@@heliocentric1756 Michael already posted the videos for the next few weeks (he’s building a new studio so obviously he won’t be available to record videos) so I went ahead and post related homeworks under unlisted videos.
the twist is the backflip, damn P.S. if a and b are equal, then the two terms cancel out and just leaving the area of the triangle, thus the shaded area of the crescent will be EQUAL to the area of the triangle beneath it. Awesome.
What if a is smaller than b? Some portion of the quarter circle will lay outside the semi circle. How should that be counted? Or what would be the green shaded area?
HOMEWORK : The square ABCD has side length 2√2. A circle with centre A and radius 1 is drawn. A second circle with centre C is drawn so that it just touches the first circle at point P on AC. Determine the total area of the regions inside the square but outside the two circles. SOURCE : Mathematical Mayhem from Crux Mathematicorum Vol. 37, No. 6
SOLUTION *8 − 2√2 − 5π/2 +9arccos((2√2)/3)* From the Pythagorean Theorem we get AC = √((2√2)² + (2√2)²) = √(8 + 8) = √16 = 4. Since CP is a radius of the larger circle, and AP = 1 we get CP = CA − PA= 4−1 = 3, therefore the radius of the large circle is 3. Now let's introduce more points : - Point E is the intersection of the circle with centre C and segment AB - Point F is the intersection of the circle with centre C and segment AD - Point G is on the circle with centre C such that the segment GDC is a radius of that circle - Point H is on the circle with centre C such that the segment HBC is a radius of that circle - Point I is the intersection of the circle with centre A and segment AB - Point J is the intersection of the circle with centre A and segment AD We're looking for the areas of the figures JPF and IPE. Using [A] to represent the area of figure A we get, [GDF] = [FCG] − [FCD] = α/2π · π(3)² - (√(3² - (2√2)²) · 2√2)/2 = 9α/2 - √2 where α = ∠FCG = arccos((2√2)/3). Next, we have [PCDF] = [PCG] − [GDF] = 9π/8 − 9α/2 + √2. Thus [JPF] = [ACD] − [APJ] − [PCDF] = 4 − π/8 − (9π/8 − 9α/2 + √2) = 4 − √2 − 5π/4 +9α/2. By symmetry [IPE] = [JPF] thus the area of the regions inside the square but not inside a circle is 2[JPE], or 8 − 2√2 − 5π/2 +9α = 8 − 2√2 − 5π/2 +9arccos((2√2)/3) ≈ 0.38 units.
Not a super difficult one but still an interesting puzzle - suppose you have a square of side length 1. Inside the square is a circle of the maximum possible radius that can fit in the square, which is shaded in. But a square has been cut out of the circle, such that it kisses the circle at 4 points, which isn't shaded in. But a circle is inside this smaller square, shaded in, with a square cut out of it, and a circle in this square, and so on to infinity. The question is, what is the total shaded area?
Depending on what you want the result for, you can then put the last two terms together and factor them using difference if two squares. That might, or might not, be a better place to stop
It's not conceptually hard, just tedious, and probably very messy. Call the intersection of the quarter circle and line AC Z. Calculate the coordinates of Z. From those you can compute the area of the triangle ZBC and the circular segment PBZ (get the angle PBZ from the dot product of vectors BP and BZ). The area of the patch is then the area of triangle ABC minus the the areas calculated. There's another geometric solution using the area of the circular cap to the right of AC, the area of triangle ABC and the quarter circle. Using calculus you can calculate the area under the circular arc PZ and the associated smaller triangle. Take your pick. I think the first one is the easiest.
Why did you go against the naming convention for the sides of a right triangle? The usual convention is that vertex C is at the right angle, side c is the hypotenuse, side a is opposite vertex A and side b is opposite vertex B. Following that convention, your ending formula be correct if you replaced your labels A, B, and C with the labels B, C, and A respectively. Or am I missing something?
Hi Michael. Did you know, a 3D body rotating under inertia describes a rotation about the axis of angular momentum which is a function of the time difference only?
By far, one of the most annoying things about youtube videos is the audio mixing. There needs to be a standard for youtube audio levels. The vocals audio levels are very low in this video, then there is blasting music that comes in at some point (flip performance). It's so annoying. RUclipsrs, please put a little more effort into making sure your audio levels are consistent across the entire video. Otherwise it makes it so that the watcher has to constantly be adjusting the volume up and down throughout the video. This is so irritating!
1:04 Frontflip
4:34 Good Place To Stop
hol up the vid was released a few minutes ago
You've chosen a good time to comment.. Basically 2 days before publishing the video.
