@@okamiexe1501oh my. This takes me back to my high school days when such half a point made me not qualify for the national math contest. Thank you for making me feel that memory all over agian
Presh Talwalker always makes it a point in his videos to highlight that if two circles are tangent, their radii are colinear and touch at the point of tangency. Simply said, if two circles touch eachother, there is a straight line that goes from both midpoints of the circles through the point where the circles touch.
Yep, that's where the solution was hidden. Bringing those radius's together, then using pythagoras to get us an equation which very easily reduced to r = √2. And then plugging that in the simple equation for the blue area which easily simplified down to π*3. Lets put that in a box. [π*3] How exciting.
It’s almost like a puzzle, I like to think of how to solve them without actually solving them. In this example I figured out that once I had that right triangle I’d be able to solve for r and get the solution. The problem is, after that, I don’t know if I’d have done the actual algebra correctly if I had a pen and paper in front of me.
at 2:14 you already have the value for r1^2. you could have just plugged it in to 3pi(r1)^2/2. No need to take the extra step of square rooting it, then plugging it back in only to square it again. Great vid BTW
I thought the same thing. But then considered, where the two circles are just touching each other, there can only be one gradient for the line of tangency, for both circles at that point. i.e the tangent lines must be equal. Try to imagine the tangent gradients somehow being different and you can see why that doesn’t work. If they share the same tangent line, then the radii have to both be perpendicular to the (same) tangent line, so the radii must be 180° (i.e. a straight line)
My logic is, assume for the sake of contradiction, that the intersection does not lie on the hypotenuse. This implies there is some path from the center of both circles which is shorter, as the shortest path will be a straight line. For this to be true, the two circles would have to overlap in some way. This violates the scenario.
Did anyone else use Trigonometry? I seen the right angle triangle of 4(adjacent), r(opposite) and 3r(hypotenuse)... So sinA=r/3r SinA=1/3 (r cancelled top and bottom) A=sin^-1(1/3) TanA=r/4 So, r=4TanA r = √2 and then continue on with Area formula for the half and quarter circles.... Love the puzzles and seeing how we differ in solution.
To save a step or two, you don't need to know the value of r{1}, you need to know the value of (r{1})^2, and you figure that out when you solve the triangle.
Theres a little where both circles are the closest to each other indicating that theyre touhcing, therefore you can draw a line through it connecting to the corner of the square
A radius will always be perpendicular to a tangent line. Since the semi circle and quarter circle are assumed to be touching at only a single point they will have the same tangent line at that point. Since each radius is perpendicular to the same line they must have the same slope and therefore make one continuous line
think of the motion of the tip of a minute hand of a clock over an hour, then simplify that back to 2D (you should be imagining a perfect circle). you know for a fact this circle is defined by the length of the minute hand (radius) you imagined and at all points on the curved line the minute hand remained constant in length. now put another "clock circle" directly beside it, but with a smaller radius and circumference. in this analogy you can imagine that if the 2 circles should overlap, then at some point in time the hands will collide with each other, and would need to be spread further apart in order to keep proper time. this example has the circles TOUCHING so imagine perfectly positioning the clock so the hands form a perfectly straight line. the proof comes from the acknowledgement that the radius is a fixed property of any natural circle.
Good problem man. An important step is missing tho: prove that the two radii form a straight line (the centers and intersection point are collinear). One way to do it: remember that radii are perpendicular to the tangent lines, and since both share the same tangent (because the circles are tangent), this implies that both radii form a straight line.
I'm not smart enough to confirm whether those answers are correct, I'm also not in the mood to pay very close attention to it all, but i like watching it, cause i see small things that'll probably help me someday, like thinking about making a triangle with that radius
Neither am I but I genuinely beleive thw answers arnt correct hes pulling out nonsense from nowhere I like watching his videos and id like to learn but it just seem true the answers
How do you know the hyphotenuse of the triangle will intersect (touch) the center of the two circles? In other word, how do you know R1 and R2 can be in straight line to form a hyphotenuse for the triangle? Thank you.
