i used to have the same problem, honestly helps just drawing things and labeling everything you think can be labeled out to start. but these are no joke! coming from an engineer!
Yeah, I second what my fellow engineer louiscatacalos6715 said, I used to struggle a lot with geometry, despite being good with algebra, Calculus, and many other fields of math, I was so bad at it that I always relied on coordinate geometry and integrals instead. If problem solving in general is sth new and difficult to you, then I suggest first doing all constructions you think might help (just mimic Andy's other circle problems constructions and you should get it), and secondly, have all geometry rules that relate in any way written in front of you (always helps to have an eye on those and I speak from experience). Overall though geometry is no easy task, it IS the origin of all math after all and may involve lots and lots of different unrelated fields of math involved in it so don't get discouraged and keep trying, you'll get there.
8 seconds: complete the big circle by mirroring it along its diameter. Now there is a ring of six little circles with room for one in the middle, and a diagonal cutting through the diameters of three circles is identical to the diameter of the large circle. Small circles is 1/3 the diameter of the large. QED.
@@tod9141 Because there are 6 small circles arranged symmetrically. I can't think of a better way to explain this but the fact of symmetry does explain it if you go the hexagon route. If you play around with seven (or more) identical coins you will see the relationships easily, one coin can only touch a maximum of 6 others in this particular arrangement, there is no closer packing of circles in a plane.
I did this one a weird way. If you make it a full circle with 6 smaller circles inside (mirroring it) then you can fit a 7th small circle in the center because that's how 7 smaller circles pack into one larger one. It's a unique construction. From there, one radius of the large circle is 3x the size of the small circle's radius (very easily shown when there's a 7th small circle in the center) Then (returning to the original construction) if you call the little circle radius "r" you have the ratio you're looking for as: [(3r)^2*(pi/2) - 3(pi(r^2)] / (3r)^2*(pi/2) divide both sides by pi and factor a little and you have: [9(r^2)/2 - 3(r^2)] / (9(r^2)/2 factor out r^2: (9/2 - 3) / (9/2) (3/2) / (9/2) 3/9 1/3 got em
Answer to the question for day 15 r = radius of the small circle with area 8 = √(8/π) = 2√(2/π) Side length of the equilateral triangle = R = radius of the large semicircle = 2(r√3) = 4√(6/π) h = the height of the left isosceles triangle = R/2 = 2√(6/π) The diameter of the left small circle = R - R/2 = R/2 = 2√(6/π) The area of the left small circle = π(√(6/π))^2 = 6
You can't just say "due to symmetry", You have to use that symmetry to prove it's a regular hexagon. Drawing all the big and small radii, then label all the lengths accordingly, then you draw a line from the center of the big circle to any tangent point of 2 small circles, this will form 3 congruent triangles with the tangent points of the diameters. Then you'll see that 90º angle of each quadrant is divided equally in 3 parts, and thus you'll find that the isosceles triangle of sides R-r and 2r has a 60 degree angle, therefore you prove it's an equilateral triangle and R-r = 2r. And with that you solve the equation at the beginning.
I got R=3r quickly without using the hexagon by just recognizing that if I connected the center point of a small circle to the center of the semicircle that would be 1/12 of a circle, making a 30-60-90 triangle. From this you can see that the hypotenuse of the triangle created from this line and the line tangent to the bottom of the small circle is going to be 2r.
@@danohana1822 3 circles in the semicircle, running a line through the center of them divides them into 6 halves in the semicircle, or 12 in the whole.
It's an elementary circle theorem: if you have 2 circles that are tangent to each other (either internally or externally), the tangent point and the centres of both circles are colinear.
tomorrow's solution is 6 the triangle inscribed in the semicircle is a 30-60-90 right angle triangle with short leg radius(R) and hypotenuse 2R making the chord R*root3 so then you can split the isosceles triangle into half to make another right angled triangle with hypotenuse R and long leg (Rroot3)/2 making the short leg R/2 so the unknown circle has a radius of R/4 so then you just jigger the above equations about until you get that the unknown circle is three quarters of the size of the 8 circle, making it 6
If you listen at 2:35 he says it without annunciating. My guess is that because his accent kinda makes it sound unclear (more like “liddle”) he wants to say it clearly (or just say it technically “correct”, even though basically everyone in America pronounces it like he does).