@@heliocentric1756 Michael already posted the videos for the next few weeks (he’s building a new studio so obviously he won’t be available to record videos) so I went ahead and post related homeworks under unlisted videos.
Me: this looks like something I can do!
Penn: does a frontflip
Me: ...
Dude that introductory backflip was dope!
It was a frontflip tho :)
length 'a' was incorrectly labelled 'b' though!
>he
@@shokan7178 he meant (1/backflip)^-1
0:54 I started working on the problem. 1:05 I started working on doing a front-flip. Nicely done. 👍🏼👍🏼👏🏼👏🏼
A front flip?!!!!!
That is the twist!
And, when you thought nothing beats the backflip in comes 1:04.
Thank you, professor.
Thank for michael penn
Good Place To Start at 1:04 this time
Hello, remember me from Physics galaxy channel
@@targetiitbcse1761 yes I do
the twist is the backflip, damn
P.S. if a and b are equal, then the two terms cancel out and just leaving the area of the triangle, thus the shaded area of the crescent will be EQUAL to the area of the triangle beneath it. Awesome.
And that is the difference between a quarter circle and a half circle (with different radii, but still).
@@57thorns radii?
That's actually a front flip, not a backflip.
and you can see it in the answer too, because if a and b are equal then the terms cancel out.
unexpectedly easy for what i thought would just be outright absurd
This was a quite easy problem for a Michael Penn video.
This was surprisingly short!
Explanation was elegant and perfect.
Made perfect sense. Nicely done Sir! Thanks for posting the video 😊😊
I’ve only seen two videos and I already love this channel
Michael you make it so easy to understand thanks for sharing the calculations are so easy to understand as well.
Great Personality Professor 👍
Love the figures on the chalkboard.
I found the case where a
Just a tip. You should be able to position your legs straight before you fall, honestly you're jumping too low. Be careful xP
What if a is smaller than b? Some portion of the quarter circle will lay outside the semi circle. How should that be counted? Or what would be the green shaded area?
"How should that be counted" seems like a definitional problem. What area are you even looking for at that point?
@@TJStellmach You could try to match the solution from the video to the problem. But the easiest would be to say that a ≥ b.
very nice slow motion demonstration of conservation of angular momentum: the more he curls up the faster he rotates.
Wow that was a very low front flip. Nice!
I think you should mention that a ≥ b, because otherwise the problem gets a bit more complicated
My answer in that case is (1/2)ab - (1/4)(b^2-a^2)arcsin(2ab/(a^2+b^2))
A frontflip! This is amazing!
HOMEWORK : The square ABCD has side length 2√2. A circle with centre A and radius 1 is drawn. A second circle with centre C is drawn so that it just touches the first circle at point P on AC. Determine the total area of the regions inside the square but outside the two circles.
SOURCE : Mathematical Mayhem from Crux Mathematicorum Vol. 37, No. 6
Michael.
@@athelstanrex Penn.
8-5pi/2
Ans 8-5π/2
SOLUTION
*8 − 2√2 − 5π/2 +9arccos((2√2)/3)*
From the Pythagorean Theorem we get AC = √((2√2)² + (2√2)²) = √(8 + 8) = √16 = 4. Since CP is a radius of the larger circle, and AP = 1 we get CP = CA − PA= 4−1 = 3, therefore the radius of the large circle is 3.
Now let's introduce more points :
- Point E is the intersection of the circle with centre C and segment AB
- Point F is the intersection of the circle with centre C and segment AD
- Point G is on the circle with centre C such that the segment GDC is a radius of that circle
- Point H is on the circle with centre C such that the segment HBC is a radius of that circle
- Point I is the intersection of the circle with centre A and segment AB
- Point J is the intersection of the circle with centre A and segment AD
We're looking for the areas of the figures JPF and IPE.
Using [A] to represent the area of figure A we get,
[GDF] = [FCG] − [FCD] = α/2π · π(3)² - (√(3² - (2√2)²) · 2√2)/2 = 9α/2 - √2 where α = ∠FCG = arccos((2√2)/3).
Next, we have [PCDF] = [PCG] − [GDF] = 9π/8 − 9α/2 + √2.
Thus [JPF] = [ACD] − [APJ] − [PCDF] = 4 − π/8 − (9π/8 − 9α/2 + √2) = 4 − √2 − 5π/4 +9α/2.
By symmetry [IPE] = [JPF] thus the area of the regions inside the square but not inside a circle is 2[JPE], or 8 − 2√2 − 5π/2 +9α = 8 − 2√2 − 5π/2 +9arccos((2√2)/3) ≈ 0.38 units.
