Nice problem! I took a general approach to get R for arbitrary half circles with radius r. It goes as follows: x² + R² = (r-R)² x² + R² = r² - 2rR + R² → x = √(r² - 2rR) y² + (r - R)² = (r + R)² y² + r² - 2rR + R² = r² + 2rR + R² y² = 4rR → y = 2√(rR) Now as y = r + x, we get 2√(rR) = r + √(r² - 2rR) 2√(rR) - r = √(r² - 2rR) 4rR - 4r√(rR) + r² = r² - 2rR 6rR = 4r√(rR) |r>0 (6/4)R = √(rR) (3/2)R = √(rR) |² (9/4)R² = rR |R>0 (9/4)R = r therefore R = 4r/9 In your case r = 18 (half of 36), so R = 4·18/9 = 8
For the people who are wondering, If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide. I got this after a while of thinking and my brain is now tired.
I always liked math and it was the one subject I was good at in school but it has been years since i sat down and solved a math problem. It was fun to solve the problem along with you. Subscribed
This was a good one. I was going down the same road with the two triangles and it was getting ugly, so I was thinking maybe it wasn't the right approach. Then I played the first few seconds of the video and you said it was a bear, so I just put my head down and pushed through. For semicircles of arbitrary radius A, the solution is R=8A/18.
This took me a very long time. I finally got it although I went around and around in a very convoluted solution. Your method was far more simple than mine. Please explain slower! I have trouble keeping up with your explanations, and it’s hard to stop the video at the right moments. Please pause a little between operations. Please! My age is the square root of 5,929.
You might consider a youtube feature under the gear icon where you can slow the video down to half or even a quarter of speed without changing the pitch of the voice (it's really an amazing algorithm, i.e. "phase vocoder" .... isn't that exciting). Apologies if you were already aware of the availability of this feature.
This puzzle was fun for me because - seeing all those tangent circles - I went off and did it with a circle inversion, which I always think is such a beautiful trick when it works. Although TBH it didn't magically make this into a quick easy problem.
What is a circle inversion? edit: ok i looked it up and it's really cool. Thank you for commenting that. One more question: Your inversions were based on what circle? I imagine you defined a new circle with center on the bottom left corner and radius equal to 36.
I was hoping someone would be curious enough to learn about it! It's pretty unusual to get a chance to use it, but it's nice when it does 'cos it's so neat@@muriloamorim2731
@@tiagoloprete It's a really nice concept, isn't it? Very fun to play around with. Glad you also enjoyed learning it! Did you try the inversion i suggested (based on a circuference centered on the bottom left corner and radius equal to 36) or any other inversion? With the inversion i suggested it was relatively easy to show that the distance between the bottom left corner and the leftmost point of the green circle is 24. Let me know if you agree/disagree or if you need help in any way with the inversion. One last thing: your name sounds Brazilian.. are you? (I am)
Hola Andy, It can be shown that for any such figure exactly 4.5 R's constitute the length of the given radius. From this we have that the length of R is 36/4.5 = 8. Thank you so much for this problem which provided a rough challenge from the directions I took. The challenge plus you enthusiasm are much welcome in problem solving departments in education these days : )
When you did draw a line from the center of the left semi circle to the collision point of the green circle and the left semi circle, how did you guarantee that the line crosses the center of the green circle?
By definition, at any point, the circumference is perpendicular to the line that connects the center to that point of the circumference. When both circumferences are tangential in one point, they are parallel in that point, so their perpendiculars in that point are collinear, and pass through the center of both circles
The semicircle and the small circle share a tangent. This means that the two radiuses (radii?) are 1) parallel (both perpendicular to the same tangent) and 2) share a point (intersection with the tangent). If they are both parallel and share a point then they belong to the same line, they are collinear.
What was the point of the yellow circle then? was it just to distract us from directly applying this method? or is there an easier way to do this problem using the yellow circle?
I’m not sure. The radius of the yellow circle is going to be some long decimal while the math for the radius of the green circle involved no decimals at all, they don’t seem too easily connected.
