well i am in class 10 and there is no chapter on intersecting chords in cbse board but i think its called power of point which derived by similar triangle (i just learned this while preparing for olympiads sooo..)
I really like all these effects you do with the squares and such. The level of detail is amazing as well! Like completing the texture on squares that aren't fully visible in the original photo.
@carmicha I feel like that would be really annoying to start and get into a rhythm, realize that you wanted the visual to do something, and then pause to make it do that, then start back up almost seemlessly.
Andy's method (with secant chords theorem) is the quickest then the smartest one. Here is a less brilliant method : a : side length of yellow square b : side length of orange square c : side length of red square d : side length of green square x : radius of the semi-circle Equation 1 : a=sqrt(18)=3*sqrt(2) Equation 2 : b+d=a=3*sqrt(2) Equation 3 : sqrt(x^2-d^2)=d+sqrt(x^2-c^2) ; (particular vertex of green and red squares are in the semi-circle) Equation 4 : sqrt(x^2-d^2)+sqrt(x^2-b^2)=a=3*sqrt(2) ; (particular vertex of green and orange squares are in the semi-circle) a,b,c,d,x are 5 unknowns and we have 4 equations. We can not determine all of the 5 unknowns : a,b,c,d,x but we will see that we can determine : area=a^2+b^2+c^2+d^2 area=a^2+b^2+c^2+d^2 area=a^2+(b+d)^2-2*b*d+c^2 area=a^2+a^2-2*b*d+c^2 area=18+18+c^2-2*b*d area=36+c^2-2*b*d Equation 4 : sqrt(x^2-d^2)+sqrt(x^2-b^2)=3*sqrt(2) sqrt(x^2-b^2)=3*sqrt(2)-sqrt(x^2-d^2) x^2-b^2=(3*sqrt(2)-sqrt(x^2-d^2))^2 x^2-b^2=18+(x^2-d^2)-6*sqrt(2)*sqrt(x^2-d^2) -b^2=18-d^2-6*sqrt(2)*sqrt(x^2-d^2) 18+b^2-d^2=6*sqrt(2)*sqrt(x^2-d^2) 18+(3*sqrt(2)-d)^2-d^2=6*sqrt(2)*sqrt(x^2-d^2) (thanks to Equation 2) 18+(18+d^2-6*sqrt(2)*d)-d^2=6*sqrt(2)*sqrt(x^2-d^2) 36-6*sqrt(2)*d=6*sqrt(2)*sqrt(x^2-d^2) 3*sqrt(2)-d=sqrt(x^2-d^2) b=sqrt(x^2-d^2) ; x^2=b^2+d^2 Equation 3 : sqrt(x^2-d^2)=d+sqrt(x^2-c^2) b=d+sqrt(x^2-c^2) (thanks to Equation 4) (b-d)^2=x^2-c^2 b^2+d^2-2*b*d=x^2-c^2 x^2-2*b*d=x^2-c^2 c^2-2*b*d=0 area=36+c^2-2*b*d area=36
I have also solved it the way you did. But I always make all equations first and simplify them to get the final equation. So i ended up with the equation A = 2a^2. And THAT was really exciting ;)
Oh my gosh! I was so sure this would finally be the video where you just troll us and say, "The answer is inconclusive. There is not enough information." Even though it uses pretty simple concepts, I don't know if I could have ever figured it out on my own. Did anybody?
Can do tgis without any algebra: basically pick sized for the inner squares at the edge cases, if tge green square is of zero legth thab the red square is of zero length and the orange square matches the yellow, thus the size is 2*18=36
So a^2 = b^2 + c^2 + d^2 means that the area of the green, red and orange squares equal the area of the yellow square, regardless of any numbers, this feels much deeper than 36 sq units.
I failed to complete it but I started with : (c + d) ^ 2 + d^2 = c^2 c + 2d = sqrt(18) There is a way to continue and solve from there without the intersecting chord theorem, right?
