Incredibly hard geometry problem from Russia

[@MindYourDecisions]

I meant to share some of the nice solutions people have sent me! Here's another solution write-up using symmetry by Bob Corns: docs.google.com/document/d/1tututTqcCr-VEkAieAyQPX5-tTIlSaD5kit8zo7KFpY/edit?usp=sharing Here's another solution from Reuven BarYehuda which is very nicely explained: kzbin.info/www/bejne/hYamZqhto8p7h9U

[@janhavishinde9999]

Very difficult to understand though

[@roqueandreeleazararroyovil8889]

Excuse me mr. Presh, I did the problem in the easiest case of a 1 by 1 square and I found that the ratio was 25/144, could you try ir that way to compare results please? Thanks a lot for your attention, I really like your videos, goodbye!

[@atasuhail]

Why to disturb the mind so much ... simply consider a circle circumscribing the octagon which will have radius approximately equal to the 1/4 of square (because it is also a || gm) side find their ratio and u will be just near the answer ....

[@shubhammalik2412]

I have easiest solution for this problem

[@djelloullezaar9851]

إخادهىؤلؤتم

[@entercherpfhalckhontralyty3542]

I'm a 9 grade russian student. At the end of the video I realised that I could solve it in a different way. In russian 8 grade class we learn that triangle medians, which always intersect each other in the same point, cut a triangle into six equal triangles. If we draw a diagonal in a quarter parallelogram, we will get a triangle, in which the lines defining the piece of the octagon are medians. So, the area of that octagon segment is two of those six little equal to each other triangles relating to the big triangle, which is a half of the quarter parallelogram. It means, that the octagon segment is one sixth of the quarter parallelogram, and the whole octagon is one sixth of the whole parallelogram as well. P.S. Let no one who is not a geometer enter.

[@adityapratapsingh2518]

Its even more easier!

[@adityapratapsingh2518]

Just consider a circle around it and find the ratio of the radius to the side

[@adityapratapsingh2518]

At 2:49 just cut it into half not 4 parts

[@artiomboyko]

Yeah, I thought about this too) P. S. Hello from St. Petersburg)

[@aaronseldes767]

Your Russian. Heres a question to see how smart you really are. If the electric filed strength was increased by 2% per second for 1minute and the area was decreased by 1% per 2 seconds and epsilon not which is 8.854x10^(-12)C^2/nm^2 . We ware measuring the area of sphere. WHat is the proportional difference between sphere and circle.

[@Dr.Druffi]

In mother russia geometry problem solves you.

[@stevethecatcouch6532]

The words of Yakov Smirnov are becoming garbled. Soon his legend will fade.

[@EmeraldEmsiron]

In mother russia, you dont overuse a joke, jokes overuse you!

[@MarkMcDaniel]

Ludites always complain about humor they don't appreciate. I like it!

[@angstygyroscope7979]

Dr. Druffi it can solve my capitalism problem

[@kekkekovich1423]

Сууука

[@disgruntledtoons]

Since the ratio is the same for any parallelogram, I applied the problem to square whose area is one, divided the octagon into eight congruent triangles which have a common vertex at the center of the octagon. The triangles have a base length of 1/4 and an altitude of 1/6. 1/2 x 1/4 x 1/6 = an area of 1/48 for each triangle, and therefore 1/6 for the octagon as a whole.

[@RA-tp4pr]

I couldn't do this if I was in the 100th grade....

[@Dontworryboutit247]

X'D

[@stevethecatcouch6532]

That's because senile dementia will have set in by then.

[@richardhead8264]

@@rishiagarwal4643 Pakistanis are brilliant!

[@dannygee_6051]

Yeah because you would be likely dead

[@RA-tp4pr]

@@dannygee_6051 clever.

[@pierreabbat6157]

Here's how I solved it: First turn the parallelogram into a square with a linear transform which preserves area. Call the side of the square 1. Four corners of the octagon are 1/4 from one side and 1/2 from the adjacent two sides. The other four corners are 1/3 from two sides. Now cut the octagon into eight equal parts which are all triangles with one corner at the center. Each triangle has base 1/4 and altitude 1/6, so its area is 1/48. The eight of them combined have area 1/6.

[@matthiasburger2315]

I have been searching through all the comments, looking for this. ;)

[@TheHuesSciTech]

Pedantically, you meant "First turn the parallelogram into a square with an *affine* transform which preserves area *ratios* ".

[@paulhofman3032]

Yeah this one seemed more logical to me to.

[@juintan3086]

Could someone draw out the pattern or describe more detail thank you

[@aaronwarwick9966]

@@TheHuesSciTech Technically it preserves the area of the overall shape and the resultant octagon.

[@wendym6227]

Ah Presh Talwalkar, you make geometry so elegant! I love your videos. Concise and to the point. PS I didn't solve this one my own either. :)

[@clyde__cruz1]

Succinct*

[@jessstuart7495]

Another approach would be to assume a square with unit area, and write equations of the lines in the upper right corner. If you use the upper right corner and make the center of the octagon (0,0) you can write equations for the two lines that cross and form one-quarter of the octagon by inspection. y=-2x+1/2 and y=(-1/2)*x+1/4 These cross at x=1/6, y=1/6 One quarter of the octogon has an area of... A/4 = Area of Upper Triangle + Area of Rectangle + Area of Right Triangle A/4 = (1/2)*(1/6)(1/4-1/6) + (1/6)*(1/6) + (1/2)*(1/4-1/6)*(1/6) A/4 = (1/6)*(1/12) + (1/36) A/4 = 3/72 A = 1/6

[@adelarscheidt]

Who else solved it by using a square instead of the parallelogram and trigonometry to find the area of the octagon?

[@adelarscheidt]

Oh wow someone in the comments pointed out that it doesn't make a regular octagon. That kinda blew my mind. You win this time, Geometry.

[@2eanimation]

That was my assumption too. I figured out the octagon's side length to be sqrt(5)/12*x where x is the square's side length. With that in mind, I followed A=a^(2)*(2+2*sqrt(2)) and got not quite 1/6 as a ratio but 0.167653... First I thought that OP was wrong, but the next day(today) I tried to figure out the outer radius of the octagon with which one can also calculate the area, and got a ratio of 0.17.... Very disappointing, but something must have gone wrong or math is broken. That's when I started calculating the length from the middle point M to one of the North-East-South-West points(S1) and another length from M to either NE, SE, SW or NW. Either will I get the same results and math is truly broken somehow or the results will differ and I can calculate the area of one of eight triangles and multiply them by 8 to (hopefully) get to 1/6. Let the square have a length of 12. M is (6,6). We need to chose S1 and S2, let them be the W and NW corners. Which can be created by two lines: g1 = m*[6,12] g2 = [0,6] + n*[12,6] We know the y-coordinate of S1, which is 6(the corner touches a horizontal line which halves the square). g1 is one of the lines forming the corner. 0.5*[6,12] = [3,6] -> S1(3,6) For S2 we need g1 = g2 [0,6] = m*[6,12] + n[-12,-6] m = 2/3; n = 1/3 m in g1: [4,8] -> S2(4,8) Vector from M to S1: S1 - M = [-3,0] = v1 Vector from M to S2: S2 - M = [-2, 2] = v2 Length of v1 and v2: |v1| = 3 |v2| = 2*sqrt(2) Yes, the lengths differ :D So we do not have a regular octagon as we can't draw a circle around it with the circle touching every corner. Just need the height of the triangle (with the M to S1 line as its base) to calculate the area. The line is at heigth(y) 6, the top point of our triangle is S2(4,8), so the height is 2(one could calculate this as well). ATriangle = |v1|*2*0.5 = 3 AOctagon = 8*ATriangle = 24 ASquare = 144 Ratio: 24/144 = 1/6 Finally 🙏🏻 That was fun :)

[@bjoernogthomas]

René Bazan Good read, solved it in the same way - however i still think the proof to why the area of the slices are the same is missing since its not a regular polygon

[@0megazeero]

I didn't even solve it lol

[@lyrimetacurl0]

@@adelarscheidt using a square is good and comes to the same answer and in that case it is also a regular octagon.

