1:44 not only is the Reuleaux polygon useful because it rolls, but the constant width feature is also important on itself. Some vending machines actually detect the width of the coin (not just will it fit the slot, but does it fill the slot as well). The old 3d coin (3 old pence) was a 12-sided polygon, with straight edges and it rolled fairly well with a nice rattling sound because 12 sides is close enough to a circle to make rolling possible. Even so, vending machine designers does not like that coin because checking it's width was dependent on orientation
In the first problem, it never struck me that we have to find the sum of the areas of the shaded region. I was busy trying to find the individual areas. I quickly realized that it is impossible because we have 3 equations and 5 variables.
Upper Region: 9π/2-9arctg(4/3)-108/25 or 1.472 Right Region: 8π-16arctg(3/4)-192/25 or 7.157 Lower Region: 25π/2-16arctg(4/3)-9arctg(3/4)-12 or 6.642 Actually you can: ●set the bottom left corner of the triangle to be the point (0,0) ●we can now describe the two circumferences by the equations x²+(y-3)²-9=0 and (x-4)²+y²-16=0 ●then we can put them in a system to find that the coordinates of their intersection (the point on the hypotenuse) are (72/25,96/25) ●now we want to calculate the upper region, and to do that we have to calculate the sector, with as its center the midpoint of the left side of the main triangle, and subtract the triangle (let it be triangle T) that has vertices the top corner of the main triangle, the midpoint of its left side, and the intersection point of the two circumferences ●so we first notice that the triangle T is isosceles, so the vertex angle is a=π-2b where "b" is the angle at the top corner of the main triangle, and we know that "a" is also the angle of the sector ●we find that b=arctg(8/6)=arctg(4/3), so the angle of the sector is a=π-2arctg(4/3) ●we can now calculate the area of the sector, A:a=πr²:2π -> A=ar²/2=9π/2-9arctg(4/3) ●now we calculate the area of the triangle T, which is b*h/2 where b is 3 and h is the x coordinate of the intersection point we caculated earlier, so T=(3*72/25)/2=108/25 ●finally the area of the upper region is A-T=9π/2-9arctg(4/3)-108/25 which is approximately 1.472 ●in the same way you can calculate the region on the right to be 8π-16arctg(3/4)-192/25 which is approximately 7.157 ●also in a similar way you can calculate the lower region, but you have to make some adjustments, you have to consider it like the sum of two region both being the difference between a sector and a triangle (like before) ●note that in this case tha triangle T is again isosceles and the formula for angle a is again π-2b, but b is no longer the arctg of the ratio of the sides of the main triangle (like in the first two cases) but instead the arctg of the ratio of the sides of the triangle that has as sides the x and y coordinates of the intersection point; fortunately arctg((72/25)/(96/25)) and arctg((96/25)/(72/25)) can be simplified respectively as arctg(3/4) and arctg(4/3) ●so we then calculate this region to be 25π/2-16arctg(4/3)-9arctg(3/4)-12 which is approximately 6.642 ●to check the results you can sum the values and find that the result is equal to the original answer of 25π/2-24 or 15.270
It's a bit silly to talk about roman vs arabic numerals in the context of decimalizing currency, because the two are basically completely unrelated things. All of Europe was already using decimal numerals _literally centuries_ before England finally decimalized their currency in 1971, for completely unrelated reasons. Heck, even India didn't convert to a decimal system for their currency until 1957 (long after many many other nations did), so they were arguably well behind the curve in this respect too...
On the coin, would there be a slight difference between the two arcs at the top, since they have different radius? Very very slight, but different 🤷♂️
You're correct that the smaller region he calculated would have a different arc than it being a circular sector in its own right. However, he calculated its area by subtraction rather than a circular arc formula. No worries there!
I don't see why anyone would draw the smalle rtrisngle pizza so can't younsolve without that nonsense isn't the area just.formula.for circular sector 1theta/2 times 5 plus whatever the last 2 segments.are..then figure out theta somehow?
Three ways you can look at why this is correct. Two require some amount of calculus, but I'll start with the non-calculus one, and then use some non-math reasoning that will set up some calculus, and then just describe the proof from calculus: (1) Consider the better known relationship for the length of an arc of a circle of radius r subtended by the angle theta. In this case the arc length is r*theta. If we wanted the full circumference, theta is 2*pi and we get an arclength of 2*pi*r. Now we might try to guess that we could do something like this for area, where the pi in pi*r^2 is theta. But that leads to theta*r^2 and a full circle requires theta = 2*pi, so this is no good. However, we can see that we're just off by a factor of 2, so we can modify our guess to be 0.5*theta*r^2. Of course, this still gives the expected area of 0 for theta = 0. We can further check other known angles to convince ourselves our modified guess works out. (2) Something you might have learned in biology, chemistry, or physics is the connection between an area and the circumference. Really, this is the relationship between any enclosed region and its boundary, so this works for surface area and volume as well. What this relationship states is that the area changes at a rate equal to its circumference (or its volume changes with a rate equal to its surface area). Note that this is not a rate in time (a speed) but a rate in size. I.e. If I want to increase the area a little bit, I'll need to increase the radius, but that means I also have to increase the circumference to accommodate that increased area, too; increasing the radius -- the size -- effects both area and circumference in a way that is connected. This is geometric relationship is what limits cellular size: as cells produce waste, they have to get rid of it, but the bigger volume a cell has, the more difficult it becomes because the volume (r^3) grows as the surface area (r^2), and removing the excess waste volume through the cell wall's surface area becomes difficult for larger and larger sizes (r). The inverse of this is also why small object cool quicker than larger objects: an object's total thermal energy depends on the objects temperature and how much material is in the object -- ie how big the object is or its volume; to give up that thermal energy, that requires the surface of the object to interact with the outside, and the surface area limits how quickly the that energy can be transferred to the outside, with smaller volumes having a lot less energy to deal with (~r^3) for a given surface area (r^2). Returning to mathematics, all of this means that the rate of change in size of the volume -- d/dr * (4/3)*pi*r^3 = 4*pi*r^2 -- is just the surface area, or the rate of change of the area -- d/dr * pi*r^2 = 2*pi*r^2 -- is just the circumference. But that works just the same for our guessed formula: d/dr * (1/2)*theta*r^2 = theta*r. If you prefer, you could also start with the familiar arc length formula theta*r and integrate it up to the area (1/2)*theta*r^2. (3) All of this can be done more generally if you're familiar with both integration and coordinate changes. We could describe a circle in terms of (x,y) coordinates, or we could use polar (r, theta) coordinates. You also need to get into the trigonometry (though there's multiple methods to go about getting from A to B, so to speak). If you do all of this, you find that an infinitesimal area dx*dy which will integrate up to some area A, is the same as saying dr*r*d(theta), where the non-infinitesimal r that is hanging out int hat is coming with the d(theta) piece in order to make sure everything still has the correct units (dx and dy have units of length, while r has units of length and theta has no units, so we need r*d(theta). Therefore, A = (integral of dA) which we can write as (integral dxdy) or (integral of r*dr*dtheta), which is really (integral r*dr)*(integral d(theta). The both are just polynomial integrals, with the first is equal to (1/2)*r^2 and the second is equal to theta. Thus, the area is (1/2)*r^2*theta.