Circles that are tangent to one another have colinear radii on the point of tangency. There's several proofs for this. The shortest distance from a point to a line is a perpendicular line. This is why lines tangent to a circle are always perpendicular to the radius; if there was a shorter distance, the line would intersect the circle. At the point of tangency of two circles, the perpendicular lines match, so they're colinear.
What a great subscription your channel is! I love these fun little easy problems, but I'd never make time for them without your new videos in my feed to remind me. The ones where your solution is quite different than mine are a special treat.
Insert filler text claiming spoilers Let the radius of the semicircle be x. The radius of the quartercircle is twice the semicircle, so its radius is 2x. Drawing a line connecting the centers of the circles, we can use the Pythagoras Theorem to derive the following equation: 4² + x² = (x + 2x)² 16 + x² = 9x² 8x² = 16 x² = 2 pi(r)² ½pi(x²) + ¼pi(2x)² = ½pi(x²) + ¼pi*4x² = ½pi(x²) + pix² = 1½pix² = 1½pi(2) = 3pi ≈9.4247
1:35 - how do we know that these two radii create a straight line to form a triangle? It looks like you just assumed it but is there a rule that explains/confirms that? How do we prove that?
How do we know for sure that the r1 + r2 hypotenuse is actually linear/the angle between the connected radii is 180 degrees? Is there some mathematical principle that if two circles are tangent to each other, the line between their centers will necessarily run through the point of tangency?
Yes. All tanget lines of a circle are perpendicular to the radius at that tanget. If two circles share the same tanget line, their radii are both perpendicular to that line and thus they must be 180 degrees.
Draw two circles of any size each, put a straight line from the center of one to the center of the other and make the circles' borders touch. You won't find any way to do it without making the line and the point of touching meet.
Bonus points to do the problem with calling the quarter circle radius r and the semicircle radius r/2. The solution is the same but the work is harder.
I look at these problems, but then I do not know how to solve them, yet the solution, if you follow rules of reason, appears. Interesting ideas about 'problems' in general. When will we follow the 'rules' for reasoning to solve our problems? And the answers are not obvious. I love the Andy Math channel.
Given that we start with the rectangle having a side of 4 cm, I would say the correct answer is in cm as the units are known. I'm not sure if this would just be nit picking on my part, but it just feels more correct.
Yeh, I did that in my head but did the triangle thing first, *then* popped it into the areas. Was "worried" at first that maybe I did the triangle thing for nothing, but that came later in the vid. Fun one!
I havent watched the full video yet but I got 3pi. I first drew a line through the centres of both circles. That would form a right triangle with radius of smaller circle (x) and the base 4. Then you can form an equation like 2xy + y² = 16. And we can clearly see that the right side of the radius of the larger circle (y) and the left side is the diameter of the smaller circles (2x). So y = 2x. Then we solve them simultaneously and find the values of x and y and use them to find the area which is 3 pi
A cool problem would be to take the Ten Penny Puzzle solution (packing of 10 circles in a square) and solve for the length of the sides of the square. Using the relationship of circles to determine the size of a square feels right up your alley.
May I ask why r1+2r1 must be a strangle line? I.e. the additional line linked centre of the r1 semi-circle and the right bottom corner of the rectangle may not cut the intercept point of the two circles. Thanks.
did this whole thing in my head in a few minutes... 3Pi (x/2)^2 + 16 = [ (x/2) + x ]^2 --> x = 2sqrt(2) blue area --> area of a circle is Pi x radius ^2 --- radius of semicircle is 2x the radius of the quarter circle do the math and you get 3Pi
So... This might be in the assumption that the edges of the rectangle is the diameter of the smaller circle and the radius of the bigger circle... But what if it isn't? Like, if the half circles are not entirely half and quarter?
Why do you assume that r1+r2 makes a triangle? What is the proof for that? It seems very convenient but how can you actually prove that the connection between large and small circles lays between the centers of both circles?