[@WhiteboardMaths]

Really cool problem, nice solution too

[@jacksonsay37]

Wanna know a problem that is even cooler? The one I posted in the comments!!!

[@MindYourDecisions]

Whiteboard Maths: Thanks! wimpykidfan37: I acknowledge your enthusiasm and have saved the problem to consider for later. Can you stop posting it in every video and referring to it comment threads? In the future, anyone can email me (find it on my blog or the "about" tab on KZbin) with suggestions. I will probably see it and reply faster than a comment.

[@jacksonsay37]

MindYourDecisions I am very sorry. Just making suggestions and stuff. Maybe I could post the puzzles on my own account.

[@BigDBrian]

since the question doesn't specify a particular parallelogram, it's safe to assume the same ratio must hold for any parallelogram, therefore you could try the problem on a square. Then you can use many identities such as for instance pythagoras that you otherwise couldn't. also there would be more symmetry. Probably easier to do it that way, maybe intended.

[@PlayTheMind]

So basically the octogon is hacking into the parallelogram

[@TheInterestingInformer]

*Pentagon

[@gamerdio2503]

FightOnGaming *Octagon

[@yawtwo]

Hahaha

[@artiomboyko]

From Russia XD

[@kordellcurl7559]

I just looked at it and it looked like it was around 1/5 - 1/6 of the parallelogram

[@dannygee_6051]

in germany we call it π×daumen

[@V1tal1t1]

All you need to know to solve this problem is the equal-area property of the medians i.e each median divides the area of the triangle in half, three medians divide the triangle into six smaller triangles of equal area (en.m.wikipedia.org/wiki/Median_(geometry)). This 1/4 of the octagon is formed by two medians in a triangle that is 1/2 of a quarter parallelogram so its area is 1/3*1/2*1/4=1/24. The same logic applies to all the other parts of the octagon, in fact all 4 of them have equal area of 1/24, so the total area is 4*1/24= 1/6.

[@sohomsaumeep5619]

"In a triangle that is 1/2 of a quarter parallelogram" - which triangle are you talking about?

[@V1tal1t1]

Sohom Saumeep If K, L, M, N are the middle points of corresponding AB, BC, CD, AD of a parallelogram ABCD, with O being its center, then I’m talking about the KLO triangle.

[@NoNameAtAll2]

Use "/" before stars to cancel making text *bold*

[@V1tal1t1]

NoName Thx a lot!

[@Icenri]

I thought this was going to be the video answer. It seemed so 8th grade obvious to me.

[@bigJovialJon]

I found an easier way. After dividing the parallelogram into quarters, divide a quarter in half (from the a corner of the octagon to the midpoint of the opposite side); giving parallelograms that are each 1/8th of the original parallelogram. Throw away the 1/8 that doesn't have any of the octagon. Note that the remaining 1/8th is divided into equal triangles by a diagonal that includes an edge of the octagon. Throw away the triangle that doesn't have any of the octagon. This leaves you with a triangle 1/16th of the original parallelogram that is mostly a quarter of the octagon. Construct a line from the center of the octagon to the corner of the octagon on the opposite side of the triangle. This divides the triangle into three triangles, two from the octagon and one other. All of these triangles have equal area (this is easy to prove if you cheat by assuming the parallelogram was a square). So 1/4 of the octagon is 2/3 of 1/2 of 1/2 of 1/4 of the parallelogram. Simplify to get the octagon is 1/6th of the parallelogram.

[@hritviknijhawan1737]

Wow, that's non-appreciated genius right there. I have to say, 3 years of a great comment having great logic and a rather easy solution, and just 6 likes and no comments. Loved your solution. 💕

[@wjgranger5]

I'm still not sure how you deduced that the trapezoid was 3/4 of the shape. How did you know the triangle was 1/4?

[@abdellahdany1689]

William Granger same here...

[@caygesinnett6474]

Because the hypotenuse of the triangle goes from the corner to halfway up the otherside. If it went from corner to corner it would be half the area, but it goes from corner to midpoint, therefore it is 1/4

[@manuelfink1613]

William Granger 5:37 look at the top left quarter: You know that the white line exactly crosses the blue line in the middle. So if you double the black triangle you will have a rectangle with 1/2 of the area of the quarter. Hope you understood, I'm not a native speaker and that makes it very hard to explain

[@manuelfink1613]

andre johan of course it goes exactly into the corner. If you zoom out you can see again that you defined the perallelogram exactly from that corner

[@stevethecatcouch6532]

+andre johan If you go back a few steps you will see that, due to similar triangles, the white line passes through the corner of the smaller para and the midpoint of its right hand side. No assumptions are required.

[@Abhisruta]

The geometry has a sense of happiness. And you@Presh Talwalker, disclosed it. That's great.

[@johnlarson505]

I had a professor in college give us this problem. It was awesome to see all the different ways people solved it. I used an algebraic approach with a little bit of calculus. It was a good demonstration of how multiple approaches will give the same answer when they're valid.

[@xenontesla122]

1/6th. For my answer I assumed the parallelogram was a unit square (knowing that all parallelogram solutions are linear transformations of each other and ratios are preserved under linear transformations) and dissected the octagon into eight triangles. I know the base of each triangle is 1/4 and I used the system of equations y=x/2=2x-(1/2) to find the triangle height of 1/6. Then I found the area of the triangle and multiplied it by 8 to get my final answer ((1/4*1/6)/2=1/48, 1/48*8= 1/6).

[@monjazebr]

Incredibly Clever Solution! I estimated to be (1/6) by chopping the parallelogram to pieces then move the pieces around to see 6 of those octagons go in it. But this solution is an actual proof. Fantastic!

[@jlittlenz]

Can't one assume that shear transformations preserve area, and go straight to a square, with its symmetries simplifying things? (Somehow nicer than deducing the same from the fact that the question implies it.) Finding the coordinates of the corner closest the square's corner is then an 8th grade problem.

[@amedjed3228]

Yeah that is how I go the answer. A good eight grader could do this.

[@MrSaemichlaus]

I did so and, not using linear functions, got 1/sqrt(32) = 1/5.66 I calculated the area of one pizza slice (eighth) of the octagon, no idea what went wrong or why I still got so close.

[@trueriver1950]

Shear goes to rectangle not to square, surely???