How do we know that all the tops of the equilateral triangles end up at the center of the big circle? I know it looks like it does, but how do we know for sure?
Tangent lines to a circle are perpendicular to its radius. Since the small circles are tangent to the big circle that means that both circles share the same tangent line and so both radii are aligned
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Some of these solutions are so beautiful.
everything looked normal in this video...
Pre-recorded maybe since it's a sponsored video.
We need a video on your thinking process because I’m having trouble just starting to solve any of the problems
i used to have the same problem, honestly helps just drawing things and labeling everything you think can be labeled out to start. but these are no joke! coming from an engineer!
Yeah, I second what my fellow engineer louiscatacalos6715 said, I used to struggle a lot with geometry, despite being good with algebra, Calculus, and many other fields of math, I was so bad at it that I always relied on coordinate geometry and integrals instead.
If problem solving in general is sth new and difficult to you, then I suggest first doing all constructions you think might help (just mimic Andy's other circle problems constructions and you should get it), and secondly, have all geometry rules that relate in any way written in front of you (always helps to have an eye on those and I speak from experience).
Overall though geometry is no easy task, it IS the origin of all math after all and may involve lots and lots of different unrelated fields of math involved in it so don't get discouraged and keep trying, you'll get there.
Andy getting PAID!
I solved the next day's puzzle on my own in the head. I'm so proud of myself! ☺️ I'm looking forward to your next video!
Yep, but the calculations themselves are really annoying, cause of the roots, especially the 1/sqrt(pi) involved here
@Z-eng0 Just leave them as they are. R1 is sqrt(8/pi). R2 is sqrt(6/pi) after multiplying R1 with sqrt(3/4).
8 seconds: complete the big circle by mirroring it along its diameter. Now there is a ring of six little circles with room for one in the middle, and a diagonal cutting through the diameters of three circles is identical to the diameter of the large circle. Small circles is 1/3 the diameter of the large. QED.
how do you know there's room for one in the middle
@@tod9141 Because there are 6 small circles arranged symmetrically. I can't think of a better way to explain this but the fact of symmetry does explain it if you go the hexagon route.
If you play around with seven (or more) identical coins you will see the relationships easily, one coin can only touch a maximum of 6 others in this particular arrangement, there is no closer packing of circles in a plane.
I love watching theses
I did this one a weird way.
If you make it a full circle with 6 smaller circles inside (mirroring it) then you can fit a 7th small circle in the center because that's how 7 smaller circles pack into one larger one. It's a unique construction.
From there, one radius of the large circle is 3x the size of the small circle's radius (very easily shown when there's a 7th small circle in the center)
Then (returning to the original construction) if you call the little circle radius "r" you have the ratio you're looking for as: [(3r)^2*(pi/2) - 3(pi(r^2)] / (3r)^2*(pi/2)
divide both sides by pi and factor a little and you have: [9(r^2)/2 - 3(r^2)] / (9(r^2)/2
factor out r^2: (9/2 - 3) / (9/2)
(3/2) / (9/2)
3/9
1/3
got em
If I had a nickle for every time Big R equals 3 LitTle r, I'd have 2 nickles 🤔
Answer to the question for day 15
r = radius of the small circle with area 8 = √(8/π) = 2√(2/π)
Side length of the equilateral triangle = R = radius of the large semicircle = 2(r√3) = 4√(6/π)
h = the height of the left isosceles triangle = R/2 = 2√(6/π)
The diameter of the left small circle = R - R/2 = R/2 = 2√(6/π)
The area of the left small circle = π(√(6/π))^2 = 6
day15
r1:r3=1:4,r2:r3=1:2sqrt3
r1:r2=sqrt3:2→A:8=3:4→A=6😊
Very well done.