At any point on a circle, there can be only one tangent at that point and it will be perpendicular to the radius. Both circles meet at one point, so both circles have their respective tangents on the same point, and both the tangents will be in the same line, then obviously the two radii will also be in a straight line.
1:16 wait, how do the 4s just magically turn into 2s? is it just because they cancel each other out so you could make them any old number and it doesn't matter?
How can you assume that is a quarter or semi circle? If you added a single pixel in illustrator there is no way to tell. It is just given in the description of the problem?
So here's what's I don't get it: There is an assumption the line going frontman bottom right corner, through the point of contact of the curves, extends to the midpoint of the left side. I don't see how that was proven, and that's important. BTW, the units were defined (cm), so the area would be 3pi cm 2
Although the correct answer is 2, you should have demonstrated that the segments are aligned with each other, as we are uncertain about where the arches intersect. But very fun and interesting content 🎉🎉🎉
people like you are great and awesome ! youtubers should follow your example for you have done right by seeking to provide good and educational content for students ! you and another good man like yourself known as "presh talwalker" are what a youtuber should aspire to be !
It's the only way to solve. The width of the rectangle is entirely dependent on the two circles touching. Without it, the 4cm that we were given wouldn't do anything.
Maybe this solution would fly in math, but not in engineering. Unit squared? Only unit I've seen is cm. Your u^2 can be any imaginary thing, unrelated to this problem.
I was working on it and got to the point where I'm not sure whether it's possible or not to draw the (r1 + r2) straight line...can you please explain how is it possible...😁
Part of the initial assumption is that the two blue areas are tangent to each other at the one point. This means that r1 and 2*r1 do not overlap. The other fun is that since the two shapes are tangent, that means the two segments connecting them will form a 180 degree angle, which is a straight line.+
the area of a circle with radius r = pi*r^2. the shape on the left is half a circle, so to get its area, we use the formula for area of a circle, but then half it. same logic applies to the right shape, which is a quarter of a circle, meaning we divide it by four.
I'm not bothered enough to know the number I just like to know the theory behind finding out.. The whole "reducing down to the exact number" I would just let the computer do.. I just need to know what numbers to plug in.
All tangent lines of a circle are perpendicular to the radius at that tangent. If two circles share the same tangent line, their radii are both perpendicular to that line and thus they must be 180 degrees.
How do you know that on 2:07 r1 + 2r1 line is straight line and not the curved one? Guess it's required to solve this task but how to prove this assumption.
Think of it this way: if it was not a straight line, ie it was bent, that would mean that you could straighten it and have it be longer. However, the start points of each line are fixed, so the point where the two arcs touch must be when the line is the longest it can be, ie when it's straight
I’m going to take off half a point from this question on the test because the unit is centimeters was included and you didn’t specify in your answer
Yes. I deserve that half point taken off!
@@AndyMath nooo you dont 😂
I just had some major secondary school flashbacks
@@okamiexe1501oh my. This takes me back to my high school days when such half a point made me not qualify for the national math contest. Thank you for making me feel that memory all over agian
@@khaelkuglerOh, aren't you the one who plays 3×10^8 ms^-1?
The rotating of r1 and r2 to form the hypotenuse blew my mind. I gasp aloud.
Lol. It's funny, I did that but I made the side x but messed myself up on the calculus
Presh Talwalker always makes it a point in his videos to highlight that if two circles are tangent, their radii are colinear and touch at the point of tangency.
Simply said, if two circles touch eachother, there is a straight line that goes from both midpoints of the circles through the point where the circles touch.
It was groundbreaking.I could almost hear the IMAX music when I watched it.
The animations are great.
Yep, that's where the solution was hidden. Bringing those radius's together, then using pythagoras to get us an equation which very easily reduced to r = √2. And then plugging that in the simple equation for the blue area which easily simplified down to π*3. Lets put that in a box. [π*3] How exciting.
yea same. I tried it before and gave up but when he rotated r1 and r2 to form the hypotenuse i literally fell off my chair.
