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If he could be as passionate about fundamentals (idk, maybe you already are) it would be perfect. Geometry and algebra are (imo) where maths gets interesting and exciting, but you need a modest amount of fundamentals before you can really sink your teeth into them, and most kids have already reached 'learned helplessness' before that, alas.
I was supposed to leave my house to go out 10 minutes ago, but instead I just watched this video to ensure I had my math question for the day. Thanks for the lesson!
"This is not a fun one" blew my mind. Didn't expect that. I thought there would be a way to use the diagonal since it runs through centers of the circle. And that radius for red circle could be calculated from pi * r2.
12 out of 15 of those minutes were for the verification of one single angle, but that angle turned out to be the lynchpin for the whole problem. 😅 That was some fantastic work.
I like math more that someone as good, and entertaining, as you can admit when a difficult problem may not be fun. Even though I would never have gotten there myself, I understand each step perfectly as you explain it. 🙌🏻
At 9:50, Andy Math has derived the equation cos(Θ) + 2cos²(Θ) = 1 + sin(Θ) + 2(sin(Θ))(cos(Θ)). We are looking for a difference of 0 between the left and right sides of the equation. Let y = f(Θ) = cos(Θ) + 2cos²(Θ) - (1 + sin(Θ) + 2(sin(Θ))(cos(Θ)) = cos(Θ) + 2cos²(Θ) - 1 - sin(Θ) - 2(sin(Θ))(cos(Θ)) = (cos(Θ))(1 + 2(cos(Θ) -sin(Θ))) - 1 - sin(Θ). (The terms with a common factor cos(Θ) have been grouped.) Then I wrote a python program to compute y over the range of Θ = 1° to 89°: import math for degrees in range (1, 90): theta = math.radians(degrees) # compute cos(theta) and sin(theta) and store result costheta = math.cos(theta) sintheta = math.sin(theta) difference = costheta*(1 + 2*(costheta - sintheta)) - 1 - sintheta print(f" {degrees:02d} {difference} ") The output follows. Note that the difference for 1° is about 2 and decreases monotonically to about -2 at 89°. The minimum is at about 30° and we assume that don't have an exact 0 because we have the residual left behind by roundoff errors. (The e-16 means that the 4.996003610813204 is raised to the 10 to the -16 power; in other words, we have a very small number.) In the video, Andy Math proved, using exact values for sin(30°) and cos(30°), that Θ = 30° is an exact solution. The computer program output implies, but does not prove, that there is only one solution. One thing missed in the video is a proof that there is only one solution. 01 1.9468866190357024 02 1.8922989068322937 03 1.8362870106122497 04 1.7789025442972033 05 1.7201985306893652 06 1.6602293420166656 07 1.5990506389125028 08 1.536719307902825 09 1.4732933974751135 10 1.408832052805517 11 1.3433954492219944 12 1.277044724482847 13 1.2098419099514597 14 1.1418498607493648 15 1.0731321849709863 16 1.003753172044541 17 0.9337777203245934 18 0.8632712640026803 19 0.7922996994232239 20 0.7209293108926789 21 0.6492266960704376 22 0.5772586910305295 23 0.5050922950835128 24 0.43279459544826454 25 0.36043269186351173 26 0.288073621229026 27 0.21578428236634695 28 0.14363136098874157 29 0.07168125496983752 30 4.996003610813204e-16 31 -0.07134680428097784 32 -0.14229406758686847 33 -0.21277728163640364 34 -0.28273259206658097 35 -0.3520968695222939 36 -0.4208077798377319 37 -0.4888038532260752 38 -0.5560245523952653 39 -0.6224103395089126 40 -0.6879027419128388 41 -0.7524444165492401 42 -0.8159792129820839 43 -0.8784522349590269 44 -0.9398099004369407 45 -0.9999999999999998 46 -1.0589717536012504 47 -1.1166758655606215 48 -1.1730645777544628 49 -1.2280917209339006 50 -1.2817127641115769 51 -1.3338848619586985 52 -1.384566900156728 53 -1.4337195386505623 54 -1.4813052527525754 55 -1.5272883720495227 56 -1.5716351170669944 57 -1.614313633648798 58 -1.6552940250114654 59 -1.694548381436876 60 -1.7320508075688772 61 -1.7677774472826895 62 -1.8017065060988244 63 -1.8338182711162414 64 -1.8640951284424698 65 -1.892521578101468 66 -1.9190842464030529 67 -1.9437718957608152 68 -1.9665754319485238 69 -1.9874879087881538 70 -2.0065045302657567 71 -2.0236226500745405 72 -2.0388417685876266 73 -2.0521635272660874 74 -2.0635917005109508 75 -2.0731321849709863 76 -2.0807929863211463 77 -2.0865842035296147 78 -2.0905180106344474 79 -2.0926086360538187 80 -2.092872339456855 81 -2.091327386225008 82 -2.087994019536823 83 -2.0828944301118386 84 -2.076052723652185 85 -2.067494886023226 86 -2.0572487462173346 87 -2.045343937147557 88 -2.0318118543205443 89 -2.016685612440704 === Code Execution Successful ===
