You are making me a genius in my own mind Andy. I couldn't do any of these a month ago. Now I am killing it. Thank you and Merry Merry Christmas to you and everyone here. This is a great Christmas present.
Answer to the next question: Quarter circle and small circle radius = R & r (2r)^2 + (12)^2 = R^2 Shaded area = (πR^2)/4 - πr^2 = π[((2r)^2 + (12)^2)/4 - r^2] = 36π
I'm thinking the next one is one of those where it doesn't matter how big the white circle is, the solution will be the same every time, since all we have is a distance of 12 at a random height. So if the area of the white circle is 0, that would make the radius of the quarter circle 12, which leads to the area being 36*pi.
Note that the problem becomes clear when we see that the right triangle whose side length is 12 and whose hypotenuse is R, the radius of the quarter circle, and whose third side is 2r, the diameter of the small circle, is, according to the Pythagorean theorem, (R)²-(2r)²=12², so (πR²)/4-π(2r)²/4=(12²/4)π, i.e. (πR²)/4-πr²=36π, which is the required area.
That’s a good question. I think if you assume the quarter circle and semicircle touch at exactly one point and fit into a square, then they must touch at the center of the semicircle. But it seems like that is something that should be established.
The right and bottom side of the square are two tangent lines of the semicircle that converge at a right angle (i.e. the lower-right corner of the square). If you connect the center of the semicircle to the point of the convergence, you divide the convergence angle equally (45° - 45°), which means this line coincides with the square’s diagonal. The upper-left corner of the square is the center of the quarter circle, So the square’s diagonal is expected to split this quarter circle equally into two.
Tomorrow looks like the area is π/4*R²-πr², and you can find a triangle with sides of 12 and 2r, and hypothenuse of R, which gives us R²-(2r)²=12, which leads to the answer being 3π
@@danohana1822 The one that consists of the radius of the bigger circle that we draw to the right side of the chord, the chord itself, and the line segment between the left side of the chord and the center of the circle, whose length happens to be exactly the diameter of the smaller circle.
Next one is easy to solve by constructing an example, but here's the proof: Radius of 1/4 circle = R, radius of small circle is r. We want ((R^2)/4 - r^2)pi. Connect the center of the 1/4 circle to the other end of the 12 length. Right triangle with legs 2r and 12, hypotenuse R. 4r^2 + 144 = R^2. Divide everything by 4 and move the r^2 term to the right to get 36 = (R^2)/4 - r^2. Since we want ((R^2)/4 - r^2)pi sub 36 in and we get 36pi.
Green radius is 2 (pir2/4), blue radius is {2} (pir2/2). The distance from where the green contacts the blue to the edge of the square is 2cos45 =2/{2}={2} Im assuming the semicircle is symmetrical around the squares axis, which means the squares sides are {2} (green vertical) + {2} (blue vertical) =2{2} Then, the area is (2{2})2= 8
I started by asking myself if the point of tangency of the quarter circle and the semicircle is at the center of the square (it sure looks like if might be, but then of course, you can’t assume that). Following along with the development in the video shows (freeze at 2:38) that it is in fact the case. Thinking some more about why this is true, I came to the conclusion that it follows from the respective areas being the same. One can deduce (by equating the areas of a quarter circle with radius s/SQRT(2) and a semicircle with radius s/2) that this common area must be equal to (pi * s^2)/8. So if the common area is given as pi, then s^2 must equal 8, as shown in the video. Summarizing: if the area of the two shapes is the same (say A), the area of the square will be 8 * A/pi, and the point of tangency will be the center of the square. Conversely, if the point of tangency is at the center of the square, the two shapes must have the same area, given by (pi * s^2)/8.
You could also multiplyed the 4 by 2(one of the triangle's height, the 4 is the base or in other way the half of the other diagonal) and it gives the same:)
I just assumed that the quarter circle crosses the centre, so its radius would be a half-diagonal, and two adjacent half-diagonals form a right triangle with hypotenuse being the side, which comes out to root 8. So the area is 8
You can also take the horizontal projection of r1 : r1x= r1 * cos45 = r1 * sqrt(2)/2. Then simply add r1x+r2 = sqrt(2)+sqrt(2) = 2sqrt(2). So the area is (2sqrt(2))^2 = 8.
Half way through the video it was found that r2 was root2. And we know r2 is half the length of the side of the square. So (2root2)^2 is area = 8. Shaves off a some working by not worrying about the diagonal or r1.