"The big ars are equal to three small arse" sorry, couldn't stop myself...haha
diametre of all 3 small circle = diameter of semicircle .........now its much easier
You can't just say "due to symmetry", You have to use that symmetry to prove it's a regular hexagon.
Drawing all the big and small radii, then label all the lengths accordingly, then you draw a line from the center of the big circle to any tangent point of 2 small circles, this will form 3 congruent triangles with the tangent points of the diameters. Then you'll see that 90º angle of each quadrant is divided equally in 3 parts, and thus you'll find that the isosceles triangle of sides R-r and 2r has a 60 degree angle, therefore you prove it's an equilateral triangle and R-r = 2r. And with that you solve the equation at the beginning.
We got a regular hexagon!!!!
I got R=3r quickly without using the hexagon by just recognizing that if I connected the center point of a small circle to the center of the semicircle that would be 1/12 of a circle, making a 30-60-90 triangle. From this you can see that the hypotenuse of the triangle created from this line and the line tangent to the bottom of the small circle is going to be 2r.
Why would it be 1/12?
@@danohana1822 3 circles in the semicircle, running a line through the center of them divides them into 6 halves in the semicircle, or 12 in the whole.
When extending the red line, don't we need to prove that it hits where the small circle touches the semicircle? Why are those points on the same line?
It's an elementary circle theorem: if you have 2 circles that are tangent to each other (either internally or externally), the tangent point and the centres of both circles are colinear.
Day 15 area is 6
Elegant.
Your other video was just posted like 10 hours ago?!?!
Yeah, I am trying to get back on pace for the Aggvent Calendar.
Yeah, recently Andy has started taking estrogen, its effects are already showing... i'm just embarrassed...
@@AndyMath Nice video! By the way, did you delete my reply? I can't see it
tomorrow's solution is 6
the triangle inscribed in the semicircle is a 30-60-90 right angle triangle with short leg radius(R) and hypotenuse 2R making the chord R*root3
so then you can split the isosceles triangle into half to make another right angled triangle with hypotenuse R and long leg (Rroot3)/2 making the short leg R/2 so the unknown circle has a radius of R/4
so then you just jigger the above equations about until you get that the unknown circle is three quarters of the size of the 8 circle, making it 6
Wowwww I lovee thissss That's Amazing
Out of curiosity, is there a particular reason you try to really annunciate the T in 'little'?
If you listen at 2:35 he says it without annunciating.
My guess is that because his accent kinda makes it sound unclear (more like “liddle”) he wants to say it clearly (or just say it technically “correct”, even though basically everyone in America pronounces it like he does).
How do we know that all the tops of the equilateral triangles end up at the center of the big circle? I know it looks like it does, but how do we know for sure?
that's how hexagons work
Tangent lines to a circle are perpendicular to its radius. Since the small circles are tangent to the big circle that means that both circles share the same tangent line and so both radii are aligned
@@radfuethanks!
1/3
Im wondering if hes a really passionate math major, or a really passionate math teacher
2r, R-2r, x
R-r, r, x
4r²=(R-2r)²+x²
=R²-4Rr+x²
x²=4Rr-R²
(R-r)²=r²+x²
R²-2Rr=+x²
R²-2Rr=4Rr-R²
2R² =4Rr+2Rr
R²=3Rr
R=3r
Area(sc)=(1/2)9πr²
Area(cir)=3πr²
Area not covered by 3 circles=
3πr²/2
(3πr²/2/9πr²/2
3/9=1/3
do you know great mathematician ramanujan
I would reckon he does
Hexagon hence 6 equilateral triangles. Yeah, we all know hexagon properties by heart...
Hi
"now we can see that R=3r", my mind literally exploded with the simplicity by which Andy arrived at that. 🤯💥💣🧨
My answer for tomorrow's question is
6 square units
Me too
Same
Same
Nothing is wrong in this one except for your pronunciation of little, but that might just be dialect!
That’s always how he has pronounced it
@ Yeah, dialect. Strange. Something must be off!
0:30 vs 2:35