Unit of rectangle side: cm
Andy: We don't use this in the US.
Man, i randomly found your channel and now i'm addicted to it... It's remind me the fun of math that i forget decades ago... Massive thanks...
🎉
It’s almost like a puzzle, I like to think of how to solve them without actually solving them. In this example I figured out that once I had that right triangle I’d be able to solve for r and get the solution. The problem is, after that, I don’t know if I’d have done the actual algebra correctly if I had a pen and paper in front of me.
I love this channel. There is zero fat on these videos. You just shred through these math problems. Keep 'em coming!
at 2:14 you already have the value for r1^2. you could have just plugged it in to 3pi(r1)^2/2. No need to take the extra step of square rooting it, then plugging it back in only to square it again. Great vid BTW
I didn't think connecting the two radiuses to form the triangle was an option because I couldn't convince myself that they would perfectly align
I thought the same thing. But then considered, where the two circles are just touching each other, there can only be one gradient for the line of tangency, for both circles at that point. i.e the tangent lines must be equal. Try to imagine the tangent gradients somehow being different and you can see why that doesn’t work. If they share the same tangent line, then the radii have to both be perpendicular to the (same) tangent line, so the radii must be 180° (i.e. a straight line)
This always happens when two curves (eg. circles) are touching at a single point
@@howmanybeansmakefive thank you man.
My logic is, assume for the sake of contradiction, that the intersection does not lie on the hypotenuse. This implies there is some path from the center of both circles which is shorter, as the shortest path will be a straight line. For this to be true, the two circles would have to overlap in some way. This violates the scenario.
Did anyone else use Trigonometry? I seen the right angle triangle of 4(adjacent), r(opposite) and 3r(hypotenuse)...
So sinA=r/3r
SinA=1/3 (r cancelled top and bottom)
A=sin^-1(1/3)
TanA=r/4
So, r=4TanA
r = √2 and then continue on with Area formula for the half and quarter circles.... Love the puzzles and seeing how we differ in solution.
Wish i had these animations and explanations 40yrs ago! 😊
To save a step or two, you don't need to know the value of r{1}, you need to know the value of (r{1})^2, and you figure that out when you solve the triangle.
Well done, but I believe the unit is cm...
Cm^2 !!
cm²
@@sammafo7131 Andy already wrote 3pi unit^2, so the unit is cm.
You bring so many interesting questions. Nice contents!
Why is the unit of the answer is u^2? Should it be cm^2 because the picture already shows 4cm?
It should be, he made a mistake
Im curious how one can find that r1 and r2 can make that triangle? is there a proof of it somewhere?
You can draw a line between the radius points of two circles that are touching and it’s going to be equal to r1+r2
Theres a little where both circles are the closest to each other indicating that theyre touhcing, therefore you can draw a line through it connecting to the corner of the square
A radius will always be perpendicular to a tangent line. Since the semi circle and quarter circle are assumed to be touching at only a single point they will have the same tangent line at that point. Since each radius is perpendicular to the same line they must have the same slope and therefore make one continuous line
think of the motion of the tip of a minute hand of a clock over an hour, then simplify that back to 2D (you should be imagining a perfect circle). you know for a fact this circle is defined by the length of the minute hand (radius) you imagined and at all points on the curved line the minute hand remained constant in length. now put another "clock circle" directly beside it, but with a smaller radius and circumference. in this analogy you can imagine that if the 2 circles should overlap, then at some point in time the hands will collide with each other, and would need to be spread further apart in order to keep proper time. this example has the circles TOUCHING so imagine perfectly positioning the clock so the hands form a perfectly straight line.
the proof comes from the acknowledgement that the radius is a fixed property of any natural circle.
@@eeple29 This is a great way to explain it. Thank you!
I just subscribed to your channel. How exciting!
Good problem man. An important step is missing tho: prove that the two radii form a straight line (the centers and intersection point are collinear). One way to do it: remember that radii are perpendicular to the tangent lines, and since both share the same tangent (because the circles are tangent), this implies that both radii form a straight line.