@@smylesg he is referring to the University of Warwick in England. The custom of most respectful people is to adopt the cultural conventions of the place referred to. Call it what you like in Rhode Island, you would be laughed at here for not knowing or impolitely ignoring the convention here.
3:10 Presh, a little pronunciation guide of British towns for our friends across the pond: Warwick: war-ick (the second W is silent) Leicester: less-ter Gloucester: Gloss-ter Derby: Dar-by Edinburgh: Edin-bruh (not berg) Birmingham: Bur-ming-hum (not HAM) There are plenty more but this is a good start. Now before the comments go off on me saying, this is a mathematics channel - well, we’re sharing knowledge in a respectful way without compartmentalising the subject matters, as Presh has wonderfully done with the history of the 50p coin.
That is a really cool history of the 50p coin, which by the way, has been reduced in size but still has those 7 sides. It never occurred to me the requirement that it would have to roll down a chute of constant width such as inside a vending machine, without getting stuck. A very clever design!
The first problem shows the two semicircles intersecting on the hypotenuse, but this technically needs to be proven, and the proof follows from the observation that drawing a perpendicular on the hypotenuse forms a right angle opposite to each circle’s diametric chord, so it must lie on both circles as a circle is the locus of points forming a right angle on any diametric chord.
Excellent! One remark though: not "perpendicular", but "altitude" . The altitude on the hypotenuse is perpendicular to the hypotenuse, but also intersects the right angle vertex opposite to the hypotenuse. (A perpendicular on the hypotenuse can be positioned anywhere on the hypotenuse; unless we specify "a line _from the right angle vertex_ of the triangle, perpendicular to the hypotenuse".)
The answer to the first question is pretty easy if you grab an inclusion exclusion style of mindset. What isn't obvious to me is why the two circles intersect on the hypotenuse of that triangle. The argument is dependent on that assumption.
It was not obvious to me either. But writing out the equations of the two circles and the hypotenuse, I was able to determine with some algebra that the circles will always intersect on the hypotenuse. Given that they do, it seems there should be a simple proof to demonstrate it, but my computation was rather tedious.
If the circle on the vertical leg has radius a and the circle on the horizontal leg has radius b, and the origin is at the lower-left vertex of the triangle, then the equations are: (y - a)^2 + x^2 = a^2 y^2 + (x - b)^2 = b^2 y = 2*a - a/b * x The circles intersect at (0, 0) and x = 2*b*a^2 / (a^2 + b^2) y = 2*a*b^2 / (a^2 + b^2) And you can verify that point is on the hypotenuse line y = 2*a - a/b * x
Ah, I think I have it. Draw the first circle and look at the point where it intersects the hypotenuse. Draw a line from the lower-left vertex of the triangle to the intersection point. That new line forms one of the legs of a new right triangle, since the angle is subtended by the semi-circle. That means the line drawn is the altitude of the original right triangle. Since there is only one possible altitude of a right triangle from the hypotenuse, the other circle must intersect at the same point on the hypotenuse.
@@XJWill1 Nice! When I thought about it more awake, I started formulating that equation based proof, but as you said, I find they rarely are the best way to prove a geometry based problem. Funny thing is I had thought about the right angle of a semicircle theorem but didn't apply it to drawing in a line from the right angle vertex to the intersection with the hypotenuse. Good thinking!
thanks for doing my question! very big thumbs up to ya. im going to a math challenge next year and ukmt 2024, and youve always been a great deal of help when it comes to geometry. -rafael, the fella that suggested the first question
PB1: the intersection point slices the bug triangle into 2 rect triangles... Then area = area of the 2 circles - area of 2 small triangles = area of the 2 circles - area of big triangle. π3²/2+π4²/2-6*8/2 = π25/2-24 Edit for extra details: just imagine slicing the "c" area in half.
That's also how I did it: by drawing a straight line between the vertices of area c . Note: the "c" area is not sliced "in half", but into two parts. The two parts are not of equal size.
When I started working out sectors, arcs, and triangles in the first one I knew I was in trouble. When you labelled the parts I knew there was something I missed. The moment you made the sections in algebraic I saw where it was going and was kicking myself! I can always use algebra, but the visuals always beat me!
I think the video's still showing a bit of a "magic trick". The big thing to consider is not working out any angles or circle sectors until you at least see that there's two semicircles so there's no need to deal with arcs. You can take both semicircles and cut out the inner triangles. Then you may realize that those inner triangles form the big triangle, so no need to calculate the area of the individual smaller triangles either. Going "I labeled these points, then I subtracted them and they just happened to turn into the equation I want" is in my opinion not a useful solution, because there's no thought behind it.
There are two thoughts, that will help you solve a lot of problems. First, they ask for the entire value of the blue area. That means, it might not be necessary to separately calculate each blue area. Second, try to find areas, that can easily be calculated and make use of them. So it's not magic, but solving this kind of challenge over and over again.
For me, it may be wrong, but the solution is simple. as there are 7 equal arcs on the outside... they average out, and you can refer to this shape as an approximate circle. Thus the area would be pi times the radius squared..or in the ballpark of 3.14. Wrong for a mathematician but 'close enough' for everyone else.