Not a super difficult one but still an interesting puzzle - suppose you have a square of side length 1. Inside the square is a circle of the maximum possible radius that can fit in the square, which is shaded in. But a square has been cut out of the circle, such that it kisses the circle at 4 points, which isn't shaded in. But a circle is inside this smaller square, shaded in, with a square cut out of it, and a circle in this square, and so on to infinity. The question is, what is the total shaded area?
This is one of the few videos of yours that I could easily do on my own
Depending on what you want the result for, you can then put the last two terms together and factor them using difference if two squares. That might, or might not, be a better place to stop
A better place to stop? What is this heresy? :-P
Nice video Dr. Penn! Very impressive flip!!!
1:02 linear isometry
One thing may be proved first: arc AC and arc PC don't crossover. If a≤b, what will happen?
My answer in that case is (1/2)ab - (1/4)(b^2-a^2)arcsin(2ab/(a^2+b^2))
Please bring in some more problems based on combinatorics, also if you could take up some calculus problems from the jee advanced exam.
The front-flip flex is why I will never be a stellar mathematician.
(1/8)(4ab+πa²-πb²)
math with parcore!
Michael Penn I like this small problem... and the big jump ;-)
3:35 Last I checked, pie are round.
The front flip is at 1:02
Easiest one 1⃣
Hmm. (ab/2) + pi(a^2+b^2)/8 - pi*b^2/4. Slightly simplify to (ab/2) + pi(a^2-b^2)/8.
Easier than what it looks ngl, great stuff!
Homework for timetravlers again :prove or disprove that pi+e is transdental
Dude, did you really Somersault standing? Wow!
Where was the big twist? At 1:03 ?
Couldn't the entire shaded figure be described as a kind of big twist?
Can every individual area be determined?
I can't find a way myself.
My dream is to have a blackboard as big as Michael's
Half circle+right triangle-quarter circle=our answer
Why you dont make a book about this
I love the chalk and the flip 😂😁
Would you like to make a video on the brachistochrone problem? I think it'd be neat :)
Hmm, how about finding the area of that small patch on the upper left side of the red right-angled triangle? lol
It's not conceptually hard, just tedious, and probably very messy. Call the intersection of the quarter circle and line AC Z. Calculate the coordinates of Z. From those you can compute the area of the triangle ZBC and the circular segment PBZ (get the angle PBZ from the dot product of vectors BP and BZ). The area of the patch is then the area of triangle ABC minus the the areas calculated. There's another geometric solution using the area of the circular cap to the right of AC, the area of triangle ABC and the quarter circle. Using calculus you can calculate the area under the circular arc PZ and the associated smaller triangle. Take your pick. I think the first one is the easiest.
Thank you Michael, geometry is good, Tnx😍❤
why the secound circle calculation becomes plz give reason
🙏
Well well well
I thought this problem required some heavy Integral calculus, I was fooled.
Motion to change QED for "And that's a good place to stop".
Seconded.
Michael Penn is gonna be the next Ronald Graham with that acrobatics.
Mathrobatics.
Your vids are just enjoyable to watch, thanks:)
Wow!
A frontal somersault in slow motion with jazzy music playing in the background?? Is there anything you can't do Michael?? :-)
Maybe he can't do organic chemistry videos
Oh yeah nice Monke Flip lol
Looks like about 5. 5 area.
Why did you go against the naming convention for the sides of a right triangle? The usual convention is that vertex C is at the right angle, side c is the hypotenuse, side a is opposite vertex A and side b is opposite vertex B. Following that convention, your ending formula be correct if you replaced your labels A, B, and C with the labels B, C, and A respectively. Or am I missing something?
I have noticed before that Michael doesn't use that convention. But that's all it is, a convention: nobody says you have to follow it
I did it the unnecessary way, with trigonometric functions. Over complicating it as usual.
where is a big twist?
Hi,
For fun:
1 "and so on and so forth".
Hi Michael. Did you know, a 3D body rotating under inertia describes a rotation about the axis of angular momentum which is a function of the time difference only?
I will use integration 😀
Awesome video
normal people: 8 him: o
o
这个问题就有点水了，但是这个前空翻。。。我靠
Penn Flip
Class 10 th mathematics in cbse
Ikr
Yep
The way u did is bit lengthy , it would have been simpler if u assign diff parts diff variables then add and subtract
EZ
Man you're such a hottie. You should open an onlyfans
By far, one of the most annoying things about youtube videos is the audio mixing. There needs to be a standard for youtube audio levels. The vocals audio levels are very low in this video, then there is blasting music that comes in at some point (flip performance). It's so annoying. RUclipsrs, please put a little more effort into making sure your audio levels are consistent across the entire video. Otherwise it makes it so that the watcher has to constantly be adjusting the volume up and down throughout the video. This is so irritating!