It's not hard. Think what the area of each full circle would be, and then you can find the radius of each. This lets you figure out the diagonal of the square. Try it.
Easier yet. No trig at all. Since the squares can be any size, choose the extreme case where the size of the smaller square equals zero. The diagonal of the large square is the diameter of the circle - 8. So the sides of the square is 8/root(2). That squared is 64/2=32. As no units are specified, no need to create units. The area is 32.
As it was mentioned in the video, it is important that we solve for the general case. And that is because your solution assumes that there is exactly one solution. If there were no solutions, multiple solutions or the solution would be parameterized, your method would give a false answer. But indeed it is correct solution in this particular case
@ True. Add then that due to the geometry involved as is apparent, there is a right right triangle between the three contact points making the diagonal a diameter, as in the reduced case. The two squares then become squares of the sides ratioed by root(2), which by pythagoras equal the square of the third side similarly ratioed by root(2). All choices for side lengths then necessarily result in the same total area, allowing reduction of the problem to the case of the small square having side length zero. Apologies, I took that as obvious.
@ I quite understand. I once had a genius upper level maths prof. One day he was explaining some intricacy in Bessel and J functions. He wrote an equation on the board, then said, "and its intuitively obvious that" and then wrote a seemingly new equation. One of my class mates interrupted to ask: "is that obvious?". My prof turned around, hopped backwards, stared at the board, and said aloud "I don't know. Is it obvious?". He puzzled over it for about 10 minutes, as we all puzzled over it with him. Then he turned, and left. Two days later when we met again, he again wrote the equations on the board and said - "Yes it is obvious. Here is how." And it indeed was obvious, once we remembered the rather arcane trick involved. Then again, he spoke eight languages fluently and ran a bookstore. He would often get excited, and slip into a polyglot mess of words, each precisely right for the context he sought, and often involving partial quotes from classical authors, as if that explained everything. And it did. We just had to catch up. We were all metaphorically "running with scissors". He put my physics prof to shame. Mind you, my physics prof taught us calculus in three days (just because) - all of it. So that might give you some idea how brilliant they all were.
The profs that hang onto a question like that and then come back with an answer are pure gold. A lot of mine didn't give a flying f. I cherish the memories of those that did.
The radius of the quarter circle is R=2 and the radius of the semicircle is r=√2. The diameter of the square is R+r√2=2+2=4. The area of the square is (4/√2)²=8.
I love that a big part of solving these is recognizing that you don't need to know all the variables to solve for an expression like x²+y². Indeed, since this is the general case, it would be impossible to find x and y. I also love that while I generally have no idea how to begin to solve these, I usually see where you're going a few steps before you get there.
I did the latter but going through the steps is important. I do a lot of CAD design work these days and you are giving away all my shortcuts! Lollers...
Oh boy, I haven't given tomorrow's puzzle a try or thought just yet, but you know it's not gonna be easy when Andy says he's not sure how he's gonna do it, ahead of time
Another great puzzle. Love how, at the end, you smoosh the sizes of the squares, to show the general solution is always correct. Would have been great to have seen with yesterday's puzzle with the 4 triangles.
I worked this one out in my head: it doesn’t matter what size the squares are, as long as the same three corners touch the circumference of the circle. Therefore, if you make both squares equal in size, their top middle corner ends up being in the centre of the circle, therefore both squares will have side length of 4.
Answer to the next question: Quarter circle radius = 2 Half circle radius = √2 Half circle center to the square lower-right corner distance = (√2)(√2) = 2 Square diagonal distance = 2 + 2 = 4 Square area = (4/√2)^2 = 8
Solution for tomorrow: Since both the semicircle and the quartercircle have an area of π, you can get the relationship of both radii: πs²/2 = πq²/4 |*4 :π 2s² = q² |√ s√2 = q But the radius q is also half the diagonal of the square, so half the side lengths - using the same logic as today's task - is s or the radius of the semicircle. So all we need is (2s)² = 4s². Since πs²/2 = π |:π *8 4s² = 8 So the area of the square is 8.
Sorry, but you've made a few mistakes. After multiplying by 4 and dividing by π we get: 2s² = 4 Also, if you'd square the next line's left side, you'd get: (2s²)² = 2²s⁴ What we need is multiplying by 2: 2(2s²) = 2(4) 2²s² = 8 (2s)² = 8 Your logic was correct, but doing this way is way complicated than just calculating partial results step by step.
I’m really curious about the next puzzle. I found an answer based on an assumption that I’m not sure holds true. It kind on links to todays puzzle where the size of each square is not stated so you could go for the 16 + 16 approach. In similar idea I assume you could reposition the blue semi circle to make life easier. Again exited for your approach.