His first line he creates starts at the middle point of the 36 line. The X side of the triangle extends up from that point. The red portion extends down from that point.
lol I actually thought for a bit he got the problem wrong after I read this cause I went to check and his center of the circle should be on the same horizontal plane as the intersection of the left and bottom circles in the middle of the pic, and his center is slightly more down. but its just the drawing that is imprecise aha
I feel like I should be eating handfuls of paste after watching this. I stared at this for 20 minutes trying to get somewhere, and Andy explains it in less than 7 minutes. If I ever feel the need to feel dumb, this is where I come.
Went to solve this on my own before watching the video, and was proud to find the answer was 9(2-sqrt2), until i came to the video and realized I found the radius of the YELLOW circle, not the green one.
Right after 42 seconds, I was able to do it... My only doubt was if it is possible for a segment from centre of the semicircle to the circumference also passing through the centre of the smaller circle
Eventhough your speed is Interstate, I am able to follow. Had you been my algebra-trig teacher and said "How exciting" a couple of times, I'd be so much more facile. But, here you are on RUclips and here I am a geometer without too much of algebra or trig, and yet curious. Let's see where this goes.
I guess this way you should imply BEFORE solving that R cannot be 0, because you cannot divide by 0. And that can be not the right way to solve this type of questions, because this way you can lose some of the answers/
@@cod3r_It's geometry there's only one real solution. When you divides like the guy up there said, you get 8R = R², you divide R in both sides and you get R = 8.
Hi good job on solving the problem but pls correct me if I’m wrong when u got 72R=9Rsquared could u of just divided by R to get 72 =9R then divided by 8 to get 9=R
No joke, I looked at it and immediately thought "well it has to be slightly smaller than a quarter of 36 so I bet it is 8" and it just so happened that my hunch was right.
Since there must be 2 values for R, since its R² (R³ would be 3 R⁴ 4 and so on) just dividing by R would make it that you no longer see one of the two values(it doesnt realy matter anyway here, but it might have so better look at all the answers)
Regarding the line at 0:30. How do we know it goes through the center of the green circle? I wouldn't have assumed that, so I'm a bit disappointed that it wasn't explained.
The straight part of the left semicircle is forming a tangent line across the leftmost point of the circle. By definition, any line drawn perpendicular to the tangent line at the point it touches a circle will pass through the centre of the circle, because the part of the circle that touches the line is parallel to the line. Been a while since i did geometry, but hope that helps at least get the broad strokes across, i think he explains it more in some of his other vids where he uses that principle, think he skimmed over it here due to the time constraint.
In the first part of this problem, how do you know the line from the tangent point of the green circle going through the green circle's center also goes through the center of the semicircle?
A line drawn from the center of a circle to its radius forms a right angle to a tangent line at any point on the circle. So two circles in contact (whether inside or outside) results in the same tangent line at the point of contact. Since the two lines from the two centers to the contact point form right angles at the contact point, both centers and the contact must be on a common straight line through the centers and the contact point.
When you draw the orange radius, length 18, how do you know it touches the tangency point of the 36 semi circle & the green circle? Or is it coincidence, or close, but meaning that it doesn't matter?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide. I got this after a while of thinking and my brain is now tired.
Hi! I throughly enjoyed the process of solving this problem with you :) I just wanted to know why did you go with a quadratic equation in the last part (9r2 - 72r = 0), couldnt we have solved it directly by : 9r2 = 72r ( cancelling 9r from both side) We get r=8.
While that works for solving this particular problem where we only want positive solutions, that won't always work. The way he did it gave us 2 solutions, 0 and 8. For this problem, 0 wasn't a valid solution, but there are lots of problems where you don't want to get rid of that solution. Be careful dividing by variables because that usually removes solutions.
I am confused with just one aspect. How do You know that first line Andy draw connect "center" of big circle with point where big circle touches green circle? And, no matter how small green circle is, If You draw line from center of vertical line to point where where both circles touch, it will always go through center of green circle?
