So what I would do to prove the second step, is to draw a third line tangent to the circle touching the other two lines, forming a triangle. Now we cut off the point parts of the triangle and are left with a series of 3 line segments, and the Arc of the circle. If we repeat this process infinitely many times, we will see that the series of line segments approaches the curve of the circle. Furthermore, due to the triangle inequality, we are always removing > 0 length from the line segments, which means that the circle must be smaller than what we started with

I used area instead of circumference, that worked out too.

@mikesteffen3378

2 күн бұрын

I did the same. You do have to use a 12 sided regular polygon instead of the inner hexagon, but it doesn’t require assumption 2.

@fangtooth-1125

2 күн бұрын

@@mikesteffen3378why do you have to use a 12 sided polygon?

@Narikvp

2 күн бұрын

The inscribed hexagon area can be calculated as C*r^2, where c is a constant. That constant turns out to be less than 3, so the inscribed hexagon area is not sufficient to justify the proof.

@acmhfmggru

2 күн бұрын

I used area and didn't get a satisfyingly correct answer, so I doubled back and used perimeter and found a satisfyingly correct answer. With area approach, I picked that the circle's area should be pi (ie: the unit circle, r = 1), the inner hexagon's area should be >= 3, and the outer hexagon's area should be

@tagore_14

Күн бұрын

Yeah same

A not so simple justification for 10:47 is that |AG| = r tan(angle(GOA)) |Arc(QG)| = r angle(GOA) (where r is the radius of the inscribed circle) So |Arc(QG)| angle(GOA), that's the case for 0

@christophniessl9279

2 күн бұрын

actually, you are already using the value of pi in calculating the length of an arc as radius times central angle. But we di not know pi yet....

@mauricereichert2804

2 күн бұрын

@@christophniessl9279 no, depends on how you define things. From the video we can assume that pi is defined in the classical geometric way with circumfrence/diameter. Then by rotation symmetry arclength is r angle(GOA), pi is just the special case where we look at a half circle. no need to know the value of pi.

@pierredumoulin527

2 күн бұрын

@@mauricereichert2804 yes but how do you know that tan(angle) > angle ? Without pi, I’m not sure there is a simple way to explain that.

@mauricereichert2804

2 күн бұрын

@@pierredumoulin527 yes that is why it is not "so simple". in my analysis course we took the route sequences -> sums -> defined sin/cos/tan by their taylor expansion -> defined pi as the first positive zero of sin. This way tan - id being positive is automatically clear (with the taylor error estimation)... For the ancient greeks idk rn. :D

@pierredumoulin527

Күн бұрын

@@mauricereichert2804 yes maybe the idea to use area instead for the big exagon is better, very simple to prove

I just solved it using areas instead of perimeters/circumference and bypassed the perimeter justification entirely. The bigger hexagons entirely encloses the circle, which entirely encloses the smaller hexagon, so the Areas must be in order of the inequality.

@Swappr

2 күн бұрын

in the actual exam it said considering perimeters but ig it works yh

@yugagrawal6102

2 күн бұрын

I tried solving it using area as well, but I calculated the area of smaller hexagon to be 3√3r²/2 which made the inequality 3√3/2

@ccost

2 күн бұрын

this works for this question but on the exam you HAD to mention perimeters (said in question)

@orchestra2603

2 күн бұрын

Using results from areas, you can also prove the similar inequality hold for perimeters too.

@DergaZuul

2 күн бұрын

@@ccost where did you hear or see perimeters mentioned in the question?

this pained me but i got it. I am pretty sure kaka is still traumitized.

@himmock

2 күн бұрын

22h ago??? Vid was 11 seconds ago??

@LegendaryBea

2 күн бұрын

​@@himmock bro got access to members only content i guess

@ronniechilds2002

2 күн бұрын

Same here. I had to reverse it and pause it a couple of times, but I got there.

@gabriql

2 күн бұрын

@@LegendaryBea nah i just know the person who sent the question in and they sent me the video before upload

@leif1075

2 күн бұрын

Thst SUCKS it pained you..why not just leave it then, if I may ask?

If we accept as prior knowledge that a circle has the smallest perimeter for a given area, then the inequality is clear because the outer hexagon even has a greater area, thus a greater perimeter than the circle.

@go_gazelle

Күн бұрын

I was having trouble connecting the dots, so I decided to rephrase it in a way that makes the most sense for me. Accepting that "a circle has the smallest perimeter for a given area" means that: A hexagon with the SAME area as a circle has a larger perimeter than the circle. Therefore: A BIGGER hexagon (one with an area LARGER than the circle's), must also have a larger perimeter than the circle.

Probably the easiest way to prove the larger hexagon has a larger perimeter is to call the original curve a1, stretch a curve towards the hexagon and call it a2, it'll be increasing the length of the line each time, and therefore when the curve matches the hexagon, you'll have a1

@dj_laundry_list

Күн бұрын

One could expand the circle to be a squiggly while still being between both, but having a larger perimeter than the outer hexagon, so this approach needs a constraint on the way the shape is expanded in order to work

It is much easier to justify the second inequality using the "other" pi from the formula for calculating the area of circle; it is clear that the area of the bigger hexagon is larger than the area of the circle; the equilateral triangles in the larger hexagon have side lngth x = 2r/sqrt(3), area A of one of the triangles is sqrt(3)*x²/4 (well known, or simple to calculate using pythagoras), so A= r²/sqrt(3). 2*sqrt(3)*r² = 6A > pi*r², hence 2*sqrt(3) > pi.

