Yeah me neither. But I think it’s pretty commmon that if you can’t solve for any variables in a problem, then either something will cancel out, or you’ll have to look out for something like this, so I guess we have to have these tricks in the back of our mind 😅
Making the identified 2 length into a triangle was very clever, that's the part I missed and got stuck on. I realized the answer would be a sum of a series of semi-circles, but I got tripped up by the 2 length not being in the centre of any of the semi-circles. How exciting
Duplicate the whole shape and rotate it 180 degrees around center point of the large semicircle. We are left with concentric circles. Also, by intersecting chords: (R+r)(R-r) = 2x2 => R² - r² = 4. Multiply everything by π: πR² - πr² = 4π. That looks important, let's put box around it. [4π] This describes the difference of the concentric circles, which is twice the area of the shaded region (from our original duplication) Therefore 4π / 2 => 2π, and that's another way to get our answer!
Based on how the question is asked, we can assume that the answer does not depend on the value of r. If we set r=0, it is straightforward that R = 2 and Area = pi/2 * R² + pi/2 * (R/2)² - pi/2 * (R/2)² = pi/2 * R² = 2*pi
You can solve this one without any algebra! As it doesn't put any constraints on the size of the smallest semicircle I just assumed it to have zero size. You can imagine shrinking it down, along with the other two semicircles connected to it, until it's just a single point in the centre of the largest semicircle, which now has radius 2. We now have a yin-yang shape made of a smaller solid semicircle on the bottom right, and a larger semicircle on the left with a chunk taken out that's equal to the smaller semicircle. Move the smaller semicircle into that gap, and you have a single, solid semicircle with radius 2, and area 2 pi.
From the largest and smallest semicircles being concentric, we can deduce the 2 medium semicircles are equal (Radius = (R+r)/2). From that there is an interesting bit of symmetry when we copy the blue area but rotated 180 degrees (having the equal medium semicircles overlap). We get the large outer circle - the small inner circle = 2x the blue area. This is more evident after you have solved it, but it is a nice way to verify the solution.
Just treat the smaller concentric circle as if it were a point circle (since its exact dimensions are immaterial to the solution), and one is left with a shaded semicircle with radius 2 (plus and minus semicircles of dimeter 2) making the answer obvious.
I have a find area challenge, classified as one of the hardest find area challenge ever made in Hong Kong public exam. Can you help solve it? 4 x 1/4 circle is placed inside a square with radius r, with center of 4 circles placed on each corner of the square can you tell me the overlapping center area? This question has been in my head for soooo long it is like : || square square square square square square square square center of 1/4 circle A center of 1/4 circle B square square square square square square square the area wanted square square square square square square square center of 1/4 circle C center of 1/4 circle D square square square square square square square square ||
There are 4 semicircles. 1. Large blue/white 2. Small blue/white 3. Large white 4. Small white, then you need all the areas including large white which was highlighted green for a bit
A technicality, but largest and smallest are grammatically superlative, implying there are at least 3 semicircles (If only the largest and smallest were semicircles, they would be larger and smaller instead of largest and smallest). There appear to be 4 semicircles, at least 3 of them must be semicircles, and the only way to solve this is to assume they all 4 are.
I understand the math’s equations and the solution but I’m struggling with the logic of the solution. The equations for the Left side and the Right side equations make sense. Individually they are the total semicircle area - the white area and therefore make the area of the blue sided area for the respective Left and Right sides. Logically if you place the two equations together the solution would be the Total size for the shaded area. However, when simplified the equation is really the big Left semicircle area - the small Right semicircle area. Aren’t you then just finding the area of the Blue shaded area on the Right? I’m sure I’m confused in some respect but I can’t wrap my head around this. Any help would be appreciated thank you!
It helps to notice the medium semicircles have the same diameter/radius/area. Let's say A=Largest semicircle, B=Medium semicircle, C=Smallest semicircle The left is A-B. The right is B-C. The total is (A-B) + (B-C) = A-B+B-C = A + (B-B) - C = A-C. The total blue shaded area is A-C. The blue shaded area on the Right is only B-C.