This is how I proved the 30° angle (it took me a long time): The diagonal parallelogram in the middle determines everything else. If the radius of the orange circles are r and the short sides of the parallelogram are labeled A, by drawing the tangent lines from the leftmost orange circle center and looking at congruent triangles, it can be shown that the long side of the parallelogram is of length 4r+A. Given this, and the fact that the long sides are spaced 2r apart, it's straightforward to show that the lower left hand angle of the parallelogram is 60°, which proves the angle of the big triangle is 30°. It can also be shown that A=4r/√3 once you know the 60° angle, so the long side is 4r+4r/√3. Then I found the length of the side D of the square in terms of the radius R of the red circles, noting that the line from the center of the red circle to the lower left corner forms a 15° angle, and using the half-angle formula for tangent(30°/2), you get D=(3+√3)R. So we then have cos(30°) = D/[long side of parallelogram], or (√3)/2 = [(3+√3)R]/(4r+4r/√3). With a bunch of arithmetic and simplifying fractions with denominators having terms with √3 in them😊, you get r = (√3)R/2,so each orange circle has an area 75% of the area of a red circle.
This reminds me of a baseball quote. It was Reggie Jackson talking about Nolan Ryan's pitching. He said, "Every hitter likes fastballs, just like everybody likes ice cream, but nobody likes it crammed down their throat a half gallon at a time." Well, I like math, but...... LOL Nice work, but this was a lot.
Brilliant explanation. I have not the energy or the level to understand all those things this night in France. Here are my calculus but i don't explain in details because i don't want to name a lot of points. x : side length of the square R : radius of one of the 2 red circles ; 24/2=12=Pi*R^2 r : radius of one of the 3 orange circles k=angle Theta of Andy's explanation Equation 1 : tan(k/2)=R/(x-R) Equation 2 : x=r*(2+4*cos(k)) (horizontal direction) Equation 3 : r=(x-x*tan(k))*cos(Pi/4-k/2)*sin(Pi/4-k/2) Nota Bene : Equation 3 could have been be replaced by another equivalent equation : x=x*tan(k)+r/tan(Pi/4+k/2)+r/tan(Pi/4-k/2) (vertical direction) Equation 3 : 2*r=(x-x*tan(k))*2*cos(Pi/4-k/2)*sin(Pi/4-k/2) 2*r=(x-x*tan(k))*sin(Pi/2-k) 2*r=x*(1-tan(k))*cos(k) 2*r=x*(cos(k)-sin(k)) then we use Equation 2 to transform Equation 3 : 2*r=r*(2+4*cos(k))*(cos(k)-sin(k)) 1=(1+2*cos(k))*(cos(k)-sin(k)) I don't explain but this equation with one variable "k" can easily be found THEN k=Pi/6 radians = 30 degree tan(k)=1/sqrt(3) ; cos(k)=sqrt(3)/2 ; sin(k)=1/2 Equation 1 : tan(k/2)=tan(15 degree)=2-sqrt(3)=R/(x-R) R=(2-sqrt(3))*(x-R) R*(3-sqrt(3))=(2-sqrt(3))*x x=R*(3-sqrt(3))/(2-sqrt(3)) x=R*(3-sqrt(3))*(2+sqrt(3)) x=R*(3+sqrt(3)) Equation 2 : x=r*(2+4*cos(k)) x=r*(2+4*sqrt(3)/2) x=r*(2+2*sqrt(3)) x=2*r*(1+sqrt(3)) R*(3+sqrt(3))=2*r*(1+sqrt(3)) sqrt(3)*R*(1+sqrt(3))=2*r*(1+sqrt(3)) sqrt(3)*R=2*r 3*R^2=4*r^2 r^2=3/4*R^2 Pi*r^2=3/4*Pi*R^2 Pi*r^2=3/4*12=9 Answer : 3*Pi*r^2=3*9=27 Total orange area = 27 u²
Question: in 12:10, how do you prove the tangent line does indeed connect the tangency point of both circles to the corner of the square? Edit: you dont need to prove it to solve, but he connected the line as if its proved to reach the corner of the screen, so I'd like to know how to prove it
Have to make a few assumptions - assume the orange circles are tangent to each other and are tangent to the diagonal lines. If so, then near the beginning of the video Andy removes the two red circles, rotates the remaining figure 90 degrees, and overlays it in the original, showing how the three circles would fit. This exercise demonstrates that there is a line starting in the upper left vertex. Further, because the image was rotated 90 degrees, the intersection of that line with either of the two original diagonal lines must also be 90 degrees. Now consider the line connecting the centers of the orange circles. That connecting line must be parallel to the two original diagonals. That means the connecting line must also be perpendicular to the diagonal drawn from upper left corner. That is, the radius of each orange circles must be perpendicular to the newly drawn diagonal from upper left. A tangent line is always perpendicular to a radius drawn from the circle center to the point of tangency. Because the radius of the orange circles are perpendicular to this diagonal, and because this diagonal was drawn with the assumption that it bounded (was tangent to) the center circles, this diagonal line MUST be tangent to both orange circles at their point of tangency.