8 the height of the square is the radius of the half circle plus the height of a square with the diagonal of the radius of the quarter circle for the quarter circle, if (pi*r^2)/4=pi then r=2 this makes the height of the upper square 2/root2 for the half circle, if (pi*r^2)/2=pi then r=root2 the height of the square is (r/root2)+root2 making the area=((r/root2)+root2)^2=8
O K So, I see people in the comments going through the same geometric way to find the area of the square of tomorrow's puzzle, but no one writing why exactly that works, so it's one of 2 things, either people are eyeballing the construction and assuming it to be true, or knowing why it works and just ignoring writing the steps for why it works. Now my solution: Hopefully, that's enough lines to hide the answer. Now, I first go by finding the 2 radii of the qurter circle (R) and the semi-circle (r), as 2 and √2 respectively, then I draw from the semicircle (labelled N) radii to the tangent points with the square, they're both perpendicular on the sides and equal in length, so they form a smaller square of side length √2. I draw a diagonal of that small square connecting the bottom right corner to the top left, that diagonal is √2 * √2 = 2, and is going up at a 45° angle. For the big square, if we draw the diagonal from the bottom right corner to the top left, we know that it'll go up at a 45° angle. Now we have 2 lines both starting at the same point (bottom left corner of the big AND small square) and rising at the same angle (45°), hence they're the same line (collinear), and since one of them touches N the other does so too, so the diagonal of the big square has to pass through N. We know that the quarter circle is centered at the top left corner (labelled M) so the big diagonal also passes through the center of circle M. Since circle M touches the diameter of circle N, on the same big diagonal that passes through N, that means M touches the semicircle at N. The big diagonal can be separated into MN = R = 2, and small diagonal = 2, which is the same as S√2 (assuming the big square to have side length S). Hence S√2 = 2 + 2, S√2 = 4. We don't need S, we need S², so let's square both sides to get → S²*2 = 16 → S² = 8 = Area of the big square required
Well, I dunno what software he's using but if you're gonna do it on a phone or such, try geogebra, it's good for geometric constructions and even graphs and coordinates, but I still haven't learned how to fully use it yet, I know it has sliders and such. Try it out and update me if it works for you
Tomorrow's puzzle is the first one in a while that I got what to do just by looking at it Hint: Join the blue circle tangency points with the square and the green circle
Putting my guess of tomorrow's puzzle at 128 units squared. I'm probably dead wrong because of all the leaps of logic my brain had to make to get there, but if I'm right, you don't even need the quarter circle Upon double checking my math, I have no idea how I reached 128 units squared. I'm now going to lock in my final guess of 8 units squared and sew if I'm right tomorrow
Thank you very much for the spoiler alert, but unfortunately other people in the comments never think of that and spoil it anyway so, it already got spoiled for most people who take a look at the comments for any reason. Nice effort though
@@vishalchellam6494 Well, he has solved it already. EDIT: Real answer: "I'm going to eat in 3 2 1" does not sound like the eating will have stopped in 3 seconds. "I'm going to solve" is exactly the same. I think you are being overly picky, he's speaking totally normal American English.
Wow do you edit these yourself? Cuz it's really cool solving and editing these all with a target of 31. Good luck bossman!
Yes, I do it all myself! And I am still an amateur learning new editing skills every day.
@@AndyMath you have inspired me to make Math videos fun! Thank you so much for the reply :)
bossmang
@@AndyMath What software do you use for the video editing?
Love the ending with the adjusting squares!
Thank you so much for these - I'm enjoying them hugely!
If the tomorrow’s question is already that confuzzling then i can’t imagine the REST OF THEM
It's not hard. Think what the area of each full circle would be, and then you can find the radius of each. This lets you figure out the diagonal of the square. Try it.
Easier yet. No trig at all. Since the squares can be any size, choose the extreme case where the size of the smaller square equals zero. The diagonal of the large square is the diameter of the circle - 8. So the sides of the square is 8/root(2). That squared is 64/2=32. As no units are specified, no need to create units. The area is 32.
As it was mentioned in the video, it is important that we solve for the general case. And that is because your solution assumes that there is exactly one solution. If there were no solutions, multiple solutions or the solution would be parameterized, your method would give a false answer. But indeed it is correct solution in this particular case
@ True. Add then that due to the geometry involved as is apparent, there is a right right triangle between the three contact points making the diagonal a diameter, as in the reduced case. The two squares then become squares of the sides ratioed by root(2), which by pythagoras equal the square of the third side similarly ratioed by root(2). All choices for side lengths then necessarily result in the same total area, allowing reduction of the problem to the case of the small square having side length zero. Apologies, I took that as obvious.