Once we notice the medium semicircles have the same diameter/radius/area, we can visualize the total (Large-Medium+Medium-Small) by putting the right side into the gap in the left side. The 2 medium sized semicircles are equal and the same shape, so it is a perfect fit. This geometric visualization is the same as the the algebraic step of canceling Medium-Medium=0. We are left with the Large semicircle from the left - the Small semicircle from the right (that was brought over to the left when we visualized canceling out the equal medium semicircles by moving the right into the left).
This sponsorship looks important. Let’s put a box around it.
How exciting
Your videos have shown me the beauty of figuring out exact values without having to solve for every missing value. Best math channel on youtube.
I agree!
How exciting
3blue1brown AHEM
Congrats on the sponsorship bro!
How exciting
Didn't expect R²-r² to be used like that with the 4
Yeah me neither. But I think it’s pretty commmon that if you can’t solve for any variables in a problem, then either something will cancel out, or you’ll have to look out for something like this, so I guess we have to have these tricks in the back of our mind 😅
This channel has made me way more interested in math than any math teacher I ever had in high school
Not too hard when he solves all of the problems for you 😂
This was a wild question, and I am very glad I got to see it. Thank you sir
Yes big R and liTTTTTTle r
He's just pronouncing the word correctly
@@danielszekeres8003 *robotically
Brilliant!!! How exciting.
Making the identified 2 length into a triangle was very clever, that's the part I missed and got stuck on. I realized the answer would be a sum of a series of semi-circles, but I got tripped up by the 2 length not being in the centre of any of the semi-circles. How exciting
This is the only channel where I'll stick around for the ad read. You earned it boss
how exiting
hoooow exciting
That's important put a box around it 😊
I love how it all jut comes together
How exciting! ❤
Sponsorship! So exciting
Duplicate the whole shape and rotate it 180 degrees around center point of the large semicircle. We are left with concentric circles. Also, by intersecting chords: (R+r)(R-r) = 2x2 => R² - r² = 4.
Multiply everything by π: πR² - πr² = 4π. That looks important, let's put box around it. [4π]
This describes the difference of the concentric circles, which is twice the area of the shaded region (from our original duplication)
Therefore 4π / 2 => 2π, and that's another way to get our answer!
How endearing!
Based on how the question is asked, we can assume that the answer does not depend on the value of r.
If we set r=0, it is straightforward that R = 2 and Area = pi/2 * R² + pi/2 * (R/2)² - pi/2 * (R/2)² = pi/2 * R² = 2*pi
You can solve this one without any algebra!
As it doesn't put any constraints on the size of the smallest semicircle I just assumed it to have zero size. You can imagine shrinking it down, along with the other two semicircles connected to it, until it's just a single point in the centre of the largest semicircle, which now has radius 2.
We now have a yin-yang shape made of a smaller solid semicircle on the bottom right, and a larger semicircle on the left with a chunk taken out that's equal to the smaller semicircle. Move the smaller semicircle into that gap, and you have a single, solid semicircle with radius 2, and area 2 pi.
From the largest and smallest semicircles being concentric, we can deduce the 2 medium semicircles are equal (Radius = (R+r)/2). From that there is an interesting bit of symmetry when we copy the blue area but rotated 180 degrees (having the equal medium semicircles overlap). We get the large outer circle - the small inner circle = 2x the blue area. This is more evident after you have solved it, but it is a nice way to verify the solution.
Just treat the smaller concentric circle as if it were a point circle (since its exact dimensions are immaterial to the solution), and one is left with a shaded semicircle with radius 2 (plus and minus semicircles of dimeter 2) making the answer obvious.
Beautiful
Nah the sponsorship was smooth and brilliantly executed with the flow
LiTTle r
How exciting and brilliant
How exciting
Loved you in smalville btw. Problem is wild)
I'm so surprised when problems like this end up having such a clean integer solution.
