I was trying to do the first bonus footage animation in my head to see if you can actually solve this problem with so little information and thought you couldn't. Actually seeing that the angle remains the same is pretty neat.
I start the video thinking 'how on earth would you figure that out'....then following along step by step, it's so simple 😂 Amazing job as always with these videos. Very fun to watch.
Litteraly the moment I saw the picture, I thought that circles could be in any position, moving them in my mind _exactly_ as in the end of the video. Though I didn't think that angle would be constant. Very cool)
Well the angle had to be constant for the question to be answerable unequivocally, otherwise the answer would be a range of angles in between the case that a circle touches two sides and the case that both circles are equal. So I expected the constant answer :)
Andy, i have found your channel by an accident. Since that tine, i always click asap on your videos. I really enjoy watching you solving all the problems. I wish you all the best! Keep up the good work!
That’s what I was thinking before the bonus footage. Nothing about the sizes of the circles was given or constrained but you got a single numerical answer rather than a formula with unknowns so it must be something that stays the same like you showed in the bonus footage.
Awesome problem and solution! I did it by joining the centers of the two circles. This creates a pentagon with three right angles and two isosceles triangles, one in each circle. If you call the head angle of one of the isosceles triangles x, the head angle of the other isosceles triangle is 270 - x. From there you can get the desired angle easily.
At 1:50, drop a perpendicular from the point of tangency with the top side of the square through the tangent circle's center. Also drop a perpendicular from the point of tangency with the right side of the square through that tangent circle's center. Extend both perpendiculars until they meet. Construct a line segment connecting the centers of the two circles. This line segment is the hypotenuse of a right triangle where the sides are segments of the perpendiculars. Both isosceles triangles constructed earlier in the video are present. The interior angles of the right triangle are exterior angles of the isosceles triangles. Each exterior angle is the sum of the two equal interior angles, 2(90° - x) = 180° - 2x for one of the triangles and 2(90° - y) = 180° - 2y for the other triangle. The sum of interior angles for the right triangle is (90°) + (180° - 2x) + (180° - 2y) = 450° - 2x - 2y and equates to 180°, so 450° - 2x - 2y = 180°, 2x + 2y = 270° and x + y = 135°. So our ? angle, which equals x + y, is equal to 135°, as Andy Math also found.
Seeing that the solution was not dependent on any dimensions was interesting! I was thinking the solution would be a line of every possible (x,y), but seeing the answer was a number independent of anything else was a surprise! Very exciting indeed.
Just prove the line joining the two centres of the circle is parallel to the 45 degree line. First join the centre of the upper circle and the upper left vertex by a line, this line bisects the upper white angle (45 degree) by property of tangent lines. Do the same thing for the big circle. The 45 degree line, the two lines joining the two centres and the upper and lower vertices and the line joining the two centres form a trapezium with both ends equal to 22.5 degree. So the line joining the two centres must be parallel to the 45 degree line. The rest is straightforward.
It always looked about 135, but if it were 123 for example, and since I am shit house at this stuff, I'd have still guessed 135 and been completely wrong. So thanks for not making me wrong 😄
Correct me if I'm wrong, but there seems to be no need for the diagonal, just circles touching eachother and different but connected sides of a sqare. The solution isn't depending on diagonal anyhow.
I love proofs that utilize the conservation of generality as a tool in their problem solving, because it feels more in line with how real math works. we know "θ" is constant, otherwise this would not be solvable, which is an artificially constructed symmetry, But there just as well could have been an inherent mathematical rule that said: for any pair of circles "C₁" and "C₂" inscribed in a 45-45-90 triangle "T" the angle T∩C₁-C₁∩C₂-C₂∩T is constant
It's also true that the circles don't have to both be tangent to the diagonal of the square. You can change the size of each circle independently as long as each one remains tangent to one side of the square and the other circle. The defined angle will always be 135 by the exact same argument as shown in the video. The square and the diagonal don't matter at all, only the right angle in the top right. And we can generalize for any angle in the top right, a. You would end up with 2x+2y+a=360, so x+y=180-a/2 The angle could be 180, in which case x+y is 90 degrees and part of a right triangle. The angle could even be greater than 180 and the same argument works.