I'm not smart enough to confirm whether those answers are correct, I'm also not in the mood to pay very close attention to it all, but i like watching it, cause i see small things that'll probably help me someday, like thinking about making a triangle with that radius
i like how well articulated you are.
@@marvinochieng6295 thanks, I've spent a lot of time practicing
Neither am I but I genuinely beleive thw answers arnt correct hes pulling out nonsense from nowhere I like watching his videos and id like to learn but it just seem true the answers
I got the same answer.
How do you know the hyphotenuse of the triangle will intersect (touch) the center of the two circles? In other word, how do you know R1 and R2 can be in straight line to form a hyphotenuse for the triangle? Thank you.
Circles that are tangent to one another have colinear radii on the point of tangency.
There's several proofs for this.
The shortest distance from a point to a line is a perpendicular line.
This is why lines tangent to a circle are always perpendicular to the radius; if there was a shorter distance, the line would intersect the circle.
At the point of tangency of two circles, the perpendicular lines match, so they're colinear.
What a great subscription your channel is! I love these fun little easy problems, but I'd never make time for them without your new videos in my feed to remind me. The ones where your solution is quite different than mine are a special treat.
Insert filler text claiming spoilers
Let the radius of the semicircle be x. The radius of the quartercircle is twice the semicircle, so its radius is 2x.
Drawing a line connecting the centers of the circles, we can use the Pythagoras Theorem to derive the following equation:
4² + x² = (x + 2x)²
16 + x² = 9x²
8x² = 16
x² = 2
pi(r)²
½pi(x²) + ¼pi(2x)²
= ½pi(x²) + ¼pi*4x²
= ½pi(x²) + pix²
= 1½pix²
= 1½pi(2)
= 3pi
≈9.4247
Ok but how do you know at first place that these are semi-circle on the left and quarter circle on the right?
1:35 - how do we know that these two radii create a straight line to form a triangle? It looks like you just assumed it but is there a rule that explains/confirms that? How do we prove that?
Ok, someone else expalained that below😉
This look important.
Let's put a square around it.
To those who wonder about the rotation part, it can be proven that the centres of the tangent circles and the point of tangency are collinear.
How do we know for sure that the r1 + r2 hypotenuse is actually linear/the angle between the connected radii is 180 degrees? Is there some mathematical principle that if two circles are tangent to each other, the line between their centers will necessarily run through the point of tangency?
Yes. All tanget lines of a circle are perpendicular to the radius at that tanget. If two circles share the same tanget line, their radii are both perpendicular to that line and thus they must be 180 degrees.
@@craigheimericks4594 Well shit, I can’t believe I didn’t think of that. Thanks man.
When did he mention 180 degrees? I've missed it, and I want to understand the question, and the answer that nice guy gave you :)
Draw two circles of any size each, put a straight line from the center of one to the center of the other and make the circles' borders touch. You won't find any way to do it without making the line and the point of touching meet.
@@AWhistlingWolf Yeah I got that, the question was how do we actually know that for certain? Another commenter already explained it.
Solved it by just staring at the thumbnail of the video w/o opening it. A nice task for a last 2-3 years of high school.
Bonus points to do the problem with calling the quarter circle radius r and the semicircle radius r/2. The solution is the same but the work is harder.
Thanks man im gonna challenge my teacher 😂 ❤❤
I look at these problems, but then I do not know how to solve them, yet the solution, if you follow rules of reason, appears. Interesting ideas about 'problems' in general. When will we follow the 'rules' for reasoning to solve our problems? And the answers are not obvious. I love the Andy Math channel.
I've been wathcing these too much, rotating the radiuses to form a hypotenuse was my first thought lol
Actually you can label it as 3pi cm^2, since they flat out give us the length in centimeters.
Given that we start with the rectangle having a side of 4 cm, I would say the correct answer is in cm as the units are known. I'm not sure if this would just be nit picking on my part, but it just feels more correct.
Yeh, I did that in my head but did the triangle thing first, *then* popped it into the areas.