The fact that the circle is there tells me I probably have to use it in some way, but given i don’t remember anything about chord theorems, is there another way using algebra and ratios? There’s gotta be no?
just make equal lines until everything is equally lined and very simply do some basic addition nvm the square behind must have a triangle shape by that thing
Here's the question. To forward you the diagram, please let me know the social media accounts to reach out to you. I appreciate if anyone answers this question. Let ABC be a scalene triangle of area 6 sq units and in-radius 2 sq units (circle is inscribed in a triangle). A1 B1 C1 are the feets of the altitudes of the triangle through the vertices A, B, C. If A2, B2, C2 are the midpoints of the sides BC, CA and AB respectively, then the length of the broken line A1 B2 C1 A2 B1 C2 A1 is?
Ok as usual my brain was clinging on my its fingernails so I probably misunderstood something obvious... but in order for the intersecting chords thing to work, don't we have to assume that the corners of the squares are touching the edge of the semicircle?
So basically, because only one measurement was given, the image was almost perfectly proportional, so I eyeballed it and guessed 36 and was pleasantly surprised it was right.
Solve for positive integers m, n less than or equal to nine such that mn + m + n = (mn)* where (mn)* is the concatenation of m and n. I suggest trying m = 6 after solving the puzzle 😉
Since there are no rules on the largest square we can move the left corner to the edge of the circle. This makes green and red square disappear and orange to match the yellow one. So the sum of areas will be double the area of largest one. Thats obviously the fastest solution but it makes a (reasonable) assumption that the puzzle CAN be answered. Not a method for all math problems but very useful for puzzles.
@@harold2 You'll have to picture the squares in your mind, or try drawing them on a piece of paper. Start with the existing setup, where the Green square is inside the Yellow Square and its top left corner is touching the semicircle. Similarly, the Red square is inside the orange square, with its upper left corner touching the semicircle. The Yellow square's width is equal to the orange square's width plus the green square's width. If you move the Yellow square over, the green square gets smaller until the green square is zero, which means the Yellow square is equal to the Orange square. Remember that as the widths change for a square, so do their heights. As the Yellow square is being moved over, it is also dragging the right side of itself and the Orange square as well, forcing the red square farther to the left as well. As the Red square gets moved to the left, its height gets moved towards zero (since the semicircle gets closer to y=0), until the red square also gets eliminated. This leaves only the Yellow square and the Orange square, both of them on top of each other. The Orange square cannot get any smaller and still remain a square (try drawing regular quadragons that have their bottom left corner at the left corner of the semicircle and their upper right corner on the semicircle's perimeter, and see how many are a square). If you expand the Yellow square beyond the halfway point of the semicircle, then the Orange square starts to go back down in size resulting in the Green and Red squares reappearing.
@@toddkes5890 that would work. However, sometimes the images are not to scale and although visually it may look like you have the answer, the real answer may be very different. Like the Amazon question on the 80ft rope.
Never even heard of the intersecting chords theorem. What a lovely, niche tool to have in one's back pocket.
We Indians have a separate chapter in math about intersecting chords theorem
@@yogindras4402 uhh no, we dont,
@@ShlokAhuja-gw7zmhe's probably talking about the circle chapter in 9th or 10th grade
well i am in class 10 and there is no chapter on intersecting chords in cbse board but i think its called power of point which derived by similar triangle (i just learned this while preparing for olympiads sooo..)
@@ikik-ko6zs I think it's present in icse
I remember studying about it
3:26 "That is... elegant." Gotta agree.. When I saw that coming together I thought "wait a minute..."
Fr bro like, Let him and her cook bro 🔥🔥🔥
@@bebektoxic2136who her
I really like all these effects you do with the squares and such.
The level of detail is amazing as well! Like completing the texture on squares that aren't fully visible in the original photo.
So true! I'm really curious how he does it behind the scenes. Does it mess with the flow of his presentation?
@@justinvance9221 There’s a lot of cuts.
@carmicha I feel like that would be really annoying to start and get into a rhythm, realize that you wanted the visual to do something, and then pause to make it do that, then start back up almost seemlessly.
Andy's method (with secant chords theorem) is the quickest then the smartest one.