So area of the top 2 shapes are 12.5π but this actually includes the shape in the middle because you are counting it twice and then minusing 1 off because we are removing the triangle as shown here (25π)/2 - 24u² And thus, this is the answer
I didn’t understand why you think the arc of big slice is equal to the arc of the small slice. The arc of the small slice is bulged out more. So the arcs don’t overlap.
Everytime I open a comment section of a yt video, there's at least 1 indian guy explaining how badly they suffered while solving integration problems in 4th grade.
BTW, the 20p coin is the same shape as a 50p coin, but smaller. and the current 50p coins (since the late 90s, IIRC) are smaller than the previous ones.
I calculated the area of the heptagon. Then I calculated the area of the round corners. Finally I added the area of the heptagon and the area of the round corners, thus obtaining the total area of the coin.
My bad, reveals my ignorance. I didn't even think to look it up because I thought it would be pronounced like Dionne Warwick. ua-cam.com/video/zphc3xcmJXk/v-deo.htmlsi=4Jd7YM3FWcEqJScG
@@ntlespino It's not a "snooty convention" it's just someone saying the word wrong. Many English words aren't spelt how they're pronounced. It's like going to another country and mispronouncing all their words and you calling them snooty for correcting your pronunciation.
"Always" is a strong assertion... and also wrong in this case! For example, the town of WICK at the far north of Scotland, which is just pronounced WICK. There's "always" an exception to every rule in English (including this rule!) :-D
I'm crap with trig questions so I'll just do the first one, though i will take some inspiration from your 50-pence story. Let the subdivisions of the semicircles/triangle figure be as follows: from top to bottom the three sections covered by the radius 3 semicircle will be I, II, and III. From left to right, the remaining two sections covered by the radius 4 semicircle will be IV and V. We need I, III, and V. To sections above are contained by the constituent polygons as follows: R3 semicircle: I + II + III R4 semicircle: III + IV + V Right triangle: II + III + IV Adding together the two semicircles and subtracting the triangle we get the following: I + II + III + III + IV + V - II - III - IV Simplifying the above, we get I + III + V, as II and IV cancel along with one of rhe IIIs, which is just what we need. R3 semicircle: A = (πr²)/2 = π(3²)/2 = 9π/2 R4 semicircle: A = (πr²)/2 = π(4²)/2 = 16π/2 = 8π Right triangle: A = bh/2 = 8(6)/2 = 48/2 = 24 9π/2 + 8π - 24 = 25π/2 - 24 ≈ 15.27
6:37 How do we know the 2 triangles formed using the line segment, are isosceles? As in, how do we know that line segment to the centre is the same length as the side of the pizza slice?
Can please someone explain the 2nd problem. Why those 2 triangles are isosceles? The lower leg is ½ but upper 2 are not ½ since it is not circular sector
The isosceles triangles are made up of radial line going from the centre of the coin. As long as a line goes from the centre to any of the 7 "corners", it's the same length.
@@AbDmitry the 2 legs are actually slightly more than 1/2 because they are inscribed from the centre of the coin to an arc from an origin on the bottom edge of the coin. Presh said, at the start of the problem, that the "pizza slice" was not a circular sector, although it looks like one. That's where the confusion is coming from.
@@AbDmitry imagine drawing 7 radial lines from the centre to each of the 7 corners on the coin edge, starting clockwise from top. Imagine measuring these lines and comparing them with each other. Would any of the lines be longer or shorter than the others?
Question 1 is quite simple. It would have been more challenging to prove that the intersection of the two semi-circle arcs is (always) indeed on the hyptenuse of the triangle.
Question 1: (π*4²)/2 + (π*3²)/2 - (2*3)(2*4)/2 = (16π)/2 + (9π)/2 - 48/2 = (25π - 48)/2 = 25π/2 - 24 Question 2: Let C be the center of the coin. Let Vᵢ for i = 1, 2, 3, ..., 7 be the vertices of the coin (counter-clockwise, starting from the top in the diagram). Let s = |CVᵢ| be the distance between C and each vertex Vᵢ . For 0 < {α , β , γ} < 180° , let α = angle(V₁CV₂) be the angle between CV₁ and CV₂ , let β = angle(V₁CV₅) = angle(V₅CV₂) be the angle between CV₁ and CV₅ , let γ = angle(V₁V₅V₂) be the angle between V₁V₅ and V₂V₅ . Then α = 360°/7 = (360/7)° = 2π/7 [rad] β = (360° - α)/2 = (360° - 360°/7)/2 = 360° * 3/7 = 6π/7 [rad] γ = 2 * (180° - β)/2 = (180° - β) = (180° - 360° * 3/7) = (7*360°/14 - 360° * 6/14) = 360°/14 = 2π/14 [rad] = π/7 [rad] Calculate s: (|V₂V₅|/2) / |CV₅| = cos(γ/2) ⇒ (1/2)/s = cos(γ/2) ⇒ s = (0.5)/cos(γ/2) Area of arc sector between V₁V₅ and V₂V₅ is: A₁ = (γ / 360°) * (π*1²) = (1/14) * π = π/14 Area of triangle V₁CV₅ is: A₂ = s * (1*sin(γ/2)) / 2 = = 0.5/cos(γ/2) * (sin(γ/2)) / 2 = 0.25 * (sin(γ/2))/cos(γ/2) = 0.25 * tan(γ/2) = 0.25 * tan(π/14) Area of coin sector between CV₁ and CV₂ is: A₃ = A₁ - 2*A₂ = = π/14 - 2 * 0.25 * tan(π/14) = π/14 - (0.5)*tan(π/14) Hence, area of coin is A = 7 * A₃ = = 7 * [ π/14 - (0.5)*tan(π/14) ] = π/2 - (7/2)*tan(π/14) Q.E.D.
1st problem I was able to work out pretty easily: Adding the areas of the 2 half-circles gave me the area we were asked to calculate + the area of the 6x8 triangle. Add them, subtract the triangle, and BAM, done. 2nd problem...I'm not ashamed to admit my whole brain just went "HUH!?!?!"
1:14 The 5 and 10 pence coins weren’t “new”. The pre-decimalisation shilling became worth 5 pence, and the 2 shilling coin became worth 10 pence, with all of these coins remaining in circulation post-decimalisation. This saved minting a whole load of new coins and having them swapped instantly on decimalisation day.
