by Pythagoras, the sum of the squares of the other two sides is constant. Since the areas of the semicircles are proportional to these squares, we have proved that the combined area is constant and can then get the answer 1/2 by considering the limiting case.
Since no information was given regarding the ratios of the two semicircles, Hence the sizes of both doesn't matter And so we can take the ratio 1:1 and solve within seconds Obviously one must learn how to solve but this was an easy way to get the answer
Quick way: look at the edge case where the radius of one of the semicircles is 0. Then the other semicircle clearly covers half the circle, thus the constant combined area is 1/2 the circle.
Doxo's Quick Way is an interesting special case. So, if you know that the area remains invariant if you change the picture, then you conclude it's 1/2. But the special case doesn't prove the invariance, therefore the general case doesn't follow immediately.
@@MozartJunior22A popular puzzle solving trick is, indeed, to jump from 'not given' to 'makes no difference', ... yes. But that rests on the assumption that the puzzle's integrity is good. From a puristic viewpoint, such an assumption is not quite satisfactory.
I did it using a different approach : construct two lines AC and BD such that they are mutually perpendicular. The line AB is such that it is parallel to the line CD. In the piece presented we can say AB is length 2*a and CD is length 2*b. We now find R such that we can construct a circle of radius R through the points A,B,C.D. With some simple manipulation we can show that R^2 = a^2 + b^2, which is sufficient to answer the question. This method uses only one theorem relating to a diameter projecting a right angle at all points on its circumference.
I solved it a much simpler way: 1. Find that given the context of the question, it's safe to assume the answer is independent of the individual size of each semicircle 2. Consider that if one semicircle has the same diameter as the full circle, it must have an area of 1/2 and the other semicircle must be reduced to 0 3. 1/2 + 0 = 1/2
By inspection I concluded the area of the two semicircles equaled one-half. I imagined the semicircles moving up and down to form one-half as his illustration showed, and he's done on other vids. Nicely done Presh! The actual solution was elegant. I like it and these are enjoyable to watch.
As soon as I finished reading the question, I realized it is a constant answer no matter the sizes of the two semicircles. So *I use the "special condition solution", where two semicircles are equally sized.* Then we can draw the "two diagonal lines", which are both the diameter of the large circle. Let the radius of the large circle = R, since the area of the large circle = pi*R^2 = 1, and since the radius of the semicircle = r = (1/√2)R, we have the whole blue area = 2*(1/2*(pi*[(1/√2)R]^2)) = 1/2(pi*R^2) = (1/2)*1 = 1/2. This is the quickest way to get the answer when you're not asked to prove "for all conditions".
Let the radius of the circle be r, the radii of the two semicircles a and b, b>=a. Let AB be the chord of length 2a, CD the parallel chord of length 2b, M the midpoint of AB, N the midpoint of CD. Let O be the centre of the circle, P the point of contact of the semicircles, let distance OP=x. By Pythagorus on the right triangle OMA, r^2 = (a+x)^2 + a^2 and similarly for triangle ONC, r^2 = (b-x)^2 +b^2 Equating the two RHS gives x=b-a Putting this back into either equation gives r^2 = 2(a^2+b^2), so the answer is 1/2 for all values of x.
I solved it much simpler but I'm not sure it's a valid proof: The blue area will always have the same ratio to the area of the entire circle. We can then construct a case where the area of the whole circle is not 1 and assign lengths to the blue area to give us easy numbers to work with. If we assume the blue area to have a radius of 1, then each semi circle will have an area of (pi*1^2)/2 = pi/2, giving us the area of the two semicircles to be pi. Using a radius for the semicircle of 1, that means the length from the cusp to the midpoint of the arc is sqrt(2). This is also the radius of the outside circle so we can plug that into the area equation and see that the circle has an area of 2pi. So, the ratio of the blue region to the entire circle is 1/2.
I figured that, since it was implied that the area would remain constant independent of the location of the chords, take it to the extreme so that one semi-circle encompasses the full diameter of the outer circle, while the second one vanishes to zero, and then you have half the area of the original circle, like the animation at the end.
ASEAN symbol has two "EQUAL" sized semicircles while the UNACADEMY symbol has "UNEQUAL" sized semicircles and here he has taken unequal semicircles only
Since this asked for "the" area, I assumed it was constant, so I considered the case where the two semicircles are equal. Then you can draw two isoceles triangles within those semicircles with hypotenuse sqrt(2/π). So the radius is 1/sqrt(2π), or the area is 1/4. Multiply by 2 to get 1/2
In order to avoid "cheating" answers about "must be same result regardless of the chosen semicercles so just consider le limit", you might ask : "find the semicercles giving the biggest shaded area" or " study the variations of the shaded area when you change the semicercles" (and that's how i'll give this problem to my students). Besides , naming C,Ca,Cb the 3 cercles, O, Oa, Ob their centers, A (resp. B) belonging both to C and Ca (resp. Cb) it's easy to justify equals the triangles O A Oa and O B Ob which gives back the a²+b²=r² formula. Anyway, thanks for the beautiful result.
Because the problem under-specifies parameters regarding the semicircles, you know right away that solving a simplified case will solve the problem. The simplest version of the problem is the one in which the semicircles are symmetric within the larger circle. In the simplified version of the problem, it is easy to see that the radius of one of the semicircles is 1/sqrt(2). Its area is pi/(sqrt(2)^2 / 2 = pi/4. Since there are two semicircles, their total area = pi/2.
Another approach: if you label points ABCD counterclockwise (|AB| = 2a, |CD| = 2b), and let P = intersection of AC and BD, then ACD = 45º, APD = 90º → so AP^2 + DP^2 = AD^2 → 2a^2 + 2b^2 = x^2. x is a chord with inscribed angle of 45º → central angle of 90º → x = √2 r. Therefore a^2 + b^2 = r^2 and the rest follows as in your solution.
