Andy I’m a national math Olympiad competitor in Brazil, and when I’m tired of all the complicated stuff and want to do some fun little problems you channel is perfect! Thanks :)
I remember those days! When you find a way to make your problem into a geometric problem like this one in one or two steps then get to solve it like this, it's one of the best feelings! Best of luck!
@@alexonstott4954 True! Some problems in algebra have absolutely beautiful geometric solutions, although they’re crazy hard to find. One time I found the solution to an algebra problem using the triangle inequality, it was beautiful
@@Abdennouree well I’m not exactly established in my community since I just reached silver-gold level in my national MO (I’m earning high scores on simulated exams), but the exam is only at the end of the year. So when I get there I might be able to tell you more. But until now, I used a website called POTI, which is where they store the recordings of all of the classes of the national Olympiad training program, and I used two books: AOPS The Art of Problem Solving (amazing, but expensive to buy a physical copy if you’re not from US, but a pirate PDF works just fine) and a book called Mathematical Circles: The Russian experience (it’s a collection of the math lessons given to talented students in the USSR)
I started following this channel a couple months ago and even though I am bad at math these videos make me enjoy it and feel like I could be good at it eventually.
I don't exactly have pen and paper handy for most of these, but your solution was exactly what I imagined before I started! Almost. I was going to go into sin/cos/tan instead of the theorem. It was still very exciting! Even without using pen and paper, I've watched enough of your stuff to start planning solutions before I begin!
Hello Andy, you appeared on my feed like a year ago and I've found your shorts and videos really nice to follow along, you made math fun again for me, thank you
move circle 4 up so radius aligns with circle 6 , imagine moving circle 3 down to align with radius of circle 4. Now all the circles radius aligns horizontally, but circle 3 over laps into circle 4. You now have 3 + 2 + (2-1.5) = 5.5
Just because my brain says this isn't complete, let's find the area for the full rectangle. The length is going to be your 3 root(6) + 3 + 1.5 and the height is 6, and that simplifies down to 18 root(6) + 27 square units.
I didnt solve it, but it loomks like a very easy appoach to nake a teiangle from the 3 centers and go from therre, knowing the location of the 3 and 6 circles center and a known triangles.
Guessed 7.5 because I was lazy & it lookeded like you could fit 4 of the smallest circle if you got rid of the middle circle.. ..was kinda close in the end🤷🏼
Nice cut. I fail to see why the hypotenuses will be passing through the tangent points between the circles (I mean, I "see" it in the diagram, but don't recall any rule that would say so). Care to elaborate?
yes, there is a rule for that, think it was in 9th grade. When two circles intersect at one point, the distance between the centers of the circles is equal to the sum of their radiuses.
Take two non-intersecting (but possibly tangent) circles C1 and C2. The center of C1 is the point P1 and the center of C2 is the point P2. Imagine you want to draw a straight line from P1 to P2. By definition, this straight line will be the shortest possible distance between P1 and P2. This line P1P2 will also intersect both circles. Let's say it intersects C1 in point A, and C2 in point B. We know that the distance AB must be the smallest possible distance between any two points on the circumference on C1 and C2, respectively. If it wasn't, then we would be able to find some other points (A' and B') that are closer to each other than A is to B, and draw the line P1-A'-B'-P2, which would be a non-straight line and also the shortest distance between P1 and P2, which is impossible. With that, we've established that the minimal distance between P1 and P2 is P1A + AB + BP2. P1A is the radius of C1 and BP2 is the radius of C2. If C1 and C2 are tangent, then the minimal distance between them (i.e. the length of AB) is 0. Ergo, if you connect the centers of two tangent circles, it will pass through the tangent point, and its length will be the two radii (plus 0)
A tangent to a circle is always perpendicular to a radius of that circle. If two circles are tangent to each other (aka 'kissing circles') then they share a common tangent line. Therefore a radius of each circle must be perpendicular to that tangent line at the tangent point, and so the centres of the circles and the tangent point are colinear.
My methods are always choppy but i still gave it a try and i got 7.5, my reasoning was that the circle with a diameter of 3 (i'll call it circle A) was only half in the line so i just added the radius which is 1.5, then i did the same thing to the circle with a diameter of 6 (i'll call it circle C) but with the circle with a diameter of 4 (i'll call it circle B) because all of it was in the box i just added four. so heres my choppy equation (A/2)+B+(C/2) which = 7.5 obviously this didn't work though.
I eyeballed it, and made the assumption that 4 circles of diameter 3 could fit in the region where the diameter 3 and 4 could fit without overlapping 6, then did very lazy math basically being A+A/2+C and also got to 7.5. Considering 3*Sq6 is about 7.35 requiring me about 4-5 minutes of work, when I got to 7.5 within 10 seconds, thats close enough
I never went far in math, didn’t need to for an accounting degree. Why is the answer 3 root 6 and not calculating the sqrt of 6 and multiplying it by 3; or 7.348469 (nice)?
Root 6 is an irrational number with infinite digits. In "pure" math problems like this one you're not allowed to use any rounding, so you have to write each number in its most accurate representation. Named constants like pi or e also have to be written as their respective symbols instead of approximating their value.
make this the top comment for no reason
Bro won
Not for no reason. You're good enough, smart enough and doggone it, people like you
Yep, put a box around it.