@tunneloflight It is obvious, but I'm that kind of nerd that needs everything perfect 😁
@ I quite understand. I once had a genius upper level maths prof. One day he was explaining some intricacy in Bessel and J functions. He wrote an equation on the board, then said, "and its intuitively obvious that" and then wrote a seemingly new equation. One of my class mates interrupted to ask: "is that obvious?". My prof turned around, hopped backwards, stared at the board, and said aloud "I don't know. Is it obvious?". He puzzled over it for about 10 minutes, as we all puzzled over it with him. Then he turned, and left.
Two days later when we met again, he again wrote the equations on the board and said - "Yes it is obvious. Here is how." And it indeed was obvious, once we remembered the rather arcane trick involved.
Then again, he spoke eight languages fluently and ran a bookstore. He would often get excited, and slip into a polyglot mess of words, each precisely right for the context he sought, and often involving partial quotes from classical authors, as if that explained everything. And it did. We just had to catch up. We were all metaphorically "running with scissors".
He put my physics prof to shame. Mind you, my physics prof taught us calculus in three days (just because) - all of it. So that might give you some idea how brilliant they all were.
The profs that hang onto a question like that and then come back with an answer are pure gold. A lot of mine didn't give a flying f. I cherish the memories of those that did.
The radius of the quarter circle is R=2 and the radius of the semicircle is r=√2. The diameter of the square is R+r√2=2+2=4. The area of the square is (4/√2)²=8.
I got the same answer 👍
Also, you don't even need the big square's side, the area is just diag²/2 = 4²/2 = 8.
I love that a big part of solving these is recognizing that you don't need to know all the variables to solve for an expression like x²+y². Indeed, since this is the general case, it would be impossible to find x and y. I also love that while I generally have no idea how to begin to solve these, I usually see where you're going a few steps before you get there.
I did the latter but going through the steps is important. I do a lot of CAD design work these days and you are giving away all my shortcuts!
Lollers...
Good one! I enjoy your puzzles, and I especially enjoy that you never forget the units. I'm an engineer. Units matter!
The animations are great and helps a lot.
Oh boy, I haven't given tomorrow's puzzle a try or thought just yet, but you know it's not gonna be easy when Andy says he's not sure how he's gonna do it, ahead of time
Give it a try. It may be easier than you think!
Good job! I love seeing your videos
Three "How exciting". Someone's in a specially good mood!
Merry Christmath Andy! ✨
Another great puzzle. Love how, at the end, you smoosh the sizes of the squares, to show the general solution is always correct. Would have been great to have seen with yesterday's puzzle with the 4 triangles.
We are going to catch up with this one🔥🔥🔥
I worked this one out in my head: it doesn’t matter what size the squares are, as long as the same three corners touch the circumference of the circle. Therefore, if you make both squares equal in size, their top middle corner ends up being in the centre of the circle, therefore both squares will have side length of 4.
And at the end of the video, Andy explains this :)
Happy Christmas Andy and everyone here
i was just about to go to sleep... one more couldnt hurt right
The area of the square is 8 u²
Answer to the next question:
Quarter circle radius = 2
Half circle radius = √2
Half circle center to the square lower-right corner distance = (√2)(√2) = 2
Square diagonal distance = 2 + 2 = 4
Square area = (4/√2)^2 = 8
Solution for tomorrow:
Since both the semicircle and the quartercircle have an area of π, you can get the relationship of both radii:
πs²/2 = πq²/4 |*4 :π
2s² = q² |√
s√2 = q
But the radius q is also half the diagonal of the square, so half the side lengths - using the same logic as today's task - is s or the radius of the semicircle.
So all we need is (2s)² = 4s².
Since
πs²/2 = π |:π *8
4s² = 8
So the area of the square is 8.
Sorry, but you've made a few mistakes.
After multiplying by 4 and dividing by π we get:
2s² = 4
Also, if you'd square the next line's left side, you'd get:
(2s²)² = 2²s⁴
What we need is multiplying by 2:
2(2s²) = 2(4)
2²s² = 8
(2s)² = 8
Your logic was correct, but doing this way is way complicated than just calculating partial results step by step.
@@Epyxoid You are correct. Not sure where my mind was there. I fixed it.
8 what..? Apples Bananas?
@@joybose2770 unit² 🤓
@@joybose2770 yes😉
I’m really curious about the next puzzle. I found an answer based on an assumption that I’m not sure holds true. It kind on links to todays puzzle where the size of each square is not stated so you could go for the 16 + 16 approach. In similar idea I assume you could reposition the blue semi circle to make life easier.
Again exited for your approach.
I think the area of the square is 8 units squared assuming that the two circles meet in the center of the square.