There's a much easier way of proving this. Start with the case where the lower semicircle fills up the top half of the circle and the upper semicircle is null. Now move the semicircles down a small amount (d for delta). By Pythagoras theorem, the radius of the bottom semicircle is sqrt(r**2 - d**2), so it's area is 1/2 pi (r**2 - d**2). The area of the top semicircle is 1/2pi (d**2), so the area of both semicircles is 1/2pi r**2, regardless of the value of d.
FYI, this is what Nestor Abad proved in another comment, by using the inscribed angle-central angle theorem (you have an inscribed angle of 45°, the central one is 90°, so that you get 2 congruent triangles with lengths a, b and r).
I used the MYD principle of "missing information is irrelevant" to solve this at first glance: we're not told the ratio of the two semicircles, therefore it doesn't matter, therefore solve for when one of them is zero.
here's my line of thought: this is a general problem, so I can Choose the semicircles as I please. as cord for the lower semicircle, I pick the diameter of the large circle. this means that the upper semi circle has a cord with length 0, and is tangent to the circle. the lower semi circle will be exactly half of the large circle. ergo the area is 1/2.
Your first statement is mathematically unjustified. Sure, you can choose the chords as you please, but nowhere have you demonstrated that the answer does not depend on how you choose your chords. If your goal is obtain the right answer without any mathematical rigor, then of course your technique is fine, but I'm not sure why you would care about getting the answer more than the math behind it.
that's exactly how I figured the answer. however, being able to prove it for every couple of semicircles, that's another story :D (i couldn't do that xD)
Henceforth from this day forward it stall no longer be known as the "Gougu " or "Pythagorean" Theorem, but rather "My Favorite" Theorem , and shall be taught that way in academia worldwide 👍👍👍 So say we all This is the way
I think a simpler way to get the equation a^2+b^2=r^2, without considering any particular values of a and b, is noticing that the triangle with vertices the centre of the white circle, the medium point of one of the two semicircles diameter and an extreme of the semicircle's diameter is a right triangle. It has hypotenuse equal to the radius of the white circle and the sides equal to b and b-(b-a)=a or equal to a and a+(b-a)=a, depending on the semicircle we chose.
I appreciate the proof-driven approach to the problem. I solved it rather quickly with two arbitrary (albeit easy-to-calculate) extremes, while you emphatically solved it for all arbitrary ratios simultaneously.
I figured this one out without the math. I just did what you did at the end, with the shifting sizes of the semicircles, in my head, and realized that if one semicircle is so small as to be basically non-existent, the other will approach 1/2 the area of the circle.
Another way of solving it would be to recognize that if two chords of a circle intersect at a right angle, each chord dividing into a, b and c, d, then for the Diameter of the circle D D² = a² + b² + c² + d² By joining the opposite ends of the top chord AB (= 2r₁) and bottom chord DC ( = 2r₂) we get two chords AD and BC intersecting perpendicularly. a = c = √2 r₁ b = d = √2 r₂ Hence D² = 2(r₁² + r₂²) and the result follows!
Solved by intutuion since we have no measures for the 2 semicircles and therefore can assume that the area is undependent of their sizes. Then just think about the extreme case where other is as large as possible and the other is the smallest (so small its not even there)
A B C D left to right intersections with large and small blue semicircle circumference circle center of radius W = square root of 1/π b small semicircle radius , B large semicircle radius AD = chord subtended by angle < ACD= 45°, corresponding angle center
A B C D left to right intersections with large and small blue semicircle circumference O center of circle of radius W = square root of 1/π b small semicircle radius , B large semicircle radius AD = chord subtended by angle < ACD= 45°, corresponding angle center
I have a more straightforward solution to this problem. Let's denote the upper chord by AB, the lower one by CD, and their midpoints (and thus the centers of the semicircles) by E and F, respectively. The point where the two semicircles touch is X, the center of the cirlce is O. AEX is an isosceles right triangle so AX=√2a. Similarly, CX=√2b. AXE∠=45°, CXF∠45°, so AXC is a right triangle. BXE∠ is also 45° so C, X, and B lie on the same line and ABC∠=45°, therefore, AOC∠=90°. This means that AC=√2r. Now apply your favorite theorem to the AXC trianble: (√2a)²+(√2b)²=(√2r)² which leads to the answer.
There's an other way to get to the answer: by using pythagorean thm in each of the following right triangles: 1. From the small semi-circle, join its vertical radius r to the radius 1/pi^.5 (hypotenuse) of the big circle. Name y the length of the third side from the center of the big circle. 2. Idem from the big semi-circle. Name x the third side and R its radius. The hypotenuse the same as the previous one. Keep going, set up equations from pythagorean thm, then notice that: x+y=r+R. Play around and sove the 3 equations for: pi/2(r^2+R^2) You end up with 1/2. Bonus: you can generalize the problem by setting the area of the big circle is k, therefore the area of the 2 inscribed semi-circles is k/2. Beautiful problem. Thank u for posting it.
If you were to guess this puzzle would work out to a constant, irrespective of a and b, as many of these puzzles are, then you could consider the case where the first cord is the diameter of a circle, such that you only get one semicircle equal to half of the original. It’s not a rigorous proof, but considering the extreme case you can infer that the answer must either be 1/2 or be dependent on a/b.
Label the endpoints of the chords in a clockwise fashion as ABCD. Connect A to C and B to D with line segments to form opposite isosceles right triangles. Rotate one of the right triangles 90 degrees about the circle's center so that A and D coincide and the chords now form a right angle. Connect B to C. BAC is a right triangle. BC is a diameter of the circle. AB^2 + CD^2 = BC^2. After this, it's just a matter of scaling to see that the area of the two shaded semicircles would be equal to half the area of the full circle.
Gosh, the intersecting chord theorem sure is a favourite. It's easier to set up circle points (a,a) and (b,-b) with the circle centre at (0,c). With large radius=1 a quick Pythagoras gives a^2+(a-c)^2=1, b^2+(b+c)=1. The difference of these is 2(a^2-b^2)-2c(a+b). Factor out 2(a+b), a-b-c=0, c=a-b. Substitute a-c=b, b+c=a so either way a^2+b^2=1 and the result 1/2 follows (since we take 1/2 of the a and b circles).