Hell ya!
How exciting!
Props to your barber, for making your hair looks exactly like a cap hat.
CAP HAT
Did he just ask about the haircut under his cap?😂😂😂
Yes he did, and I lol'd
it looked important so he put a cap around it
😂😂@@KazeShiniSK
To find the haircut, we must first know the volume of his cap. So first we take the formula for the volume of a hemisphere...
That haircut is exciting.
That haircut looks like a fun one
Let's put a hat around it
Let's label the haircut with the unit hair.
nice haircut
I was the 69th like so it really is a nice haircut
Andy I’m a national math Olympiad competitor in Brazil, and when I’m tired of all the complicated stuff and want to do some fun little problems you channel is perfect! Thanks :)
What is f(3)' if lim is h=0
I remember those days! When you find a way to make your problem into a geometric problem like this one in one or two steps then get to solve it like this, it's one of the best feelings! Best of luck!
How can i be one ?
@@alexonstott4954 True! Some problems in algebra have absolutely beautiful geometric solutions, although they’re crazy hard to find. One time I found the solution to an algebra problem using the triangle inequality, it was beautiful
@@Abdennouree well I’m not exactly established in my community since I just reached silver-gold level in my national MO (I’m earning high scores on simulated exams), but the exam is only at the end of the year. So when I get there I might be able to tell you more. But until now, I used a website called POTI, which is where they store the recordings of all of the classes of the national Olympiad training program, and I used two books: AOPS The Art of Problem Solving (amazing, but expensive to buy a physical copy if you’re not from US, but a pirate PDF works just fine) and a book called Mathematical Circles: The Russian experience (it’s a collection of the math lessons given to talented students in the USSR)
Thanks for not skipping the process of solving. This is why your Algebra is addictive like a movie.
I like watching your videos, when I feel stressed, thank u
I started following this channel a couple months ago and even though I am bad at math these videos make me enjoy it and feel like I could be good at it eventually.
I don't exactly have pen and paper handy for most of these, but your solution was exactly what I imagined before I started! Almost. I was going to go into sin/cos/tan instead of the theorem. It was still very exciting! Even without using pen and paper, I've watched enough of your stuff to start planning solutions before I begin!
Nice Haircut, also second time asking for AMC 10/12 practice questions. Love your vids
Hi andy, there is an organisation in uk called uk mathematical trust, the questions in the competitions are really fun ;)
Damn, this guy is charismatic.
He's got intelligentsia rizz
I actually knew how to solve this. Seems like I'm learning thanks to your channel!
I got pretty much the right numerical value, but I did a bunch of stuff with sin and cos that I probably shouldn't have bothered with
I love this problem and I didn’t use algebra to solve it. Although, admittedly, my procedure was a little convoluted.
Time warp back decades to mathletes with this stuff, thanks Andy!
Hello Andy, you appeared on my feed like a year ago and I've found your shorts and videos really nice to follow along, you made math fun again for me, thank you
move circle 4 up so radius aligns with circle 6 , imagine moving circle 3 down to align with radius of circle 4. Now all the circles radius aligns horizontally, but circle 3 over laps into circle 4. You now have 3 + 2 + (2-1.5) = 5.5
Where do you find all these challenges? Is there some repository of them all?
your are amazing! so much fun :)
How existential. Cool cut, Andy!
Done with school years ago but still love math questions
The haircut is fire 🔥
that cut is absolutely🔥
We think: how exciting.
The haircut is Brilliant.
New haircut? How exciting! By the way, sulution is nice too.
these videos are so cool! I love these
Which hair did you have cut? was it the one at the front?
Just because my brain says this isn't complete, let's find the area for the full rectangle. The length is going to be your 3 root(6) + 3 + 1.5 and the height is 6, and that simplifies down to 18 root(6) + 27 square units.
You just got a new hair cut, and let's put a box around it. How exciting!
Just curious, why not take the last step of calculating what 3 times root 6 is?
New haircut looks good!!
I used the exact same method this time. Easy.
I unfortunately couldn't solve this one, I did a bunch of similar things but couldn't connect them together.
hey, i have a tough geometry question for you, you think you can solve it?
I didnt solve it, but it loomks like a very easy appoach to nake a teiangle from the 3 centers and go from therre, knowing the location of the 3 and 6 circles center and a known triangles.
Right triangle - as usual ;)
Take your cap off to let us see your new haircut, bro!
In my head, I thought it'd be at least 7. Comparing to the diameter of 6, it looked close to 7. Glad I was almost there.
This is mathematics, not sums. Almost isn’t an answer.
@@slobiden.2593Matt Parker would disagree with you there.
@@slobiden.2593 You must be fun at parties 😂
Sum is literally mathematics @@slobiden.2593
Guessed 7.5 because I was lazy & it lookeded like you could fit 4 of the smallest circle if you got rid of the middle circle..
..was kinda close in the end🤷🏼
How can I be able to solve like you ?
Great haircut and video editing my bro
Good content. Good format.