Wow, complex! I found the question confusing because it wasn't clear if both red circles were the same size. Not that I would have been able to solve it had this been stated anyway. Also, at 0:37 your premise that the same circles fit diagonally makes sense, but I don't see how to prove it. If the middle circle was exactly in the centre of the square, then rotating the square 90 degrees would do it; but we don't have this information about it exactly in the centre.
Wow! You made an easy puzzle look hard The angle can only be 30 if the tangents of the orange circles originate in the corners of a square and the vertical tangents of the left and right circles are sides of the square. The symmetry puts the centre of the centre circle in the centre of the square. Now draw another 4 orange circles to form a ring around the centre circle….and the rest is obvious
How do you know for sure that the tangent line between the orange circles properly meets the corner of the square? Your entire solution was based on the idea that that formed a triangle with the side of the square as its hypotenuse, but it's not clear to me that it necessarily is one.
If it doesn't meet the corner then you wouldn't have any particular answer. Imagine a scenario where the slopes of the lines tangent to the red circles was very close to 0. Then the area in between the lines would be squished and you would not have room for any orange circles.
Hey Andy - this was fun! Reminds me of my years tutoring math and how none of my students liked it when I said it was going to be fun. I think you can assume/prove that the center of the middle orange circle is in the center of the square, otherwise you wouldn't be able to rotate to get the image of the 5 orange circles. Then construct two lines from the corner of the square to the center of the middle orange circle and the center of the bottom right orange circle that you added. Also connect the centers of those two circles. This gives you three triangles where the sum of the angles at the corner of the square is 45 degrees and each angle must be equal to the others (because of congruent triangles). Each small angle must be 15 degrees and then the angle we want is then 30 degrees. Does that prove it?
Phew! That was a lot of work to prove 30 degrees. Couldn't you have used something about the fact the two Red Circle hypotenuses (hypotenii? :D) have to be parallel, since the orange circles are tangent to them? I feel like there's some magic formula that somehow would prove that.
I thought you would have just gone with the two triangles being proportionate, and working from there. The exact values of the other two angles in a right triangle would not have mattered.
Ok, so for for the angles thing, imagine an equilateral side length 2 (with the bottom side horizontal) and a line descending from the top point. Imagine it swinging. At what distance between the two bottom corners does the descending line cross the bottom side at 90°? Only at the midpoint (2:1 ratio). If you don't start with an equilateral triangle, you don't get the 2:1 ratio when the descender hits 90°, or you don't get 90° when the ratio is there. I don't know how you'd prove prove all this, but yes, 2:1 ratio and 90° has to be 60° and 30° for the other angles.
Was it not possible to use tangent properties to find theta? More specifically this property: Lengths of tangents to circles are equal if they come from a common point.
So I feel like I know a way to prove it's 30 all around in a different way, but I'm not sure how valid it would be, and I can't really put it in the comments, but if you're interested let me know
To prove that theta is 30 degrees at around 3:00, could you not use trigometric inverses? Since you have a triangle where the side opposite theta (R) is half of the hypotenuse (2R), sine inverse of 1/2 when theta is between 0 and pi over 2 is pi over 6 or 30 degrees.
This only works if the intersection of the long line equals half the length of the square. It's true in this case, but if the angle was anything else the circle wouldn't line up with the corner of the square. The problem is proving that it does.
Dear Andy, @JoyDaBeast asked a good question: in 12:10, how do you prove the tangent line does indeed connect the tangency point of both circles to the corner of the square? It's the same type of geometric difficulties that stopped your first two proof attempts, isn't it?
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now @s4873djfia also replied to @JoyDaBeast original comment. His comment might have some additional insight.
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now in the comments @s4873djfia also replied to @JoyDaBeast similar comment about this question. His comment might have some additional insight.
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now in the comments @s4873djfia also replied to @JoyDaBeast similar comment about this question. His comment might have some additional insight.