Is the 135 degree related to the fact that the square is bisected at 45 degrees? 135+45=180..? If it was the diagonal of a rectangle how would it change?
The old man brain in me was screaming at the beginning "what do you mean what is the angle? they are just arbitrary angles...." a bit later I realize that the square box surrounding the whole thing was also an important part of information... this is why I do physics not math 🤣🤣
Why didn’t my high school teacher teach us math this way? It would have made so much more sense using visual proofs! Not to mention that this method is way more FUN!!!
Hey there, i saw this online. I wonder if its solvable with relatively short solution or it's just troll The question (20%) / (5/8) * 15 ^ 2 + 60! - 2 =
I started by looking at the longer line of the angle from within the circle on the right and drawing a line the runs between the angle but makes itself perpendicular the the longer one creating a 90 degree angle. Then by this point I could tell there wasn't much to the rest of the angle so I made an educated guess that the angle on the left of my dividing line was around 45 degrees and by adding that onto 90, I finish with an answer of 135. This is certainly not a question you'd find on any exam papers 'cause that angle was way too literal and not abstract enough.
OK, without watching the video: if the problem can be solved, then the answer should be the same regardless of how we choose the circles. Which means, we can choose them equal-sized, which would make the other diagonal their common tangent. And it would divide the angle in question in two equal halves. Each one of those halves would be an angle of an isosceles triangle with an apex being half of the direct angle. So, the apex angle is pi/4, and our angle is 2*(pi - pi/4)/2 = 3*pi/4. Come to think of it, that's the solution: draw a common tangent until it intersects both sides of the square; our angle would be the sum of two angles of different isosceles triangles, and their peaks, while different, would sum up to pi/2 anyway, because they have a common side. So, our angle would be (pi - a)/2 + (pi - b)/2 = pi - (a+b)/2 = pi - pi/4 = 3*pi/4. OK, watching now.
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Did you like the bonus footage?
Best plot twist.
Yesss✨✨ please do more with this format. Love ur vids
Yes very much 🎉👍
I was trying to do the first bonus footage animation in my head to see if you can actually solve this problem with so little information and thought you couldn't. Actually seeing that the angle remains the same is pretty neat.
Yep
Bro makes the best math question vids
A sausage triangle.
I can hear that too. I'm not native speaker and don't know all the english maths terminology. So for me it is i-sausage triangle. How exc...
@@wiader666 haha, same for me, I was thinking I was the only one hearing that^^
Haha that's so funny, He's actually saying isosceles triangle if you guys don't know
You have destroyed math
from now on, when you are suppose to say isosceles triangle, say: Eye-sausage-Lee's triangle. It might hurt saying that, but try it.
I start the video thinking 'how on earth would you figure that out'....then following along step by step, it's so simple 😂 Amazing job as always with these videos. Very fun to watch.
Litteraly the moment I saw the picture, I thought that circles could be in any position, moving them in my mind _exactly_ as in the end of the video. Though I didn't think that angle would be constant.
Very cool)
Well the angle had to be constant for the question to be answerable unequivocally, otherwise the answer would be a range of angles in between the case that a circle touches two sides and the case that both circles are equal. So I expected the constant answer :)
@@z000ey yes, it had to be constant for the question; I meant that I'd say it isn't constant without the question)
Andy, i have found your channel by an accident. Since that tine, i always click asap on your videos. I really enjoy watching you solving all the problems. I wish you all the best! Keep up the good work!
Bonus insights were awesome. More please.
I really enjoy watching you solve stuff, especially when you highlight which theorems are involved.
I loved the bonus footage. It might have been a quicker way to solve the problem. I really liked this problem.
I try to do these or even guess what the approach would be, but I was completely stumped and subsequently impressed at how eloquently you solved that.