[@wwoods66]

@True River -- Yes, but after transforming the parallelogram into a rectangle without changing the areas, you can scale one dimension up or down to match the other without changing the _ratio_ of the areas. So the result you get from analyzing the case of a square will apply to any parallelogram.

[@robertbeach7942]

do 8th graders know that shear transformations preserve area?

[@AirWick141]

1/6 is wrong. The ratio is ✓2/2. I made the octagon's radius equal to r, and the pollygons radius equal to 2r. Then solved for their general Area, and rationalized them.

[@sergiokorochinsky49]

in 5:55 I thought he will reapply the method infinite times and then solve the series... :-p

[@ericsun1006]

same

[@squeekycheese]

My solution was a little less dizzying. Using a square I split the area of the parallelogram minus the octagon into 2 sets of four similar triangles. Using geometry and simple linear algebra calculated that four of them had an area of 1/8 the area of the parallelogram and four of them had an area of 1/12 the area of the parallelogram. A little addition and bada bing badda bang. This was a tough one. Enjoyed it. thanks! Keep them coming.

[@bckzilla]

Given that the question asks for a solution for any parallelogram, it may help to consider a square. Dividing a square (say a 6 x 6 square) in the same way it is done in the video makes it quite easy to (intuitively at least) understand the ratios of the triangles compared to the square. I don't think 8-graders know about lemmas or proofs.

[@rosomak8244]

In Russia they know.

[@vinayv8791]

This problem is a perfect example of the "search for the truth". Everything appears to be random, only when we focus on the problem, definitive patterns appear and we will be able to look at the actual picture. Indeed it is a great problem.

[@JLPicard1648]

I took the engineer's solution: Hey, there's a similar parallelogram framing the octagon in the center. Those angles look similar. I'd bet if I took the time I could prove they were similar parallelograms. Let's see... It looks about 1/2 of the height of the big one and 1/3 the width, so the parallelogram is like 1/6. Then you have to cut the corners off, which'll take it down a bit, but not quite to 1/7. Since 1/7 is 2/14, let's call it 2/13. Got it within 0.013 without any difficulty or hard math

[@JLPicard1648]

Don't call my comment imbecilic when it is you who has made the idiotic and uninformed claims. I work on a daily basis building robots (although, I'll freely admit, as a student) and hobby rockets. 0.013 is a perfectly acceptable margin of error. I frequently have to work with parts of tolerances of 1" or greater. In fact, I can cite several industrial manufacturing companies using waterjets and laser cutting devices that have 0.01 tolerances with state of the art equipment ( www.bystronicusa.com/en/news/case-studies/archive-2014/140210_Bending_Better.php and rapidmanufacturing.com/rapid-machining/cnc-machining-services/waterjet-cutting/?gclid=Cj0KCQjw1JbPBRCrARIsAOKj2PkPyBd0knis-LCcE0ePKazLaSPAVU7uc83C_9H75Xqc_U-fTFzNh2UaAib2EALw_wcB for instance ). Most engineering, at least in the field of robotics, doesn't need to be super precise. That's why PID exists. The only parts I can think of that I've worked with in 3 years (again, not too long, but bear with me) that a 0.013 margin of error is unacceptable for are bearings and electronics--and there are even bearings with margins of error of 0.005" ( www.mcmaster.com/#1291n1/=19uvq0c ) that are dependable and regularly used for prototypes and flight robots. And furthermore, how are you thinking this is going to be applied? I decided to pursue that method because A. I suck at math and B. I wanted to prove a point that in the real world not everything needs to be so precise, a point which flew right over your head, apparently due to lack of relevant knowledge and sense of scale. Finally, if our hypothetical application is that important and if peoples' lives depend on the margin of error being really small, then sketch it out and use a meter stick with the method I described and the margin of error all but disappears. The only time you need to calculate something to that degree of precision is when you're doing theoretical math or you're taking a math test in Russia, imbecile.

[@Perserus]

The lesson here is, don't get on Ben Hamiltons bad side.

[@SnapstickGamer]

Alf Mikael 👍

[@aajjeee]

Ben Hamilton you seem to ba assuming the sape is not the size of a building

[@JLPicard1648]

If it were, then 0.013 is still just 1% of its size, not really going to make a difference

[@sir_prize]

I did this one a little differently, although it's difficult to describe in words without drawing a diagram. I also divided the parallelogram into four quadrants, but I did not subdivide it any further. I (eventually) noticed that the portion of the octagon in each of the four small parallelograms was contained in a triangle that had an area of 1/4 the small parallelogram. (Half of the base, same height, with an extra factor of 1/2 since it's a triangle.) It remained to show the ratio of the octagon portion's area to the triangle's area. The remainder of each triangle (excluding the octagon portion) was a smaller triangle that (with some difficulty) I proved to have an area of 1/3 that of the larger triangle. (The base of the small triangle was 2/3 that of the larger one, and it was 1/2 the height. 2/3 x 1/2 = 1/3.) Thus each of the 4 portions of the octagon had an area of 2/3 that of a large triangle whose area was 1/4 of a small parallelogram with an area of 1/4 that of the large parallelogram. 4 x (2/3) * (1/4) * (1/4) = 1/6. The difficult part (besides drawing an accurate diagram) was showing that the small triangle had an area that was 1/3 that of the larger triangle. I can provide details if anyone is interested, but it's hard without being able to refer to a diagram where I can label the points.

[@tiripoulain]

I guess it made the thing more concise, but the truth is nothing was “proven” at all. No mathematical principles were used to SHOW these assumptions. Unless proven mathematically, you can't really consider graphical analysis to be proof, can you? Maybe it would've been a good idea to at least orally explain that some of the steps were supported by the properties of parallelograms.

[@pengin6035]

Thierry Poulin But this channel explained the idea of the solution, I think that's good enough already

[@_inabox]

2(3/4)(p/64) pulled out of a hat

[@durnominator]

3/4 because the line goes from corner to the half of a side of those p/64 trapezoids. Divide that trapezoid by half and you have one full trapezoid and one half full trapezoid, divided from corner to corner by the line, thus 3/4.

[@TheVictoryHawk]

you have to prove it goes halfway, just because it looks like it is not enough

[@durnominator]

It does go halfway, just look at which sub-trapezoids of the top right quadrant it goes corner to corner from. Halfway from corner to corner goes halfway. MYD skipped all that explaining, though. He also skipped showing that all four quadrants of the octagon have same area. Only two of them are of same shape and go by symmetry directly. So two pairs by symmetry, but what about the other pair? Since it doesn't depend of the trapezoid's corner angle, it is thus the same?

[@copperdragon9286]

I can not help it, but I come to different solution: 1 : 4*sqrt(2) My solution is dead-simple: Assume the parallelogram to actually being a square with side 1. Then you immediately see that the octagon has a diameter of 1/2 (corner - to - corner) which means its radius is 1/4. One 8th of the octagon is a triangle with base 1/4 and height (1/4)/sqrt(2). This gives a total area of 8*(1/4)*((1/4)/sqrt(2))*(1/2). And the ratio is 1 : 4*sqrt(2).