I was thinking about the second problem: if it's a shape of constant width, wouldn't it's area form / map to a rectangle over a 180 degree roll? It's width would be 1, and it's length would be half the perimeter of the coin. Is this not right?
Apparently not. (Width of coin is 1 , perimeter of coin is 7 * (2π/14) = π , so the area of the rectangle you're looking at is width * half perimeter = 1 * π/2 = π/2 which is _not_ the area of the coin.) By the way, your formula isn't even true for a circle, so why would it be true for this coin?
It is only possible if you know the angle. Obviously the curve can be drawn from close to a straight line to almost making a full circle with the others. So knowing about the construction of the shape is crucial. Without it you only can calculate the heptagon and tell tha the area of the hole shape is bigger than the area of the heptagon.
@@robertveith6383 Well, I did not watch the video. So, sorry for my misunderstanding. After watching the video part with the construction it is rather simple. But I don't do it, because it is not interesting for me anymore.
At 7:39 , the blue-colored area is a _circular sector_ (in other words, a _sector_ of a circle) with a radius of r =1 . (Note: that circle isn't drawn here, but its area is approximately 4 times as large as the coin's area.) The area of a circle sector is calculated by the formula (r²/2)*θ , where θ is the angle of the sector. This formula follows from the fact that the area of a circle is πr² , so the area of just a sector of angle θ, equals πr² * (θ/(2π)) = (r²)*θ/2 . So, for example, if θ = 60° = π/3 , the area of the sector would be (1/6) of the full circle area. In our case, r = 1, and θ = 2c , because θ is the angle at the bottom vertex of the blue area (where the two "flat" isosceles triangles meet, and the angle at the corner of either isosceles triangle is c ; so θ = c+c = 2c). I hope that helps.
Shorter answer: the blue area at 7:39 is a "pizza slice" with a radius of r=1 and an angle of _theta_ = 2c . The area of a pizza slice is given by the formula r^2 * (theta)/2 , where r is the radius of the pizza, and _theta_ is the angle of the slice. ( _theta_ is also called "central angle" because the point of the slice corresponds with the _center_ of the original pizza.)
Funny how in question 2, you calculate angle b as (360 - a)/2 , _just as I did_ ; instead of simply noticing that b = 3a . In question 1, we actually don't need b and d. Instead, we can simply divide area c into c1 + c2 (with c1 to the left/on top of c2) by drawing a straight line between the two vertices of area c . This line also cuts the main triangle into two (triangular) parts. The required answer is then easily seen as (a + c2) + (c1 + e) with (a + c2) = {half circle with radius 3} - {top/left part of main triangle} , (c1+e) = {half circle with radius 4} - {bottom/right part of main triangle} , hence a+c2+c1+e = {half circle with radius 3} + {half circle with radius 4} - {whole main triangle}
Anyone else ghik its kind of poor wordong st 2:07 to say because the WIDTH ks cistsnt on the shape the HEIGHT doesnt change..just because width doesn't change as you roll something doesnt mean height will stay constant..
I have to admit that, at least here in the US, when working out geometry problems, we wouldn’t think of expressing angles in radians. I never did until I was comfortable memorizing common radian-degree correspondences.
There's a much easier solution to the first problem. You just need to add the two hemicircles and subtract the triangle. How you know to get this is by writing a "+" on all the regions of the diagram that you are adding. Then you write a "-" to all the regions you are subtracting. Naturally you do this until the parts you don't want to calculate the area of has both a "+" and a "-", cancelling each other out, while the part you DO want the area of has a net 1 "+". It sounds more complicated than it actually is, it's a lot easier when you draw it out.
or you just realize that bc the two semicircles share a region that is shaded and within the triangle, subtracting the triangle's area removes the area of one semicircle for the shared region, leaving behind the area for the other semicircle.
The solution to the first problem really floored me! So simple, quick and elegant! Great job, man!!!
Man I spent like 2 hours doing the 1st question using cos rule and sectors, only to find such a simple solution. At least I got the right answer.
1:44 not only is the Reuleaux polygon useful because it rolls, but the constant width feature is also important on itself. Some vending machines actually detect the width of the coin (not just will it fit the slot, but does it fill the slot as well).
The old 3d coin (3 old pence) was a 12-sided polygon, with straight edges and it rolled fairly well with a nice rattling sound because 12 sides is close enough to a circle to make rolling possible. Even so, vending machine designers does not like that coin because checking it's width was dependent on orientation
But how tonactusly dope the problem without what I think is a cheat Presh does.
Another thing about those polygons is that they need to have an odd number of sides
In the first problem, it never struck me that we have to find the sum of the areas of the shaded region. I was busy trying to find the individual areas. I quickly realized that it is impossible because we have 3 equations and 5 variables.
Upper Region: 9π/2-9arctg(4/3)-108/25 or 1.472
Right Region: 8π-16arctg(3/4)-192/25 or 7.157
Lower Region: 25π/2-16arctg(4/3)-9arctg(3/4)-12 or 6.642
Actually you can:
●set the bottom left corner of the triangle to be the point (0,0)
●we can now describe the two circumferences by the equations x²+(y-3)²-9=0 and (x-4)²+y²-16=0
●then we can put them in a system to find that the coordinates of their intersection (the point on the hypotenuse) are (72/25,96/25)
●now we want to calculate the upper region, and to do that we have to calculate the sector, with as its center the midpoint of the left side of the main triangle, and subtract the triangle (let it be triangle T) that has vertices the top corner of the main triangle, the midpoint of its left side, and the intersection point of the two circumferences
●so we first notice that the triangle T is isosceles, so the vertex angle is a=π-2b where "b" is the angle at the top corner of the main triangle, and we know that "a" is also the angle of the sector
●we find that b=arctg(8/6)=arctg(4/3), so the angle of the sector is a=π-2arctg(4/3)
●we can now calculate the area of the sector, A:a=πr²:2π -> A=ar²/2=9π/2-9arctg(4/3)
●now we calculate the area of the triangle T, which is b*h/2 where b is 3 and h is the x coordinate of the intersection point we caculated earlier, so T=(3*72/25)/2=108/25
●finally the area of the upper region is A-T=9π/2-9arctg(4/3)-108/25 which is approximately 1.472
●in the same way you can calculate the region on the right to be 8π-16arctg(3/4)-192/25 which is approximately 7.157
●also in a similar way you can calculate the lower region, but you have to make some adjustments, you have to consider it like the sum of two region both being the difference between a sector and a triangle (like before)
●note that in this case tha triangle T is again isosceles and the formula for angle a is again π-2b, but b is no longer the arctg of the ratio of the sides of the main triangle (like in the first two cases) but instead the arctg of the ratio of the sides of the triangle that has as sides the x and y coordinates of the intersection point; fortunately arctg((72/25)/(96/25)) and arctg((96/25)/(72/25)) can be simplified respectively as arctg(3/4) and arctg(4/3)
●so we then calculate this region to be 25π/2-16arctg(4/3)-9arctg(3/4)-12 which is approximately 6.642
●to check the results you can sum the values and find that the result is equal to the original answer of 25π/2-24 or 15.270
@@gabrielel.9615 Lot's of respect to you bro!