Good question. The green circle and the large semicircle are touching on the edge. That means the green circle and the large semicircle have the same tangent line at that point. The black line Andy drew is perpendicular to this tangent line of both circles. Since the tangent line is the same among both circles, and the black line is drawn perpendicular to it, it will by definition extend through the center of both the green circle and large semicircle.
How do you know that the centre of green circle, Center of semi circle, and point where curved surfaces of green circle and semicircle touched are all in the same line?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide. I got this after a while of thinking and my brain is now tired.
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide. I got this after a while of thinking and my brain is now tired.
I have only one doubt. Please help me. How do we know that the red part below x is equal to the radius of the circle? The x may or may not overlap with the radius. I believe don't know that for sure.
@@ajamessssss he made a line starting from the center of the semi circle which became the hypotenuse of the triangle with one side as x. Therefore, the red part below the x starts from the center down to the edge of the semi circle and is equal to the radius. Is that what you were asking?
I did not solve this. Could be that I try solving them in my head, could be that I did not know how to approach this and would need a lot of exploratory algebra.
This may sound like philosophical bullshit, but your math videos actually help me cope with my executive dysfunction from ADHD - I "just" need to break down every task/problem into small enough parts that I can solve. Everything can be broken down into smaller steps and at some point the step is small enough for me to know the solution. Thank you :-)
Clever! How, though, do you know that the line segments starting at the origins of the two large semi-circles go through the center of the green circle. I'm glitching on that point. Thanks!
Good question , the answer may be above me , however with the algebra solution done , IMHO we can indulge in some reverse engineering abstractly . Notice that units are not mentioned , could be miles or meters , it does not matter , because everything in the problem exist as ratios . So R as 8 gives us the big circle diameter as a ratio of 38/8 to R or 4.5 R . This is after proof , but the answer also exists in terms of R , not solved numerically until we give some unknown an actual value . So that 4.5R diameter can be any one choice from a an infinite range . This includes zero as the scale of the figure zooms away towards zero. Zero is a funny one but it is somewhere .
Hey at 72R=9R^2 how come you just didnt divide both sides by 8R? You would have arrived to the conclusion that R=8. That is instead of subtracting both sides by 72R, then factoring out an “R,” coming to two conclusions for R.
I could not see an algebra solution. Did it graphically. The center of the circle is found as you start at zero for R and increse it. You have 3 curves that eventually meet at a single point as R approaches 8. The 3 curves are x=R, the upper curve circle as the radius is 18-R and the lower circle as the radius is 18+R. Did it quick and dirty on a TI-84. Equation 1, the upper circle ... sqrt((18-R)squared-X squared)) +18.... equation 2, the lower circle .... sqrt(18+R)squared-(X-18)squared).... equation 3 is just a vertical line for x=R ..... (X-R)*10000. Think of as R gets bigger the alowable center position of the circle is restricted by needing to be R distance from the y axis and R distance from the upper circle and R distance from the lower circle. The allowable center point is inside the 3 curves and as R increase you get one allowable point as R gets its maximum value of 8.
How are you able to assume that the line equal to r+18 when solving for the y variable is indead a straight line simply bevause its a tangent for both curves? Im having trouble figuring out how you would know that.
It's a basic circle theorem: when you have 2 circles that are tangent to each other, either externally or internally (aka 'kissing circles'), the centres of the circles and the tangent point are colinear.
Not good at math, so can someone explain how can we be sure that Radius at 0:35 Goes through the middle of green circle and hits tangent of both circles in the same point? is there some sort of law that confirms it always happens?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide. I got this after a while of thinking and my brain is now tired.
In the very first step, how come you can assume that all those points are collinear? Couldn't it be that that those points are not on the same line, so you cannot connect them?
in the very first step of his solution (at 0:27), he draws a single line passing through a) the centre of the large semicircle on the left b) the centre of the green circle c) the point of contact between the green circle and the large semicircle How does he know for sure those points can be connected with a straight line of length 18?