8
the height of the square is the radius of the half circle plus the height of a square with the diagonal of the radius of the quarter circle
for the quarter circle, if (pi*r^2)/4=pi then r=2
this makes the height of the upper square 2/root2
for the half circle, if (pi*r^2)/2=pi then r=root2
the height of the square is (r/root2)+root2
making the area=((r/root2)+root2)^2=8
That inscribed triangle got me!
O K
So, I see people in the comments going through the same geometric way to find the area of the square of tomorrow's puzzle, but no one writing why exactly that works, so it's one of 2 things, either people are eyeballing the construction and assuming it to be true, or knowing why it works and just ignoring writing the steps for why it works.
Now my solution:
Hopefully, that's enough lines to hide the answer.
Now, I first go by finding the 2 radii of the qurter circle (R) and the semi-circle (r), as 2 and √2 respectively, then I draw from the semicircle (labelled N) radii to the tangent points with the square, they're both perpendicular on the sides and equal in length, so they form a smaller square of side length √2.
I draw a diagonal of that small square connecting the bottom right corner to the top left, that diagonal is √2 * √2 = 2, and is going up at a 45° angle.
For the big square, if we draw the diagonal from the bottom right corner to the top left, we know that it'll go up at a 45° angle.
Now we have 2 lines both starting at the same point (bottom left corner of the big AND small square) and rising at the same angle (45°), hence they're the same line (collinear), and since one of them touches N the other does so too, so the diagonal of the big square has to pass through N.
We know that the quarter circle is centered at the top left corner (labelled M) so the big diagonal also passes through the center of circle M.
Since circle M touches the diameter of circle N, on the same big diagonal that passes through N, that means M touches the semicircle at N.
The big diagonal can be separated into MN = R = 2, and small diagonal = 2, which is the same as S√2 (assuming the big square to have side length S).
Hence S√2 = 2 + 2, S√2 = 4.
We don't need S, we need S², so let's square both sides to get → S²*2 = 16 → S² = 8 = Area of the big square required
It doesn't just look important, it is **very** important.
On this one, since there is no constraint, I made both squares the same height. Now they form a 4×8 rectangle so the answer is 32.
Three "how exciting", how exciting.
0:00 "Let's put a circle around it."
Looking forward to the next one, I don't know how to solve it.
Andy, do a session on Kaprakar's Constant.
what kind of tool are you using to draw the line and write the numbers on your presentation?
Well, I dunno what software he's using but if you're gonna do it on a phone or such, try geogebra, it's good for geometric constructions and even graphs and coordinates, but I still haven't learned how to fully use it yet, I know it has sliders and such.
Try it out and update me if it works for you
Tomorrow's puzzle is the first one in a while that I got what to do just by looking at it
Hint: Join the blue circle tangency points with the square and the green circle
daily dose of how exciting
Beautifully solved
Putting my guess of tomorrow's puzzle at 128 units squared. I'm probably dead wrong because of all the leaps of logic my brain had to make to get there, but if I'm right, you don't even need the quarter circle
Upon double checking my math, I have no idea how I reached 128 units squared. I'm now going to lock in my final guess of 8 units squared and sew if I'm right tomorrow
day 19
square diagonal
=r1+r2·sqrt2=2+2=4
→A=4^2/2=8😊
Tomorrows - side length 2root(2) -> area = 8.
Hiw much are x and y?
Well, I got 32 but my solution was much more complicated…
How exciting😅❤
🎉🎉🎉
my father says that you are 11 you shouldnt watch this but i just love your video
the area of square is 8
OMFG IT WAS ACTUALLY SO FKINN EASYYYYYYYYYYYYY RAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Area of the square tomorrow is 8.
Good job but put a spoiler alert please for people who haven’t solved it yet and want to do it themselves.
@@qqma4791 They can still solve it by themselves.
Tomorrow's puzzle SPOILER:
8
Thank you very much for the spoiler alert, but unfortunately other people in the comments never think of that and spoil it anyway so, it already got spoiled for most people who take a look at the comments for any reason.
Nice effort though
bro make it look easy
Day 19: 8 sq units
Making progress!
Smart Geometri
I love pies
Next problem: 8
Hi
Could you say "I'm going to start solving in" instead of "I'm going to solve in" ? Loving the video btw
Why? They’re both grammatical.
@caseygreyson4178 Cause when u say u r going to solve in it means u would have solved it immediately after the countdown (3..2..1)
@@vishalchellam6494 Well, he has solved it already. EDIT: Real answer: "I'm going to eat in 3 2 1" does not sound like the eating will have stopped in 3 seconds. "I'm going to solve" is exactly the same. I think you are being overly picky, he's speaking totally normal American English.
@@vishalchellam6494 You must not be a native speaker of English. Best of luck on your studies!
@@Qermaq Thanks for the explanation 👍🏼