I did that exam, I cried a little when I saw this. Remember we have like 2 minutes to read, solve and write each problem.

@CoolRubiksCube

2 күн бұрын

No we don't! It's 80 marks and we get quite a while. For a 4-marker, we have more than 5 minutes at least.

@acre4406

2 күн бұрын

@@CoolRubiksCube oh sorry, it is 3½ exactly. Chill

@CoolRubiksCube

2 күн бұрын

@@acre4406 wdym chill? I'm not mad. Also, the test is 90 minutes, 80 marks, that's 4 and a half minutes for a 4-marker

@acre4406

2 күн бұрын

@@CoolRubiksCube With all the interruptions we had, it was actually 80 minutes, plus, the marks are not evenly distributed, there are 22 exercises to cover. Read, understand, solve, write, review. 22 times. Some of them are fast to do, others are not. You have a couple of minutes to do each. This wasn't even solvable. They know they have put a impossible exam. Why do you think they are passing us with only 15 marks?

@CoolRubiksCube

2 күн бұрын

@@acre4406 I really don't think that the exams are impossible; I usually end in around 30 minutes and then have time to check. On my most recent mock I got full marks on paper one and lost 5 in total.

As someone who actually did the paper this question was interesting roughly about 5 people in the entire exam hall got it .but it wasnt really a hard question it was a weirdly formatted question that people were not used to

@Rouja1

2 күн бұрын

I remember the absolute terror that I got when I read the question for the first time😭

@bjorneriksson2404

Күн бұрын

How old are the students taking this test?

@Rouja1

Күн бұрын

@@bjorneriksson2404 16 but from all abilities

I think the easiest justification would use the argument of areas. The area of OAG > area of OQG for obvious reasons (it just lies fully in there). Then, the area of OAG is 0.5 * r * r / sqrt(3) and area of OQG is 0.5* pi/6 * r^2. From here we should grasp that pi Then, we switch to the lengths of the half of the side AG and the length of the arc QG. length(AG) = r/sqrt(3) and length(QG) = r * pi/6 length(QG) * 6 / r = pi And finally, length(QG) If we multiply by 12 both sides, we arrive at the same inequality for the perimeters

@Rouja1

2 күн бұрын

In the actual question it said "use perimeters" which probably made it easier because you know where to start

this video gives me GCSE flashbacks

UK Maths GCSE exams "Prove π is between these values", US maths exams "Multiple choice to make marking easier 😀".

@bazdarinothebizier9085

2 күн бұрын

UK Maths GCSE exams "Prove that 1L of acid will cause more damage than 5 stabbings with this knife oi"

@Tommy_007

2 күн бұрын

Danish version: Here is the answer. How does it make you feel?

@Kero-zc5tc

2 күн бұрын

Chinese version: neighbour already solved the exam 1 second in! Why you a failure!!!

@bazdarinothebizier9085

2 күн бұрын

@@Kero-zc5tc Chinese version: *steals answer from superior neighbour* See, we're smart just like you 👲

@nakultiwari5162

2 күн бұрын

​@@Kero-zc5tcyou haven't see the Indian version my boy

For the length of the side of the external hexagon and call it _s,_ we use the _apothem_ (the line OG joining the centre to the mid way point of a hexagon side), and call that _a._ Note also that GA-AH is the same length as the side s. A little bit of trig (and knowing angles are pi/3 (60deg), we get that the length of s = 2a/tan(pi/3). The length of the inscribed arc of the circle GH is simply a*(pi/3). Let a be 1. Then s is 1.1154 and the arc is 1.0472. So the circumference of the hexagon is greater than the circumference of the circle.

@chillywilly5882

2 күн бұрын

The question is to prove that the value of pi is between 2 numbers, so you can't use the actual value of pi to work that out!

This was mental maths. Keep them coming

Interesting points first about circle vs hexagon. 1. Intuitively we know the outer hexagon is larger than the inner hexigon. 2. The length AB and arc QR on their own cannot be intuitively see as >, =, or , =, or

I’d say to show that the outer hexagon is bigger than the circle is to think about triangles and rope or string. The sum of the two sides is greater than the third. In order to make your way to the full size of the two added sides, you pull the string towards the corner, stretching the string. Not exactly mathematically rigorous, but it shows that it can be justified. It turns into a calculus problem of gaining length to equal the two lengths of an isosceles triangle. If you split the base into two, and move the lines towards the vertex where the two equal sides meet, you start with a right triangle and get increasing lengths as you move one of them towards that vertex. Since we’ve already shown the arc to be greater than the straight inner line, it can be shown that the two outer lines are greater than the arc due to there being a point along the path of moving the two split parts of the base of the isosceles that it will equal the arc. Which can be seen by noting that the arc and the two connected outer lines connect at their endpoints. Another thing to notice is that a circle (or arc) is like a constantly changing line, like it’s “straightening out”. As we move along the arc, it gets more like a straight line. Effectively giving a shorter length of a curve than you would with just the two straight lines to get to the other end.

My first thought was to rotate one of the hexagons 30⁰ so the two are tangent at the vertices of the smaller and have the circle between them.