I did it differently. The top left point is r from the center. Using Pythagoras, r^2 = a^2 + (square of the distance of midpoint of the small chord to the center). That distance is simply a + b - √(r^2-b^2)). You simply need to solve for π(a^2 + b^2)/2, given that πr^2 = 1. Very conveniently most things cancel out and you get 1/2
Wow. I can't believe I solved it correctly without seeing the answer. I am very amazed, although I did not solve it that way, but I assumed that the semicircles could change dimension and make points at the ends of their diameters form a square. Also I didn't realize that it could also be minimized and maximized, conserving the total area, until it was clearly seen that it was half the circle. I'm from Latin America, and I love to challenge my mind and that of my friends with math problems like these. Thank you for continuing to encourage us to live mathematics.
Here's how I did: first of all I centered the circle in cartesian system, so the circumference was {x^2+y^2=1/pi} Then I named "h" the ordinate of the point where the two semi-circles meet. Then through some easy geometrical constructions I figured out there are two perpendicular lines that link the opposite points where the circumference meets the semi-circles, and these two lines are: s1: y=x+h, s2: y=-x+h . For the rest of the problem I only needed one of them, so I went for s1. I then found the two intersections of s1 with the circumference and, solving for x, i got the two radiuses. Probably not the most elegant way, but it worked
An alternative approach is to apply analytic geometry. Take the origin to be the center of the semicircle with radius b and the x-axis along its diameter then the points of intersection of the two semicircles with the large circle: A(b,0), B(-b,0), C(a,a+b) and D(-a, a+b). The mid point of line segment AC is ( (a+b)/2, (a+b)/2) and the line perpendicular to AC passing through the mid point should intersect y-axis at the center of the large circle which can be found to be (0,a). Therefore the radius of the large circle is given by sqr((a-0)^2 + (a+b-a)^2)) = sqr(a^2+b^2) from which follows the required result.
the fact that this question was asked without specifying the sizes of the small circles made me conclude that its independent of them, so picking a nice extreme case like in the end solves the puzzle. there are a lot of puzzles like that, where from the mere fact that an answer is expected you can already conclude it.
Here both the semi circles touch each other at the centre of the circle, then perpendicular to the two parallel diameters pass through the centre of the circle ( cause perpendicular from the centre of the circle to the chord divides it into two equal parts, I have used r^2 = 2r_1^2 = r_2^2 Required area = 1π/2(r1^2+r2^2) =1π/2 (r^2) =1/2
Cheat method to solve: when a problem is phrased like this, it suggests that you will get the same answer no matter what values are used for A and B. So you can pick convenient values. You could set A = 0, and then B fills up exactly half the of the circle. Or you can set A = B, and it's a very convenient shape to look at (both semi-circles fit into a square in the middle of the circle).
That’s what I realized too.
Thanks man
Yep
First I realised that the area must be constant, then I adjusted my view and realized when one semicircle was infinitesimal the others base must be the diameter of the circle.
Then, for some reason I didn't just say "oh that means it's half" but instead realised if I went the other way and made both semicircles equal in size they would fit the square that fits in the circle.
Then I messed up the maths and ended up with a nonsensical answer.
Yes, but it would be lacking proper mathematical problem development, therefore you would only be answering for two particular cases, not solving as a whole (although here it's indifferent).
Many teachers would not accept that as a correct answer.
I just thought of the animation you showed at the end, and figured it must be 1/2....nice proof though....
?
How is this comment written 5 days ago?
How 5 days ago lol ?
Dis dude is from da PAST
@@utsab2010 He's a Patreon Supporter and has early access to the vids
I really don’t understand what people do not like about math. It’s so cool to discover these sorts of things
Indians be like, he got this idea from Unacademy logo🤣🤣
Lol same thought 😅
True
👍👍👍
Name this unacedmy problem
@@Ksraghu1c Gaaon Gaaon mein baccha baccha solve kar sakta hai🤣
I figured it out by imagining the case when a = 0 and got to 1/2
same..
Same here.
Intuition is not proof though; although the question heavily implies it, initially we do not know that the area is the same regardless of the radii of the semi-circles.
There's the assumption that as no lengths are given for a & b and it's solvable that it holds for all a & b >= 0, which means you can use a limit condition of a (or b) = 0.
@@cigmorfil4101 no, you cannot prove it that way. You can "guess" the result and then try to prove it regardless of the radii; but the question could be wrong in what it implies and makes you assume wrong conclusions. For example, "given an ellipse with a and b as its long and short axes, what is its perimeter and area?" You would reason that since a circle's area is pi*r*r and its perimeter is pi*(r+r), and given that a a and b have the "same role" as the circle's radius, the logical answers would be pi*a*b (true) and pi*(a+b) (false). It does works for particular cases : a point, where a=b=0, a circle, where a=b=r, and a parabola, where a= infinity... but it is not true for the general case.
@@takix2007
I didn't say it proved it, but using the limiting conditions makes those assumptions.
There is a beautiful alternative trigonometry solution.
You express both radii a and b of the semicircles as r*cos(alpha) and r*cos(beta), where r is the radius of the big circle. With symmetry you can show that cos(beta)=sin(alpha). With the formula sin(alpha)^2+cos(alpha)^2=1 you get
Area = 1/2pi*a^2 + 1/2pi*b^2
= 1/2pi*r^2 * (sin(alpha)^2 + cos(alpha)^2)
=1/2
My solution, using the inscribed angle-central angle theorem: ibb DOT co SLASH 9YqGHTC (please replace DOT and SLASH by . and / respectively; it seems that RUclips doesn't like links in the comments.)
As the central angle is 90º, the two shaded triangles must be congruent (they have same hypothenuse R and same angles).