@@dylanjayabahu2878 I think the diagram is to scale. That's why he did it. I get your point though that those points might not be colinear but I dot see many ways to solve it otherwise.
Touching circles theorem, if two circles touch each other (internally or externally), then the point of contact will lie on the line joining the two centres. Same thing again at 2:27.
I love that unuseful yellow circle
That's me
bro forgot the word useless 😭
@@contentlacking5950 i forgor
It's just like me fr
@@contentlacking5950why is it called orange juice? It's not yellow at all, it's orange.
I am not a math person, but I love Andy's attitude and could watch his videos all day.
You are not a math person that's why you don't know why he took so much time 😂😂
@@shivakrishna4754 what is bro yapping about
How exciting indeed
GOOD GOD, EVERYTHING WAS A TRIANGLE ALL ALONG
Always is.
@@A_Loyalist Always was...
Always has been.
Always will be...@@Heronoobie
the type of questions you bring perfectly match my grade level. thanks. i can solve some tough probs now
which grade r u tho?
@@detroitstudios397 10th India
@@epic_divyanshu well im 8th india
@@detroitstudios397 nice. icse?
I'm in grade 12 , can't say about the calculations but the observations are really good
Nice problem! I took a general approach to get R for arbitrary half circles with radius r.
It goes as follows:
x² + R² = (r-R)²
x² + R² = r² - 2rR + R² → x = √(r² - 2rR)
y² + (r - R)² = (r + R)²
y² + r² - 2rR + R² = r² + 2rR + R²
y² = 4rR → y = 2√(rR)
Now as y = r + x, we get
2√(rR) = r + √(r² - 2rR)
2√(rR) - r = √(r² - 2rR)
4rR - 4r√(rR) + r² = r² - 2rR
6rR = 4r√(rR) |r>0
(6/4)R = √(rR)
(3/2)R = √(rR) |²
(9/4)R² = rR |R>0
(9/4)R = r therefore R = 4r/9
In your case r = 18 (half of 36), so R = 4·18/9 = 8
Yeah
Generally it might be a bit hard to understand but it is simpler if you understand your terms!🤌🏼🤌🏼🤌🏼
bro is so underrated, keep up the good work man appreciate it a lot
Man I love this guys videos! Stumbled upon them a few days ago now I cant stop lol
For the people who are wondering, If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide.
I got this after a while of thinking and my brain is now tired.
I was wondering exactly for this explanation, than you!!
I wish you were my math teacher, you are so patient with the explainations and it is crystal clear, thanks !
I always liked math and it was the one subject I was good at in school but it has been years since i sat down and solved a math problem. It was fun to solve the problem along with you. Subscribed
This was a good one. I was going down the same road with the two triangles and it was getting ugly, so I was thinking maybe it wasn't the right approach. Then I played the first few seconds of the video and you said it was a bear, so I just put my head down and pushed through. For semicircles of arbitrary radius A, the solution is R=8A/18.
This took me a very long time. I finally got it although I went around and around in a very convoluted solution. Your method was far more simple than mine. Please explain slower! I have trouble keeping up with your explanations, and it’s hard to stop the video at the right moments. Please pause a little between operations. Please! My age is the square root of 5,929.
Hey! Me too! 1947 was a good year, huh?
You might consider a youtube feature under the gear icon where you can slow the video down to half or even a quarter of speed without changing the pitch of the voice (it's really an amazing algorithm, i.e. "phase vocoder" .... isn't that exciting). Apologies if you were already aware of the availability of this feature.
This puzzle was fun for me because - seeing all those tangent circles - I went off and did it with a circle inversion, which I always think is such a beautiful trick when it works. Although TBH it didn't magically make this into a quick easy problem.
What is a circle inversion?
edit: ok i looked it up and it's really cool. Thank you for commenting that.
One more question: Your inversions were based on what circle?
I imagine you defined a new circle with center on the bottom left corner and radius equal to 36.