@RAG981

Күн бұрын

Yes if you draw it with the vertices of the inner hex at the tangent points of the outer, it becomes completely obvious.

@wyattstevens8574

Күн бұрын

@@RAG981 Of course, you *would* need to know the perimeter formula.

I wouldn’t call this a proof, but to build my intuition for the second step, I would do the following: I would start with 2 position vectors one, with tail at O, head at A and the other, with tail at O and head at Q. We see that OQ

That was fun! I performed the calculations using the diameter rather than radius, and I already knew enough about hexagons to be able to calculate their perimeters, so I was able to miss out a few of the steps in the demonstrated proof. It provided a great opportunity to for me to do a bit of mental gymnastics to start my day anyway. Thanks for sharing.

I would try to prove that the external hexagon being longer by 1. taking the section and rotating it by having the corner on the top 2. cutting the shape into (infinite) vertical slices 3. as the hexagon angle is always larger than circle one (expect in the endpoints), the hexagon section length is also longer.

It is interesting to try to show the perimeter of the circumscribed hexagon is larger than the perimeter of the circle. I would try to argue this way. Start by noticing there is a bigger circle with radius 2 pi/3 such that this hexagon is the inscribed hexagon. The perimeter of this hexagon is smaller than the perimeter of the larger circle. Next, find the regular 12-gon that circumscribes the original circle (and has a tangent at 12:00). The perimeter of the circumscribed 12-gon is smaller than the perimeter of the circumscribed hexagon. This 12-gon is the inscribed 12-gon for a larger circle. Keep doing this for 3*2^n-gons and observe they keep having smaller perimeters while also being inscribed inside circles that keep getting smaller. Thus, the perimeters of the 3*2^n-gons are decreasing and converge to 2 pi r because the larger circles have radii that converge to r (this step actually has several parts than need justification but intuitively is right). Since the perimeters were decreasing, the perimeter of the first circumscribed hexagon must be the largest of them all, and bigger than the limiting value of 2 pi r.

To prove the actual problem you used perimeters. And u ‘assumed’ the perimeter of outer circle will be larger and of inner circle will be smaller. This case can be used vice versa without assumptions but solid statements. As you know the correct formulae for perimeters of circle and hexagons and also for their sides, u can use the perimeter formulae to show how they are smaller or greater to each other

The circle is defined as the line which has a constant distance to a point, its origin point. If a circle is inscribed in a polygon all points of the poligon are further from the origin than the point of the circle at the same angle. A simply inscribed circle is a circle where all points of the poligon boundry (the line) are not inside the circle area, but may be on the circle's boundry, and at least one point us. A maximally inscribed circle is one where the points of the poligon boundry that are also points on the circle boundry are maximised. For regular polygons this happens when the origin of the circle coincides with the center of the poligon and where the radius of the circle is the same as the distance between the center of a side of the poligon and the center of the poligon. With this definition in mind and using radial coordinates with the origin of the circle as the origin of the coordinate system, we can write two formulas: R_circle(a) = r and R_poly(a) >= r for all a, where R is the function that gives a point's radius for the angle a. If R_poly where to be less than R_circle, the point of the poligon would be inside the circle, which is not allowed by definition of the inscribed circle. If R_poly = R_circle for all a we are back to the definition of the circle, so there must be at least one point where R_circle is strictly less than R_poly. (In fact, since poligons need to be continuous, there must be a segment of the poligon where this is true) The line integral in polar coordinates is S(√(R(a)^2 + dR/da^2)da). For the circle dR_c/da is 0 and R_c is constant. Since R_c

9:57 - Surely it is easy to calculate the arc length HQG = (pi/3)xr, and the short leg AH of the right triangle AHO = r/(root3).

Your problem of justifying only exists because the original diagram is drawn badly. Have the vertices of the inner hex at the centres of the outer hex still traps the circle, but the order of distances is obvious.

I think we can justify the 2nd step with area arguments and calculus. We can show that for a fixed area of a convex polygon that the circumference is minimised in the limit of the polygon tending to a circle. Therefore if we took area arguments we can show that the inscribed hexagon has less area than the circle which in turn has less area than outer hexagon. From then on we can argue that for the outer hexagon to have bigger area has to be due to bigger circumference than the circle.

Construct HG segment and connect its midpoint (Call it X) to A In the triangle AXG the base is R/2 (=.5000 R) the Curve is (PI/6)R (~.5236R) and the hypotenuse is (1/Sqrt(3))R (~.5774R) Since the error in calculating PI is =/- .0013, the error in PI/6

A convex shape A is inscribed in another convex shape B then, 1). Area of A 2). Perimeter of A < Perimeter of B

Thank you

To justify the circumscribed polygon's larger perimeter, you can just successively approximate the circle by adding segments that shortcut across the polygon's vertexes. Since you are always cutting an angle of less than 180 you are always replacing two sides of a triangle with one side so it is indeed shorter, and since you only created more angles of less than 180 you can repeat and induce, and since you are approaching both the position and the tangent of the circle you avoid the staircase paradox

My attempt: Draw the line IJ tangent to the circle at point Q, where I is a point between HA and J is a point between AG. Since IJ is a straight line, IA+AJ > IJ, so HA+AG > HI+IJ+JG. We can repeat this for arc HQ < lines HI+IQ. As we approach infinite tangent lines, we approach the arc of the circle. Since with each additional tangent line, the total distance is less than the previous one, therefore arc HG < HA+AG.

