Since the positions of A, B, C, and D relative to the center of the red circle P are not given, we can assume they don't matter, and thus we can devolve the figure down to where the right hand quarter circle radius is equal to zero, since the area value will hold until that point. This places O on the circumference of the red circle, and by Thales' Theorem, any internal 90° angle on a circle will connect to two ends of a diameter. This means that AB = 2r = 4 (radius was resolved same as in video), which means that OA = OB = 2√2. In this devolution of the problem, the blue area is now equal to just the left hand quarter circle, and now we have its radius so the solution is as follows: Aʙ = πr²/4 = 8π/4 = 2π Note that this solution also holds for the opposite situation where the left and right quarter circles are congruent, and is even easier to calculate, as the radii of the red circle and the blue now semicircle are the same. Indeed, the area of the blue area will always be 1/2 the area of the red circle as long as A, B, C, and D (or whatever points they merge into, if devolved) are on the circumference of the red circle, the basic layout of two side-by-side quarter circles (or one and a zero-sized one), and OP being at 45° to the horizontal is maintained.

I agree. Method 1 is more fun!

Inutile de faire de grands calculs. Faisons glisser les "coins" du premier (celui qui est à gauche) quart de cercle sur la circonférence rouge. Rien ne nous l'interdit : c'est donc qu'il n'y a pas de variation de l'aire bleue. On obtient alors un demi-cercle bleu de rayon 2 (rayon du cercle rouge)...! Il suffit alors de diviser par 2 ! 4pi/2=2pi !!!

I solved it this way. Assuming that a solution is possible, the large circle is not defined in terms of the large quarter circle - it could be anywhere. You need three points ion the circumference to define a circle. So the solution -if there is one- will hold if the radius of the small semicircle becomes zero - in other words the red circle passes through the N and W points of the large quarter circle and the center. It thus becomes trivial. The straight line distance from the two corners of the quarter circle are a diameter of the red circle because it subtends a right angle. The red circle has area 4pi so r^2 = 4 and r=2, d=4. Applying Pythagoras to the quarter circle R^2 + R^2 = 16, so R = √8. Area of the quarter circle will be 1/4 x pi x (√8)^2 or 2pi. I'll post this then look at the answer.

@Darisiabgal7573

3 күн бұрын

Just looking I will say you’re right. So just looking when three points of the Square touch the circle then thus is an inscribed perfect quadrilateral. The chords are 90° for the radii which means length is SQRT (2) on the unit scale. The area of a circle is pi r^2 and the quarter circle is pi/4 r^2 so if r = SQRT(2) then on the unit square pi/4 * 2 = pi/2 now we need scale. Our area scale is 4pi/pi so 4 * pi/2 = 2 pi.

I liked your method 1 better. I tried solving by using your method 2, calculating the Blue Area when the quarter circle is minimum size (radius = red circle radius) >> Blue Area = 2π and then calculating the Blue Area when the quarter circle is maximum size (radius = (red circle diameter)/√2, point O is on circumference of red circle, OC = 0) >> Blue Area = 2π Since both minimum and maximum radii gave the same answer I figured it must be the same for all radii in between.

@ThePhantomoftheMath

3 күн бұрын

Nice thinking! Clever!

@BruisedReedofTas

3 күн бұрын

I think the second method is way more elegant and this comment adds to the strength of that opinion

Method 2 relies upon the assumption that the blue area has the same value independent of the construction of the quarter circle. I find it more interesting to find it out without this assumption or hint.

@ThePhantomoftheMath

Күн бұрын

Me too! 😀

You only can apply the second method if and only if you know that the blue area is constant. Since you gave no such argument the second method is incomplete. Alternatively, you could use the intersecting chords theorem: (0.5 (R+r))*(0.5 (R+r)) = (r_red - 0.5*(R-r))*(r_red + 0.5*(R-r)) 1/4 (R+r)^2 = 2^2 - 1/4*(R-r) 1/4 ((R+r)^2 + (R-r)) = 4 1/4 (2 R + 2 r^2) = 4 1/4 (PI R + PI r) = 2 PI A_blue = 2 PI

@ThePhantomoftheMath

8 сағат бұрын

Hi and thank you for the comment. In the explanation of the second method, I did mention that point C is an arbitrary point. That is more than enough to prove that the area is, in fact, constant. To be more clear: If you divide any given area into two parts using an arbitrary point, the overall area of the two parts will remain constant and equal to the original area. This is because the total area of the original shape is simply being partitioned into two smaller areas, and the sum of these two smaller areas must always add up to the area of the original shape. For example, if you have a rectangle with an area of 20 square units and you place an arbitrary point within this rectangle, no matter how you divide the rectangle using this point, the sum of the areas of the two resulting regions will always be 20 square units. Hope this clarifies things.