Plus I have another solution to this question I wanted to confirm If we look at the figure of 2:08 containing 5 orange circle, we can just draw 4 lines passing through the centers of each surrounding in a way that they are tangent to the middle circle from that we can say that the hypotunouse is 4r and side is 2r which also proves the angle to be 30-60-90 pair
At the part where you evaluated the sin and cos equation at 30 degrees, you also needed to do a quick proof that no other distinct angle would satisfy the equation. Otherwise you could have accidentally evaluated something that would simplify down to like sin(x) = sin(x) which is true for all angles and tells you nothing.
I agree. I had already checked myself to make sure they didn't simplify to the same thing, but I did not mention that in the video. I should have done that or your idea in the video.
Hey here's a puzzle that's not quite as hard. Construct a quadrilateral such that: * the base AB has length 8 * the left side AD is perpendicular to the base and has length 2x * the right side BC is also perpendicular to the base and has length x. (The top CD is slanty, and hint: don't need it.) M is the midpoint of AB. Connect CM and DM. They meet at 45 degrees. What's x?
did you know, there’s an alternative way to solve this with the help of-get this-an image editing program. select the areas of the orange circles, add a solid color layer to fill the areas and use the eyedropper to pick the red from the other circles. now, do the opposite color with the originally red ones. I know you see now where this is going… by looking up the number given in the question, it is proven that the answer is 24.
I feel like everyone would be math geniuses if more teachers were even half as enthusiastic about math as Andy. Truly the GOAT.
Thank you!
@@AndyMath thank you for the amazing content!
If he could be as passionate about fundamentals (idk, maybe you already are) it would be perfect. Geometry and algebra are (imo) where maths gets interesting and exciting, but you need a modest amount of fundamentals before you can really sink your teeth into them, and most kids have already reached 'learned helplessness' before that, alas.
15 minutes of Andy Math? In one video? How exciting!
Probably a great way to cure insomnia!
@@AndyMath actually made it harder to sleep.
If Andy says it was NOT a fun one, you know it wasn't a fun one
I didn’t even try once I heard that. I knew this would be a leaning video.
This 15 minute video from Andy Math is perfect to watch while having lunch. Thank you and how exciting.
Thank you, I was worried it was too long, but once commited to the problem, I had to finish.
I usually watch them during my breakfast, everyone have their own habits 😁
I was supposed to leave my house to go out 10 minutes ago, but instead I just watched this video to ensure I had my math question for the day. Thanks for the lesson!
This was intense! The proof that the angle was 30° blew my mind
"This is not a fun one" blew my mind. Didn't expect that. I thought there would be a way to use the diagonal since it runs through centers of the circle. And that radius for red circle could be calculated from pi * r2.
12 out of 15 of those minutes were for the verification of one single angle, but that angle turned out to be the lynchpin for the whole problem. 😅 That was some fantastic work.
I like math more that someone as good, and entertaining, as you can admit when a difficult problem may not be fun.
Even though I would never have gotten there myself, I understand each step perfectly as you explain it. 🙌🏻
If bro said "It's not a fun one" then I'm finished 💀🙏
He couldnt even pug a box around it to make it exciting
Always a weird feeling when you know something is probably true but can't directly prove it.
At 9:50, Andy Math has derived the equation cos(Θ) + 2cos²(Θ) = 1 + sin(Θ) + 2(sin(Θ))(cos(Θ)). We are looking for a difference of 0 between the left and right sides of the equation. Let y = f(Θ) = cos(Θ) + 2cos²(Θ) - (1 + sin(Θ) + 2(sin(Θ))(cos(Θ)) = cos(Θ) + 2cos²(Θ) - 1 - sin(Θ) - 2(sin(Θ))(cos(Θ)) = (cos(Θ))(1 + 2(cos(Θ) -sin(Θ))) - 1 - sin(Θ). (The terms with a common factor cos(Θ) have been grouped.) Then I wrote a python program to compute y over the range of Θ = 1° to 89°:
import math
for degrees in range (1, 90):
theta = math.radians(degrees)
# compute cos(theta) and sin(theta) and store result
costheta = math.cos(theta)
sintheta = math.sin(theta)
difference = costheta*(1 + 2*(costheta - sintheta)) - 1 - sintheta
print(f" {degrees:02d} {difference} ")
The output follows. Note that the difference for 1° is about 2 and decreases monotonically to about -2 at 89°. The minimum is at about 30° and we assume that don't have an exact 0 because we have the residual left behind by roundoff errors. (The e-16 means that the 4.996003610813204 is raised to the 10 to the -16 power; in other words, we have a very small number.) In the video, Andy Math proved, using exact values for sin(30°) and cos(30°), that Θ = 30° is an exact solution.