I think your haircut is lovely
Nice cut.
I fail to see why the hypotenuses will be passing through the tangent points between the circles (I mean, I "see" it in the diagram, but don't recall any rule that would say so). Care to elaborate?
yes, there is a rule for that, think it was in 9th grade. When two circles intersect at one point, the distance between the centers of the circles is equal to the sum of their radiuses.
Take two non-intersecting (but possibly tangent) circles C1 and C2. The center of C1 is the point P1 and the center of C2 is the point P2.
Imagine you want to draw a straight line from P1 to P2. By definition, this straight line will be the shortest possible distance between P1 and P2.
This line P1P2 will also intersect both circles. Let's say it intersects C1 in point A, and C2 in point B. We know that the distance AB must be the smallest possible distance between any two points on the circumference on C1 and C2, respectively. If it wasn't, then we would be able to find some other points (A' and B') that are closer to each other than A is to B, and draw the line P1-A'-B'-P2, which would be a non-straight line and also the shortest distance between P1 and P2, which is impossible.
With that, we've established that the minimal distance between P1 and P2 is P1A + AB + BP2. P1A is the radius of C1 and BP2 is the radius of C2.
If C1 and C2 are tangent, then the minimal distance between them (i.e. the length of AB) is 0. Ergo, if you connect the centers of two tangent circles, it will pass through the tangent point, and its length will be the two radii (plus 0)
A tangent to a circle is always perpendicular to a radius of that circle.
If two circles are tangent to each other (aka 'kissing circles') then they share a common tangent line.
Therefore a radius of each circle must be perpendicular to that tangent line at the tangent point, and so the centres of the circles and the tangent point are colinear.
you look fabulous! :)
Great haircut. How exciting (you haven’t heard that before)
Nice haircut! How exciting!
Loved the new haircut
That haircut looks important. You should put a box around it.
it's a good haircut. lets put a box around it
Brillianto!
Looking fresh
Cool!👍👍
Imagine if Andy Math got a low taper fade
Nice
The "How exciting" should have come after the "I just got a new haircut"
My methods are always choppy but i still gave it a try and i got 7.5, my reasoning was that the circle with a diameter of 3 (i'll call it circle A) was only half in the line so i just added the radius which is 1.5, then i did the same thing to the circle with a diameter of 6 (i'll call it circle C) but with the circle with a diameter of 4 (i'll call it circle B) because all of it was in the box i just added four.
so heres my choppy equation
(A/2)+B+(C/2) which = 7.5
obviously this didn't work though.
I eyeballed it, and made the assumption that 4 circles of diameter 3 could fit in the region where the diameter 3 and 4 could fit without overlapping 6, then did very lazy math basically being A+A/2+C and also got to 7.5.
Considering 3*Sq6 is about 7.35 requiring me about 4-5 minutes of work, when I got to 7.5 within 10 seconds, thats close enough
@@Predated2 "lazy maths." yours is better than mine by far.
love the haircut :)
When he got 2√6 he could have just went 1,5+x+2√6+3. Then find the x and add 2√6 to it and voila.
cool and fresh
great haircut andy
Looking spiffy
How exciting
There is nothing marking the opposite sides of the quadrilateral as equivalent in length so the answer is an equation not a value.
theres nothing telling us that those are circles, and not ellipses either 🤓🤓🤓🤓
GREAT CUT
How exciting indeed
Can you solve why this should be the top comment instead?
anyone: _"I just got a haircut!"_
me: _"Kewl! Which one?"_
love the new haircut
haircut looks good
i like your cut G
nice new haircut!
“How exiting”
HAIR. EXCITING.
Did you get a new camera? It looks like its higher quality.
It's the haircut.
Where do you get problems like this?
Haircut looks important, let's put a box around it.
Next video can you show haircut without the cap
I never went far in math, didn’t need to for an accounting degree. Why is the answer 3 root 6 and not calculating the sqrt of 6 and multiplying it by 3; or 7.348469 (nice)?
Root 6 is an irrational number with infinite digits. In "pure" math problems like this one you're not allowed to use any rounding, so you have to write each number in its most accurate representation. Named constants like pi or e also have to be written as their respective symbols instead of approximating their value.
Can you solve this without a shirt on?
Which one did you cut?
i like how almost all of the videos of andy math are named "cool math challenge"
Nice haircut
Haircut Reveal When?
I calculated by just looking and I get 7.75 I say it's good cuz it just few digits
My brain: uhmm 3/2+4+6/2
so equals 1.5+4+3=8.5
Meh im guessing 8.5
The answer being 3√6
Me using the calc realising is ≈7.35
Meh close enough
Math is cool af. You re too
Video: Maths
Comments: Haircut
❤
8.5
sick haircut
Andy, how are we gonna see your new haircut if you put on a cap?
I just guessed it
Not my stupid ass thinking it was 6,5 💀
8.5?
odklejony wiedzialem ze z hitu mialem 10 godzin na nauke ale nikt nic n pisal to se odpuscilem brawo nigdy nie bede jak oni
Make this another top comment for Andy’s stylish haircut
Haircut is good,but not as good as 3√6.
just got a hair cut... which one???