Just a heads up but you do not need to pronounce the second w in Warwick
2nd problem is a nice one. Enjoy it.
The first question is a 10th or 9 th standard question in India.
Yes
Yes
Same in Bangladesh
I assume the inclusion/exclusion method is taught? That would make the question 10x easier
What do you mean by "10th standard"?
The second one I could not get right
bro i solved the first problem the exact same way, using the same variables for each part.
It's a bit silly to talk about roman vs arabic numerals in the context of decimalizing currency, because the two are basically completely unrelated things. All of Europe was already using decimal numerals _literally centuries_ before England finally decimalized their currency in 1971, for completely unrelated reasons. Heck, even India didn't convert to a decimal system for their currency until 1957 (long after many many other nations did), so they were arguably well behind the curve in this respect too...
Thanks man you gained my respect
8 more days to prepare...
On the coin, would there be a slight difference between the two arcs at the top, since they have different radius? Very very slight, but different 🤷♂️
There's only one arc. You're subtracting two triangles from the sector with radius 1 to find the remaining area.
You're correct that the smaller region he calculated would have a different arc than it being a circular sector in its own right. However, he calculated its area by subtraction rather than a circular arc formula. No worries there!
I don't see why anyone would draw the smalle rtrisngle pizza so can't younsolve without that nonsense isn't the area just.formula.for circular sector 1theta/2 times 5 plus whatever the last 2 segments.are..then figure out theta somehow?
Can some one please explain me about how he took the area of sector to be(( r^2)/2)*(theta)
Area of circle is pi*r^2. So for a sector's area multiply by the angle of the sector and divide by (2*pi). The pi cancels out
Three ways you can look at why this is correct. Two require some amount of calculus, but I'll start with the non-calculus one, and then use some non-math reasoning that will set up some calculus, and then just describe the proof from calculus:
(1) Consider the better known relationship for the length of an arc of a circle of radius r subtended by the angle theta. In this case the arc length is r*theta. If we wanted the full circumference, theta is 2*pi and we get an arclength of 2*pi*r. Now we might try to guess that we could do something like this for area, where the pi in pi*r^2 is theta. But that leads to theta*r^2 and a full circle requires theta = 2*pi, so this is no good. However, we can see that we're just off by a factor of 2, so we can modify our guess to be 0.5*theta*r^2. Of course, this still gives the expected area of 0 for theta = 0. We can further check other known angles to convince ourselves our modified guess works out.
(2) Something you might have learned in biology, chemistry, or physics is the connection between an area and the circumference. Really, this is the relationship between any enclosed region and its boundary, so this works for surface area and volume as well. What this relationship states is that the area changes at a rate equal to its circumference (or its volume changes with a rate equal to its surface area). Note that this is not a rate in time (a speed) but a rate in size. I.e. If I want to increase the area a little bit, I'll need to increase the radius, but that means I also have to increase the circumference to accommodate that increased area, too; increasing the radius -- the size -- effects both area and circumference in a way that is connected. This is geometric relationship is what limits cellular size: as cells produce waste, they have to get rid of it, but the bigger volume a cell has, the more difficult it becomes because the volume (r^3) grows as the surface area (r^2), and removing the excess waste volume through the cell wall's surface area becomes difficult for larger and larger sizes (r). The inverse of this is also why small object cool quicker than larger objects: an object's total thermal energy depends on the objects temperature and how much material is in the object -- ie how big the object is or its volume; to give up that thermal energy, that requires the surface of the object to interact with the outside, and the surface area limits how quickly the that energy can be transferred to the outside, with smaller volumes having a lot less energy to deal with (~r^3) for a given surface area (r^2).
Returning to mathematics, all of this means that the rate of change in size of the volume -- d/dr * (4/3)*pi*r^3 = 4*pi*r^2 -- is just the surface area, or the rate of change of the area -- d/dr * pi*r^2 = 2*pi*r^2 -- is just the circumference. But that works just the same for our guessed formula: d/dr * (1/2)*theta*r^2 = theta*r. If you prefer, you could also start with the familiar arc length formula theta*r and integrate it up to the area (1/2)*theta*r^2.
(3) All of this can be done more generally if you're familiar with both integration and coordinate changes. We could describe a circle in terms of (x,y) coordinates, or we could use polar (r, theta) coordinates. You also need to get into the trigonometry (though there's multiple methods to go about getting from A to B, so to speak). If you do all of this, you find that an infinitesimal area dx*dy which will integrate up to some area A, is the same as saying dr*r*d(theta), where the non-infinitesimal r that is hanging out int hat is coming with the d(theta) piece in order to make sure everything still has the correct units (dx and dy have units of length, while r has units of length and theta has no units, so we need r*d(theta). Therefore, A = (integral of dA) which we can write as (integral dxdy) or (integral of r*dr*dtheta), which is really (integral r*dr)*(integral d(theta). The both are just polynomial integrals, with the first is equal to (1/2)*r^2 and the second is equal to theta. Thus, the area is (1/2)*r^2*theta.
The second w in warwick is silent - the word is pronounced “worrick”
Yes, he has definitely never heard of Warwick. 😂
In Rhode Island, Warwick's second w is sounded.