So, the result follows quickly.
Thanks for sharing!
Oh good
How 5 days ago if this is 30 seconds ago ?
@@manveshdwivedi9329 ikr ?!!?
@@manveshdwivedi9329 he's a patreon supporter so he has early access to this video
@@h3xhexagonvn211 ohhhh
Plot Twist : Unacedmy sponsored this video 😂😂
Whole comments section is filled with unacademy
Exactly
@@targetiitbcse1761 You're 2023 aspirant na?
@@sparshgupta499 yes
@@targetiitbcse1761 All the best
1:33 Presh Talwalkar for Smash Bros. confirmed.
*Presh Talwalkar MINDS his decisions!*
Hi, please find below another way for solving. If we consider A1A2 and B1B2 the parallel diameters of the two semicircles, X the tangent point and O the center of the circle of radius R that contains them, we will have:
-the angle A1B2B1 equals 45 degrees (from the isosceles triangle XB2B)
-this angle with the tip on the circle subscribes the same chords as the center angle A1OB1 so this will be 90 degrees.
-The right triangles A1XB1 and A1OB1 have the common side A1B1 (hypotenuse). Thus (A1B1) exp2 = 2 (a) exp2 + 2 (b) exp2 = 2 (R) exp2.
Result: (a) exp2 + (b) exp2 = (R) exp2. So the area of the two semicircles is half the area of the circle that contains them
Very cool solution 👍
My favourite right angle theorem! The comments have influenced him😂
..... and it's a far more preferable name for it!
Ikr
@@piman9280 Agreed, atleast it stopped the incessant argument about the theorem's name
When he got to this part, I was ready with dislike button - but it was a surprise
@@satishchaudhary7978 i'm outta the loop. whats the argument? pythagoras or the earlier indian dude?
My answer is 1/2, I Imagined that one of the semicircles is just a single dot, so the other will have an area of 1/2
Fast way of solving it!
Nice way of thinking.
My strategy was the same, but by moving in this direction we did an assumption: "area doesn't depend on the form of inscribed semicircles". This assumption comes implicitly from the task description. I think those assumptions can be made ONLY for educational tasks, but NOT FOR research/real-world tasks
I did that too but I was unhappy with that solution. not because I thought it was wrong but because it assumes that there is a static solution to be found. imo if the problem doesn't explicitly instruct the solver that there is a unique solution regardless of (in this case) semicircle size then that should not be assumed. the solution the video presents does not assume that there is a solution and is a way more satisfying solution imo
@@robinlindgren6429 I used this method also, but then also thought about the case where both semicircles are the same size.
This then creates a square in the middle of the large circle where two of the sides of the square are the diameter of the semicircles. The diagonal of that square is the diameter of the large circle. That famous theorem that Presh likes to use shows that the sides of the square are 1/sqrt(2) the size of the diameter of the circle.
Meaning that in this case also the area of the semicircles is half the large circle. (1/sqrt(2))^2 =1/2)
The question implies a static answer or an answer dependent on the size of one of the semicircles. both edge cases show the same answer, making a answer dependant on the size of the semicircles unlikely.
Now I will watch the rest of the video.
@@robinlindgren6429 Yeah, this problem would actually be pretty cool, if it said "radius of top semicircle is A and radius of second one is B, go solve". Only for us later to find out it was all but misdirection :D
I was waiting for the animation at the end and that's what intrigues me the most. 👍👍👍👏👏👏
Connect the two bases of the semicircles with two diagonal chords and two outer chords (length called p). Inscribed angle of an outer chord is 45 deg, so central angle of an outer chord is 90 deg. So p^2 = 2.r^2 = (a+b)^2+(a-b)^2; r^2 = a^2 + b^2; 0.5.pi.(a^2+b^2) = 0.s.(pi.r^2); area of semicircles is half of area of full circle...
That was my approach too. I went to the bother of making a diagram, but can't see a way to put it in comments.
Legend know this question was promotion of unacademy 😂😂
LoL yeah
Yah
Legends hoga because plural verb use ki h na know
Greatt😂😂
LMAO😂
The distance between centers of small circles is "a + b". Using Pythagoras theorem, we can write this distance also as sqrt(r^2 - a^2) + sqrt(r^2 - b^2) . Now rearrange this equation with terms containing "a" and "b" on same side and square both sides and simplify. You will have your answer.
Hey, one I got right! Though I did it more through deduction than proof. Namely, that because there was no numbers given for the two semicircles, it must be that the sum of their sizes is conserved for the problem to be solvable.
Then simply imagine how one semicircle would grow as the other vanishes (essentially the animation at the end of the video, played in my head). Because the two diameters are parallel and the semicircles are tangent, the only possible result for the unvanished semicircle is that it must cover exactly the half of the larger outer circle.
There is a simpler way to the solution. Let's x be the distance of the full circle center to the diameter of bottom semicircle. Then x^2 + b^2 = R^2. Also, (a+b-x)^2 + a^2 = R^2. Combining these two we first derive that x=a. Therefore a^2 + b^2 = R^2
@George Xomeritakis
Neat way of doing it. Plus it also gives an insight about the center of the big circle in relation to the semicircles.
2:12 "my _favorite_ right-triangle theorem" -- *YES!!*
As to the problem, I intuited from the title that it wouldn't matter what the relative sizes of the semicircles would be, so I let one go to zero ... in which case the other one filled half the circle.
That's what I did as well
Amazing!!
No matter what are the radiuses of the two semicircles, they will always have togheder an area of half of the big circle!!
The GRAPHIC illustration at the end of the video is great!!
Consider one semicircle infinitely small, then the other semicircle is limited to exactly 1/2 of the original circle.
Or in other words with zero radius:)
Yes, the answer is obvious once one realizes that _if the problem didn't have a unique answer it wouldn't have been asked!_
Same answer if you assume the point where the blue circles touch each other is the centre of the bigger one.