I was hoping someone would be curious enough to learn about it! It's pretty unusual to get a chance to use it, but it's nice when it does 'cos it's so neat@@muriloamorim2731
@@muriloamorim2731 i'm curious too, hoping we'll get an answer :p found out about inversion today and spent half an afternoon learning about it
@@tiagoloprete It's a really nice concept, isn't it? Very fun to play around with.
Glad you also enjoyed learning it!
Did you try the inversion i suggested (based on a circuference centered on the bottom left corner and radius equal to 36) or any other inversion?
With the inversion i suggested it was relatively easy to show that the distance between the bottom left corner and the leftmost point of the green circle is 24. Let me know if you agree/disagree or if you need help in any way with the inversion.
One last thing: your name sounds Brazilian.. are you? (I am)
Thanks for the work you put in to fashion each clear and concise attack you document. You're doing very good work indeed.
It was great, Thanks for solving these kind of problems 👍
Hola Andy,
It can be shown that for any such figure exactly 4.5 R's constitute the length of the given radius.
From this we have that the length of R is 36/4.5 = 8.
Thank you so much for this problem which provided a rough challenge from the directions I took.
The challenge plus you enthusiasm are much welcome in problem solving departments in education these days : )
How exciting! Where do you find these problems?
I wanna know too!
Math books, math guides, 8th grade onwards, try IMO olympiad guides as well
His website
on 5:40 you could divide both sides by 9R to get 8=R
When you did draw a line from the center of the left semi circle to the collision point of the green circle and the left semi circle, how did you guarantee that the line crosses the center of the green circle?
Exactly my question as well
hold on i just realized you're right
By definition, at any point, the circumference is perpendicular to the line that connects the center to that point of the circumference. When both circumferences are tangential in one point, they are parallel in that point, so their perpendiculars in that point are collinear, and pass through the center of both circles
@@toni9810 Ahh, indeed! That makes total sense. Thanks!
@toni9810 i really wanna understand this can someone explain more simpler 🥲
Where do you get these interesting geometry problems?
How do you know the center of the semicircle, the center of the green circle, and the point of tangency between those two circles are colinear points?
The semicircle and the small circle share a tangent. This means that the two radiuses (radii?) are 1) parallel (both perpendicular to the same tangent) and 2) share a point (intersection with the tangent). If they are both parallel and share a point then they belong to the same line, they are collinear.
@@dimitrisdimitriadis4913 thank you!
The line joining the centres of two circles (that touch each other) pass through the point of contact.
@@rudrodeepchatterjee got it
I would love to know your process of thinking of a way of solving. Itd be great if you can explain how to approach these mathematical problems!
Where the f do you get these beautiful questions????
True
What was the point of the yellow circle then? was it just to distract us from directly applying this method? or is there an easier way to do this problem using the yellow circle?
I’m not sure. The radius of the yellow circle is going to be some long decimal while the math for the radius of the green circle involved no decimals at all, they don’t seem too easily connected.
I was totally stumped by that first part, solvin for X in terms of R
4:14 how do you know the red portion is half of 36?
His first line he creates starts at the middle point of the 36 line. The X side of the triangle extends up from that point. The red portion extends down from that point.
radius
lol I actually thought for a bit he got the problem wrong after I read this cause I went to check and his center of the circle should be on the same horizontal plane as the intersection of the left and bottom circles in the middle of the pic, and his center is slightly more down. but its just the drawing that is imprecise aha
I feel how my classmates feel when I explain something and they say “I don’t understand”
Tried to think of a solution in my head:
Couldn’t 💀
I did know we were going to be constructing triangles, does that get me any points?
I feel like I should be eating handfuls of paste after watching this. I stared at this for 20 minutes trying to get somewhere, and Andy explains it in less than 7 minutes. If I ever feel the need to feel dumb, this is where I come.