Pretty exciting use of angles!
Your videos are always fun to watch.
Love it!!! Keep it up and EXCITING 😊
That’s what I was thinking before the bonus footage. Nothing about the sizes of the circles was given or constrained but you got a single numerical answer rather than a formula with unknowns so it must be something that stays the same like you showed in the bonus footage.
Awesome problem and solution! I did it by joining the centers of the two circles. This creates a pentagon with three right angles and two isosceles triangles, one in each circle. If you call the head angle of one of the isosceles triangles x, the head angle of the other isosceles triangle is 270 - x. From there you can get the desired angle easily.
How exciting
Bonus footage was exciting brother
A double how exciting... How exciting that is
At 1:50, drop a perpendicular from the point of tangency with the top side of the square through the tangent circle's center. Also drop a perpendicular from the point of tangency with the right side of the square through that tangent circle's center. Extend both perpendiculars until they meet. Construct a line segment connecting the centers of the two circles. This line segment is the hypotenuse of a right triangle where the sides are segments of the perpendiculars. Both isosceles triangles constructed earlier in the video are present. The interior angles of the right triangle are exterior angles of the isosceles triangles. Each exterior angle is the sum of the two equal interior angles, 2(90° - x) = 180° - 2x for one of the triangles and 2(90° - y) = 180° - 2y for the other triangle. The sum of interior angles for the right triangle is (90°) + (180° - 2x) + (180° - 2y) = 450° - 2x - 2y and equates to 180°, so 450° - 2x - 2y = 180°, 2x + 2y = 270° and x + y = 135°. So our ? angle, which equals x + y, is equal to 135°, as Andy Math also found.
Seeing that the solution was not dependent on any dimensions was interesting! I was thinking the solution would be a line of every possible (x,y), but seeing the answer was a number independent of anything else was a surprise! Very exciting indeed.
Alright I subbed. I really enjoy this niche content
Very clever stuff there.
Nice job! Thank you!
This is doubly exciting 😊
That really was a fun one.
Really good graphics!!!
a bonus exciting? how exciting!
This is a great video.
bro is a genius❤️❤️
Just prove the line joining the two centres of the circle is parallel to the 45 degree line. First join the centre of the upper circle and the upper left vertex by a line, this line bisects the upper white angle (45 degree) by property of tangent lines. Do the same thing for the big circle. The 45 degree line, the two lines joining the two centres and the upper and lower vertices and the line joining the two centres form a trapezium with both ends equal to 22.5 degree. So the line joining the two centres must be parallel to the 45 degree line.
The rest is straightforward.
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| LET'S PUT A BOX AROUND IT |
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How. Exciting.
That was slick
Indeed...exciting af
Great!!
awesome!
How exciting :)
Do we get extra credit for the bonus footage?
It always looked about 135, but if it were 123 for example, and since I am shit house at this stuff, I'd have still guessed 135 and been completely wrong. So thanks for not making me wrong 😄
How exciting.
Correct me if I'm wrong, but there seems to be no need for the diagonal, just circles touching eachother and different but connected sides of a sqare. The solution isn't depending on diagonal anyhow.
I love proofs that utilize the conservation of generality as a tool in their problem solving, because it feels more in line with how real math works.
we know "θ" is constant, otherwise this would not be solvable, which is an artificially constructed symmetry,
But there just as well could have been an inherent mathematical rule that said:
for any pair of circles "C₁" and "C₂" inscribed in a 45-45-90 triangle "T" the angle T∩C₁-C₁∩C₂-C₂∩T is constant
Yeh, I figured that without either/both radii specified, the angle will always be a constant no matter the ratio of r1:r2.
It's also true that the circles don't have to both be tangent to the diagonal of the square. You can change the size of each circle independently as long as each one remains tangent to one side of the square and the other circle. The defined angle will always be 135 by the exact same argument as shown in the video. The square and the diagonal don't matter at all, only the right angle in the top right.