I did it the other way around, with no math at all...just from the thumbnail. I started by thinking, "If the problem's solvable, then the size ratio of the two semicircles doesn't matter - so in the version where the upper semicircle's area is zero then the lower semicircle will have the full circle's diameter and its area will be 1/2...so the total area is always 1/2." Done.
Great channel, Presh! I've been enjoying it for months while quarantined!
This was a fun one to solve and the first one i was able to prove in a different way than than Mr. Talwalkar. I could tell it was 1/2 pretty quickly by shrinking the length of a to 0 but it took a bit of time to find a proof. It might be easier to follow along with my solution if you draw it out.
Lets call the left,midpoint, and right points of the diameter of the upper semicircle A, B and c, and we'll call the left, midpoint, and right points of the diameter of the lower semicircle D, E, and F. We'll also label the the point that the semicircles touch as T and label the point at the center of the circle as P.
I inscribed a right isosceles triangle ATC inside the upper semicircle. Angle BCT is 45° and doing the same thing on the bottom semicircle means that angle TDE is also 45°. The similar angles and the fact that AC is parallel to DF means that the points D T and C are collinear.
So knowing that points C D and F lie on the large circle and that the angle TDE is 45°, we can use the inscribed angle theorem to know that an angle CPF with the vertex at the center of the circle P is 90°.
Now consider the angles EPF, FPC, and CPB. They must sum up to be 180° degrees because EPB forms a straight line. If you also consider the angles EPF, FEP, and PFE, they must sum up to 180° because they form a triangle. We have shown that angle FPC is 90°, and we know that angle FEP is perpendicular to the parallel lines AC and DF so it must be 90° as well. This means that angles CPB and PFE are congruent.
Now it gets interesting. Knowing that PF and PC have the same length because they are both radii of the large circle, we can use angle-angle-side rule to show that CBP and PEF are congruent right triangles. Line segment BC, which is given a length 'a' in the video, has the same length as line segment PE. So now we have a right triangle PEF which has leg EF = b from the video, leg PE = a, and hypotenuse PF which is equal to the radius of the large circle. Now using the right angle theorem we know that a²+b²= r² and we can finish the solution the same way Mr. Talwalkar did.
I liked the hidden bit of trivia that this solution showed along the way that the distance of the bottom line segment to the center of the circle is always equal to the radius of the top circle, and vice versa. Also its nice that it holds true without needing to constrain the puzzle such that b>= a. All in all a very fun problem.
*EDIT* After looking at you website, I see that used the same method used by the person on cut-the-knot website.
Very cool problem! I solved it in a totally different fashion, using only Pythagorean Theorem. I drew lines from the center of the big circle to one corner of each semi-circle, as well as a vertical line connecting the centers of the two horizontal lines and passing through the center of the big circle. This made two right triangles both with hypotenuse equal to the big circle's radius. I applied Pythagorean Theorem to both triangles and combined the two equations. I plugged the results into the area formula for the 2 semi-circles and got 1/2.
Amazing result.
(I wish there were less comments about the name of THAT theorem, and about the problem itself.)
Here's another sneaky way to do it:
(1) Replace the two half circles with isosceles triangles of width 2a and height a (and width 2b and height b) The base angle of both these triangles is 45%.
(2) Swing around one of these triangles maintaining its bases contacts with the circle until it kisses the other triangle.
(3) We now have an angle of 45+45= 90 between the two chords (bases of the triangles). Draw a line between the other ends of the chords which will give us a hypothenuse which is also a diameter of the circle.
(4) Bring back the two semicircles but this time place them on the outside of the chords (just for neatness sake). We already have a semicircle on the hypothenuse which is half the original circle.
(5) Finally we have our proof using the area of the semicircle on the hypothenuse is the sum of the areas of the semicircles on the other two sides - yes Pythagoras doesn't just work with squares- honest.
Pretty smart way of doing.👍
Who else had no idea what was gonna happen but guessed it was gonna be half
Yeah
Naia
Here's a less algebraic way to see that the combined area doesn't depend on the sizes of the semicircles:
Consider the triangle formed by the right-hand vertex of each semicircle together with their point of tangency. This is right-angled because the two edges inside the semicircles are both at 45° to the parallel diameters. The length of the hypotenuse doesn't depend on the sizes of the semicircles, since it's always subtended by a 45° angle from the left-hand vertices of the semicircles. So by Pythagoras, the sum of the squares of the other two sides is constant. Since the areas of the semicircles are proportional to these squares, we have proved that the combined area is constant and can then get the answer 1/2 by considering the limiting case.
This is a brilliant and intuitive way of solving this.🎉
So, presh is promoting unacademy
😄😄
Since no information was given regarding the ratios of the two semicircles,
Hence the sizes of both doesn't matter
And so we can take the ratio 1:1 and solve within seconds
Obviously one must learn how to solve but this was an easy way to get the answer
Quick way: look at the edge case where the radius of one of the semicircles is 0. Then the other semicircle clearly covers half the circle, thus the constant combined area is 1/2 the circle.
Doxo's Quick Way is an interesting special case. So, if you know that the area remains invariant if you change the picture, then you conclude it's 1/2. But the special case doesn't prove the invariance, therefore the general case doesn't follow immediately.
@@benvanrensburg4261 It actually does follow immediately because the radiuses of the semi-circles were not given
@@MozartJunior22A popular puzzle solving trick is, indeed, to jump from 'not given' to 'makes no difference', ... yes. But that rests on the assumption that the puzzle's integrity is good. From a puristic viewpoint, such an assumption is not quite satisfactory.
I did it using a different approach : construct two lines AC and BD such that they are mutually perpendicular. The line AB is such that it is parallel to the line CD. In the piece presented we can say AB is length 2*a and CD is length 2*b. We now find R such that we can construct a circle of radius R through the points A,B,C.D. With some simple manipulation we can show that R^2 = a^2 + b^2, which is sufficient to answer the question.
This method uses only one theorem relating to a diameter projecting a right angle at all points on its circumference.