The computer program output implies, but does not prove, that there is only one solution. One thing missed in the video is a proof that there is only one solution.
01 1.9468866190357024
02 1.8922989068322937
03 1.8362870106122497
04 1.7789025442972033
05 1.7201985306893652
06 1.6602293420166656
07 1.5990506389125028
08 1.536719307902825
09 1.4732933974751135
10 1.408832052805517
11 1.3433954492219944
12 1.277044724482847
13 1.2098419099514597
14 1.1418498607493648
15 1.0731321849709863
16 1.003753172044541
17 0.9337777203245934
18 0.8632712640026803
19 0.7922996994232239
20 0.7209293108926789
21 0.6492266960704376
22 0.5772586910305295
23 0.5050922950835128
24 0.43279459544826454
25 0.36043269186351173
26 0.288073621229026
27 0.21578428236634695
28 0.14363136098874157
29 0.07168125496983752
30 4.996003610813204e-16
31 -0.07134680428097784
32 -0.14229406758686847
33 -0.21277728163640364
34 -0.28273259206658097
35 -0.3520968695222939
36 -0.4208077798377319
37 -0.4888038532260752
38 -0.5560245523952653
39 -0.6224103395089126
40 -0.6879027419128388
41 -0.7524444165492401
42 -0.8159792129820839
43 -0.8784522349590269
44 -0.9398099004369407
45 -0.9999999999999998
46 -1.0589717536012504
47 -1.1166758655606215
48 -1.1730645777544628
49 -1.2280917209339006
50 -1.2817127641115769
51 -1.3338848619586985
52 -1.384566900156728
53 -1.4337195386505623
54 -1.4813052527525754
55 -1.5272883720495227
56 -1.5716351170669944
57 -1.614313633648798
58 -1.6552940250114654
59 -1.694548381436876
60 -1.7320508075688772
61 -1.7677774472826895
62 -1.8017065060988244
63 -1.8338182711162414
64 -1.8640951284424698
65 -1.892521578101468
66 -1.9190842464030529
67 -1.9437718957608152
68 -1.9665754319485238
69 -1.9874879087881538
70 -2.0065045302657567
71 -2.0236226500745405
72 -2.0388417685876266
73 -2.0521635272660874
74 -2.0635917005109508
75 -2.0731321849709863
76 -2.0807929863211463
77 -2.0865842035296147
78 -2.0905180106344474
79 -2.0926086360538187
80 -2.092872339456855
81 -2.091327386225008
82 -2.087994019536823
83 -2.0828944301118386
84 -2.076052723652185
85 -2.067494886023226
86 -2.0572487462173346
87 -2.045343937147557
88 -2.0318118543205443
89 -2.016685612440704
=== Code Execution Successful ===
How interesting!
-"Is that some kind of pirate movie you are watching there?!"
-"Nope! There are just many, many radiuses.."
This is how I proved the 30° angle (it took me a long time):
The diagonal parallelogram in the middle determines everything else. If the radius of the orange circles are r and the short sides of the parallelogram are labeled A, by drawing the tangent lines from the leftmost orange circle center and looking at congruent triangles, it can be shown that the long side of the parallelogram is of length 4r+A. Given this, and the fact that the long sides are spaced 2r apart, it's straightforward to show that the lower left hand angle of the parallelogram is 60°, which proves the angle of the big triangle is 30°. It can also be shown that A=4r/√3 once you know the 60° angle, so the long side is 4r+4r/√3.
Then I found the length of the side D of the square in terms of the radius R of the red circles, noting that the line from the center of the red circle to the lower left corner forms a 15° angle, and using the half-angle formula for tangent(30°/2), you get D=(3+√3)R. So we then have cos(30°) = D/[long side of parallelogram], or (√3)/2 = [(3+√3)R]/(4r+4r/√3). With a bunch of arithmetic and simplifying fractions with denominators having terms with √3 in them😊, you get r = (√3)R/2,so each orange circle has an area 75% of the area of a red circle.
This reminds me of a baseball quote. It was Reggie Jackson talking about Nolan Ryan's pitching. He said, "Every hitter likes fastballs, just like everybody likes ice cream, but nobody likes it crammed down their throat a half gallon at a time." Well, I like math, but...... LOL Nice work, but this was a lot.
You are a great teacher how exciting!
When I saw the thumbnails I was like there ain’t no way I know how to solve this 😂
Very fun video
Brilliant explanation.
I have not the energy or the level to understand all those things this night in France.