[@Geigenzaehler]

Some remarks... 1. Yes, the four quarters of the octagon have the same area. But this is not trivial at all (the shapes have different angles). I don't know why you assume that in the beginning. This assumption does make the proof insufficient. 2. Throughout the proof you always assume, that some line segments are perfectly cut in half. They certainly are, but you do not say why.

[@HopUpOutDaBed]

The 4 quarters of the octagon have the same area due to symmetry.

[@Geigenzaehler]

No, there is no symmetry. Just like i said, the parts of the octagon have different angles. I know how to prove, that they are equal. But as there is no symmetry this must not include symmetrical arguments.

[@HopUpOutDaBed]

He is first assuming the parallelogram is a square, then saying each part is the same due to symmetry. It is true if the original parallelogram is a square. I see what you're saying now though, he needs to explain why it's still true for all parallelograms, not just squares. Each section isn't "technically" the same area, but will be the same ratio when added up. Seems pretty intuitive to me, but you're right they aren't all the exact same.

[@Geigenzaehler]

Yes, you are right. His mistake in the proof was mentioning the x/4 in the first place. What he needed to do was adressing one of the smaller parallelograms as the new problem. Since the yellow area in the four parallelograms is equally constructed, you can apply the ratio to the other three.

[@MrHatoi]

The video would be way too long if he explained every single line being cut in half; and it would get pretty boring too. In subjects like math and hard sciences sometimes you just have to accept that something is true without proving absolutely everything because it makes it long and complicated to explain.

[@emisor9272]

"Each quarter of the octagon is x/4" i feel like you ought to prove that

[@thalesogoncalves1]

Hey Presh Talwalkar, which program do you use for these amazing presentations?

[@えへへ-y6d]

Your solution is so beautiful that I was impressed!

[@Conserpov]

Solution in the video involves way too many subdivisions to my taste. I did it once and then turned to triangles: ibb.co/dY3P1R There are 2 pairs of triangles that share altitude and have equal base (same area), and 2 pairs of similar triangles on vertical angles with 2x base (2x ratio = 4x area). Also, A = B, since 3 medians in a triangle divide it into 6 triangles of equal area.

[@beng4186]

I think this is most elegant solution I've seen.

[@qbert8695]

Hello from Russia. Using Menelaus's theorem, I've found that segment connecting parallelogramm vertex and the middle of the opposite side is divided by the other segmets in a ratio 12:3:5:4:6 counting from the longest. Knowing that, it's pretty easy to find the desired area by subtracting ones from anothers. 1 / 6 is a correct answer and GeoGebra thinks the same way.

[@soheilghane7910]

Even the teachers are crazy in Russia :/

[@Kitulous]

agree. mine is a real demotivator and stupidness-spreader.

[@knocewwmckskdmdo976]

Soheil Ghane of course they are

[@yesofcourse8002]

Can approve. "Uh? Exams are 3h 55min long? I think, that you'll be able to solve 21 out of 25 problems in 40 minutes. What? You said, that there's not enough time for you? Ok, here goes your "3" (One step better from F. I don't really know about American marks system, but 3 is not "didn't even know what's going on" and not "mainly good, but made a minor mistake" somewhere between)

[@mradvocat]

@@yesofcourse8002 But there are 19 problems in RSE) and if you mean OGE(sorry my english) first part is too easy, and really hard problem is only 25th other are not so hard. Anyway all tasks in math exam much easier than writing in russian

[@davidellis1929]

I may have found something simpler. Draw a centered nested parallelogram one quarter the size of the original, bounding the octagon, then draw its diagonals. The octagon is equal to the smaller parallelogram minus 16 tiny triangles, each (1/3)(1/2)(1/8) of the inner parallelogram. If you arbitrarily assign the area of the original parallelogram to be 16, then the inner parallelogram is 4 and each tiny triangle is (1/12). This makes the area of the octagon 4 minus (16)(1/12), or (8/3), which is (1/6) of 16.

[@danmimis4576]

dude, why complicate yourself? Since it's independent of the parallelogram angles (either implied by the text of the problem OR it can be demonstrated) it means that there is the same ratio for a square (pause at 2:08). So we have a perfect octagon with the diagonal of 1/2 the side of the square and from here is a walk in the park ...

[@CarolHaynesJ]

Dan Mimis Just about to say exactly that ... since we can't make any assumptions about the initial shape the ratio must be a fixed value for any parallelogram so just choose the simplest - a square.

[@adelarscheidt]

+Dan Mimis I thought the exact same, I used 10 for the size of the square, therefore the radius of the octagon is 5. But try finding the area of the octagon from its radius. Not a walk in the park, for sure.

[@adelarscheidt]

+Dan Mimis oh and apparently it isn't a regular octagon, as someone pointed out in the comments. The lines don't make a regular octagon, just looks like it.

[@danmimis4576]

in a rectangle it HAS to be a regular octagon due to the symmetry of the lines that generates it, it just can't be otherwise. So IT IS a walk in the park :) But now I'm watching Unsolved Mysteries with Dennis Farina and decided that this problem doesn't deserve more attention (except for, eventually, demonstrating that the ratio is a constant 1/6 for ANY parallelogram)

[@stevethecatcouch6532]

+Dan Mimis A walk in the park in the dark. The octagon is not regular. The angles alternate between 126.9 degrees and 143.1. A regular octagon with a diagonal of 0.5 * s would have resulted in a ratio of ~ 0.177, not 1/6.

[@rupasarkar8276]

Really nice way to solve a problem which appears to be difficult.

[@sergiokorochinsky49]

all the assertions become apparent from the graph, but they are far away from being proven... how do you know that white lines intersect blue lines at midpoints??

[@MrHatoi]

These assertions do have proofs; these are things that you learn in Geometry class.

[@stevethecatcouch6532]

Due to similar triangles. Any 8th grader faced with this problem would be so familiar with that property of parallelograms that it would be second nature to her.

[@zaqmko0]

Woot... I was able to solve this. Because the problem asked for THE ratio of the octagon's area to that of the parallelogram, I assumed for the time being that starting with ANY parallelogram would yield the same answer. If you start with a 12x12 square, it's pretty easy to determine the vertices of the octagon. From the vertices, it was easy to determine that the area was 24 *. 24/144 = 1/6. Once I worked it out for the case of a square, it was fall-down easy to see that the ratio would work out to be the same for a rectangle. And from there, it was only a little more effort to see that skewing the rectangle into a parallelogram would not change the are of either the quadrilateral or the octagon. * Area of an Irregular Polygon (where you're given the points' Cartesian coordinates): Area = abs(1/2 * sum(x[i] * y[(i+1)%N] - x[(i+1)%N] * y[i]) ) where i ranges from 0 to N-1. This is quick and easy to do on paper by listing the points in a column (including a copy of the first point below the last) and then doing the two multiplications for each pair of adjacent points in an X pattern. Add each column of products, subtract the two sums and divide by two.

[@sanskartiwari2996]

I doubt that it is a 8th grade question

[@Tektus]

As a Russian, I can confirm we did something similar to this back then.

[@seangrand3885]

I cant confirm that, even though i am russian as well

[@human3507]

sanskar tiwari As a non-russian Our class did this kind of geometry problem quite often in 8th grade

[@entercherpfhalckhontralyty3542]

As an actual 9 grade russian student, I want to say that we learned in the previous grade, that triangle medians cut a triangle into six equal to each other triangles. I don't remember that I'd face such a task that time, but with all the knowledge I have solved it now.