Here is a less brilliant method :
a : side length of yellow square
b : side length of orange square
c : side length of red square
d : side length of green square
x : radius of the semi-circle
Equation 1 : a=sqrt(18)=3*sqrt(2)
Equation 2 : b+d=a=3*sqrt(2)
Equation 3 : sqrt(x^2-d^2)=d+sqrt(x^2-c^2) ; (particular vertex of green and red squares are in the semi-circle)
Equation 4 : sqrt(x^2-d^2)+sqrt(x^2-b^2)=a=3*sqrt(2) ; (particular vertex of green and orange squares are in the semi-circle)
a,b,c,d,x are 5 unknowns and we have 4 equations.
We can not determine all of the 5 unknowns : a,b,c,d,x but we will see that we can determine : area=a^2+b^2+c^2+d^2
area=a^2+b^2+c^2+d^2
area=a^2+(b+d)^2-2*b*d+c^2
area=a^2+a^2-2*b*d+c^2
area=18+18+c^2-2*b*d
area=36+c^2-2*b*d
Equation 4 :
sqrt(x^2-d^2)+sqrt(x^2-b^2)=3*sqrt(2)
sqrt(x^2-b^2)=3*sqrt(2)-sqrt(x^2-d^2)
x^2-b^2=(3*sqrt(2)-sqrt(x^2-d^2))^2
x^2-b^2=18+(x^2-d^2)-6*sqrt(2)*sqrt(x^2-d^2)
-b^2=18-d^2-6*sqrt(2)*sqrt(x^2-d^2)
18+b^2-d^2=6*sqrt(2)*sqrt(x^2-d^2)
18+(3*sqrt(2)-d)^2-d^2=6*sqrt(2)*sqrt(x^2-d^2) (thanks to Equation 2)
18+(18+d^2-6*sqrt(2)*d)-d^2=6*sqrt(2)*sqrt(x^2-d^2)
36-6*sqrt(2)*d=6*sqrt(2)*sqrt(x^2-d^2)
3*sqrt(2)-d=sqrt(x^2-d^2)
b=sqrt(x^2-d^2) ; x^2=b^2+d^2
Equation 3 :
sqrt(x^2-d^2)=d+sqrt(x^2-c^2)
b=d+sqrt(x^2-c^2) (thanks to Equation 4)
(b-d)^2=x^2-c^2
b^2+d^2-2*b*d=x^2-c^2
x^2-2*b*d=x^2-c^2
c^2-2*b*d=0
area=36+c^2-2*b*d
area=36
I have also solved it the way you did. But I always make all equations first and simplify them to get the final equation. So i ended up with the equation A = 2a^2. And THAT was really exciting ;)
you should do some more calculus videos
Yes, more calculus
Yes more calculos
Yes more calculus
DUDE your videos cure my brainrot love your approaches
Wow, your solution really snuck up there! Was not expecting it to be that easy. Love your work!
how exciting 🗣️🗣️
Oh my gosh! I was so sure this would finally be the video where you just troll us and say, "The answer is inconclusive. There is not enough information." Even though it uses pretty simple concepts, I don't know if I could have ever figured it out on my own. Did anybody?
I definitely knew you had to use the diagonals to solve the question, but would never have known what formulas to use or how to use them lol
How. Brilliant.
Oooh first time I hear about the Intersecting Chords Theorem. Good to know. Thank You.
Thank you Andy!
Can do tgis without any algebra: basically pick sized for the inner squares at the edge cases, if tge green square is of zero legth thab the red square is of zero length and the orange square matches the yellow, thus the size is 2*18=36
Dude. Yes, I am excited. Rock on
Obrigado por incendiares os meus neurónios!... 😀 How exciting!
This is an elegant solution.
How exciting
So a^2 = b^2 + c^2 + d^2 means that the area of the green, red and orange squares equal the area of the yellow square, regardless of any numbers, this feels much deeper than 36 sq units.
This one was soooo satisfying
3:30 From the figure is doesn’t look like a = b+d. Is the figure incorrect?
He's talking only about the sides and not the area, when you square the sides you'll get the area. So the figure is correct.
Math is amazing!
All of these Katerina problems are great.
