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keeping the people maths ready (GCSE geometry problem, Reddit r/GCSE)
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- Опубликовано: 14 дек 2023
- Here's a fun GCSE (General Certificate of Secondary Education) geometry problem that I found on Reddit. We have three circles tangent to each other and we will find the length of the band that goes around the circles.
Original post on Reddit: / ydo9w6ssnw
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#math #algebra #mathbasics
What’s i^-1 = ?
Answer here ruclips.net/video/vpgF42O7lsA/видео.htmlsi=t1EIYh-jCO13cj9w
I know the standard way would be an easier conjugate bc there’s no other terms but “i” in the bottom, but what would happen if we make this e^ln(i^-1) = e^(-ln(i))?
What would evaluating a power series on this look like, would we plug in -ln(i) for everything that is normally x? I tried writing it out but I’m getting nowhere 😂
I find it weird some extra credit in school classes is n/0 with n being the extra credit points received.
@@johndickinson82 if only it were 0+. Then you’d be getting infinite extra credit
wouldn't that just be like, -i
cause i^-1 = (-1)^(-1/2) = -i
you could also just say (1/i)*(-i/-i) = -i
probably super obvious to be fair, video linked probably talks about fancy stuff that I haven't learn cause am only in highschool
0:03 How do you know the 3 circles are equal? I don't see it mentioned in the question. Just by sight? What if one of them is a fraction of a % smaller/bigger than the other 2?
The good thing about geometry is that everything is either triangles or circles if you squint hard enough
Where there's matter there's geometry. So yea, you can pretty much decode the universe with geometry and other math domains
if you look at how a 3d engine works, all objects are made from a mesh, and the mesh is guess what... made up entirely of triangles
in fact you can make a circle on your computer screen by adding a lot of small triangles together side by side
Even topology?
@@wesamzxc you can get pretty close to a circle but it'll never be a true circle
@@wesamzxcor just in Art, in general, everything is broken down into basic shapes.
Customer: I need a band (4π+12)(2-√3) long please.
Well, that's irrational!
@@nulltan 😆
All these years later even after being well acquainted with calculus and differential equations, geometry still lurks in the darkness
for real, geometry is my weakest point
Kind of a shame really, feels like geometry has so much potential to give an intuitive sense for why stuff works but then it just isn’t used in much of higher maths
@@cosmologicalturtle9528 Cause its hard
I'm the complete opposite. If I can put it in geometric terms it makes way more sense to me
@@NFBartos How do know what to manupilate/branch off in geometry while solving questions?
Don’t forget that you were given a unit! It is that many meters! (There’s a reason I study engineering and not maths)
Yes, you got me lol.
Units matter especially if you live in America 😞
in>cm
@@d4rk_1egendIn what way
@apdj94 because your have to use the imperial system?
Just of note that this is not actually a GCSE question, this is much harder than what you would find in a GCSE paper. This question is from the UKMT maths challenge, likely the intermediate level, as it is in r/GCSE, and is set to 13-16 year olds in the UK. This challenge has easier and harder questions, the question being from the C section of the paper it is one of the harder ones. Most students do not reach these questions in the allocated time, and the maths challenge is particularly aimed at high flying students who if they do well can go be selected by the UKMT to compete at higher levels.
O my gooooodddddd
I found it
Even though m a biology student o my godddddd
I remember taking one of these in my maths class in year 8, because I was in the advanced class for my school. I struggled incredibly
If this is harder than gcse I’m getting 100%
~6.58m, for anyone who was curious
Thanks
I think this is terrible question with terrible answer - because, for schoolchildren, it is preferable to have simple integers in answers, not just bunch of equations as it.
Yep. Was pretty easy. Solved it in my head in a few seconds.
@@Koba_78 here's your attention you asked for
@@scubasteve6175 Thank you, though your time would be better spent improving your math if you found this question difficult.
I'm a machinist, I have to deal with crap like this every day. LOL
tell us some stories of applied math Robert!
thx 👍🏻
Omg I thought geometry was useless for the longest time
@@m4nman I am an engineer currently installing a ventilated facade, I use geometry every day! Very fun stuff.
@@jesusalexisovallesgiuseppe5002what kind? Interested in Chemical or Industrial myself.
Engineers can't choose a useful measurement to save their lives, you always have to work back from whatever they choose to give you.
Then there are engineers, who draw that in cad in 30s and measure it there
Dimensions
I remember doing a question from rotational dynamics which had this same diagram.