[@СуперканалВлада]

In Russia there are 11 grades, so in American standard, it should be cathegorized as a 9th grade problem, so that it wouldn't be as much unbelievable...

[@AyushGupta-bg4bw]

Sir this solution was epic. Hats off boss!!

[@DavidsKanal]

How the hell is the area of that cornerpiece of one of the quarters of the octagon x/64? It seems arbitrary and off to me. Edit: nvm

[@brawlgod4366]

Concept of similarity in quadrilaterals😂

[@steam_bigh8899]

In a square you can draw an equal sided octagon. Within the octagon you can draw another smaller square by drawing 4 lines which split the big square into 3 slices from up to down and left to right which leads you into 9 equal sized quadrants. The inner square has the area of 1/9 or 4/36. To figure out the rest parts of the octagon just split again into half each quadrant and you'll find out another 8/36 around the inner square but where is only 1/4 occupied by the remaining octagon triangles. So that means the inner square is 4/36 + (8/4)/36 = 6/36

[@gameofday5299]

6:14 how you said that last piece is scaled down by 4 times sir??

[@machivaldemon]

yeah i dont quite understand that. need someone to elaborate

[@wolffang21burgers]

From looking at the lines in this pattern. |\/\/| |/\/\| it is easy to deduce the shape /\ \/ has area p/4. Similarly, the shape < > has area p/4 The star formed by the union of these 2 shapes has area p/2 - x. The remainder of the parallelogram has area p/2+x. Colouring this remainder, and connecting four big parallelograms together (and looking at the middle), we can see the coloured region forms an octagon similar to the initial one. (e.g. each coloured corner is similar to a quarter of the initial octagon with the same ratio) By similar triangles, (sides are double in length, area is 4 times) one can deduce this octagon has area 4x. p/2+x=4x So x/p = 1/6 .

[@harryandruschak2843]

Ahem....at 3:08, how does one know the 4 areas of the octagon are equal? Lemma?

[@FryGuy1013]

Look at the x/4 polygon. It's bounded by the left wall, the line from the midpoint of the left to the bottom right corner, the line from the midpoint of the bottom to the top left corner, and the bottom wall. It's the same for the x/64 polygon.

[@stephenhousman6975]

a parallelagram has 2 sets of parallel lines. if you make a line between a pair of parallel lines at the midpiont and cut along it you effectively cut the shape in half . if you do it with both pairs you cut it in 1/4. Since these cuts go though the center of the shape all the sides and angles are the same (for the octagon). therefore the octagon has also been cut into 1/4.

[@pengin6035]

Or you just apply the principle of Cavalieri or divide the shape into more triangles and you will see that the areas of the subtriangles are equal

[@Geigenzaehler]

@Stephen Housman No this is a non sequitur. Just because the parallelogram is cut in fourth, doesnt say, that the octagon is as well. The crucial difference is, that while the four parallelograms are congruent, the four octagonparts are not.

[@Geigenzaehler]

@Kevin Hjelden Yes, they are bounded by the same principle. But they have different angles. You need a lemma that proofs that the area is the same for adjacent angles. This is as nontrivial as: "If a parallelogram is devided by its diagonals, the four parts have equal area"

[@simonstark2948]

I just used a tilt coordinate-system which makes the parallelogram a de-facto rectangle, gave the parallelogram height and width 2 (and argued that stretching one side doesn't affects the ratio) then argued via symmetry that each quarter of the parallelogram is exactly like the others (after rotating in the tilt coordinate-system), gave both lines which were bordering the area in this quarter a function, calculated the intersection-point and then integrated the area. The area was 1/6 which is the solution since the area of this quarter is normed to 1 and all quarters are equal in terms of the fraction of the searched area.

[@xnevernikx]

This is how I did it in my mind (and after that I created this applet): geogebra.org/m/ZNb7u3S7 The idea is basically the same as yours in the beginning: P(octagon) / P(parallelogram) = P(tetragon) / P(1/4 of parallelogram), but then I used what I saw at your drawing without drawing any new object. On the applet you can see my thoughts. Remark: The applet is interactive, which means you can move points A, B and C (independent objects). The rest of the objects can not be moved, because they are constructed using points A, B and C (dependent objects).

[@dashohoxha]

The most easy and clean solution.

[@flarbear]

This solution is much more intuitive and concise. The Pklj = Pglj/3 step wasn't as obvious, and would only take a couple of lines to demonstrate (I wasn't aware of that property off-hand and had to diagram it out to see it, but it wasn't too hard to show).

[@flarbear]

Actually, it might be even simpler to calculate Pijk and Pilk since they are formed from median triangles and so are 2/6 of Pifg which is 1/2 of Pifcg. Thanks for turning me on to GeoGebra - I tried to illustrate this simpler solution in a diagram here: ggbm.at/GtmpCXr6 (It's not as pretty as your diagram since I am new to the site, but I think the derivation which was based on the comment from Виталий Удовенко is simpler.)

[@OmegaMusicYT]

The area of any triangle that shares its base and height with the parallelogram is half the area of the paralellogram. The area of the octagon is the common area for the four triangles shown in the figure

[@sayanmondal4570]

Russians 🙇 respect

[@fernandopizarrovillagarcia6992]

Step 1. Take the big triangle formed by the two vertixes at the bottom and the top vertix. That is 1/2 of the total. Step 2. Halve the octagon and take the bottom triangle. That is 1/4 of the triangle, i.e. 1/8 of the total. Step 3. Connect each vertix of the octagon and the triangle with the center of the octagon. It forms 6 identical smaller triangles. As the yellow area has 4 of the 6 small triangles, it means 2/3 of the 1/8 of the total, or 1/12 of the total area. Step 4. It is only the bottom half of the octagon, so the ratio is the double of this value. This is, 1/6 of the total area. Final answer: 1/6

[@wongwh7375]

I dont understand x/64 :/

[@voselideas5517]

me neither

[@teeheejejo]

Congruency. The area bounded by x/4 shares the same exact internal angles with the one he labelled x/64. Thus by similarity/proportion, as proven with scaling, the sides (linear) are 1:4, so squaring it will give you the ratio (area) 1:16. Finally, 1/16th of the x/4 is x/64. This works on any polygonal area scaled smaller or larger.

[@maartenofbelgium]

If you have an area of x/4 and a congruent figure of which both sides are scaled 1/2, then the area is scaled (1/2)²=1/4. The area of the scaled-down area should thus be x/16.

[@JianJiaHe]

I solved the problem by using the analytical method. First, we can project the diagram on to a paper in such a way that the parallelogram becomes a square, and the ratio of the areas doesn't change. Then, I put the diagram on to a Cartesian coordinate system, and solved for the side length of the octagon. After that, calculate the area of the octagon, now we have the ratio.

[@merubindono]

Is this one is those questions that Russian universities use to keep Jewish students from getting in? "See, even 8th graders can solve this problem, so you can't say we were being discriminatory." (Look up the video "simple math problems to fool the best")

[@Conserpov]

Jewish professors in Russian universities keep Jewish students from getting in? Yep, makes sense /sarcasm

[@Theo0x89]

*Sowjet universities used to. (Basically affirmative action for non-Jewish Russians.)