Look like Asean Logo 😁
Oh yeah
But the wheat is tightened a bit much...
But it is little bit hard
Unacademy hai😂
Unacademy logo
For me its look like a bowl upside down
This is one of the best puzzle video, I have seen on MindYourDecisions.
Lol everybody send this to Unacademy teachers
I solved it a much simpler way:
1. Find that given the context of the question, it's safe to assume the answer is independent of the individual size of each semicircle
2. Consider that if one semicircle has the same diameter as the full circle, it must have an area of 1/2 and the other semicircle must be reduced to 0
3. 1/2 + 0 = 1/2
Whenever he says, "Mind your decision!, I'm Presh Talwalker"
I feel like i said something that offended him🙄😳😳
😂🤣
By inspection I concluded the area of the two semicircles equaled one-half. I imagined the semicircles moving up and down to form one-half as his illustration showed, and he's done on other vids. Nicely done Presh! The actual solution was elegant. I like it and these are enjoyable to watch.
We're going to get awfully confused if your favourite right-angled triangle theorem changes!
As soon as I finished reading the question, I realized it is a constant answer no matter the sizes of the two semicircles.
So *I use the "special condition solution", where two semicircles are equally sized.*
Then we can draw the "two diagonal lines", which are both the diameter of the large circle.
Let the radius of the large circle = R, since the area of the large circle = pi*R^2 = 1,
and since the radius of the semicircle = r = (1/√2)R,
we have the whole blue area = 2*(1/2*(pi*[(1/√2)R]^2)) = 1/2(pi*R^2) = (1/2)*1 = 1/2.
This is the quickest way to get the answer when you're not asked to prove "for all conditions".
As soon as he said semicircles are curious I'm like: "it's gonna be half"
Very good. And enlightenment in the last seconds ...
Hii! You may have to visit this channel for more harder questions #onlymathlovers .....
@@madhukushwaha4578 yes, yours is nice too. I can subscribe to anything related to math.
This is so beautiful--
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
@@madhukushwaha4578 yep I'm a regular viewer this channel provides amazing analysis and intuitive ways to solve problems
Let the radius of the circle be r, the radii of the two semicircles a and b, b>=a.
Let AB be the chord of length 2a, CD the parallel chord of length 2b, M the midpoint of AB, N the midpoint of CD.
Let O be the centre of the circle, P the point of contact of the semicircles, let distance OP=x.
By Pythagorus on the right triangle OMA, r^2 = (a+x)^2 + a^2 and similarly for triangle ONC, r^2 = (b-x)^2 +b^2
Equating the two RHS gives x=b-a
Putting this back into either equation gives r^2 = 2(a^2+b^2), so the answer is 1/2 for all values of x.
Nice one. Exactly the way I did it.
Question sponsored by unacademy😂😂
I solved it much simpler but I'm not sure it's a valid proof: The blue area will always have the same ratio to the area of the entire circle. We can then construct a case where the area of the whole circle is not 1 and assign lengths to the blue area to give us easy numbers to work with. If we assume the blue area to have a radius of 1, then each semi circle will have an area of (pi*1^2)/2 = pi/2, giving us the area of the two semicircles to be pi. Using a radius for the semicircle of 1, that means the length from the cusp to the midpoint of the arc is sqrt(2). This is also the radius of the outside circle so we can plug that into the area equation and see that the circle has an area of 2pi. So, the ratio of the blue region to the entire circle is 1/2.
Please say what your favourite theorem is, it would make my day. I know it’s become a “thing” now, but you can do it.
Gougu theorem
Gougu theorem
Gougu theorem
Please don't encourage him to "Go ugly" again!
Grogu, a.k.a. Baby Yoda
I figured that, since it was implied that the area would remain constant independent of the location of the chords, take it to the extreme so that one semi-circle encompasses the full diameter of the outer circle, while the second one vanishes to zero, and then you have half the area of the original circle, like the animation at the end.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
Everyone be like UnAcaDEmY
Me: ASEAN but the wheat isn't having a good time
Same
@@nzf-kx2qol1g12 same bro
ASEAN symbol has two "EQUAL" sized semicircles while the UNACADEMY symbol has "UNEQUAL" sized semicircles and here he has taken unequal semicircles only
@@ujjwaldwivedi1226 ok, but one of the possible arrangements is equal
Since this asked for "the" area, I assumed it was constant, so I considered the case where the two semicircles are equal.
Then you can draw two isoceles triangles within those semicircles with hypotenuse sqrt(2/π). So the radius is 1/sqrt(2π), or the area is 1/4. Multiply by 2 to get 1/2
Mind your decisions!
In order to avoid "cheating" answers about "must be same result regardless of the chosen semicercles so just consider le limit", you might ask : "find the semicercles giving the biggest shaded area" or " study the variations of the shaded area when you change the semicercles" (and that's how i'll give this problem to my students).
Besides , naming C,Ca,Cb the 3 cercles, O, Oa, Ob their centers, A (resp. B) belonging both to C and Ca (resp. Cb) it's easy to justify equals the triangles O A Oa and O B Ob which gives back the a²+b²=r² formula.
Anyway, thanks for the beautiful result.
Master your engineering knowledge here❤❤❤
Because the problem under-specifies parameters regarding the semicircles, you know right away that solving a simplified case will solve the problem. The simplest version of the problem is the one in which the semicircles are symmetric within the larger circle.
In the simplified version of the problem, it is easy to see that the radius of one of the semicircles is 1/sqrt(2). Its area is pi/(sqrt(2)^2 / 2 = pi/4. Since there are two semicircles, their total area = pi/2.
Hii! You may have to visit this channel for more harder questions #onlymathlovers .....
The simplest version of the problem is when one semicircle has size zero and the other fills half the big circle.
@@OscarCunningham That IS simpler! I like it.
After Edited: One of the Most beautiful curve is a Circle.