9:47 ✅Yes I have solution just connect line OH and OG. Now we see two Areas sector OHG and quadrilateral OHAGO as we see area of OHAGO MUST BE greater than sector OHG from fig or we can prove it too. So since OH and OG length are same the sum of length AH and AG must be greater inorder to have a of large area of OHAGO 😊 Hope this could be the simplest explanation.

People did do the problem with areas instead and are getting: 3 * sqrt(3) / 2 The original problem did say perimeter, but a follow up explaining the mistake would be nice. Also a way to prove the perimeter homework you gave us. Assume: the arc |HG| use the Arc Length Formula to find: arc |HG| = 2 * pi * r / 6 Construct Shape HAGO Given |HO| = r, and |GO| = r, (r is the radius, as defined in the prompt) Angle H = 90° and angle G = 90° because they are perpendicular to the hexagon Construct |AO| to make 2 triangles AOH and AOG Using hypotenuse leg theorem, the two triangles are congruent This means |HA| = |AG| Angle HOG was previously found to be 60°, since the triangles are congruent, angles HOA or GOA = 30° Knowing that the interior angles of a triangle add to 180, these are 30, 60, 90 triangles using the 30 60 90 rules, or law of sines, etc. |HA| and |AG| = r / sqrt(3). Plug into the original assumption to see: 2 * pi * r / 6 < 2r / sqrt(3)

@Synthesized05

2 күн бұрын

Since this involves "back in the day" level of math skills, this proof uses hypotenuse leg theorem to prove the triangles are congruent, which may not have been discovered yet and will extend the proof. also if they where using this to estimate pi, having pi and r (essentially 2 variables) in the solution is probably not acceptable and the proof will need to be continued or a new proof will need to be crafted.

If this guy was my math teacher as a kid, I would've been so much better at it today..

very interesting historical inside to Pi. Now I'll have to go back and check the various books on Pi to see whether they too reference this. Thanks for sharing.

Draw the circle with a compass. The outer hexagon could be drawn with the compass too if the angle varies. The angle is alwas larger than for cricle. That makes the distance travelled by the compass tip greater than the circle.

Here's a little challenge I found in a journal: Find two mathematically correct equations that use all ten numbers (0 to 9) as well as the four operation symbols (+, *, /, -). Knowing that if a symbol or a number is used in one of the equations, it cannot be used in the other (except the =). How many pairs of equations are there and what methods are used to find them?

This was actually the method I used my Sophomore year in HS to estimate π.

I knew the top half (any half) of the circumference was πr, so I knew if I calculated the top half of the other two perimeters, the r's would drop off at the end. I knew the 30-60-90 side ratios, so I knew to divide r by sqrt(3) and multiple by 2. Then since 2r/sqrt(3) is expressed as 2sqrt(3)r/3, it was easy multiplying by 3 (to get half the perimeter). I don't know why we need to prove the outer hexagon has a larger perimeter if we have already calculated it to be larger. But, this might be an accepted proof: arcVQ = arcQR = arc(midpointVQ to midpointQR) Now rotate the outer hexagon (and circle if you wish) by 30 degrees so that all three distances being compared have the same endpoints. From here, it should easy to see and conclude that the distance around the outer hexagon is greater than the distance around the circle. (Compared to the path along the circle, the path along the outer hexagon is stretched further from the straight line of the inner hexagon.)

my justification might not be right but this is how I analyzed it. taking the kite shape, I draw a tangent at the point where the vertex intersects the circle. this gives symmetry, even though it is not needed. now this tangent intersects the kite at 2 points, forming 2 new vertexes. now in the triangle formed, we know that one side of the triangle is always lesser than sum of the other 2. so the new line is a shorter distance than going along the vertex of hexagon. but now we have 2 more smaller kites formed where the tangent intersected with the sides of the hexagon. we can repeat the process and realize that there are infinitely small triangles we can keep creating all on the external side of the arc of the circle . Meaning we keep finding shorter paths, until becomes the arc itself. hence arc HG is shorter than the HAG path

You can avoid having to justify that the perimeter of the outer hexagon is greater than the circumference of the circle by simply calculating the area instead. You end up getting the same inequality.

@yugagrawal6102

2 күн бұрын

I tried solving it using area as well, but I calculated the area of smaller hexagon to be 3√3r²/2 which made the inequality 3√3/2

@Ninja20704

2 күн бұрын

@@yugagrawal6102you can use area to prove it for only for the right side of the inequality. The two parts can be proven independantly and then combined together.

Here's a weaker but more familiar version of Archimedes's assumption 2: among all curves enclosing a given area, the circle has the smallest perimeter. If you're willing to assume this, you can finish the proof as follows. Clearly, the area of the circle is between the areas of the two hexagons. Therefore, there exists a third regular hexagon, lying between the given two, which has the same area as the circle. By the assumption, this has a larger perimeter than the circle. But it's smaller than the outer hexagon, so the outer hexagon has even larger perimeter, as claimed. A lot of the comments I'm seeing here argue in one way or another that since the circle is inside the hexagon, the hexagon much have a larger perimeter. If (like Archimedes) you are careful about convexity, this is valid. But if you try to apply the same principle to concave shapes, then you run into contradictions like the famous false proof that pi = 4, by squeezing a circle between a square and a rectilinear approximation of the circle.