Was "worried" at first that maybe I did the triangle thing for nothing, but that came later in the vid.
Fun one!
I havent watched the full video yet but I got 3pi. I first drew a line through the centres of both circles. That would form a right triangle with radius of smaller circle (x) and the base 4. Then you can form an equation like 2xy + y² = 16. And we can clearly see that the right side of the radius of the larger circle (y) and the left side is the diameter of the smaller circles (2x). So y = 2x. Then we solve them simultaneously and find the values of x and y and use them to find the area which is 3 pi
I dont like this kind of math questions because tou have to assume those are circles. Better if it was stated
Got this one...but with a slightly different approach. How exciting.
A cool problem would be to take the Ten Penny Puzzle solution (packing of 10 circles in a square) and solve for the length of the sides of the square. Using the relationship of circles to determine the size of a square feels right up your alley.
im not flexing here, but i never did math at school and even before you explained the answer i KNEW i had no idea
u did an extra step, u didn't need to find r1, u just need r1^2
which software do you use to animate and present the equations???
May I ask why r1+2r1 must be a strangle line? I.e. the additional line linked centre of the r1 semi-circle and the right bottom corner of the rectangle may not cut the intercept point of the two circles. Thanks.
did this whole thing in my head in a few minutes... 3Pi
(x/2)^2 + 16 = [ (x/2) + x ]^2 --> x = 2sqrt(2)
blue area --> area of a circle is Pi x radius ^2 --- radius of semicircle is 2x the radius of the quarter circle do the math and you get 3Pi
So... This might be in the assumption that the edges of the rectangle is the diameter of the smaller circle and the radius of the bigger circle... But what if it isn't? Like, if the half circles are not entirely half and quarter?
I can see it, its right there
Why do you assume that r1+r2 makes a triangle? What is the proof for that? It seems very convenient but how can you actually prove that the connection between large and small circles lays between the centers of both circles?
Is there a good book I can buy with these kinds of problems?
How did you assume that the the center of smaller circle, the center of larger circle and their point of contact lie on the same line?
I agree.
At any point on a circle, there can be only one tangent at that point and it will be perpendicular to the radius.
Both circles meet at one point, so both circles have their respective tangents on the same point, and both the tangents will be in the same line, then obviously the two radii will also be in a straight line.
try to make two circles NOT line up that way
How do you know the lower right corner is the center of the large circle?
1:16 wait, how do the 4s just magically turn into 2s? is it just because they cancel each other out so you could make them any old number and it doesn't matter?
Making the triangle is very creative I didn’t think of that good video
I'm assuming there should be no assumptions in STEM fields
How can you assume that is a quarter or semi circle? If you added a single pixel in illustrator there is no way to tell. It is just given in the description of the problem?
Can you just assume the right circle is a perfect quarter circle? I don't think you can solve it otherwise, but I don't think it is explicitly stated.
So here's what's I don't get it:
There is an assumption the line going frontman bottom right corner, through the point of contact of the curves, extends to the midpoint of the left side. I don't see how that was proven, and that's important.
BTW, the units were defined (cm), so the area would be 3pi cm 2
That was a really clever way to solve it.
Although the correct answer is 2, you should have demonstrated that the segments are aligned with each other, as we are uncertain about where the arches intersect. But very fun and interesting content 🎉🎉🎉
circles are always going to meet between their centers
I am not capable to understand such level of math
I found it. There’s a bit on the left and a bigger bit on the right.
Can you please tell me what software you are using?? Want to follow in your footsteps
How do we know the three dots in 1:37 make a straight line though?
I'm wondering the same thing
it is implied BECAUSE there is a dot, similar to how lines are often implied to be parallel by dashes through them
Straight line is shortest distance between two point.
people like you are great and awesome !
youtubers should follow your example for you have done right by seeking to provide good and educational content for students ! you and another good man like yourself known as "presh talwalker" are what a youtuber should aspire to be !
This was the first one i knew how to do
How can you ensure that the two rays will be aligned and form the hypotenuse?