Here are my calculus but i don't explain in details because i don't want to name a lot of points.
x : side length of the square
R : radius of one of the 2 red circles ; 24/2=12=Pi*R^2
r : radius of one of the 3 orange circles
k=angle Theta of Andy's explanation
Equation 1 : tan(k/2)=R/(x-R)
Equation 2 : x=r*(2+4*cos(k)) (horizontal direction)
Equation 3 : r=(x-x*tan(k))*cos(Pi/4-k/2)*sin(Pi/4-k/2)
Nota Bene : Equation 3 could have been be replaced by another equivalent equation : x=x*tan(k)+r/tan(Pi/4+k/2)+r/tan(Pi/4-k/2)
(vertical direction)
Equation 3 :
2*r=(x-x*tan(k))*2*cos(Pi/4-k/2)*sin(Pi/4-k/2)
2*r=(x-x*tan(k))*sin(Pi/2-k)
2*r=x*(1-tan(k))*cos(k)
2*r=x*(cos(k)-sin(k)) then we use Equation 2 to transform Equation 3 :
2*r=r*(2+4*cos(k))*(cos(k)-sin(k))
1=(1+2*cos(k))*(cos(k)-sin(k))
I don't explain but this equation with one variable "k" can easily be found THEN k=Pi/6 radians = 30 degree
tan(k)=1/sqrt(3) ; cos(k)=sqrt(3)/2 ; sin(k)=1/2
Equation 1 : tan(k/2)=tan(15 degree)=2-sqrt(3)=R/(x-R)
R=(2-sqrt(3))*(x-R)
R*(3-sqrt(3))=(2-sqrt(3))*x
x=R*(3-sqrt(3))/(2-sqrt(3))
x=R*(3-sqrt(3))*(2+sqrt(3))
x=R*(3+sqrt(3))
Equation 2 :
x=r*(2+4*cos(k))
x=r*(2+4*sqrt(3)/2)
x=r*(2+2*sqrt(3))
x=2*r*(1+sqrt(3))
R*(3+sqrt(3))=2*r*(1+sqrt(3))
sqrt(3)*R*(1+sqrt(3))=2*r*(1+sqrt(3))
sqrt(3)*R=2*r
3*R^2=4*r^2
r^2=3/4*R^2
Pi*r^2=3/4*Pi*R^2
Pi*r^2=3/4*12=9
Answer : 3*Pi*r^2=3*9=27
Total orange area = 27 u²
1=(1+2*cos(k))*(cos(k)-sin(k))
Let's call y=cos(k) THEN sin(k)=sqrt(1-y^2)
1=(1+2*y)*(y-sqrt(1-y^2))
1/(1+2*y)=y-sqrt(1-y^2)
sqrt(1-y^2)=y-1/(1+2*y)
1-y^2=(y-1/(1+2*y))^2
1-y^2=y^2+1/(1+2*y)^2-2*y/(1+2*y)
2*y^2-1+1/(1+2*y)^2-2*y/(1+2*y)=0
(2*y^2-1)*(2*y+1)^2-2*y*(2*y+1)+1=0
(2*y^2-1)*(4*y^2+4*y+1)-4*y^2-2*y+1=0
(8*y^4+8*y^3-2*y^2-4*y-1)-4*y^2-2*y+1=0
8*y^4+8*y^3-6*y^2-6*y=0
2*y*(4*y^3+4*y^2-3*y-3)=0
2*y*(y+1)*(4*y^2-3)=0 ; y=cos(k) and 0 < k < 45 degree
y=cos(k)=0 impossible OR y=cos(k)=-1 impossible OR 4*y^2-3=0
4*y^2-3=0
y^2=3/4
y=cos(k)=-sqrt(3)/2 impossible OR :
answer : y=cos(k)=sqrt(3)/2 then k = 30 degree = Pi/6 radians
Question: in 12:10, how do you prove the tangent line does indeed connect the tangency point of both circles to the corner of the square? Edit: you dont need to prove it to solve, but he connected the line as if its proved to reach the corner of the screen, so I'd like to know how to prove it
Have to make a few assumptions - assume the orange circles are tangent to each other and are tangent to the diagonal lines. If so, then near the beginning of the video Andy removes the two red circles, rotates the remaining figure 90 degrees, and overlays it in the original, showing how the three circles would fit. This exercise demonstrates that there is a line starting in the upper left vertex. Further, because the image was rotated 90 degrees, the intersection of that line with either of the two original diagonal lines must also be 90 degrees.
Now consider the line connecting the centers of the orange circles. That connecting line must be parallel to the two original diagonals. That means the connecting line must also be perpendicular to the diagonal drawn from upper left corner. That is, the radius of each orange circles must be perpendicular to the newly drawn diagonal from upper left.
A tangent line is always perpendicular to a radius drawn from the circle center to the point of tangency. Because the radius of the orange circles are perpendicular to this diagonal, and because this diagonal was drawn with the assumption that it bounded (was tangent to) the center circles, this diagonal line MUST be tangent to both orange circles at their point of tangency.