Assume HA, AG are metal plates hinged at A and HQG is a thread tied to the ends H and G. Now, open up HAG and make H A and G colinear, now Q and A should coincide if their lengths are equal. Rotate the plates back to the original position and we should not see the curve if the lengths are equal, if we see a curve it means HQG < HAG

Please make video on Gödel's incompleteness theorem

Did it in my head in far less time than dude painstakingly did.

Gonna take a stab at the 2nd assumption: - We draw lines HQ, QG, and AQ. We now have a similar diagram. We know that HA and AG are shortest path from H to A to G, same for HQ, QG, and H to Q to G. - We draw hexagon GHIJKL from all the midpoints of ABCDEF. We can prove it's the same as QRSTUV (same "radius", side length, etc). - You can prove that AH is longer than QH by comparing triangles AHG and QHG (same base, different height), and/or extending line AQ to intersect with HG. That proves the 2nd assumption (at least, for regular polygons arranged like this, but I think it could be proven for any triangle AHG and QHG, given that point Q is inside of AHG). - (I think?) you then prove the last step is true on a regular shape with N sides, approaching a circlular arc. You could also look at the circle that goes around the outer hexagon. You can prove that circle's circumference is longer than the inner circle's (longer radius), and longer than the path around the outer hexagon (straight lines). Might be way off, and I guess it's a bit recursive either way, but if you combine those ideas, I think you can prove length of lines HA + AG is greater than the length of arc HG. - Look at our circle again. We know arc HQ is longer than line HQ, which we proved is shorter than AH.

Daniel Rubin made a video explaining why the second part works. It has something to do with convexity.

The outer hexagon is larger than the circle because you can reduce its circumference by cutting off bits that are outside of the corner. As long as those cuts are straight it is obvious the circumference gets smaller, as the starting and ending point of each cut are connected with a straight line where in the original shape the line between those points was bent. Any part protruding outside of the circle can be cut off by a tangent straight line, this can be repeated at infinitum. Finally nothing lies outside of the circle; the remaining shape is that circle.

This made me work out that the length of a chord from angle θ is 2r*sin(θ/2). Which I feel like I should have already known.

Here's my answer: From the diagram, it can be easily seen that the area of the smaller hexagon > the area of the circle > the area of the larger hexagon. The area of the circle (pi*r^2) = pi * r^2 The smaller hexagon may be divided into six equilateral triangles (ORS, OST, OTU, OUV, OVQ, OQR) with a side length of r; the area of each of these triangles is r^2 * 1/4 * sqrt (3) [I believe GCSE students are expected to memorise the formula for the area of an equilateral triangle] therefore the area of the smaller hexagon is 6/4 r^2 sqrt3. The same idea applies to the larger hexagon (OBC, OCD, ODE, OEF, OFA, OAB) except this time, the _height_ of the equilateral triangle is r, not the side length. Since the height of the equilateral triangle is r/2 sqrt3 times the side length, the "radius" of the larger hexagon (OA, OB, OC, etc.) is r/(0.5sqrt3) or (2r/sqrt3). Therefore, the area of the larger hexagon is 6/4 * (2r/sqrt3)^2 * sqrt3 = 6/4 *r^2 * sqrt3 * 4/3 = 2sqrt3.* r^2 From this, we have 2sqrt3 > pi > 1.5sqrt3 (I'm skipping the step where the r^2s cancel out). Which gets us part of the way there. Unfortunately, 1.5sqrt3 is not 3... However, from the above, we know that QR, RS, etc = r, so the perimeter of the hexagon is 6r; the perimeter of the circle is 2r*pi, since both are convex shapes, the perimeter of the hexagon is less than the perimeter of the circle, so 6r 3 Which gives us 2sqrt3 > pi and pi > 3, which combine to make 2sqrt3 > pi > 3. QED. I think

"Given that r the is the radius of circle O" is irrelevant to the question. The size of the circle is irrelevant. Also, O is not a circle, O is a point. The real question is: "show that pi is between 3 and 2 times square root of 3" and "Hint: consider the inscribed and circumscribed regular hexagons of a circle, and here is a diagram to save you time".

@Kero-zc5tc

2 күн бұрын

On the original paper, the question specifically asked to prove it with the diagram. They give you r is the radius of circle o so that point o is shown to be centre to circle and hexagons (though it didn’t really need stating, it helps with people not looking at diagram with eyes tho). They say r is radius because one of the solutions (most commonly used by students) was to split the small hexagon into 6 triangles with 2 sides o and work out info then do similar (more work needed this time though) to the large one so you get an equation where all the r’s cancel out.this question is more to show a way we derive pi from the bounds of 2 shapes (with more sides more accuracy) so your real question is not the question

10:47 I think I got an approach that works, but please let me know if you find an error. For the sake of not confusing anyone, I am going to try to use the same verbiage as the video. Firstly, by definition, it must be true that |HA|+|AG| must be either less than, greater than or equal to the length of arc HG. If we suppose that the sum is less than, then that would mean that the Perimeter of the Hexagon ABCDEF would be less than that of the Circumference of Circle O. This would contradict the fact that we defined that hexagon as a circumscribed not an inscribed hexagon, thus it cannot be less than. If we suppose that they are equal, then that would mean that |OG|=|OB| as they would both need to be on the circle to be the same distance. This however again is a contradiction, thus leaving the only option left to be the Perimeter of the Hexagon ABCDEF is greater than the Circumference of Circle O. Again, I could be talking out my butt here, but I think that I might be onto something here thought to.