@@smylesg he is referring to the University of Warwick in England. The custom of most respectful people is to adopt the cultural conventions of the place referred to. Call it what you like in Rhode Island, you would be laughed at here for not knowing or impolitely ignoring the convention here.
3:10 Presh, a little pronunciation guide of British towns for our friends across the pond:
Warwick: war-ick (the second W is silent)
Leicester: less-ter
Gloucester: Gloss-ter
Derby: Dar-by
Edinburgh: Edin-bruh (not berg)
Birmingham: Bur-ming-hum (not HAM)
There are plenty more but this is a good start.
Now before the comments go off on me saying, this is a mathematics channel - well, we’re sharing knowledge in a respectful way without compartmentalising the subject matters, as Presh has wonderfully done with the history of the 50p coin.
Not to forget Cholmondeley: Chum-lee
@@josephfredbillOr Milngavie (just up the road from me) in Scotland, pronounced Mill-guy.
Or Trottiscliffe in Kent... Pronounced "Troz-lee"
To simplify, Presh should have said "Worrick".
*All of the names of those places need to spelled much differently if they are expected to be pronounced so differently.*
That is a really cool history of the 50p coin, which by the way, has been reduced in size but still has those 7 sides. It never occurred to me the requirement that it would have to roll down a chute of constant width such as inside a vending machine, without getting stuck. A very clever design!
The first problem shows the two semicircles intersecting on the hypotenuse, but this technically needs to be proven, and the proof follows from the observation that drawing a perpendicular on the hypotenuse forms a right angle opposite to each circle’s diametric chord, so it must lie on both circles as a circle is the locus of points forming a right angle on any diametric chord.
Excellent!
One remark though: not "perpendicular", but "altitude" . The altitude on the hypotenuse is perpendicular to the hypotenuse, but also intersects the right angle vertex opposite to the hypotenuse. (A perpendicular on the hypotenuse can be positioned anywhere on the hypotenuse; unless we specify "a line _from the right angle vertex_ of the triangle, perpendicular to the hypotenuse".)
The answer to the first question is pretty easy if you grab an inclusion exclusion style of mindset. What isn't obvious to me is why the two circles intersect on the hypotenuse of that triangle. The argument is dependent on that assumption.
It was not obvious to me either. But writing out the equations of the two circles and the hypotenuse, I was able to determine with some algebra that the circles will always intersect on the hypotenuse.
Given that they do, it seems there should be a simple proof to demonstrate it, but my computation was rather tedious.
If the circle on the vertical leg has radius a and the circle on the horizontal leg has radius b, and the origin is at the lower-left vertex of the triangle, then the equations are:
(y - a)^2 + x^2 = a^2
y^2 + (x - b)^2 = b^2
y = 2*a - a/b * x
The circles intersect at (0, 0) and
x = 2*b*a^2 / (a^2 + b^2)
y = 2*a*b^2 / (a^2 + b^2)
And you can verify that point is on the hypotenuse line y = 2*a - a/b * x
Ah, I think I have it. Draw the first circle and look at the point where it intersects the hypotenuse. Draw a line from the lower-left vertex of the triangle to the intersection point. That new line forms one of the legs of a new right triangle, since the angle is subtended by the semi-circle. That means the line drawn is the altitude of the original right triangle. Since there is only one possible altitude of a right triangle from the hypotenuse, the other circle must intersect at the same point on the hypotenuse.
@@XJWill1 Nice! When I thought about it more awake, I started formulating that equation based proof, but as you said, I find they rarely are the best way to prove a geometry based problem. Funny thing is I had thought about the right angle of a semicircle theorem but didn't apply it to drawing in a line from the right angle vertex to the intersection with the hypotenuse. Good thinking!
@@XJWill1This is an elegant proof!
"Warwick" is pronounced "Worrick", not "War-wick".
these are way too smart
the solutions are so presentable and elegant and the figures are so clean looking, great job on these videos
thanks for doing my question! very big thumbs up to ya. im going to a math challenge next year and ukmt 2024, and youve always been a great deal of help when it comes to geometry.
-rafael, the fella that suggested the first question
update: i got gold for UKMT!!!!
PB1: the intersection point slices the bug triangle into 2 rect triangles...
Then area = area of the 2 circles - area of 2 small triangles = area of the 2 circles - area of big triangle.
π3²/2+π4²/2-6*8/2
= π25/2-24
Edit for extra details: just imagine slicing the "c" area in half.
That's also how I did it: by drawing a straight line between the vertices of area c .
Note: the "c" area is not sliced "in half", but into two parts. The two parts are not of equal size.
When I started working out sectors, arcs, and triangles in the first one I knew I was in trouble. When you labelled the parts I knew there was something I missed. The moment you made the sections in algebraic I saw where it was going and was kicking myself! I can always use algebra, but the visuals always beat me!
I think the video's still showing a bit of a "magic trick".
The big thing to consider is not working out any angles or circle sectors until you at least see that there's two semicircles so there's no need to deal with arcs. You can take both semicircles and cut out the inner triangles.
Then you may realize that those inner triangles form the big triangle, so no need to calculate the area of the individual smaller triangles either.
Going "I labeled these points, then I subtracted them and they just happened to turn into the equation I want" is in my opinion not a useful solution, because there's no thought behind it.
There are two thoughts, that will help you solve a lot of problems.
First, they ask for the entire value of the blue area. That means, it might not be necessary to separately calculate each blue area. Second, try to find areas, that can easily be calculated and make use of them.
So it's not magic, but solving this kind of challenge over and over again.
For me, it may be wrong, but the solution is simple. as there are 7 equal arcs on the outside... they average out, and you can refer to this shape as an approximate circle. Thus the area would be pi times the radius squared..or in the ballpark of 3.14. Wrong for a mathematician but 'close enough' for everyone else.
So area of the top 2 shapes are 12.5π but this actually includes the shape in the middle because you are counting it twice and then minusing 1 off because we are removing the triangle as shown here
(25π)/2 - 24u²
And thus, this is the answer
I didn’t understand why you think the arc of big slice is equal to the arc of the small slice. The arc of the small slice is bulged out more. So the arcs don’t overlap.
I haven't done the math, but can problem 2 be generalized for all Rouleaux polygons with n sides by substituting n for 7 and 2n for 14?
Everytime I open a comment section of a yt video, there's at least 1 indian guy explaining how badly they suffered while solving integration problems in 4th grade.