The question was framed using the basics of pure rolling.
This was very fun to try solving! Also I think the area inside the band is (28-16sqrt(3))pi+120-68sqrt(3)
I got {[4(π+6+√3)]÷[7+4√3] }cm² which got same value as yours and which is : 3.1227699026 cm² 😊 [edit : I forgot to add unit ]
I was like how tf did you calculate it, but it's actually obvious from his drawing. Too long; didn't calculate, though
great way of solving the r that differs from my way (i didnt expect it!)
my way is like:
first, make a triangle with the points of the center of the right bottom circle, the point at which the bottom circles touch each other, the center of the top circle.
then apply pathagorous theorem to the triangle we made → r^2 + (2m-2r)^2 = (2r)^2
expand r^2 + 4m^2 - 8mr + 4r^2 = 4r^2
beautify r^2 - 8mr + 4m^2 = 0
solve r = 4m - 2m√3 (4m + 2m√3 doesnt satisfy the condition)
(i dont know if the m of 2m means meter or just a variable, if so lets remove the m from all of the formula in this explanation)
I was able to figure out that the perimeter would be 6r+2(pi)r but wasn’t able to find out how to get the length of the part in the middle. I failed to see the 30 60 90° right triangle in the middle, so as soon as you pointed it out I figured out how to do the rest.
I really like how you keep the colors consistent, thank you for showing us this one
Damn you made it look so easy! Now i want to go through some problems myself
4:13 You are not calculating the hypotenuse. You are referring to one of the legs of the right triangle. √3*r is still correct for that leg though.
You can make it the hypotenuse if you just rotate the triangle. That just makes the math easier.
Edit: Shit scratch that I just woke up
@@goldencreeper2551what is people maths exactly?
@@Quadratic4mula I dont know this means
He just skipped over the Pythagorean theorem to find the ratio, and shouldn't be calling it the hypotenuse. Here's the solution when r=1
1²+x²=(1+1)²
1+x²=4
x²=3
x=√3
@@goldencreeper2551 put it at the back of your head that the ratio of the sides of a 30,60,90 triangle have side ratios of 1,sqrt(3), and 2 respective of the angle's opposite.
The parts of the band that tightly hug the circle appear to be 1/3 of a circle's perimeter if we cut the band when it is tangent with the inner circles. So I'm going to hazard a guess and say that a close approximation would be the circumference of one of those circles plus the perimeter of a triangle formed by the leftover lengths.
Edit: I swear to god I did not watch the video and this was made before he started drawing the schematic. Somehow my guess was more than just a guess!
I thought the same before watching. Didn't bother trying to find r though
Another way to find the angle to be 120° is to extend the tangent lines until they connect. This creates a triangle from the tangent points to the new point. The obtuse triangle will then have angles 30, 30, 120.
Or just realize the line goes equally around 360 degrees. If there are only 3 (identical) circles, each circle's arc must be 120 degrees.
That's just more work than the problem asked for
@@shruggzdastr8-facedclown It was more intuitive when I did it than in writing 😅.
There's no way in icy hell that this is a damn GCSE question 😂 easily an a level question at least.
It's from the gcse level version of the UKMT, so it should be reasonably doable by gcse students, but it is probably still slightly harder than an average gcse question.
Many years ago we had questions like this on the higher GCSE maths paper, but on the higher paper the questions got progressively harder. Idea being the first questions were C grade, and final questions were A*. They do it differently nowadays but back in the day complex geometry questions like this were at the very back of the paper and worth a lot of marks to reflect their difficulty
Very precisely explained, appreciate it a lot👏👏👏
What is people maths?
This is my favorite part about math, you are just using non complex methods, the core problem becomes a logic puzzle
Drawing different radii from circle centers to useful points on edges (usually tangent lines or vertices of polygons) is a good way to start Geometry problems!
Thanks
It is interesting that the band still has the same length of contact (2*pi*r) with the circles (or rolls) when compared to wrapping a band around 1 circle. It seems this will hold true for 2 rolls and 4 rolls as well. It might be nice to formulate a proof that this holds for any (if this is true) amount of rolls.
How do you know these things are rectangles? I get when you drop the radii to the tangent point you get 90 degrees. But it's not clear to me (without invoking symmetry handwaving!) that when you connect the centres of two circles that they intersect the radii at 90 degrees. But then it's been 45 years since i did my GCSEs!