Before Edited:
The most beautiful curve is a Circle ⭕ 🎉 ❤️
your vocabulary of curves is too low
@@adithyan9263 🤔
@@AnandhaKrishnanPT list all the curves you know
now don't google it. 😑
@@adithyan9263 Open Curve and Closed Curve.
@@adithyan9263 Simple Open and Simple Closed Curve.
Another approach: if you label points ABCD counterclockwise (|AB| = 2a, |CD| = 2b), and let P = intersection of AC and BD, then ACD = 45º, APD = 90º → so AP^2 + DP^2 = AD^2 → 2a^2 + 2b^2 = x^2. x is a chord with inscribed angle of 45º → central angle of 90º → x = √2 r. Therefore a^2 + b^2 = r^2 and the rest follows as in your solution.
One thing Presh forgot to add :
Sponsored by #Unacademy.
#UnacademyJee
Hi Presh, I appreciate all of these interesting problems!
Hii! You may have to visit this channel for more harder questions #onlymathlovers .....
Sir, what animation software do you use to make your videos.
these are simple animations, you can definitely do in GeoGebra, not sure what the author used though
There's a much easier way of proving this. Start with the case where the lower semicircle fills up the top half of the circle and the upper semicircle is null. Now move the semicircles down a small amount (d for delta). By Pythagoras theorem, the radius of the bottom semicircle is sqrt(r**2 - d**2), so it's area is 1/2 pi (r**2 - d**2). The area of the top semicircle is 1/2pi (d**2), so the area of both semicircles is 1/2pi r**2, regardless of the value of d.
You need to prove that the d is necessarily the same for both semicircles. That's not obvious at all!
FYI, this is what Nestor Abad proved in another comment, by using the inscribed angle-central angle theorem (you have an inscribed angle of 45°, the central one is 90°, so that you get 2 congruent triangles with lengths a, b and r).
mY fAvOuRiTe rIgHt tRiAnGlE tHeOrEm11!!1!1!!
I thought it'd be... Gougu... 😹
He didn't use gugu theorem! My whole life is a lie😬😭
I used the MYD principle of "missing information is irrelevant" to solve this at first glance:
we're not told the ratio of the two semicircles, therefore it doesn't matter, therefore solve for when one of them is zero.
Nice question,the question that I can't solve is always nice question.😂😂
here's my line of thought:
this is a general problem, so I can Choose the semicircles as I please. as cord for the lower semicircle, I pick the diameter of the large circle.
this means that the upper semi circle has a cord with length 0, and is tangent to the circle.
the lower semi circle will be exactly half of the large circle.
ergo the area is 1/2.
Your first statement is mathematically unjustified. Sure, you can choose the chords as you please, but nowhere have you demonstrated that the answer does not depend on how you choose your chords. If your goal is obtain the right answer without any mathematical rigor, then of course your technique is fine, but I'm not sure why you would care about getting the answer more than the math behind it.
that's exactly how I figured the answer. however, being able to prove it for every couple of semicircles, that's another story :D (i couldn't do that xD)
what is the area of unacademy logo.....
1/2. Square unit
Really enjoyed doing this puzzle! Thank you for sharing it with us all!
Henceforth from this day forward it stall no longer be known as the "Gougu " or "Pythagorean" Theorem, but rather "My Favorite" Theorem , and shall be taught that way in academia worldwide 👍👍👍
So say we all
This is the way
Oh no!
We now have three names for the theorem.
What say that we compromise on calling it the
Mythagougu Theorem
From the intersecting chords we also have 2(a + b)^2 + 2(a - b)^2 = 4r^2.
look like unacademy logo i think this is unacademy promotion
I think a simpler way to get the equation a^2+b^2=r^2, without considering any particular values of a and b, is noticing that the triangle with vertices the centre of the white circle, the medium point of one of the two semicircles diameter and an extreme of the semicircle's diameter is a right triangle. It has hypotenuse equal to the radius of the white circle and the sides equal to b and b-(b-a)=a or equal to a and a+(b-a)=a, depending on the semicircle we chose.
I'm glad you've stopped saying Gougu theorem.
I appreciate the proof-driven approach to the problem. I solved it rather quickly with two arbitrary (albeit easy-to-calculate) extremes, while you emphatically solved it for all arbitrary ratios simultaneously.
2:13 The Grogu Theorem!
No no no, it's right triangle theorem!
I figured this one out without the math. I just did what you did at the end, with the shifting sizes of the semicircles, in my head, and realized that if one semicircle is so small as to be basically non-existent, the other will approach 1/2 the area of the circle.
Why does it look like unacademy logo😂😂😂😂😂😂😂
Another way of solving it would be to recognize that if two chords of a circle intersect at a right angle,
each chord dividing into a, b and c, d, then for the Diameter of the circle D
D² = a² + b² + c² + d²
By joining the opposite ends of the top chord AB (= 2r₁) and bottom chord DC ( = 2r₂)
we get two chords AD and BC intersecting perpendicularly.
a = c = √2 r₁
b = d = √2 r₂
Hence D² = 2(r₁² + r₂²) and the result follows!
Plot twist
This video is sponsored by unacademy
Nice animation at the end. That's what I imagined and guessed ½ at the start of the video.
People in the comments: "i SoLvEd It In My HeAd"
Solved by intutuion since we have no measures for the 2 semicircles and therefore can assume that the area is undependent of their sizes. Then just think about the extreme case where other is as large as possible and the other is the smallest (so small its not even there)
@@shortestway2home888 ya, but that isn't really a rigorous proof. I though of that first, but I had no way to prove that what I thought was correct.
@@loneranger4282 yeah I completely agree with you and actually made a mistake using the word "solved". Should have said "got the right answer".
Really like the music, just replaying the last 25s over and over...
I had my answer in just 2 sec by intuition just by seeing the question...
So did I
So you guessed the answer? Okay.