I’m terrible with math but I have a crush so I watch😂
Oh, boy.
Catriona Agg and Andy Math.
It's one of those nights.
Oh, boy. 😊
Brilliant!!!
I failed to complete it but I started with :
(c + d) ^ 2 + d^2 = c^2
c + 2d = sqrt(18)
There is a way to continue and solve from there without the intersecting chord theorem, right?
✨Magic!✨
Good one.
The fact that the circle is there tells me I probably have to use it in some way, but given i don’t remember anything about chord theorems, is there another way using algebra and ratios? There’s gotta be no?
God bless, man
just make equal lines until everything is equally lined and very simply do some basic addition
nvm the square behind must have a triangle shape by that thing
Hey Andy, i have got a Maths especially geometry question for you. Kindly let me know how to reach out to you?
Here's the question. To forward you the diagram, please let me know the social media accounts to reach out to you. I appreciate if anyone answers this question.
Let ABC be a scalene triangle of area 6 sq units and in-radius 2 sq units (circle is inscribed in a triangle). A1 B1 C1 are the feets of the altitudes of the triangle through the vertices A, B, C. If A2, B2, C2 are the midpoints of the sides BC, CA and AB respectively, then the length of the broken line A1 B2 C1 A2 B1 C2 A1 is?
that was fun!
Ok as usual my brain was clinging on my its fingernails so I probably misunderstood something obvious... but in order for the intersecting chords thing to work, don't we have to assume that the corners of the squares are touching the edge of the semicircle?
So basically, because only one measurement was given, the image was almost perfectly proportional, so I eyeballed it and guessed 36 and was pleasantly surprised it was right.
Done without intersecting chords.
18. That was too easy.
Solve for positive integers m, n less than or equal to nine such that
mn + m + n = (mn)*
where (mn)* is the concatenation of m and n.
I suggest trying m = 6 after solving the puzzle 😉
Since there are no rules on the largest square we can move the left corner to the edge of the circle. This makes green and red square disappear and orange to match the yellow one. So the sum of areas will be double the area of largest one. Thats obviously the fastest solution but it makes a (reasonable) assumption that the puzzle CAN be answered. Not a method for all math problems but very useful for puzzles.
I dont understand
@@harold2 You'll have to picture the squares in your mind, or try drawing them on a piece of paper. Start with the existing setup, where the Green square is inside the Yellow Square and its top left corner is touching the semicircle. Similarly, the Red square is inside the orange square, with its upper left corner touching the semicircle. The Yellow square's width is equal to the orange square's width plus the green square's width.
If you move the Yellow square over, the green square gets smaller until the green square is zero, which means the Yellow square is equal to the Orange square. Remember that as the widths change for a square, so do their heights. As the Yellow square is being moved over, it is also dragging the right side of itself and the Orange square as well, forcing the red square farther to the left as well. As the Red square gets moved to the left, its height gets moved towards zero (since the semicircle gets closer to y=0), until the red square also gets eliminated.
This leaves only the Yellow square and the Orange square, both of them on top of each other. The Orange square cannot get any smaller and still remain a square (try drawing regular quadragons that have their bottom left corner at the left corner of the semicircle and their upper right corner on the semicircle's perimeter, and see how many are a square). If you expand the Yellow square beyond the halfway point of the semicircle, then the Orange square starts to go back down in size resulting in the Green and Red squares reappearing.
@@toddkes5890 that would work. However, sometimes the images are not to scale and although visually it may look like you have the answer, the real answer may be very different. Like the Amazon question on the 80ft rope.
@@ajaxfire That's why you always do the math.
2bd or not 2bd. How elegant.
That's too good of a joke XD
You ate that
the total area of all 4 squares is 18
brilliant
The answer is 18 as they all overlap
POV u did all that perfectly only to fail 2nd grade math and get 34 from 18+18 (something I would do)
neat
Damn
Nahh bro you're terrible the answer is 34 the smallest square is as 2x as small than you think it is
I would say the total area is 18. All of the squares are contained within the 18 unit area.
I didn't understand and was like what squares at the start
But isn't the red one the biggest