Loved the ending thought process! Great work
As an engineer I can confidently state that R < 36
Love these videos. I subscribed, its worth it
GOod esplanations with out jumping crucial steps.. Good job - I enjoyed it :)
I'm thinking the yellow circle is a separate problem (much easier). It's radius is 18*(1-1/√2).
why center of a green circle, "endpoint" and point of a vertical half radius is on the same line?
really nice. I tried to do myself got stuck of course. thanks.
how would you solve for the radius of the other circle?
Good question for my 15 year old students exam
U deserve a lot subscribers
Keep going
Went to solve this on my own before watching the video, and was proud to find the answer was 9(2-sqrt2), until i came to the video and realized I found the radius of the YELLOW circle, not the green one.
At 72R = 9R^2 you could just divide by 9R on both sides, resulting in 8 = R
Right after 42 seconds, I was able to do it... My only doubt was if it is possible for a segment from centre of the semicircle to the circumference also passing through the centre of the smaller circle
Eventhough your speed is Interstate, I am able to follow. Had you been my algebra-trig teacher and said "How exciting" a couple of times, I'd be so much more facile. But, here you are on RUclips and here I am a geometer without too much of algebra or trig, and yet curious. Let's see where this goes.
If I watch all these videos a million times I’ll be able to do them too without even thinking about it.
I made a start but I just went off at a tangent.
How do you know the radius you drew of the semicircle goes through the center of the green circle?
When you are at 72R = 9R^2, why not divide both sides by 9R? You end up getting R=8 and the whole thing is a little simpler.
I guess this way you should imply BEFORE solving that R cannot be 0, because you cannot divide by 0.
And that can be not the right way to solve this type of questions, because this way you can lose some of the answers/
@@cod3r_It's geometry there's only one real solution. When you divides like the guy up there said, you get 8R = R², you divide R in both sides and you get R = 8.
have to be careful as if the solution is zero you cant divide by the variable
@@ellazychavito9222 yeah, but you can't have 0 meters (I forgot how you guys say the general form of unities of size). My comment still correct.
@@ellazychavito9222 the point here is that you need your answer to be positive because it is geometry.
I did it different. I used sohcahtoa, pythagoras theorem, b²-4ac, and the quadratic formula without using Y and I got a similar amswer to you
6 min is "potentially too long" for people? Amazing we're still a competitive country at this point.....
Hi good job on solving the problem but pls correct me if I’m wrong when u got 72R=9Rsquared could u of just divided by R to get 72 =9R then divided by 8 to get 9=R
45 years ago when I was 17, I could have solved this problem. Now I am just old and stupid. Nice work youngster!
Could someone please do "how exciting" compilation 😂
How do you know the red portion is starts at the midpoint of the semicircle?
x is line between center of semicircle and outer center of green circle.
No joke, I looked at it and immediately thought "well it has to be slightly smaller than a quarter of 36 so I bet it is 8" and it just so happened that my hunch was right.
now this was sooo good... and it took youtube only 7 months to recommend it to me 😂😂
How do you know that the second hypotenuse is R+18? How do you know it bisects the exact point that the circles are tangent to each other?
Same question
5:41 you could just divide by 9R and so you will get that R = 8 because R can't be 0
When you got to 36(2R)=9R^2, why did bother subtracting 72R and factoring? Just divide both rides by R to get 72=9R, which gives R=8.
Since there must be 2 values for R, since its R² (R³ would be 3 R⁴ 4 and so on) just dividing by R would make it that you no longer see one of the two values(it doesnt realy matter anyway here, but it might have so better look at all the answers)
I'm looking at the diagram, but it doesn't say it is drawn to scale. How do we know those are semicircles?
Outstanding.
i dunno why but when you said the radius was r+18 i laughed saying ah the problem is rated R :P
Regarding the line at 0:30. How do we know it goes through the center of the green circle? I wouldn't have assumed that, so I'm a bit disappointed that it wasn't explained.
The straight part of the left semicircle is forming a tangent line across the leftmost point of the circle. By definition, any line drawn perpendicular to the tangent line at the point it touches a circle will pass through the centre of the circle, because the part of the circle that touches the line is parallel to the line.