I absolutely hated doing these loooong geometry problems in school but man, Andy makes it entertaining to me 30+ years later!
Thank you Andy, very well done video. I have a maths degree and love your videos and how you explain things 😊
This one was the best of this month, cheers!
Thanks Andy I wait for your posts
Thank you for the journey 😊
Someone should make Andy a math teacher
8:21
Is actually is a teacher
Officially my favorite math teacher.
How exciting.
This problem was insane
"This looks like a fun one. It's not a fun one." Giggled quite a bit at that
Waiting for that 1m sub count...
Cut out circles. Weigh reds. Weigh oranges. Plug the data in this formula: Orange area=orange weight/(red weight/24)
Wow, complex!
I found the question confusing because it wasn't clear if both red circles were the same size.
Not that I would have been able to solve it had this been stated anyway.
Also, at 0:37 your premise that the same circles fit diagonally makes sense, but I don't see how to prove it. If the middle circle was exactly in the centre of the square, then rotating the square 90 degrees would do it; but we don't have this information about it exactly in the centre.
I don't know what's going on but it's 🔥🔥🔥
Wow! You made an easy puzzle look hard
The angle can only be 30 if the tangents of the orange circles originate in the corners of a square and the vertical tangents of the left and right circles are sides of the square.
The symmetry puts the centre of the centre circle in the centre of the square.
Now draw another 4 orange circles to form a ring around the centre circle….and the rest is obvious
How do you know for sure that the tangent line between the orange circles properly meets the corner of the square? Your entire solution was based on the idea that that formed a triangle with the side of the square as its hypotenuse, but it's not clear to me that it necessarily is one.
If it doesn't meet the corner then you wouldn't have any particular answer. Imagine a scenario where the slopes of the lines tangent to the red circles was very close to 0. Then the area in between the lines would be squished and you would not have room for any orange circles.
Hey Andy - this was fun! Reminds me of my years tutoring math and how none of my students liked it when I said it was going to be fun.
I think you can assume/prove that the center of the middle orange circle is in the center of the square, otherwise you wouldn't be able to rotate to get the image of the 5 orange circles. Then construct two lines from the corner of the square to the center of the middle orange circle and the center of the bottom right orange circle that you added. Also connect the centers of those two circles. This gives you three triangles where the sum of the angles at the corner of the square is 45 degrees and each angle must be equal to the others (because of congruent triangles). Each small angle must be 15 degrees and then the angle we want is then 30 degrees. Does that prove it?
"That sometimes clear . . . and sometimes vague stuff . . . which is . . . mathematics."
Imre Lakatos, 1922-1974
Phew! That was a lot of work to prove 30 degrees. Couldn't you have used something about the fact the two Red Circle hypotenuses (hypotenii? :D) have to be parallel, since the orange circles are tangent to them? I feel like there's some magic formula that somehow would prove that.
"How exciting"🗣🗣🔥🔥💯💯🔛🔝
Que excitante! 😀 Thanks for sharing.
How exciting! 😊 Obrigado por compartilhar.
I thought you would have just gone with the two triangles being proportionate, and working from there. The exact values of the other two angles in a right triangle would not have mattered.
I think we needed the angles to get the ratio of the side lengths.
Would it even have mattered if the angle were different than 30? The second part would still have worked.
Holy crap! Nice solve!!!
you are great man
Ok, so for for the angles thing, imagine an equilateral side length 2 (with the bottom side horizontal) and a line descending from the top point. Imagine it swinging. At what distance between the two bottom corners does the descending line cross the bottom side at 90°? Only at the midpoint (2:1 ratio).
If you don't start with an equilateral triangle, you don't get the 2:1 ratio when the descender hits 90°, or you don't get 90° when the ratio is there. I don't know how you'd prove prove all this, but yes, 2:1 ratio and 90° has to be 60° and 30° for the other angles.
At 12:10, I'm not sure how you know that the tangent definitely intersects the corner of the square. Am I overlooking something obvious?
Today , Iam certained that I'm Stupid
Something tells me you could have found the ratio of radii, but i have no idea how.
Was it not possible to use tangent properties to find theta?
More specifically this property:
Lengths of tangents to circles are equal if they come from a common point.
This video seems important. I put a box around it.
0:56 My poisioned mind got a little worried there😅
??
i wonder how can you acually solve for θ
Every problem can be solved with the 30-60-90 right triangle.
At 2:57 how do you know that the 2 circles will not overlap
Was it difficult? I couldn't solve it, and I felt a bit disappointed in myself.
How can be square of the ratio of sides is equal to ratio of inscribed circle area ??