A B C D left to right intersections with large and small blue semicircle circumference
circle center of radius W = square root of 1/π
b small semicircle radius , B large semicircle radius
AD = chord subtended by angle < ACD= 45°, corresponding angle
center
A B C D left to right intersections with large and small blue semicircle circumference
O center of circle of radius W = square root of 1/π
b small semicircle radius , B large semicircle radius
AD = chord subtended by angle < ACD= 45°, corresponding angle
center
As an Asian, I used 1 sec to get the answer, 0.5 😂
I have a more straightforward solution to this problem. Let's denote the upper chord by AB, the lower one by CD, and their midpoints (and thus the centers of the semicircles) by E and F, respectively. The point where the two semicircles touch is X, the center of the cirlce is O. AEX is an isosceles right triangle so AX=√2a. Similarly, CX=√2b. AXE∠=45°, CXF∠45°, so AXC is a right triangle. BXE∠ is also 45° so C, X, and B lie on the same line and ABC∠=45°, therefore, AOC∠=90°. This means that AC=√2r. Now apply your favorite theorem to the AXC trianble: (√2a)²+(√2b)²=(√2r)² which leads to the answer.
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There's an other way to get to the answer: by using pythagorean thm in each of the following right triangles:
1. From the small semi-circle, join its vertical radius r to the radius 1/pi^.5 (hypotenuse) of the big circle. Name y the length of the third side from the center of the big circle.
2. Idem from the big semi-circle. Name x the third side and R its radius. The hypotenuse the same as the previous one.
Keep going, set up equations from pythagorean thm, then notice that:
x+y=r+R.
Play around and sove the 3 equations for: pi/2(r^2+R^2)
You end up with 1/2.
Bonus: you can generalize the problem by setting the area of the big circle is k, therefore the area of the 2 inscribed semi-circles is k/2.
Beautiful problem. Thank u for posting it.
If you were to guess this puzzle would work out to a constant, irrespective of a and b, as many of these puzzles are, then you could consider the case where the first cord is the diameter of a circle, such that you only get one semicircle equal to half of the original. It’s not a rigorous proof, but considering the extreme case you can infer that the answer must either be 1/2 or be dependent on a/b.
Hii! You may have to visit this channel for more harder questions #onlymathlovers ..
設上半圓半徑為a，下半圓半徑為b，將右上跟左下與大圓的接觸點相連，可以得到 根號2倍(a+b)=2R，R為大圓直徑，簡單
Label the endpoints of the chords in a clockwise fashion as ABCD. Connect A to C and B to D with line segments to form opposite isosceles right triangles. Rotate one of the right triangles 90 degrees about the circle's center so that A and D coincide and the chords now form a right angle. Connect B to C. BAC is a right triangle. BC is a diameter of the circle. AB^2 + CD^2 = BC^2. After this, it's just a matter of scaling to see that the area of the two shaded semicircles would be equal to half the area of the full circle.
Pythagoras Theorem --> Gougu Theorem --> My favourite right-angled triangle theorem
Gosh, the intersecting chord theorem sure is a favourite. It's easier to set up circle points (a,a) and (b,-b) with the circle centre at (0,c). With large radius=1 a quick Pythagoras gives a^2+(a-c)^2=1, b^2+(b+c)=1. The difference of these is 2(a^2-b^2)-2c(a+b). Factor out 2(a+b), a-b-c=0, c=a-b. Substitute a-c=b, b+c=a so either way a^2+b^2=1 and the result 1/2 follows (since we take 1/2 of the a and b circles).
I did it differently. The top left point is r from the center. Using Pythagoras, r^2 = a^2 + (square of the distance of midpoint of the small chord to the center). That distance is simply a + b - √(r^2-b^2)). You simply need to solve for π(a^2 + b^2)/2, given that πr^2 = 1. Very conveniently most things cancel out and you get 1/2
This throws it all in the air..which is best ...the simple or the complex answer...
Wow. I can't believe I solved it correctly without seeing the answer. I am very amazed, although I did not solve it that way, but I assumed that the semicircles could change dimension and make points at the ends of their diameters form a square. Also I didn't realize that it could also be minimized and maximized, conserving the total area, until it was clearly seen that it was half the circle.
I'm from Latin America, and I love to challenge my mind and that of my friends with math problems like these. Thank you for continuing to encourage us to live mathematics.
Wow! Never heard Presh so excited in the epilogue!
Here's how I did: first of all I centered the circle in cartesian system, so the circumference was {x^2+y^2=1/pi}
Then I named "h" the ordinate of the point where the two semi-circles meet.
Then through some easy geometrical constructions I figured out there are two perpendicular lines that link the opposite points where the circumference meets the semi-circles, and these two lines are: s1: y=x+h, s2: y=-x+h . For the rest of the problem I only needed one of them, so I went for s1.
I then found the two intersections of s1 with the circumference and, solving for x, i got the two radiuses.
Probably not the most elegant way, but it worked
An alternative approach is to apply analytic geometry. Take the origin to be the center of the semicircle with radius b and the x-axis along its diameter then the points of intersection of the two semicircles with the large circle: A(b,0), B(-b,0), C(a,a+b) and D(-a, a+b).
The mid point of line segment AC is ( (a+b)/2, (a+b)/2) and the line perpendicular to AC passing through the mid point should intersect y-axis at the center of the large circle which can be found to be (0,a). Therefore the radius of the large circle is given by sqr((a-0)^2 + (a+b-a)^2)) = sqr(a^2+b^2) from which follows the required result.
the fact that this question was asked without specifying the sizes of the small circles made me conclude that its independent of them, so picking a nice extreme case like in the end solves the puzzle. there are a lot of puzzles like that, where from the mere fact that an answer is expected you can already conclude it.
All your vids are mine-blowingly simple
Here both the semi circles touch each other at the centre of the circle, then perpendicular to the two parallel diameters pass through the centre of the circle ( cause perpendicular from the centre of the circle to the chord divides it into two equal parts, I have used r^2 = 2r_1^2 = r_2^2
Required area = 1π/2(r1^2+r2^2) =1π/2 (r^2) =1/2