Sorry, I mean straight after when the diagonal of the small square is confirmed as 2, same size as r1. Then you know r2 is half length of the side of the large square.
@@juanignaciolopeztellechea9401At 2.30 or so he erases the smaller square and calculates the length of the side of the large square by using the newly found long diagonal and pythagoras. But we already know r2 is half length of large square side. That's the only step that can be skipped. It's a very minor point. At 2.30 visually you can see the smaller square is exactly a quarter the size of the large square with the numbers written on the image at that point (the long diagonal is exactly r1 is same length as short diagonal).
Can one just assume that the radius of the green quarter circle and diagonal of smaller square are a straight line? It's easy enough to prove if you extend some lines and fill in angles. Otherwise, nice video. I got the same answer in the same manner.
A lot of assumptions are made with this question. The first assumption is that the question is presenting basic shapes, i.e. a quarter and semi-circle (which should answer your question). The second assumption is the circle shapes are placed symmetrically in the square, which means the quarter-circle radius must intersect the semicircle at a perfectly diagonal and symmetrical 45 degree angle.
An immediate observation is that if pi times something equals pi, that something equals 1. But the first question I asked was, is it true that those given squares are equal, so I would have probably started with proving it. (Unfortunately I'm not a mathematician to go for it.)
36pi took me way too long to figure out how to prove it area of the quarter circle is piR^2 area of the circle is pir^2 you can make a right angled triangle with hypotenuse being R and the other sides being 12 and 2r making R^2=12^2+(2r)^2 so the shaded area=(pi(12^2+(2r)^2))/4-pir^2 a=144pi/4+(4pir^2)/4-pir^2 a=36pi+pir^2-pir^2 a=36pi
i'm half way through the video and i stopped just to say that i think you only need r2 since it is half the side of the square so the area of the square would be 2*(2*sqrt(2)) which would be 8... did i got it right? yei i did 😂
Hey I was right! Though I absolutely used the wrong(or at least different) equations 😅 I just figured out the radius of the semi circle to be sqrt(2), and kinda assumed that the center of the diameter was the center of the square(as the 2 points whrere the semi circle touches the square look to be in the center, and also drawing lines inward from them look like the exact center and create a 90⁰ angle), and then I dou led it for the length of the square, being 2sqrt(2), and then squared if to get 8 Completely wrong method, but hey, it worked
If you calculate the radius of the green circle, this confirms that the base of the blue semi circle lies on the diagonal since its radius matches the diagonal of the smaller square formed by connecting the three tangent points.
@@grahamkay4034 Sorry, what I really mean is that there's no need to calculate r1. Alternatively, complete the blue semicircle into a full blue circle. The diameter of that is a side of the square.
You are making me a genius in my own mind Andy. I couldn't do any of these a month ago. Now I am killing it. Thank you and Merry Merry Christmas to you and everyone here. This is a great Christmas present.
You can do it Andy, get all the puzzles done in 31 days!
I applaud the effort and dedication 👏
Where do you solve your problems?
On a piece of paper.
No, power point.
He probable made all the slides and just clicks through them
@@grahamkay4034 This has the same energy as
"- where do you work out"
"- in the library"
In the ring
Probably in his room.
Answer to the next question:
Quarter circle and small circle radius = R & r
(2r)^2 + (12)^2 = R^2
Shaded area = (πR^2)/4 - πr^2 =
π[((2r)^2 + (12)^2)/4 - r^2] = 36π
I'm thinking the next one is one of those where it doesn't matter how big the white circle is, the solution will be the same every time, since all we have is a distance of 12 at a random height. So if the area of the white circle is 0, that would make the radius of the quarter circle 12, which leads to the area being 36*pi.
12 is less than, not equal to, the radius of the quarter circle
12 specifies the size of the small circle.
@@cyruschang1904 As the area of the small circle approaches 0, the radius of the quarter circle approaches 12.
@@Haver2000000 OK. I see what you’re saying 👌
6 days - 12 puzzles! Let's go Andy we believe in you!
Even if it goes over into the new year, it’s ok.
We appreciate Andy’s great work! 🙂
@JLvatron very well put!
Note that the problem becomes clear when we see that the right triangle whose side length is 12 and whose hypotenuse is R, the radius of the quarter circle, and whose third side is 2r, the diameter of the small circle, is, according to the Pythagorean theorem, (R)²-(2r)²=12², so (πR²)/4-π(2r)²/4=(12²/4)π, i.e. (πR²)/4-πr²=36π, which is the required area.