Instead of an Octagon, if you use a Hexadecagon, you get an even more fascinating inequality!

Imagine the curve HQ and the line segment HG construct an angle QHG, the curve QG and the line segment HG construct an angle QGH. We see angle AHG > angle QHG ; angle AGH > angle QGH So AH + AG > QH + QG Then AH + AG > HG

I'm never comfortable that the outer hexagon has a perimeter greater than the circumference of the circle! I always work with area! Similarly with the limit of sin(x)/x as x goes to 0

I remember being so scared and then so happy when it came in the exam and I solved it

How very Archimedes… I haven’t watched the video yet, but here goes… 1) Assuming the circle has a radius of 1, the inscribed hexagon has a perimeter of 6. 2) Using the circle to measure proportionality, the dimensions of the larger hexagon are 2/(sqrt3) times those of the smaller hexagon. 3) using these proportions, the larger hexagon has a perimeter of 12/(sqrt3) = (4*sqrt3*sqrt3)/(sqrt3) = 4*(sqrt3). 4) it can be observed visually that the perimeter (circumference) of the circle is greater than that of the smaller hexagon and less than that of the larger hexagon. 5) thus, dividing the perimeters of both hexagons and the circle by 2r gives 3

@michaeldakin1474

2 күн бұрын

Having now watched the video, i propose a proof for the second step (perimeter of the larger hexagon is greater than the circumference of the circle)… 1) let r=1, and consider the angles in radians. 2) mark any point external to the circle and draw the two tangent lines to the circle, as well as the radii at the tangent points. 3) let ‘a’ be the angle (in radians) between the two radii. 4) draw a third line from the centre of the circle to the external point, making two identical (albeit mirrored) right triangles. 5) now, noting that ‘a’ is equal to the length of the arc between the tangent points, compare that length with 2*tan(a/2) - whatever value you enter for ‘a’, I think that the comparison will effectively show that the perimeter of the circumscribed polygon is always greater than the circumference of the circle.

@michaeldakin1474

2 күн бұрын

Just to simplify, I think this basically amounts to saying that, provided the adjacent side has a length equal to 1, the tangent of an angle is greater than the numerical value of that angle in radians. Thus tan(theta)>theta. I’m no expert in trigonometry, so I cannot offer further proof that this is always true (with the condition that the adjacent side is equal to 1), but I dare say such proof exists. For the specific example of a circumscribed hexagon, it is certainly true that 2*tan(Pi/6)>(Pi/3).

Consider the lines FA and AB. They are tangents to the circle, and thus perpendicular to the radii at the points of tangency (H and G respectively). Thus, looking only at one side, tangent FA is perpendicular to radius OH, thus angle OHA must be 90°, with the same reasoning for OGA. Now draw the quadrilateral AGOH. We just established OHA and OGA are each 90°. Angle GOH is 60° (as seen in the video). The sum of the interior angles of a quadrilateral is 360°, and we know three of the angles: 90° + 90° + 60° = 240°, therefore angle HAG is 120°. Of course, this is standard for a regular hexagon, but we'll pretend we didn't know that already. Now draw triangle AOH, trivially shown as a 30-60-90 triangle. OA is the hypotenuse, OH the "long side" (and also the radius of the circle) and AH the "short side". In that triangle, if you define the hypotenuse as 2x, the short side AH is x and OH is x√3. Since HA is the same length as AG, the total length of those two lines is 2x which is, conveniently, the same length of the hypotenuse. They are also 1/6 the perimeter of the hexagon. So, calculate the hypotenuse. OH is x√3, which is also equal to the radius of the circle, and the radius of a circle is the circumference (circ) divided by 2π. Thus x√3 = circ/2π. Divide both side by √3, and you have x = circ/(2π√3). The hypotenuse, and conveniently the combined lengths of lines HA and AG is 2x, therefore has a length of circ/π√3. Which means the perimeter of the hexagon is 6circ/(π√3), or 6/(π√3) times the circumference of the circle. π√3 is less than 6, which means 6/π√3 is greater than 1, which means the perimeter of the hexagon must be greater than the circumference of the circle.

If i add a line inside the corner, splittling the 2 lines, i now created a path that is smaller then the original one The more we do that the closer we get to the circle, showing the circle is not the optimal path, but smaller then the 2 outside lines.

From 5:00 mins, work out the base AB instead of the hypotenuse. Let M be the midpoint of AB, and let AB = 2x, then AM = x. In the right-triangle OMA, tan 60° = r/x. tan 60° = √3, so r/x = √3 => x = r√3 / 3. Area of triangle OAB = 1/2 (2x) r = xr = r² √3 / 3. So area of hexagon = 6 r² √3 / 3 = r² 2√3. Area of circle = πr². Area of inscribed circle πr² π First part can also be proven using areas.

i made a 1 on my sat.

@mmburgess11

2 күн бұрын

i made a smudge on mine.

My justification for the missing step: We know that a circle has the least perimeter for a given area. The circle's area is less than the outer hexagon. So whatever the perimeter of the outer hexagon is, it would be more than a circle with the same area. And our circle has less area than that, so the perimeter of our circle is even lower.