[@Conserpov]

Let me be clear - this is a moronic, entirely made up, fake news myth, from Cold War era Zionist anti-Soviet propaganda hogwash about "oppression of Jews" in USSR. There was something remotely similar to "affirmative action" - but for ethnic minorities (very "ethnic") - and those were conditioned, more like educational benefits US military provides.

[@bh8642]

You assume - with NO proof whatever - that the professors behind those difficult problems are Jewish. Your total lack of logic is appalling.

[@bh8642]

Oh, and you AGAIN state nonsense WITHOUT proof - WHERE does it say that the majority of Russian university professors are Jewish? In your fervid imagination? Get a life, little child.

[@owem6511]

I used a different method. First, I labeled the sides of the parallelogram as: height=2 length=4 Next, I found the length of the hypotenuse of the triangle that had the top length for the base and the left side for the height, extending the hypotenuse from the top left corner to the right side midpoint. Since it extended half of the length of the parallelogram's left side, i knew the height would be 1. As well as that, since it extended the full length of the parallelogram's top base, it had a length of four.Plug in the numbers and you get 1^2+4^2=c^2 1+16=17 the square root of 17 is 4.12310562562 Then, I realized that the triangles that lined each side of the octagon had a base length that, if discovered, could grant me the information needed in order to solve for the area of the octagon. I also realized that since one of the base lengths that lined the octagon was on line with the hypotenuse of the triangle whose length I had discovered, I could divide that length until I reached the size of it, giving me the information needed. The length of the base of one of the tiny triangles whose base covered one of the sides of the octagon was one eighth the size of the hypotenuse on the triangle I had just discovered, so I divided 4.12310562562 by 8 and got 0.5153882032, giving me the size of one of the lengths of the octagon. I then plugged in the length of one of the octagon's sides into an equation to solve for the area of an octagon as shown A=2(1+√2)a^2 A=2(1+√2)0.5153882032^2 A=1.282550955 I then divided the area of the octagon by the area of the parallelogram, being 8, giving me the decimal: 0.16031886937 This fraction roughly converts to 4/25, but they are asking for a ratio, implying the probably want some single fraction(e.g. 1/2, 1/8, 1/23, ect.), so I went with the closest thing, being less than 6 thousandths off, 1/6. This is the first one of these geometry videos that I got right, so I'm feeling pretty proud of myself, not to bad for a 7th grade education!

[@cr1216]

The easy way to do this: imgur.com/a/I0koX No algebra required, only 8th grade geometry is needed (parallel triangles and area formula)

[@stevethecatcouch6532]

You start by stating that QM/MD = PM/MB = PQ/DB = 1/2 as if it were obvious. PQ/DB = 1/2 by similar triangles, but it took me a bit of work to convince myself that QM/MD = PM/MB = PQ/DB. Is it obvious and I'm just missing it?

[@cr1216]

Hi Steve. This is a pretty trivial pattern if you are familiar with geometry problems involving parallel lines with "X" shape. There are two similar triangles one on the top and one on the bottom, rather than the more common case where one triangle is inside another. But as long as the lines stay parallel these two cases are essentially the same thing.

[@stevethecatcouch6532]

+Manuel Cano +C R Thanks guys. I was looking for similar triangles in the wrong places. The more I look at this, the more fascinating it becomes.

[@lanat1512]

I couldn't see very clearly from the picture. Could you please tell me where the h1 and h2 are? I'm an 8th grader myself, and can't figure that out.

[@GurwinderSingh-hf3te]

I did it in a different way and arrived at the same answer. What i did is this. Since it is the ratio that we are looking for, I took a square and assigned sides to it, and then made those lines through the midpoints. which gave me an octagon. After that I started to exclude the small triangles areas until I am left with the octagon itself. Then i divided that by the area of the square to give me the ratio. 0.1666

[@xVitOSx]

Горжусь росией 🇺🇦

[@olegshtolc7245]

Naruto runner слава Украине 🇷🇺

[@madpad759]

I think there is a simpler way to prove this: 1) Connect the opposite corners of the parallelogram 2) Connect the middle points of the opposite edges of the parallelogram 3) This will split the parallelograms into 8 triangles. It also split the octagon into 8 small triangles 4) We can prove that the area of each small triangle is 1/6 of the larger triangle which it resides in. Proof: (take the west-north triangles as example) a) The bottom side of the small triangle is 1/2 of the bottom side of the larger triangle (horizontal line) b) The height of the small triangle is 1/3 of the larger triangle (The centroid cuts the median in the ratio of 2:1.) -> the area of the smaller triangle is 1/6 of the larger triangle. Also note that the 8 larger triangles have the same area. But this is not used in the proof. Please forgive my crappy English. It is not my first language.

[@nilanshur]

This is the most simplest and provable proof I think. Good job!

[@ClammySparkz]

THE ANSWER 1/6 IS WRONG!!! (OR IT DOESN'T WORK FOR EVERY PARALLELOGRAM) IT'S 1/5.656854249 (1/(4*sqrt(2)) 1. Let's suppose we're solving the problem for a square with side length a=4, which makes it's area a^2=16 2. Then looking at the octagon in the middle, we can see that the diameter of the circumcised circle (the bigger circle touching the corners) is 2 or half a. This means the distance from the center to the corner of the octagon is 1 or a/4. 3. To get to the value of one of the sides of the octagon, we first split it into 8 equal triangles (like pieces of cake). We take one of those isosceles triangles and split it into down the middle into 2 to get a right angle triangle with hypotenuse 1 (a/4) and the smallest angle equaling 22.5 degrees (pi/16). 4. Taking the sinus of the corner we get that half the of the side of the octagon equals sin(22.5) or 0.3826834324, so one side is equal to 2*sin(22.5) 5. To get to the area of the octagon we put this value in to the formula for a right octagon (taken from Wolfram Alpha): A = 2*(1 + sqrt(2))*a^2. 6. The area of the octagon is therefore: 2*(1 + sqrt(2))*(2sin(pi/8))^2 = 2.82842712474619 7. Dividing this value by the area of the square, we get the fraction: 2.82842712474619/16=0.17677669 8. Consequently the ratio of the octagon area to the square area is 1 : 1/0.17677669 or 1 : 5.656854249 PLEASE VOTE UP SO PRESH CAN SEE OR TELL ME WHERE I WENT WRONG

[@Geigenzaehler]

@Jose Use Geogebra

[@ClammySparkz]

Oh shit, didn't notice that it wasn't a regular octagon, i thought it was to easy to be true, thanks

[@MichaelOnines]

Klemen Iskra, I fell into the same trap. Once you realize the octagon isn't regular you can go back and easily calculate the height of one of the triangles perpendicular to the side with length s/4, and find it is s/6. (1/2)(s/4)(s/6)*8 = s*s/6.

[@Stratelier]

I asked the same thing ... but eventually realized that in the case of a square, the lines of the octagon are _not_ angled +/- 22.5deg from either axis as a regular octagon would be ... they are at _slopes_ of +/- 0.5 and +/- 2.0 ... an error of about 20%.