Merry Christmas Andy. I have enjoyed following you, you make it look so easy.
How do we know the green circle touches at the midpoint of the blue semicircle’s base
it just happens to be
That’s a good question. I think if you assume the quarter circle and semicircle touch at exactly one point and fit into a square, then they must touch at the center of the semicircle. But it seems like that is something that should be established.
Every shape is symmetrical in this composition. What's your argument against the symmetry?
The green a blue circle could be any size and it would still work, pie is just very nice cause it cancels out with the circle area fotmula
The right and bottom side of the square are two tangent lines of the semicircle that converge at a right angle (i.e. the lower-right corner of the square). If you connect the center of the semicircle to the point of the convergence, you divide the convergence angle equally (45° - 45°), which means this line coincides with the square’s diagonal.
The upper-left corner of the square is the center of the quarter circle, So the square’s diagonal is expected to split this quarter circle equally into two.
Tomorrow looks like the area is π/4*R²-πr², and you can find a triangle with sides of 12 and 2r, and hypothenuse of R, which gives us R²-(2r)²=12, which leads to the answer being 3π
Almost. It's R² - (2r)² = 12², which leads us to 36π.
@Epyxoid oh dang. Shame on me for missing that!
What triangle?
@@danohana1822 The one that consists of the radius of the bigger circle that we draw to the right side of the chord, the chord itself, and the line segment between the left side of the chord and the center of the circle, whose length happens to be exactly the diameter of the smaller circle.
Happy Christmas Andy and everyone here
Next one is easy to solve by constructing an example, but here's the proof:
Radius of 1/4 circle = R, radius of small circle is r. We want ((R^2)/4 - r^2)pi. Connect the center of the 1/4 circle to the other end of the 12 length. Right triangle with legs 2r and 12, hypotenuse R. 4r^2 + 144 = R^2. Divide everything by 4 and move the r^2 term to the right to get 36 = (R^2)/4 - r^2. Since we want ((R^2)/4 - r^2)pi sub 36 in and we get 36pi.
I love how you say "How exciting" 2 times in the video. How exciting!
Green radius is 2 (pir2/4), blue radius is {2} (pir2/2).
The distance from where the green contacts the blue to the edge of the square is 2cos45 =2/{2}={2}
Im assuming the semicircle is symmetrical around the squares axis, which means the squares sides are {2} (green vertical) + {2} (blue vertical) =2{2}
Then, the area is (2{2})2= 8
Tomorrow's is pi/4*r1^2-pir2^2
Making a triangle with the displayed segment, (2r2)^2+(12)^2=r1^2
So the area is pi/4*4r2^2+pi/4*144-pir2^2=36pi
You could also use the area formula for a rhombus! (d1*d2/2) is all you need once you get the diagonal of 4! (Not 24 haha)
ooo love that solution
I started by asking myself if the point of tangency of the quarter circle and the semicircle is at the center of the square (it sure looks like if might be, but then of course, you can’t assume that). Following along with the development in the video shows (freeze at 2:38) that it is in fact the case. Thinking some more about why this is true, I came to the conclusion that it follows from the respective areas being the same.
One can deduce (by equating the areas of a quarter circle with radius s/SQRT(2) and a semicircle with radius s/2) that this common area must be equal to (pi * s^2)/8. So if the common area is given as pi, then s^2 must equal 8, as shown in the video.
Summarizing: if the area of the two shapes is the same (say A), the area of the square will be 8 * A/pi, and the point of tangency will be the center of the square. Conversely, if the point of tangency is at the center of the square, the two shapes must have the same area, given by (pi * s^2)/8.
You could also multiplyed the 4 by 2(one of the triangle's height, the 4 is the base or in other way the half of the other diagonal) and it gives the same:)
I just assumed that the quarter circle crosses the centre, so its radius would be a half-diagonal, and two adjacent half-diagonals form a right triangle with hypotenuse being the side, which comes out to root 8. So the area is 8
You can also take the horizontal projection of r1 : r1x= r1 * cos45 = r1 * sqrt(2)/2. Then simply add r1x+r2 = sqrt(2)+sqrt(2) = 2sqrt(2). So the area is (2sqrt(2))^2 = 8.
i got the answer as 36pi . But it seems very unintuitive and now my brain is not braining . HOW EXCITING.
You didn't explain why the diagonal of the small square is perpendicular on the blue semicircle's diagonal. Happy Holidays!
Half way through the video it was found that r2 was root2. And we know r2 is half the length of the side of the square. So (2root2)^2 is area = 8. Shaves off a some working by not worrying about the diagonal or r1.