Hi!!
So I feel like I know a way to prove it's 30 all around in a different way, but I'm not sure how valid it would be, and I can't really put it in the comments, but if you're interested let me know
so early it feels weird
To prove that theta is 30 degrees at around 3:00, could you not use trigometric inverses? Since you have a triangle where the side opposite theta (R) is half of the hypotenuse (2R), sine inverse of 1/2 when theta is between 0 and pi over 2 is pi over 6 or 30 degrees.
This only works if the intersection of the long line equals half the length of the square. It's true in this case, but if the angle was anything else the circle wouldn't line up with the corner of the square.
The problem is proving that it does.
Dear Andy, @JoyDaBeast asked a good question: in 12:10, how do you prove the tangent line does indeed connect the tangency point of both circles to the corner of the square? It's the same type of geometric difficulties that stopped your first two proof attempts, isn't it?
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now @s4873djfia also replied to @JoyDaBeast original comment. His comment might have some additional insight.
Phew!
Any solution using inversion?
12:07 how did you figure the tangent between the first two orange circles would reach the exact corner?
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now in the comments @s4873djfia also replied to @JoyDaBeast similar comment about this question. His comment might have some additional insight.
Whew!
12:10 how do we know that the tangent passes through the vertex of the square?
I agree, I could have dug deeper on that. I'll try to type it out. It is the same logic as 0:50. if the 3 circles fit in the square in the given orientation, (which does intersect the corner) they should also fit the same way when rotated (and intersect the other corner). The side of the triangle is a portion of the rotated 3 circles/2 lines. I see now in the comments @s4873djfia also replied to @JoyDaBeast similar comment about this question. His comment might have some additional insight.
Coordinate geometry makes it easy
I want to try the problem out... can you give some more necessary facts? Are the sides of the square bisected? Thanks!
The sides are not bisected, the only things you know are that the outer two of three circles are tangent to the sides of the square
Physicists will just get a ruler out and measure it.
0:56 you technically could have solved it right here by seeing the right angle
Is it possible for such question always?
Plus I have another solution to this question I wanted to confirm
If we look at the figure of 2:08 containing 5 orange circle, we can just draw 4 lines passing through the centers of each surrounding in a way that they are tangent to the middle circle from that we can say that the hypotunouse is 4r and side is 2r which also proves the angle to be 30-60-90 pair
Liddle r, not lit-tle r?
At the part where you evaluated the sin and cos equation at 30 degrees, you also needed to do a quick proof that no other distinct angle would satisfy the equation. Otherwise you could have accidentally evaluated something that would simplify down to like sin(x) = sin(x) which is true for all angles and tells you nothing.
I agree. I had already checked myself to make sure they didn't simplify to the same thing, but I did not mention that in the video. I should have done that or your idea in the video.
Ngl I just assumed they were 30-60-90 triangles and did most of it in my head
This mf is practicing be spiderman, he´ll defenetly could calculate the route balancing
👏👏👏👏👏(square)
Hey here's a puzzle that's not quite as hard.
Construct a quadrilateral such that:
* the base AB has length 8
* the left side AD is perpendicular to the base and has length 2x
* the right side BC is also perpendicular to the base and has length x.
(The top CD is slanty, and hint: don't need it.)
M is the midpoint of AB. Connect CM and DM. They meet at 45 degrees.
What's x?
08:40 isn't (2r / (cos / (2r)) = (4r^2 / cos)?
the most bottom line is a large line. he's dividing the nominator
"Hey guys this looks like a fun one!"😊😊😊
"It is NOT a fun one"😡😡😡
🔴🟠🟠🟠🔴
2:30 yes all remaining 3 corners of the inner square makes angle of 30 degrees. i dont know why do u think about proof bcz its visible
4 min to find the area, 11 min just to prove that dang angle was 30 degrees
Math doesn't fuck around lol
Hey guys it looks like a fun one 😊
But it isnt a fun one 😧
daha iyi bi 15 dakika geçiremezdim
ne kadar da, heyecan verici !
@@clexist mskdmw aynen how exciting!
@@im_zinc kişilik tipin ne
@@clexist hiç hatirlamiyorum ya pek ilgili degilim
@@clexist neden sordun
Second
did you know, there’s an alternative way to solve this with the help of-get this-an image editing program. select the areas of the orange circles, add a solid color layer to fill the areas and use the eyedropper to pick the red from the other circles. now, do the opposite color with the originally red ones. I know you see now where this is going… by looking up the number given in the question, it is proven that the answer is 24.
You know things are bad when he says it isn't a fun one.
I lost it at 5 minutes. It's not for me, no
x²=x^x
x=?
Pin my message bro its my birthday and its september