If you connect the centers of the bottoms two circles, you get a horizontal segment, which is parallel to the very bottom one. Thus that’s a rectangle. And by symmetry, the other two are the same too. 😃
@@bprpmathbasics Forcing me to ask: how do we know it's parallel?
@@silver6054 Let's say u have a line AB on the ground and let's say AB is 20 m. At point A, you have a stick going up vertically. Since the stick is not slanted, the stick and the line AB will be 90 degree. Let's make the stick 8 m. U do the same at point B. And please remember to use an 8 m stick at point B as well.
At the top of the 2 sticks, u tied a 20 m bar. The bar at the top, the line AB on the ground and the 2 sticks will form a rectangle and that is how they made the goal posts at a soccer game.
@@chinchang5117 That's not quite the model I see here though! Again, we drop two radii to the tangent points and join them. We have a line (outside the circles) of an unknown length at this point, and two perpendicular lines of radius R come out, and now we join the top of the radii with a line length 2R. I see in the real world those lines will complete the fourth side of a rectangle, and showing (as in the video) that the original line is 2R. But how do we know this centre connecting 2R line doesn't make say 105 degrees at one radius and 75 at the other? I am assuming something like congruent triangles will show this, but could only get half way with this.
Let us name it Quadrilateral ABCD, where AB connects the 2 centres, AB = 2r
BC = AD = r
angle C = angle D = 90°
We know that the opposite sides of this Quadrilateral are equal (BC = AD = r)
BC || AD as angle C + angle D = 180°
Therefore ABCD is a ||gm (1 set of opp sides are equal and parallel)
A ||gm with any angle as 90° is considered a rectangle.
Hence we prove that ABCD is a Rectangle
i had this all except the 60 30 90 triangle in the middle lol, the r root 3 was the bit i was missing
Very useful with exams coming up, thank you!
Wow thats beautiful. Simple and elegant solution.
how is that elegant?!?!?
The problem was very interesting and fun!
I just saw the thumbnail, and just tried it going from that. Luckily, I actually got it right ^^
(And for anyone inputting the numbers, I think it should be around 6,58 m for the solution.)
Thanks for the upload!
Not luck it’s skill
Before watching the video:
We assume that the band is stretched taut (0 slack) going to the Circumference + 6 * the radius.
It's touching each of the circles for 1/3 of their circumference, three circles so we can multiply 1/3 by 3 to get 1 circumference.
It travels the radius before it becomes its neighbor's domain where it travels another times the radius to get to the part where it's taut, so (r+r)* 3 for 6r.
From there we determine the radius with 3r (one whole circle + half the bottom 2 circles) + that empty bit in the middle = 2.
Finally we can solve for that empty bit (x) by creating a right angle triangle from
- where the two bottom circles touch
- the center of the top triangle
- the center of either left/right triangles.
And that's gonna be (r+x)^2 = r^2 + (2r)^2
Rewrite that for x and we've got x = √(r^2 + (2r)^2) - r
Back to our height we've got 2r + √(r^2 + (2r)^2) = 2.
Then just plug everything back into 6r+c and you're golden.
Edit: oh yea I'm also assuming the circles are even, which I assume the problem wants me to do even if they aren't technically marks.
After watching the video I think I was close enough to count. I had the same logic but I just forgot my special right triangles.
Btw the green angle is 90° because it connects a tangent line to the circle's center.
You can just set the height to 2r + r√3 due to the height formula of the equilateral triangle: h= a√3/2, where a = 2r and therefore h = r√3.
he did exactly that except he didnt use the formula because just doing the geometry is trivial enough
okay so band goes around the circles and then straidhg horizontal or in angle of 60 degrees top.
The length of the straight is 2 * of circle radius, also along circles it makes the "full" circle splitted on 3 different parts. So its length is 6* circle radius+ 1 length of the circle
2m consists of 2 radius + 2 radius * sin 60 degree,
so
2 = 2(r+√3/2)
r = 1-√3/2
length = 6(1-√3/2) +2* pi (1-√3/2)
One thing I really enjoy about maths is seeing cameos for previous maths topics which make an appearance for these questions.
Circle theorems, Pythagoras, Arc lengths, Solving for X and a sneaky appearance of Quadratics at the end. Really fun to see all this knowledge from previous topics.
Okay, I did it exactly the same way as you did lol. That usually never happens to me when solving math problems in youtube
This is even more obvious of comparing the form with those created with 2 or 4 (or more) circles:
The lenght of the band is always the circumference + 2r times the number of circles
Man I just smiled wider and wider as the picture came into focus what a genuinely fun problem!