How do you know that it's half the length of the square's side?
It is, but that isn't given in the image nor in the text.
You would have to assume that because it wasn’t known until he confirmed the diagonal was 4.
Sorry, I mean straight after when the diagonal of the small square is confirmed as 2, same size as r1. Then you know r2 is half length of the side of the large square.
@@MartB1979 but to get that you had to calculate r1 and r2. You didn't skip anything
@@juanignaciolopeztellechea9401At 2.30 or so he erases the smaller square and calculates the length of the side of the large square by using the newly found long diagonal and pythagoras. But we already know r2 is half length of large square side. That's the only step that can be skipped. It's a very minor point. At 2.30 visually you can see the smaller square is exactly a quarter the size of the large square with the numbers written on the image at that point (the long diagonal is exactly r1 is same length as short diagonal).
Can one just assume that the radius of the green quarter circle and diagonal of smaller square are a straight line? It's easy enough to prove if you extend some lines and fill in angles. Otherwise, nice video. I got the same answer in the same manner.
How do you know that the green circle has its centre on the vertex of the square?
A lot of assumptions are made with this question. The first assumption is that the question is presenting basic shapes, i.e. a quarter and semi-circle (which should answer your question). The second assumption is the circle shapes are placed symmetrically in the square, which means the quarter-circle radius must intersect the semicircle at a perfectly diagonal and symmetrical 45 degree angle.
An immediate observation is that if pi times something equals pi, that something equals 1. But the first question I asked was, is it true that those given squares are equal, so I would have probably started with proving it. (Unfortunately I'm not a mathematician to go for it.)
Love these Andy. Make it dark instead of white background, helps when watching at night.
wow that was... exciting.. havea good christmas.
What kind of whiteboard or presentation tool do you use to make demonstration clear like this?
36pi
took me way too long to figure out how to prove it
area of the quarter circle is piR^2
area of the circle is pir^2
you can make a right angled triangle with hypotenuse being R and the other sides being 12 and 2r
making R^2=12^2+(2r)^2
so the shaded area=(pi(12^2+(2r)^2))/4-pir^2
a=144pi/4+(4pir^2)/4-pir^2
a=36pi+pir^2-pir^2
a=36pi
Could have done it another way:
The width of the square is:
√2 + √2 = 2√2
So the area of the square is:
2√2 x 2√2 = 8 sq.units
This was probably the only one i couldve done by myself so far
I love these Aggvent math puzzles.
Merry Christmas 2024!
You missed the rhombus area equation.
Can’t wait for the next problem
i'm half way through the video and i stopped just to say that i think you only need r2 since it is half the side of the square so the area of the square would be 2*(2*sqrt(2)) which would be 8... did i got it right?
yei i did 😂
Merry christmas andy🫂🧑🏻🎄🧑🏻🎄💫
Needs more triangles
I ALMOST get this one. I was so close T-T
Two Pi’s just in time for Xmas Eve dessert.
Radii crew represent!
Catching up
Day 20: 36pi sq units
For the next day I got:
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36*pi
Hey I was right! Though I absolutely used the wrong(or at least different) equations 😅
I just figured out the radius of the semi circle to be sqrt(2), and kinda assumed that the center of the diameter was the center of the square(as the 2 points whrere the semi circle touches the square look to be in the center, and also drawing lines inward from them look like the exact center and create a 90⁰ angle), and then I dou led it for the length of the square, being 2sqrt(2), and then squared if to get 8
Completely wrong method, but hey, it worked
If you calculate the radius of the green circle, this confirms that the base of the blue semi circle lies on the diagonal since its radius matches the diagonal of the smaller square formed by connecting the three tangent points.
Why do we need the green circle? If r2 = sqrt(2), then the length of one side is sqrt(2)+sqrt(2), giving us an area of (2*sqrt(2))^2 = 8.
You would need something else to define where the top side and left side of the square are located if you don’t have that green quarter-circle.
Don't you need the green quarter circle to establish that the semi circle lies on the square's diagonal?
@@grahamkay4034 Sorry, what I really mean is that there's no need to calculate r1. Alternatively, complete the blue semicircle into a full blue circle. The diameter of that is a side of the square.
day20
(2r)^2+12^2=R^2
→r^2+6^2=R^2/4
→R^2/4-r^2=36→A=36丌😊
option A=丌(12/2)^2=36丌😊
Next problem: 36*pi
Nice
hi
recycle these notes??
...
fourth ig