BTW, the 20p coin is the same shape as a 50p coin, but smaller. and the current 50p coins (since the late 90s, IIRC) are smaller than the previous ones.
I calculated the area of the heptagon.
Then I calculated the area of the round corners.
Finally I added the area of the heptagon and the area of the round corners, thus obtaining the total area of the coin.
A 50p is what it is, no more, no less! Has Your question any application to real life?
I dont see WHY ANYONE would draw a point at the center at 5:35 like that ao why solve this way..isnt there another way to solve?
Why the reluctance in calling it by its known name - the Arabic numeral system?
You are "solving the world's problem two problems with one video at a time"
it is a pity that you didn't offer an visual explanation for the simple solution of 1.
Just Amazing 😍!!
The English WICK is always pronounced ICK. Saying War Wick is a social faux pas on a level with saying Ar Kansas.
My bad, reveals my ignorance. I didn't even think to look it up because I thought it would be pronounced like Dionne Warwick. ua-cam.com/video/zphc3xcmJXk/v-deo.htmlsi=4Jd7YM3FWcEqJScG
I look forward to when these hidden snooty conventions fade into obscurity/no longer become anything worth remarking over
@@ntlespino It's not a "snooty convention" it's just someone saying the word wrong. Many English words aren't spelt how they're pronounced. It's like going to another country and mispronouncing all their words and you calling them snooty for correcting your pronunciation.
Not really a faux pas though is it?
"Always" is a strong assertion... and also wrong in this case! For example, the town of WICK at the far north of Scotland, which is just pronounced WICK.
There's "always" an exception to every rule in English (including this rule!) :-D
I'm crap with trig questions so I'll just do the first one, though i will take some inspiration from your 50-pence story.
Let the subdivisions of the semicircles/triangle figure be as follows: from top to bottom the three sections covered by the radius 3 semicircle will be I, II, and III. From left to right, the remaining two sections covered by the radius 4 semicircle will be IV and V. We need I, III, and V.
To sections above are contained by the constituent polygons as follows:
R3 semicircle: I + II + III
R4 semicircle: III + IV + V
Right triangle: II + III + IV
Adding together the two semicircles and subtracting the triangle we get the following:
I + II + III + III + IV + V - II - III - IV
Simplifying the above, we get I + III + V, as II and IV cancel along with one of rhe IIIs, which is just what we need.
R3 semicircle:
A = (πr²)/2 = π(3²)/2 = 9π/2
R4 semicircle:
A = (πr²)/2 = π(4²)/2 = 16π/2 = 8π
Right triangle:
A = bh/2 = 8(6)/2 = 48/2 = 24
9π/2 + 8π - 24 = 25π/2 - 24 ≈ 15.27
NOICE
Your comment says translate to English underneath lol.
@@mike1024. And when you click on it, the text changes into "NOISE" .😅
6:37 How do we know the 2 triangles formed using the line segment, are isosceles? As in, how do we know that line segment to the centre is the same length as the side of the pizza slice?
I'm pretty sure it's because the point marked as center is equidistant from each quasi-vertex along the edge.
Can please someone explain the 2nd problem. Why those 2 triangles are isosceles? The lower leg is ½ but upper 2 are not ½ since it is not circular sector
The isosceles triangles are made up of radial line going from the centre of the coin. As long as a line goes from the centre to any of the 7 "corners", it's the same length.
@@robair67but if they are isosceles, then the 2 legs are ½ each and the third is 1, which is impossible for triangle
@@AbDmitry the 2 legs are actually slightly more than 1/2 because they are inscribed from the centre of the coin to an arc from an origin on the bottom edge of the coin. Presh said, at the start of the problem, that the "pizza slice" was not a circular sector, although it looks like one. That's where the confusion is coming from.
@@robair67 I know, that’s why we can’t state that the triangles are isosceles
@@AbDmitry imagine drawing 7 radial lines from the centre to each of the 7 corners on the coin edge, starting clockwise from top. Imagine measuring these lines and comparing them with each other. Would any of the lines be longer or shorter than the others?
The second W in Warwick is silent
Great Video
This ill have to watch again. Very Difficult.
How about Susan B Anthony next?
I actually solved the 1st question in a mock paper before the UKMT so it was pretty easy during the exam.
Question 1 is quite simple. It would have been more challenging to prove that the intersection of the two semi-circle arcs is (always) indeed on the hyptenuse of the triangle.
Question 1:
(π*4²)/2 + (π*3²)/2 - (2*3)(2*4)/2 = (16π)/2 + (9π)/2 - 48/2 = (25π - 48)/2 = 25π/2 - 24
Question 2:
Let C be the center of the coin.
Let Vᵢ for i = 1, 2, 3, ..., 7 be the vertices of the coin (counter-clockwise, starting from the top in the diagram).
Let s = |CVᵢ| be the distance between C and each vertex Vᵢ .
For 0 < {α , β , γ} < 180° ,
let α = angle(V₁CV₂) be the angle between CV₁ and CV₂ ,
let β = angle(V₁CV₅) = angle(V₅CV₂) be the angle between CV₁ and CV₅ ,
let γ = angle(V₁V₅V₂) be the angle between V₁V₅ and V₂V₅ .
Then
α = 360°/7 = (360/7)° = 2π/7 [rad]
β = (360° - α)/2 = (360° - 360°/7)/2 = 360° * 3/7 = 6π/7 [rad]
γ = 2 * (180° - β)/2 = (180° - β) = (180° - 360° * 3/7) = (7*360°/14 - 360° * 6/14) = 360°/14 = 2π/14 [rad] = π/7 [rad]
Calculate s:
(|V₂V₅|/2) / |CV₅| = cos(γ/2) ⇒
(1/2)/s = cos(γ/2) ⇒
s = (0.5)/cos(γ/2)
Area of arc sector between V₁V₅ and V₂V₅ is:
A₁ = (γ / 360°) * (π*1²) = (1/14) * π = π/14
Area of triangle V₁CV₅ is:
A₂ = s * (1*sin(γ/2)) / 2 =
= 0.5/cos(γ/2) * (sin(γ/2)) / 2
= 0.25 * (sin(γ/2))/cos(γ/2)
= 0.25 * tan(γ/2)
= 0.25 * tan(π/14)
Area of coin sector between CV₁ and CV₂ is:
A₃ = A₁ - 2*A₂ =
= π/14 - 2 * 0.25 * tan(π/14)
= π/14 - (0.5)*tan(π/14)
Hence, area of coin is
A = 7 * A₃ =
= 7 * [ π/14 - (0.5)*tan(π/14) ]
= π/2 - (7/2)*tan(π/14)
Q.E.D.