[@michaelempeigne3519]

you rounded the numbers so you in essence proved that it is 1 / 6.

[@wwoods66]

Okaaay. 1) By inspection, we see that the ratio will be the same for any parallelogram. So let's simplify by making it a square, with a side length of 4 units. 2) We divide the octagon into eight pizza wedges, each triangles of equal area. Let's figure the area of the 8-9 o'clock slice. 3) Putting the origin at the bottom left, the center is at (2,2). Another corner is obviously on the horizontal center line, halfway to the center, at (1,2). The third.... 4) The third is at the intersection of the diagonal down from the top left (0,4) to (2,0) and the diagonal from (0,2) to (4,0). The equations are y = -2x + 4 y = -1/2 x + 2 Solving gets x = y = 4/3. So the point is 2/3 below the center line, and the length of the side of the triangle along the centerline is 1. 5) The area of the triangle is 1/2 * 1 * 2/3 =1/3. 6) The area of the octagon is 8/3. 7) The ratio of areas is 8/3 / 4^2 = 1/6. Huh, I figured it out!

[@ShinySwalot]

Haha, he couldn't solve it! PreshTallWalker, more like FreshWallTalker! ((I couldn't solve it either....)

[@jacksonsay37]

But can you solve the puzzle I posted in the comments?

[@s888r]

One way you can do it is by taking a square (a square is a parallelogram) and drawing the lines. You get a regular octagon. Since one of these line segments divide a side in half and it's opposite side is also divided in half, the line segments dividing these opposite sides intersect at their mid-points. This only means that the vertex of the octagon facing a side of the square is a quarter side length away from it and longest diagonals of the octagon have length equal to half the length of any of the square's sides. Since it's a regular octagon, you can find it's area by a formula.

[@manishamotwani782]

In india we solve such problems almost everyday lol

[@awesomedude8812]

MANISHA MOTWANI looks similar but this is more harder especially for students of grade 8th.

[@pranjalverma3501]

I bet no such level of problems are taught in India,you just shut up.

[@DennisOya]

I have solved it using Area formulas for parallelogram S=a°Ha, triangle S=1/2a°Ha and for Trapezium S=(a+b)°Ha/2, where a and b are relevant horizontal sides and Ha is perpendicular to side a. If you draw horizontal line from left midpoint to right midpoint you divide octagon to 2 equal shapes. Then you can see each of those shapes consist of one Trapezium and one Triangle. And you can calculate it's areas by abovementioned formulas. If we mark horizontal side of parallelogram as 'a' and perpendicular to it as 'Ha' then it's Square will be S=a°Ha. For Trapezium S=5/72a°Ha, as one side of it = 1/2a, other side = 1/3a and perpendicular = 1/6Ha. For Triangle S=1/72a°Ha as it's side = 1/3a and perpendicular = 1/12a. Now combined area of half of octagon = 1/12a°Ha, so full octagon area will be 1/6a°Ha, which is 1/6 of parallelogram's area. I skip the proofs why sides of Trapezium and Triangle have those ratios to the side of parallelogram and as well for perpendicular ratios. It can be solved using Similarity properties and crossing angles. You can check it yourselves.

[@krauser_]

Lol, I am from Russia. We don't have such difficult shit in school at all. What a click bait

[@antonantonov8747]

check description - entrance exam in a special school

[@YamiSuzume]

So... you say that every school in your country has the exactly same math problems? And even if thats true and your country has the exact same education everywhere.. Why is this a clickbait? Because it has Russia in it? Are you so addicted with your own country that the word "russia" makes a video a must-click for you? wtf is wrong with you?

[@MuantanamoMobile]

He's not from Russia, just your regular Trump troll, as his youtube channel shows.

[@laincoubert7236]

well, 8th grade in russia isn't high school so maybe there is some mistake

[@laincoubert7236]

just wanted to write the same thing lol

[@vankrelian]

Wow, that was a very roundabout way to solve this question. There's a much simpler and intuitive way to solve this. If you trace a line from the center of the parallelogram to each vertex of the octagon you'll see that it it splits in 8 triangles, each of these small triangles extends all the way to the parallelogram. Due to some symmetries you only need to find the ration between one of the smaller triangles to its bigger one and it'll be the same as the octagon to the parallelogram, that being 1/6.

[@markwilliams1555]

Such a beautiful solution

[@willemkelles7942]

Hi, there is an easier solution: if we say that P is the size of the parallelogram, we can calculate the size of the diamond in the middle as P - 2*P/4 - 2*P/8 (2*P/8 being the size of 2 triangles with halve the base and half the hight). This gives P/4 for the diamond with the top and bottom equal to the top and bottom of the octagon and the left and right points on the original parallelogram. We still have to extract the size of the left and right extensions of the diamond to cover the size of the octagon only. These extensions can be seen as 4 small triangles. For each of these triangles there is a bigger triangle of the same shape with size 3*P/16 (a triangle with base 3/4 of the long side of the parallelogram and hight the short side/2) . As the small triangle has 1/3 the hight and base its size becomes 9 times smaller. As we have 4 of them the size of the extensions left and right become 4*3*P/16*1/9=P/12. So we have P/4 - P/12 = P/6 as the area of the octagon.

[@mikhailplotkin684]

I solved this in a simpler way by using the fact that medians in a triangle are divided in 2:1 ratio in their intersection point. If you consider a (top right) quarter of the parallelogram, the 1/4 of the octagon area can be found by subtracting areas of two triangles. First triangle is OPM where O is the center of parallelogram, P is the middle of the top side of the parallelogram and M is "east" vertice of the octagon. Second triangle is PNK, where N is the "north" vertice of the octagon and K is the "north-east" vertice of the octagon. The area of the first triangle is obviously 1/16 p. The area of the second triangle is 1/48 p, it can be found if we remember that (from the mentioned medians property) PK : PM = 2 : 3 (I can give more details if needed). So the quarter of the octagon area equals (1/16 - 1/48) p = 1/24 p. So the total area of the octagon has area 4 * 1/24 p = 1/6 p

[@bluey3575]

I love these seemingly hard math puzzle that have elegant simple solution 😁👍

[@jarikosonen4079]

It looks there is error, even the final result is correct. The areas marked as "x/4" actually are different sized. The left top area would be A1=(12*ah+dxh)*bh/72 and right top area would be A2=(12*ah-dxh)*bh/72, where ah=a/2 (base/2), bh=b/2 (height/2) and dxh is the half the sideways shifting of the base. The solution however seems to work for dxh0 cases also!

[@RobinsonDanieldosSantos]

Fantastic solution!

[@Champ-rb2yw]

I just turned it into a square, since that is a parallelogram. Then, I found that from one corner of the octagon to the opposite corner is 1/2 the side length of the square, because of the parallel lines the form the sides of the octagon. Finally, I used this length along with properties of a regular polygon from geometry class to find out the answer.