Nice video, thanks!
Kinda happy as I worked the answer out from the thumbnail and I haven’t studied maths for 20 years. Thanks for a neat video :)
Rationalizing the denominator works out nicely
Before watching (and many years out of school), if the circles had r = 0,5m this would be simple. But I am looking at the graphic and it seems like the measurements are not that neat.
Well anyway, we can see that the band goes around each circle for 120 degrees so the length is just one perimeter of these identical circles, plus 3 times the tangent to tangent line. Which you would like to be 3m but I don't know yet.
I fear no man.... But that *thing*
(Geometry)
It scares me
mathematicians after spending years of their life learning to calculate shapes only to find out about tape measurers
theory vs experimentalism 👍🏻
Ideal and the real
It is also the distance a circle of radius r would travel if rolled around the central equilateral triangle 1 time
calculating the resulting formula gives you 7 m (taking into account the right amount of significant figures)
I wasn't expecting sig. figs on a maths channel.
well played my friend!
Man, I have been working on a farm too long. I looked at the question and said 7.5m instantly because that's about how long a strap or banding would have to be to go around those.
I figured out the perimeter steps, but didn’t want to try and find r. This problem reminds me a lot of those packing kinds of questions from my properties/structures of materials class
I started with R + R + cos(30)R + cos(30)R = 2
To me, that is 2R + 2cos(30)R = 2
Then in my head it became 2 / 2 / 2 = R + cos(30)R
For 2/2/2/cos(30) = 2R
But clearly it was super wrong. Even getting close i kept getting 0.57735, which is also wrong.
How was i so far off? What part of my approach was flawed?
Even with 2R + cos(30)2R = 2,
2/cos(30) should equal 4R, but it doesn't.
I feel really stupid, could someone point out what fundamental mistake i'm making?
I found it. One cannot cancel out by dividing individual terms, it has to be the entire side of the equation. Imagine understanding trig but missing that! Stay in school!
Thanks for the video, it indirectly forced me to learn stronger fundamentals
I was able to solve it before watching the video, with a slightly modified (worse) calculation path and taking more time.
And this is GCSE level? My god
it looked like he was about to summon a demon with that breakdown
I was confused about the lines forming a rounded triangle but then realised you don’t need it and
The way my brain started making triangles the second I saw this is distressing. Engineering classes have broken me.😊
This was fun remeinded me of my high school geometry good times thank you
Bprp has given up to technology 😭
That’s the only way for me to make geometry videos 😃
I lost you where you factored out the r. No idea what maths logic was applied there
Ah I missed the equilateral triangle in the middle. Nice one
the hypoteneuse isn't sqrt(3)*r, but the opposite to the 60º angle is.
something that bugs me about this question, we have to assume all 3 circles are all equal. the problem in the picture doesn't make any mention of the radius of the 3 circles being the same.
Even in University it's easy to forget that you can just make a triangle... From a triangle...
Before watching.
I know that the solution is going to be a circle plus the triangle which ends up being
Pi r^2 +6r
But I can't figure out how to get the radius from the given 2m, I tried bisecting the object to make a 1x2 triangle, but that's not the same angle as the 4r I was trying to find.
When I was at school this was a slightly different question. We were asked to determine the height of a stack of cannonballs.
Fun task that I actually managed. 🙂👍
4:15 - I'm pretty sure the hypotenuse is 2r 🙂 but I agree that the side being referenced is √3r
Ok.... but how you apply this to the real world?
I love math but this is what drives me crazy about algebra: this crazy non actual answers ... to me is not done until I see a number and no symbols or letters next to it.
That said, love your content, thank you.
I remember this question, it's actually much easier than it looks.
This would work great for changing lawnmower belts of an unknown size. Too bad I couldn't do a math problem correctly if my life depended on it...
So because I wanted to see how good I am I tried it myself, had a different approach where I made it more difficult for myself.
One of the things I saw here was a triangle one leg going top to bottom and the hypotenouse going 15° from the top through the touching point of two circles to the lowest point of the bottom circle. So I calculated the hypotenouse with 2÷cos(15). The I devided that by 4 because I would have basically the same triangle inside one circle and this would be one leg, then I diveded that by cos(15) again and had r. Multiplied it by 6 and another time with 2×pi. My calculation was:
2÷cos(15)÷4÷cos(15)×6+2÷cos(15)÷4÷cos(15)×2×pi
Which gives the same result as you got when I punched it into my calculator.