Find the blue area
Me: there
Don't take sq root of 2 then square it substitute directly
not too bad of a question. was able to solve in double the time
I found the blue area, it's towards left and bottom right. Inside a rectangle
How can you be sure the pythagoras aligns nicely?
I did it in my head! Watching your videos made me become a *god*
the idea for rotating the radii, thats genius
It's the only way to solve. The width of the rectangle is entirely dependent on the two circles touching. Without it, the 4cm that we were given wouldn't do anything.
@@tomdekler9280 i know it is, but i wouldve just went around in circles (lol) and never wouldve gotten anywhere
@@lolok6439 yeah the fact that circles in tangent have colinear radii is a huge part of most circle-related geometry problems.
Well that was easy, it is a good problem definitely
Maybe this solution would fly in math, but not in engineering.
Unit squared? Only unit I've seen is cm.
Your u^2 can be any imaginary thing, unrelated to this problem.
I was working on it and got to the point where I'm not sure whether it's possible or not to draw the (r1 + r2) straight line...can you please explain how is it possible...😁
The units were given, so the answer is in cm2
Love from India 🇮🇳 sir
it's not unit square tho, cause one of the rectangles length is literally 4cm
1:39 i had to puse to relize why it's true (both are vertical to the tangent of the circuls in that points, with needs to be the same)
Why these 2 sectors don't overlap when the hypotenuse equals to r1 + 2*r1? Why r1 and 2*r1 must bond to a straight line?
Part of the initial assumption is that the two blue areas are tangent to each other at the one point. This means that r1 and 2*r1 do not overlap. The other fun is that since the two shapes are tangent, that means the two segments connecting them will form a 180 degree angle, which is a straight line.+
Funny symbols you have in there, magic man.
You must be really young then
@@aarusharya5658 last year in highschool
@@aarusharya5658 +ngl it does look very confusing
Is it just me or does him saying "r sub-one" sound like "arse of one"?
Thank you, I was hearing Arse Of One and couldn't figure out what he was actually saying!
What and which University have you studied in Andy?
How exciting!
How can you justify r1 and 2r1 form straight line
Anyone here can reccomend books that have similar nature to this video like solving interesting math problems?
01:39 It's not obvious. U'd prove it.
This is not solvable without making an assumption that there are half and quarter circles.. They look like they may be, but it is an assumption.
where does those divided by 2 and 4 come from? I just curious im not a math geni...
the area of a circle with radius r = pi*r^2. the shape on the left is half a circle, so to get its area, we use the formula for area of a circle, but then half it. same logic applies to the right shape, which is a quarter of a circle, meaning we divide it by four.
Proud to have solved this in my mind
What a wonderful channel . Thank you so much
Looks Hard
Simple after seeing him solve
Yet How Exciting❤💯
That's the blue area 🎉
I'm not bothered enough to know the number
I just like to know the theory behind finding out..
The whole "reducing down to the exact number" I would just let the computer do..
I just need to know what numbers to plug in.
I am in 9th grade and average at math. Idk how but I solved this first try. Am I good at math now???
How do you know that r1 and r2 form a straight line at 1:36 ?
By using your brain
All tangent lines of a circle are perpendicular to the radius at that tangent. If two circles share the same tangent line, their radii are both perpendicular to that line and thus they must be 180 degrees.
@@kaiji2542 ohh I see. Thanks!
How exciting? Really exciting!
How do i send you questions?
How do you know that on 2:07 r1 + 2r1 line is straight line and not the curved one? Guess it's required to solve this task but how to prove this assumption.
Think of it this way: if it was not a straight line, ie it was bent, that would mean that you could straighten it and have it be longer. However, the start points of each line are fixed, so the point where the two arcs touch must be when the line is the longest it can be, ie when it's straight
It's that the ratio of A series paper?
I'm here within 3 minutes of uploading!
How. Exciting.
How exciting indeed!
I tried so hard to find the triangle in this and couldnt' do it :(