And we can generalize for any angle in the top right, a. You would end up with 2x+2y+a=360, so x+y=180-a/2
The angle could be 180, in which case x+y is 90 degrees and part of a right triangle. The angle could even be greater than 180 and the same argument works.
I was just about to think the radius of the circles doesn't matter cause the r was irrelevant to find x+y (in the equation ofc)
Kare veya köşegene ne gerek var? Cevabın 135° olması için sağ üst köşedeki açının 90° olması yeterli
is it posibole to solve most of your questions and get 600 in sat?
Damn you!
how exciting
So now we have double exciting
W video again.
Pure brain nourishment
Is the 135 degree related to the fact that the square is bisected at 45 degrees? 135+45=180..? If it was the diagonal of a rectangle how would it change?
Dude aged a lot between recording and editing the video 💀
Btw what software you use for doing the equations
Always 135 degrees? Wow! I did not expect that.
I wait for your videos
My guy must have schizophrenia or something cause he sees triangles everywhere haha.
Your videos are amazing
W vid
But mama i'm in love with a mathematician
The old man brain in me was screaming at the beginning "what do you mean what is the angle? they are just arbitrary angles...." a bit later I realize that the square box surrounding the whole thing was also an important part of information... this is why I do physics not math 🤣🤣
clever
Why didn’t my high school teacher teach us math this way? It would have made so much more sense using visual proofs! Not to mention that this method is way more FUN!!!
Is it significant that the ? angle is equal to 180 - 45?
Gg
wow he aged
triple exciting in this video
Just let the circles be the same size, then the angle = 180° - 2*(45°/2) = 135°.
The angle of the dangle.
Motion of the ocean
Eye sauce less triangle
I'm not a native speaker and I kept hearing "I-sausage triangles", guess it's time to go eat :p
why the hell does my gut start raising when im looking at this image as if im falling in my sleep
I have a question, why x is not equal to y?
Sisyphus reference
Actually solved it through the bonus process, getting 2*(alpha+beta)=360-90
2:30 Is that really a legal move to do? (2x + 2y) / 2 = x + y?
2:24 but why is the graph so angry?
Anyone else see an angry chameleon face at 2:04?
oof
Hey there, i saw this online. I wonder if its solvable with relatively short solution or it's just troll
The question
(20%) / (5/8) * 15 ^ 2 + 60! - 2 =
Looks 135 degrees but lemme try it out and edit after.
Edit: I swear it was a coincidence.
I don't want to be that guy but I decided to guess the angle by eyeballing it...
And got the answer right ✅ 👌 😏
I started by looking at the longer line of the angle from within the circle on the right and drawing a line the runs between the angle but makes itself perpendicular the the longer one creating a 90 degree angle. Then by this point I could tell there wasn't much to the rest of the angle so I made an educated guess that the angle on the left of my dividing line was around 45 degrees and by adding that onto 90, I finish with an answer of 135.
This is certainly not a question you'd find on any exam papers 'cause that angle was way too literal and not abstract enough.
No 69 involved, but still nice
i was waiting for your feet as extra footage…
OK, without watching the video: if the problem can be solved, then the answer should be the same regardless of how we choose the circles. Which means, we can choose them equal-sized, which would make the other diagonal their common tangent. And it would divide the angle in question in two equal halves. Each one of those halves would be an angle of an isosceles triangle with an apex being half of the direct angle. So, the apex angle is pi/4, and our angle is 2*(pi - pi/4)/2 = 3*pi/4.
Come to think of it, that's the solution: draw a common tangent until it intersects both sides of the square; our angle would be the sum of two angles of different isosceles triangles, and their peaks, while different, would sum up to pi/2 anyway, because they have a common side. So, our angle would be (pi - a)/2 + (pi - b)/2 = pi - (a+b)/2 = pi - pi/4 = 3*pi/4.
OK, watching now.
Without stating the angle measurement is in radians, only partial marks.
How edgy.
How exciting