I'm happy with myself that I was still able to do this.
If you draw out the outer triangle, it is better provable that the circle slices are 120 degree.
I wish these types of questions were on American tests, they look to fun!
Sign your high school up for University of Waterloo CEMC - a set of fun computer and math contests.
They have stuff from Grade 7 on up, and advanced students can try the 7th grade one as early as Grade 3.
Nice problem, thanks
lol "the answer isn't pretty"
welcome to engineering.
It never occurred to me to reduce it down symbolically. I saw the height of the inner triangle is 2r × sin(60°) and just plugged it into a calculator:
r = 0.54 m
L = 2r(3+π) = 6.63 m
or 7 m if you're into significant digits 😆 (chemical engineer, here).
If just trying to find length of band around shape, why not just use arc lengths of 3 120 degree portions, plus 3 times the base of rectangle, 3r(theta) + 6r?
I absolutely love watching videos I don’t understand
I have no idea how you infer the rectangles. @2:07 you have a bottom green 90° angle drawn. How did you infer that ? How do we demonstrate the band is parallele to the bottom leg of the equilaterale triangle ? I see it, but cant manage to prove it. It's not as simple as saying the normal radii in red are vertical and 90° to the tangent. Ok but the blue r is 90° to the red r ? Since when ? Because i say so ?
I would just try to approximate to the triangle with 2m height
engineering: compute r, its r~0.536, and if you estimate the total height as being about 4r, its about the same as r~0.5
If you think it’s 4r then you are totally wrong. It would be the case with 4 pipes. But with 3 the one on top goes somewhat between the lower 2
Ah yes, it'll be great asking if they have a (4pi+12)(2-sqrt3) length belt in stock. I hate these kinds of questions because real life doesn't work like this. Figuring it out was fun though.
I had a minor brain fart by using 60 deg for each circle segment instead of 120, but other than that my derivation was correct. Feels good to be reassured I’m still capable of working out a problem like this lol.
actually something like this was in my book as a challenge question but what was only given was the diameter of the circles and the solution is similar to what you said in the video but the answer was 0.5(pi) + 3. Also it was Edexcel International GCSE Mathematics B
I got stuck realising that I could factor out r from "2r+sqrt(3)r"
I did rough calculations and estimated and i got within .3 meters
The language of the question is ambiguous.
Class 10th India's question... For Olympiad lvl...
Isnt the hypotenuse the side across from the right angle? Meaning the 2r side?
This problem is so easy bru I saw it in my mind in like a minute , no show off it’s just basic
but what if the 2m line is meant to be from outside band to outside band?
assume the band is infinitely thin. ✅
lol
4:17 that is not the hypotenuse
Sir you are a genius
The length of the band will be 6.5825 meters.
It's not a particularly difficult problem... *_if_* you have adequate time to ponder it. The difficulty comes from doing under the time pressure of 10 mins per question for GCSE...
...I could do it given an hour or so...
I have a real concern: nothing shows on the diagram that the circles have the same diameter no?
Do videos on ukmt maths challenge
I did it! Just needed to look up height formula for triangle 😆
I just make a Solidworks part and it calculates everything...
The problem doesn’t specify that all three circles have the same diameter. They look that way as drawn, but usually, if not specified, we shouldn’t assume.
True but if they weren't the same size it wouldn't be possible to solve
Actually, the paper requires us to make that assumption, or the question is unsolvable. Also,given they ARE all the same size on the paper, who are we to assume they're "meant" to have different diameters? Unless specified that wouldn't make any sense
first time ive enjoyed watching a math video
Why didn’t you say the length of the band is 3 multiplied by the circumference each 120 degree sector covers + the length of each rectangle
Therefore 3x2r + 3x120/360 (2πr)
Which simplifies to 6+2π
h=2
h = 2r+ H
2 = d + H
h = r + r + u + r
h = 3r + u
2 = 3r + u
i really don't know how to measure u :(
Excellent, thanks.
4:23 , the hypotenuse cant be sqrt 3r because we know the hypotonuse is 2r. did u perhaps say it on accident or am I not following?
before watching:
4(2-sqrt(3))(pi+3)
Math can be so elegant sometimes
Instead of getting it in in roots and pies, i did some approximation along the way and got approximately 6m as the answer, hahaha