@derwolf7810

5 сағат бұрын

@@ThePhantomoftheMath No, that point C is arbitrary is not enough to prove that the area is constant. Take for example the same problem, but replace the small quarter circle, with a three quarter circle of the same (small) radius (which you labeled a) and you get a varying area of 2 𝜋 + 0.5 𝜋 a^2, despite point C still beeing an arbitrary point. The argument (in your second paragraph) only shifts the goalposts, because we don't (and can't) initially know whether or not its premise is true. Hence its consequence might be true (like in the example of the video) or false (like in the example i gave above). Having an arbitrary point C allows both possibilities, so we can't deduce anything from its existence.

@ThePhantomoftheMath

Сағат бұрын

@@derwolf7810 Hi again friend. I understand your point, and I agree, but the problem is that would then be a completely different problem. To rephrase better: You would be completely right if I had never mentioned precisely how the quarter circles were constructed, which I did at the beginning of the video. I said that both the radius of the larger and the smaller quarter circle must be horizontally aligned at their bottom sides. That provides enough information for the second method later. If the problem were, for example, as your example suggests: to construct one larger quarter circle and 3/4 of the smaller circle, and if we again put point C all the way up, the radius labeled 'a' would be the same as the one labeled 'b', you would get a full circle! This is because point C dictates the length of both radii 'a' and 'b'.

I followed all but 1:49 when the smallest quarter circle is extended, it cuts line OA at the same point (D) as the red circle cuts OA. This does not seem obvious to me, needs to be proved? Thanks.

@osmanfb1

4 күн бұрын

looks like we can show that they cut at the same point. Assume the smallest circle cuts the OA line at a different point D'. Now based on the red circle, OC*OB = OD * OA (power of circle?). But since OA=OB , OC = OD. But we know the smallest circle's radius, OC = OD'. So OD = OD'. D and D' are coincident. Maybe this is obvious in some other way of looking at this problem? It has been more than 50 years since I was in high school studying synthetic geometry.

@ThePhantomoftheMath

4 күн бұрын

@@osmanfb1 Hi! Thank you for the question. I will try to clarify. To understand why the smaller quarter circle intersects line OA at the same point D as the red circle, let’s consider the geometric properties involved: Square and Diagonal Concept: Imagine a square with points A,O, and B, where AO=OB. Drawing the diagonal from point O to the opposite corner of the square divides it into two identical isosceles right triangles. Symmetry and Chord: Any line perpendicular to this diagonal will create two symmetrical isosceles right triangles.In our problem, the line DC represents a chord of the red circle, and the symmetry of the circle implies that the diagonal from point O is the perpendicular bisector of this chord. Intersection Point D: Because the diagonal OD bisects the chord DC of the red circle, OD and OC are equal. This equality means that D is the same point where the red circle intersects OA. Therefore, lines OD and OC must be the same because they are the legs of two identical isosceles right triangles formed by the diagonal and the perpendicular bisector. This confirms that the smaller quarter circle, when extended, intersects OA at the same point D as the red circle.

@RAG981

4 күн бұрын

ODxOA = OCxOB by intersecting chord properties . But OA = OB so OD = OC = a.

I haven't understood the second method. Even initially you didn't mention the arc of a larger quarter circle passes through the centre of a red circle. Is it obvious? Am I missing something?

@ThePhantomoftheMath

6 сағат бұрын

Hi, thank you for the question. The arc of the larger quarter circle does not pass through the center of the red circle. It can, but it doesn't change anything; therefore, it is neither a condition of the problem nor mentioned in the problem. The second method only relies on the arbitrary point C and the property that any given area divided into two parts using an arbitrary point will remain constant and equal to the original area. This is because the total area of the original shape is simply being partitioned into two smaller areas, and the sum of these two smaller areas must always add up to the area of the original shape.

@vcvartak7111

4 сағат бұрын

@@ThePhantomoftheMath i admit it was my mistake that the arc need not pass through centre of the circle. But still not grasp the second method. If C is arbitrary point that can divide the blue area but actually larger quarter -circle shrink and smalller quarter-circle enlarge to match that. May be cut paste( larger quarter to smaller quarter)

@ThePhantomoftheMath

Сағат бұрын

@@vcvartak7111 Hi friend. Just think about it this way: Point C dictates the length of both radii (labeled 'a' and 'b') of the small and large quarter circles. Since C is not constrained by any condition, we can adjust it along the intersection line as we please. So, if we move it all the way up, we will increase radius 'a' but decrease radius 'b'. And since both circular segments are horizontally aligned at the bottom, their overall area will not change. This means that if we slide point C all the way down, the small quarter circle would completely disappear, and we would be left with only the larger quarter circle with the maximum possible radius. ☺️

As writing is on the figure, very difficult to follow also it is difficult to understand the speach.