Been a while since i did geometry, but hope that helps at least get the broad strokes across, i think he explains it more in some of his other vids where he uses that principle, think he skimmed over it here due to the time constraint.
In the first part of this problem, how do you know the line from the tangent point of the green circle going through the green circle's center also goes through the center of the semicircle?
A line drawn from the center of a circle to its radius forms a right angle to a tangent line at any point on the circle. So two circles in contact (whether inside or outside) results in the same tangent line at the point of contact. Since the two lines from the two centers to the contact point form right angles at the contact point, both centers and the contact must be on a common straight line through the centers and the contact point.
Why can't you replace x² with 36(9-R) for the first equation? I'm not trying to be critical, I'm just curious
When you draw the orange radius, length 18, how do you know it touches the tangency point of the 36 semi circle & the green circle? Or is it coincidence, or close, but meaning that it doesn't matter?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide.
I got this after a while of thinking and my brain is now tired.
How does a line from centre of semicircle to green circle is coincidental to centre of green circle
Hi! I throughly enjoyed the process of solving this problem with you :)
I just wanted to know why did you go with a quadratic equation in the last part (9r2 - 72r = 0), couldnt we have solved it directly by :
9r2 = 72r ( cancelling 9r from both side)
We get r=8.
dang you smarter than him
While that works for solving this particular problem where we only want positive solutions, that won't always work. The way he did it gave us 2 solutions, 0 and 8. For this problem, 0 wasn't a valid solution, but there are lots of problems where you don't want to get rid of that solution. Be careful dividing by variables because that usually removes solutions.
I am confused with just one aspect. How do You know that first line Andy draw connect "center" of big circle with point where big circle touches green circle? And, no matter how small green circle is, If You draw line from center of vertical line to point where where both circles touch, it will always go through center of green circle?
Good question.
The green circle and the large semicircle are touching on the edge. That means the green circle and the large semicircle have the same tangent line at that point. The black line Andy drew is perpendicular to this tangent line of both circles. Since the tangent line is the same among both circles, and the black line is drawn perpendicular to it, it will by definition extend through the center of both the green circle and large semicircle.
Am I right to assume that mostly these are minimal length heuristics , with no equals .
I wouldn't have the nerves. I would just draw and measure with a ruler...
Its like a 14 15 year old Knowledge that's needed but a few l'oreille years to find the way to get into it
This was a very nice problem!! I enjoyed it A LOT!!
How do you know that the centre of green circle, Center of semi circle, and point where curved surfaces of green circle and semicircle touched are all in the same line?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide.
I got this after a while of thinking and my brain is now tired.
This is why i was a mathlete. Yes...how exciting
x is not equal to y - 18. Because both horizontal tangents are not the same. So y is less than x+18.
Me being a doctor watching this at 2Am. This is just awesome.
How do you prove the hypotenuse is R+18? What if there's a small gap between radius R and semicircle?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide.
I got this after a while of thinking and my brain is now tired.
I have only one doubt. Please help me. How do we know that the red part below x is equal to the radius of the circle? The x may or may not overlap with the radius. I believe don't know that for sure.
0:26
@@sykroza what's that?
@@ajamessssss he made a line starting from the center of the semi circle which became the hypotenuse of the triangle with one side as x. Therefore, the red part below the x starts from the center down to the edge of the semi circle and is equal to the radius. Is that what you were asking?
When he drew the line from the center of the green circle to the bottom semi-circle
@@sykroza Yes. Thank you.
I did not solve this. Could be that I try solving them in my head, could be that I did not know how to approach this and would need a lot of exploratory algebra.
Loved it
The next level question, Find the radius of the yellow circle
Its radius is 18*(1-1/√2)
Why at 0:26 he think that line from center of semicircle goes through center of green circle? I can’t see reason why it must be.