Hey bro
I love your content.
What tool do you edit with?
I would like to make content like yours.
How exciting :)
Que emocionado
how exciting!
How Brilliant
这个图案和一种☯️图很像，先天太极图
Love it once again :)
how exciting
Insane how you got the area without finding r nor R first
Another sponsor!!
this channel is a big W
please andy i need to know where you find theese problems😩😩
What software are you using for the presentations? Is it just a slideshow that you are working through?
Cooking to Betty Crocker and Taylor Swift is life.
Did I miss the "cuz I'm about to solve it in 3, 2, 1"?
Do we assume that they are all 4 semi-circles?
LITOU 🗣🗣🗣🗣
GYATTTT 😍
It looks like the little circle on the left side has radius (R-r), not (R+r)
Just not true.
It has the diameter of (R+r).
3:21 to skip the ad
I have a find area challenge, classified as one of the hardest find area challenge ever made in Hong Kong public exam.
Can you help solve it?
4 x 1/4 circle is placed inside a square with radius r, with center of 4 circles placed on each corner of the square
can you tell me the overlapping center area? This question has been in my head for soooo long
it is like :
||
square square square square square square square square
center of 1/4 circle A center of 1/4 circle B
square square
square square
square square
square the area wanted square
square square
square square
square square
center of 1/4 circle C center of 1/4 circle D
square square square square square square square square
||
*"The sky is pickles"*
Sorry if it's a silly question. How can you assume those are all semi circles when it's not stated?
It is actually stated in words above the initial diagram. Though it could be clearer.
how can i send questions for u to do ?
i don't get the logic behind getting the diameter of the smaller green one, can anyone elaborate?
There are 4 semicircles. 1. Large blue/white 2. Small blue/white 3. Large white 4. Small white, then you need all the areas including large white which was highlighted green for a bit
Who up andying they math rn
is it assumed that all the halfcircles are in fact circles and not only the largest and smallest?
A technicality, but largest and smallest are grammatically superlative, implying there are at least 3 semicircles (If only the largest and smallest were semicircles, they would be larger and smaller instead of largest and smallest). There appear to be 4 semicircles, at least 3 of them must be semicircles, and the only way to solve this is to assume they all 4 are.
box
Lituhl r squared
Hi
#oddlysatisfying
stop annunciating the T in little.
I understand the math’s equations and the solution but I’m struggling with the logic of the solution.
The equations for the Left side and the Right side equations make sense. Individually they are the total semicircle area - the white area and therefore make the area of the blue sided area for the respective Left and Right sides.
Logically if you place the two equations together the solution would be the Total size for the shaded area. However, when simplified the equation is really the big Left semicircle area - the small Right semicircle area. Aren’t you then just finding the area of the Blue shaded area on the Right?
I’m sure I’m confused in some respect but I can’t wrap my head around this. Any help would be appreciated thank you!
It helps to notice the medium semicircles have the same diameter/radius/area.
Let's say A=Largest semicircle, B=Medium semicircle, C=Smallest semicircle
The left is A-B. The right is B-C.
The total is (A-B) + (B-C) = A-B+B-C = A + (B-B) - C = A-C.
The total blue shaded area is A-C. The blue shaded area on the Right is only B-C.
Once we notice the medium semicircles have the same diameter/radius/area, we can visualize the total (Large-Medium+Medium-Small) by putting the right side into the gap in the left side. The 2 medium sized semicircles are equal and the same shape, so it is a perfect fit. This geometric visualization is the same as the the algebraic step of canceling Medium-Medium=0.
We are left with the Large semicircle from the left - the Small semicircle from the right (that was brought over to the left when we visualized canceling out the equal medium semicircles by moving the right into the left).
Waw
i hate when he doesn't count down
458th viewer 😂
1,472 here!
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Try Vegan Please
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No
how exciting!
How exciting