Draw radii OS, OT, and OM, where M is the midpoint of EF. Also draw OE and OF. As OS = OT = r and QRSTUV is a regular hexagon, then ∠SOT = 60°, which means ∆SOT is equilateral and ∠TSO = ∠OTS = (180°-60°)/2 = 60°, and ST = r as well. By symmetry, as QRSTUV is regular, QR = RS = TU = UV = VQ = r as well. This means the perimeter of QRSTUV = 6r. As QRSTUV is inscribed inside circle O, that means the perimeter of QRSTUV must be less than the circumference of circle O, which is 2πr: 6r 2r(3) 3 As EF is tangent to circle O at M, then ∠FMO = ∠OME = 90°. As ABCDEF is a regular hexagon, ∆EOF is an equilateral triangle (shown above with ∆SOT), then by rule, OM, as a line from thd opposite vertex intersecting the midpoint at 90°, bisects ∆EOF, forming two congruent right triangles ∆FMO and ∆OME. As ∠EOM = ∠MOF = 60°/2 = 30° and ∠OFM = ∠MEO = 60°, ∆FMO and ∆OME are 30-60-90 special right triangles. cos(30°) = OM/OE √3/2 = r/OE √3OE = 2r OE = 2r/√3 As ∆EOF is equilateral, OF = FE = OE = 2r/√3. As ABCDEF is a regilar hexagon, then AB = BC = CD = DE = FA = 2r/√3 as well. This means the perimeter of ABCDEF is 6(2r/√3) = 2√3(2r) = 4√3r. As circle O is inscribed inside ABCDEF, then the circumference of circle O is less than the perimeter of ABCDEF. 2πr 2r(π) π < 2√3

I did by comparing areas rather than perimeter, and I got the answer

I don't know why they bother to say the radius is r.

I had to quickly google the unit circle for a refresher, but I got it just from looking at the thumbnail. It felt good.

@limitbreak2966

2 күн бұрын

Yea It’s been so long that I had no clue what it wanted, and I don’t ever think I did inequalities? So that may be why I had no clue what I was supposed to do with the information given since they literally gave you nothing besides knowing the angles, like I felt if I even had the length of side RS I could have done SOMETHING, but since I don’t have a good memory I had forgotten about those known lengths to triangles with like 2PiR or whatever it was.

@majordude83

2 күн бұрын

Yeah, it did take a minute to think, "I guess it's OK to assume it's a shorter distance around the inner hexagon, because straight line between same points". I remembered pi is ratio of radius to circumference, and then went from there. Presh used a "special triangle", once he had the angle and needed the hypotenuse for the outer hexagon; the unit circle is kinda like a bunch of the special triangles in one diagram. ​@@limitbreak2966

You can also do this question without any trig by dividing the radius by the inner radius of the smaller hexagon, then multiplying the radius by this scaling factor for the outer radius, getting one segment to then obtaining the perimeter. At least that's how I did it out on paper before the video

Now take the limits of the inscribed and circumscribed n-agons as n→∞

If we draw a straight line between points HG, the arc HG is never further (always closer or almost always, deprnding on if we count the „points” exactly at H and G) than the „double line” HAG (u know what i mean), so arc HG is shorter im not sure if its ok, u might have to proove this thing with lines being closer meaning theyre shorter

When I first saw the thumbnail, my idea was to "simply" use the relation between the area of the hexagons and the circle. I grabbed my notes and did the maths, until I found out i got that 3 < π < 4 😭 We can't guess we should use the perimeter instead of the area :/

AWESOME SOLUTION, PRESH !! But, I thought that the second part was needless bcoz it is "obvious".. 🙂🙂🙂

I actually did this exam, I got it right during the exam! Hoping to get a 9 (A*) On the exam, it said it was required to do a method involving the perimeter, so it was easier to find how to solve it.

I used perimeter to show that pi is greater than 3 and the area to show that pi is less than 2*sqrt(3)

Yooo this was in my gcse!!!

It's rather an unfair question, as you have to pull the original inequalities out of the air. It could be improved by allowing Euclid's definition of a straight line as the shortest distance between two points -- both inequalities can be proved from this, though rigorously only as less-than-or-equal. EDIT: This doesn't work with the second inequality (between the outer hexagon and the circle). I'll leave it up in case anyone wants to fix it.

@limitbreak2966

2 күн бұрын

I feel it’s unfair cause it requires perfect memorization of the triangle side length thingys, like unless something was given like unit circle or something like that; then i feel it’s not fair to students who are bad at memorizing random bits of information, but could still do the problem easily if they had the relevant bits

@chillywilly5882

2 күн бұрын

@@limitbreak2966 Don't need to memorise it. Equilateral triangle, say side length 2. Drop a perpendicular from a point to the base, so half the base is 1. So you have a right-angles triangle with hypotenuse 2, side 1, so other side has to be 2 squared - 1 = 3; and take the square root of that, so √3 (pythagoras).