This may sound like philosophical bullshit, but your math videos actually help me cope with my executive dysfunction from ADHD - I "just" need to break down every task/problem into small enough parts that I can solve. Everything can be broken down into smaller steps and at some point the step is small enough for me to know the solution. Thank you :-)
Clever! How, though, do you know that the line segments starting at the origins of the two large semi-circles go through the center of the green circle. I'm glitching on that point. Thanks!
What's the significance/meaning of the R=0 solution though?
Good question , the answer may be above me , however with the algebra solution done , IMHO we can indulge in some reverse engineering abstractly . Notice that units are not mentioned , could be miles or meters , it does not matter , because everything in the problem exist as ratios . So R as 8 gives us the big circle diameter as a ratio of 38/8 to R or 4.5 R . This is after proof , but the answer also exists in terms of R , not solved numerically until we give some unknown an actual value . So that 4.5R diameter can be any one choice from a an infinite range . This includes zero as the scale of the figure zooms away towards zero. Zero is a funny one but it is somewhere .
Typoo 36/8
Just so exciting
How exciting! I love this guy!
That was just great
Hey at 72R=9R^2 how come you just didnt divide both sides by 8R? You would have arrived to the conclusion that R=8. That is instead of subtracting both sides by 72R, then factoring out an “R,” coming to two conclusions for R.
you could have done that because you know that R is not 0, but I think he just wanted to factor and show it more clearly in another way.
Why don't you use in your explanations the binomial formulas?
wen you got to 72R=9R^2 m first tougt was to just divide both sides by R, giving 72=9R, so R=72/9=8
thinking outside the box..err.. circle, one this one
What's the radius of the orange circle? 😛
I could not see an algebra solution. Did it graphically. The center of the circle is found as you start at zero for R and increse it. You have 3 curves that eventually meet at a single point as R approaches 8. The 3 curves are x=R, the upper curve circle as the radius is 18-R and the lower circle as the radius is 18+R. Did it quick and dirty on a TI-84. Equation 1, the upper circle ... sqrt((18-R)squared-X squared)) +18.... equation 2, the lower circle .... sqrt(18+R)squared-(X-18)squared).... equation 3 is just a vertical line for x=R ..... (X-R)*10000. Think of as R gets bigger the alowable center position of the circle is restricted by needing to be R distance from the y axis and R distance from the upper circle and R distance from the lower circle. The allowable center point is inside the 3 curves and as R increase you get one allowable point as R gets its maximum value of 8.
Correct me if I'm wrong.. but I think you made a mistake... That looks like a yellow circle not an orange one 🤔
How are you able to assume that the line equal to r+18 when solving for the y variable is indead a straight line simply bevause its a tangent for both curves? Im having trouble figuring out how you would know that.
It's a basic circle theorem: when you have 2 circles that are tangent to each other, either externally or internally (aka 'kissing circles'), the centres of the circles and the tangent point are colinear.
Not good at math, so can someone explain how can we be sure that Radius at 0:35 Goes through the middle of green circle and hits tangent of both circles in the same point? is there some sort of law that confirms it always happens?
If you look at the point of contact between the green circle and the 36circle's circumference, there can be a perpendicular radius of the green circle drawn there. There can also be a perpendicular radius of the 36circle that can be drawn, since both are perpendicular to the same point, they coincide.
I got this after a while of thinking and my brain is now tired.
Both's radius are perpendicular to the same point.
In the very first step, how come you can assume that all those points are collinear? Couldn't it be that that those points are not on the same line, so you cannot connect them?
what points?
in the very first step of his solution (at 0:27), he draws a single line passing through
a) the centre of the large semicircle on the left
b) the centre of the green circle
c) the point of contact between the green circle and the large semicircle
How does he know for sure those points can be connected with a straight line of length 18?
@@dylanjayabahu2878 I think the diagram is to scale. That's why he did it. I get your point though that those points might not be colinear but I dot see many ways to solve it otherwise.
Sameee
Touching circles theorem, if two circles touch each other (internally or externally), then the point of contact will lie on the line joining the two centres. Same thing again at 2:27.
bro almost forgot to say how exciting
Pretty good