1st problem I was able to work out pretty easily: Adding the areas of the 2 half-circles gave me the area we were asked to calculate + the area of the 6x8 triangle. Add them, subtract the triangle, and BAM, done.
2nd problem...I'm not ashamed to admit my whole brain just went "HUH!?!?!"
"Reuleaux polygons" named after the German engineer Franz Reuleaux.
Coin problem seems easier if if the coin was made of 3 arcs or 4 or 6.
It cannot be made of 4 or 6 arcs. A Reuleaux Polygon has always an _odd_ number of vertices and an _odd_ number of sides/circle arcs.
5:33-35 was the key to the solution. I didn't know this.😮😂
1:14 The 5 and 10 pence coins weren’t “new”. The pre-decimalisation shilling became worth 5 pence, and the 2 shilling coin became worth 10 pence, with all of these coins remaining in circulation post-decimalisation. This saved minting a whole load of new coins and having them swapped instantly on decimalisation day.
This sounds like the real cause, why the penny farthing finally died 😉
War-wick you mean Worrick!
I was thinking about the second problem: if it's a shape of constant width, wouldn't it's area form / map to a rectangle over a 180 degree roll? It's width would be 1, and it's length would be half the perimeter of the coin. Is this not right?
Apparently not.
(Width of coin is 1 , perimeter of coin is 7 * (2π/14) = π , so the area of the rectangle you're looking at is
width * half perimeter = 1 * π/2 = π/2
which is _not_ the area of the coin.)
By the way, your formula isn't even true for a circle, so why would it be true for this coin?
@@yurenchu because it sounds right lol
25 pi -24
It is only possible if you know the angle. Obviously the curve can be drawn from close to a straight line to almost making a full circle with the others. So knowing about the construction of the shape is crucial. Without it you only can calculate the heptagon and tell tha the area of the hole shape is bigger than the area of the heptagon.
* whole shape
@@robertveith6383 Well, I did not watch the video. So, sorry for my misunderstanding. After watching the video part with the construction it is rather simple. But I don't do it, because it is not interesting for me anymore.
Didn't get why the central angle is 2c, in r^2(2c)/2??
At 7:39 , the blue-colored area is a _circular sector_ (in other words, a _sector_ of a circle) with a radius of r =1 . (Note: that circle isn't drawn here, but its area is approximately 4 times as large as the coin's area.)
The area of a circle sector is calculated by the formula (r²/2)*θ , where θ is the angle of the sector. This formula follows from the fact that the area of a circle is πr² , so the area of just a sector of angle θ, equals πr² * (θ/(2π)) = (r²)*θ/2 . So, for example, if θ = 60° = π/3 , the area of the sector would be (1/6) of the full circle area.
In our case, r = 1, and θ = 2c , because θ is the angle at the bottom vertex of the blue area (where the two "flat" isosceles triangles meet, and the angle at the corner of either isosceles triangle is c ; so θ = c+c = 2c).
I hope that helps.
Shorter answer: the blue area at 7:39 is a "pizza slice" with a radius of r=1 and an angle of _theta_ = 2c . The area of a pizza slice is given by the formula r^2 * (theta)/2 , where r is the radius of the pizza, and _theta_ is the angle of the slice. ( _theta_ is also called "central angle" because the point of the slice corresponds with the _center_ of the original pizza.)
@@yurenchu Thank you soo much for the awesome details!
@@JosueMartinez-ww1vj You're welcome! I'm glad to have been of help.
Funny how in question 2, you calculate angle b as (360 - a)/2 , _just as I did_ ; instead of simply noticing that b = 3a .
In question 1, we actually don't need b and d. Instead, we can simply divide area c into c1 + c2 (with c1 to the left/on top of c2) by drawing a straight line between the two vertices of area c . This line also cuts the main triangle into two (triangular) parts. The required answer is then easily seen as
(a + c2) + (c1 + e)
with
(a + c2) = {half circle with radius 3} - {top/left part of main triangle} ,
(c1+e) = {half circle with radius 4} - {bottom/right part of main triangle} ,
hence
a+c2+c1+e = {half circle with radius 3} + {half circle with radius 4} - {whole main triangle}
Problem 1: 容斥原理 - 小圓形+大圓形-三角形
Of course i can.
1st area is on the top left.
2nd area is on the bottom left.
3rd area is on the right.
Why in the second exercise, you assume that a is equal to one seventh of the 2π? Without any prove
Anyone else ghik its kind of poor wordong st 2:07 to say because the WIDTH ks cistsnt on the shape the HEIGHT doesnt change..just because width doesn't change as you roll something doesnt mean height will stay constant..
You should go back in and edit your many spelling/typo errors.
I have to admit that, at least here in the US, when working out geometry problems, we wouldn’t think of expressing angles in radians. I never did until I was comfortable memorizing common radian-degree correspondences.
There's a much easier solution to the first problem. You just need to add the two hemicircles and subtract the triangle. How you know to get this is by writing a "+" on all the regions of the diagram that you are adding. Then you write a "-" to all the regions you are subtracting. Naturally you do this until the parts you don't want to calculate the area of has both a "+" and a "-", cancelling each other out, while the part you DO want the area of has a net 1 "+". It sounds more complicated than it actually is, it's a lot easier when you draw it out.
"You just need to add the two hemicircles and subtract the triangle" This is exactly what he did. 4:30
Yeah, that's basically what he did, but he was really thorough in showing why it worked.
or you just realize that bc the two semicircles share a region that is shaded and within the triangle, subtracting the triangle's area removes the area of one semicircle for the shared region, leaving behind the area for the other semicircle.
First comment
who?
Congratulations on your truly astounding accomplishment. :-P
The kwekt answer is that he shows a 2017 shorty award nominee screen and he thinks he’s one of the best UA-camrs 🤡