@gabriql

12 сағат бұрын

@@chillywilly5882 @limitbreak2966 this is exactly how I solved it in the exam w/o any unit circle or anything.

i used areas instead of perimeter, i got really close to the answer though. instead of 3, i got 3 x 3/2sqr3

why do you want AG+AH which is actually side AB of outer hexagon. That is larger than 1/6 of the circumference of circle which is πr/3( r is radius of circle) and side AB is 2r/√3 is greater than πr/3

Is it possible to use induction proof for the remaining assumption? Consider all possible polygons that can circumscribe a circle and show that the most simple polygon, p_3 has larger circumference than the circle. Then show that when comparing two consecutive polygons p_n and p_[n+1], the higher order polygon has smaller circumference. And the limit of the circumference, when n→∞ will be the circumference of the circle, but all other polygons will have larger circumference than the circle. This is just a hunch, I have not a clue if it holds true mathematically, or if an induction proof is sufficient...

Did this using areas. Got the same answer.

Prove the formula of area of an oval

ctrl F "96" no hits

how can you say that the perimeter of the exterior hexagon is more than the circle?

As we can see that it's challenging to prove the Circumference 3 Alternate approach: Compare areas of Circle and ABCDEF Area Circle pi * r * r ==> pi Combine equations (1) and (2): 3 Give me the credit, if u find this useful. -Rasika Srimal from Sri lanka

Just curious tell me if i am wrong but if Archimedes know that circumference is 2πr then why did not is manual calculated the radius and circumference of a circle manual then put in the equation of 2πr=circumference he would get the value of π did i miss anything or he used another method to prove this please some one tell me

But those are true in euclidean space, can we make that proof in non-euclidean space? By the way, I thik the only way to prove that the str8 line is the shortest distans is in those definitions of spaces. I had something about it on my math course at university but it was some time ago.

I thought to compare the area LOL.

if we assume that the tangent points g and h are directly in the middle of the outer hexagon's sides |AB| and |AF|, we can calculate that length to be (2r√3)/3 = ~1.155r now then, the arc length S = rθ that gives us the length πr/3 radians = ~1.047r 1.047r ∴ circle this feels a little like circular reasoning so i'm probably missing something, right?

@chillywilly5882

2 күн бұрын

The question is to prove that the value of pi is between 2 numbers, so you can't use the actual value of pi to work that out!

@wham_sandwitch

Күн бұрын

@@chillywilly5882 right of course, that makes sense

Do you need the perimeter on the larger hexagon or could you use area instead?

@Rouja1

2 күн бұрын

In the actual question they asked for perimiter

I can do better than this 2pi = 360° The chord 180° = 2 the tangent at the bisector of the circle is infinity The chord 120 ° = SQRT(3) x 3 is 3 SQRT(3)/2 and the triangle outside is 3 SQRT(3) The chord 90° = SQRT(2) multiplied by 4/2 = 2.828 the outscribed square is 4 The chord 60° = 1 multiplied x 6/2 = 3.0 and The outside hexagon is 2/SQRT(3) x 6/2 = 12/2 * SQRT(3) = 3*2/SQRT(3) = 2*SQRT(3) Thus 3 But there is another bit of probabilistic data encoded If we look at each value So let’s do that 2 2.598 2.828 3

no way this is GCSE...

man I'm just tryna pass

i did this in the exam lol

takes too long. use symmetry to reduce the time

Great video. Can anyone help me by confirming if my answer to the following question is correct: Two fair dice (red/blue) are rolled, what is the probability that one of them will show a number greater than 3 while the other will show an even number? I calculated the answer to be 7/18

Nahhh now just use π < 2√3 to prove the second part😂

You gloss over the main difficulty at 1:45 - How do we KNOW that the perimeter of the outer hexagon > the circumference of the circle. The rest comes from junior high Trig, but Proving (from Postulates) that Circum-circle < Perim-Outer-hex is lacking. And I'm not sure how to prove it. Intuition is not proof.

@Grizzly01-vr4pn

Күн бұрын

Easily proven using a 30-60-90 triangle whose 60°-opposite leg = 1, and thus the 30°-opposite leg is √3/3, compared with a circular sector of radius 1 and subtended angle of 30° which will therefore have an arclength of π/6. √3/3 > π/6, and so multiplying each by 12 gives the respective perimeter/circumference, for which the inequality also holds.

@nealcarpenter3093

Күн бұрын

@@Grizzly01-vr4pn Sorry, but you're missing my point. Yes, finding the perim. of the outer hex. is easy. But how do we PROVE it has to be greater than the circum. of the circle?

@Grizzly01-vr4pn

Күн бұрын

@@nealcarpenter3093 How would you prove that any given number is bigger or smaller than any other given number?

Come on, this is way too easy to be featured here.

@Rouja1

2 күн бұрын

The exam it came from was 2.5 weeks ago so he just wanted to solve it for the people that did the exam (like me)

@gabriql

12 сағат бұрын

this was a pretty huge exam that hundreds of thousands of people sat a few weeks ago (including myself). Do remember that this exam has to be sat by people who aren't gifted at maths at 16 years old, and also is not an optional exam at all (every 16 year old took it or a varient). I understand that it is quite easy for this channel but knowing one of the people who sent this question in personally, we both just wanted a nice clear proof and for other people to see the question ( I think it is a very good question).

the only way for the curve to be longer would be to be straighter than a straight line

A vague idea on how to prove |HAG| > |arcHQG| would be to use calculus to prove the area in between is always positive. Instead of just corner A, subdivide into points A1 and A2 (etc) and state the length is equal (or less than) the arc's. Then show this would result in a negative area, proving by contradiction it must be greater in length.