[@WorBlux]

Yay!, I go the answer. I did it a little simpler way. I looked at an interior parralelagram of 1/4 the area of the original entirely containing the octagon. There's three of them, you want to consider only those shapes that use two of the original midpoints as a vertex. The you can draw a line between those midpoints, and you will end up with 4 triangles of equal size in the parallelogram, but not inside the octogon. You know the length of this line is 1/4 of the parallel dimension of the original figure as the intersection of two the midpoint-vertex lines is on one side, and the end of the figure on the other. You can use some similar triangle magic with the triangle that shares a vertex and whose base is midpoint (one of the same in the last step)- vertex (line parallel to the 1/4 dimension one) to find out the other dimesion of the triangle is 1/6 or the original. So you then take 1/4{area of reduced figue} - 4{small triangles}(1/2(1/4*1/6){area formula}) = 1/4-1/12 = 1/6

[@bryanw8310]

Consider the special case of a square, then divide the big square into four smaller squares. The centre of the octagon is now at one corner of the small square, draw the diagonal line from this corner. Now the small square should be divided into 6 triangles, it is easy to notice that all the four small triangles are of the same area and the two big triangles are of the same area and 4 times to each small triangle. Now, the ratio of the original question can be conjectured at (1 + 1) : (1 + 1 + 1 + 1 + 4 + 4) = 1 : 6.

[@someonespadre]

I did it empirically in Excel. A parallelogram skewed 20° right, 5 units vertically (along skewed line), 10 units left and right. Area by Coordinates. Divide octagon by parallelogram 0.166666666…. I calculated the coordinates in my Land Surveying software and CSVd them into excel. Software will return area but not precisely enough.

[@ivarangquist9184]

I solved it in a much simpler way: The nice thing about parallelogram is that you can morph them into a square by applying a linear transformation to the space. And since all linear transformations preserve the ratio of areas, you only need to find the solution of the special case when the parallelogram is a 1x1 square, and it will apply to all cases. And I solved that simply by drawing the shape on a piece of paper, dividing it into triangles and adding them all up. It took less than 20 seconds.

[@RiotGearEpsilon]

This is a great problem! One of the most engaging yet. :D

[@mstmar]

I found the solution in a different way. I started with the same parallelogram as at 3:50, only keeping lines that intersect the octagon. Then draw a rotated version, so you end up with another octagon slice in the top right corner of the parallelogram. Then draw both diagonals of the parallelogram. You end up with 12 smaller triangles, each having the same area. This can be proved with same base + height -> same area, and similar triangles (to each eighth of the octagon). Since the octagon is made up of 2 triangles, the ratio is 2:12 = 1:6

[@JBroadElf]

If it were a square the ratio would be (sqrt2)/8. But a square is a parallelogram so why isn't it 1/6?

[@sebastiansimon7557]

This is how I solved it: I immediately realized, that it applies to a square as well, so I used that. Then I divided the square into four quadrants, since the area ratios don’t change from doing that. So the reduced question was about the ratio between the area of a quarter octagon to a quarter square. I tried to divide the quarter octagon into two triangles. There are two ways of doing that. In any case, I tried to figure out, where the corner points are in the quadrant (let’s range it from 0 to 1 horizontally, and 0 to 1 vertically; I took the lower right quadrant). The first three points are obviously at (0, 0.5), (0, 1), (0.5, 1). For the last point I observed where the two lines crossed and how the two lines looked like. They looked like f(x) = 2x and g(x) = 0.5x + 0.5; so I set f(x) = g(x) to figure out where they intersected, and received x = 1/3; so the last point, due to symmetry, was (1/3, 2/3). From there, I could calculate the areas of the two triangles, add them together, and divide by the area of the quadrant.

[@danijelandroid]

I devided the octagon in one parrallelogram + 4 triangles with equal areas. The width of the original parrallelogram is a height is b. The base of all the triangles is 1/3a or 1/3b. The height is (1/2 - 1/3)/2 =1/12 b or a. The octagon parrallelogram is 1/3a *1/3b=1/9ab. Triangles are ether 1/2* 1/3a*1/12b=1/72ab or 1/2*1/3b*1/12a=1/72ba Area octagon: 1/9ab + 4*1/72ab = 1/6 ab

[@travisbaskerfield]

I used a program called GeoGebra which allows one to drag points. In every case, the area of the octogon was 1/6 so I knew the answer but never would have thought of this approach. I liked that even PT had to be given the technique to solve it.

[@taflo1981]

I got 1/6 by a slightly different argument. As in the video, consider a quadrant of the parallelogram, say the one on top-right. The triangle formed by the centre of the parallelogram, the topmost corner of the octagon, and the midpoint of the right-hand edge of the parallelogram comprises one quarter of the area of the quadrant. Analogously, the same holds for the triangle formed by the centre, the rightmost corner, and the midpoint of the the top edge. Thus, in total these triangles have half the area of the quadrant. Furthermore, the triangles cover the corresponding quadrant of the octagon twice. In addition, they also cover two small "blue" triangles. But each of these two small triangles has the same area as half of the quadrant of the octagon (same lengths of base lines and same heights). In other words, the total area of the two triangles we started with is three times the area of the quadrant of the octagon. Therefore, the quadrant of the octagon has 1/6 of the area of the quadrant of the parallelogram. This holds for all quadrants, proving the area of the octagon to be 1/6 of the area of the parallelogram.

[@mjones207]

I solved this using coordinate geometry, with vertices of the parallelogram (0, 0), (4a, 0), (4a + 4b, 4c), and (4b, 4c). The line y = 2c chops the octagon in half, forming two congruent compound shapes that each can be separated into a trapezoid and a triangle with horizontal lines. I also gave this problem to two of my more gifted students (an 8th grader and a 10th grader) and am curious to find what they do with it this week.

[@willbishop1355]

Start with a square of area 1. Diagonals from each set of two opposite midpoints form a diamond. The intersection of those diamonds is the octagon. The area of each diamond is 1/4. The area of the union of the two diamonds is 1/3. Therefore the area of the octagon must be 2*1/4 - 1/3 = 1/6. That ratio applies to any parallelogram because linear transformations scale all areas the same.

[@QuakeJoz]

Here's my solution: 1) Start with the parallelogram. Chop off the top and bottom around the octagon (1/2 area left) 2) chop off the corners of the remaining parallelogram, stopping at the centre of each line. This leaves a diamond in the middle (1/4 area left) 3) Draw in the octagon, and lines from the centre to each point of the octagon/diamond. This makes 12 triangles, 4 of which we want to remove. They are all the same area (see below), so we have 8/12 * 1/4 = 1/6 left, which is the answer. Each of the formed trianges have equal area because they either share a base and height directly, or can easily be split into two sets of triangles which share a base and height.

[@Acid31337]

1. Cuz result same for every parallelogram(ortho-projection save squares blabla). We calculate it for square. 2. cuz it's now symmetric(8-axis!1) just throw away 7/8 of this. 3. we got small yellow triangle in transparent one. look at horisontal/vertical side of small triangle it has 1/4 of square side length, because point is intersection of 1x0.5 rect diagonals. 4. small yellow triangle has same square as another one, with same height and also 1/4 bottom length. 5. small yellow triangle has same shape as third big one, and 1/2 one linear size and 1:4 of square. so parts are 1+1+4 by square. in result 1/(1+1+4)=1/6.

[@aatmansupkar8747]

I don't care about the main question, but it surprises me that there are a whole lot more properties this figure shows.

[@hj8607]

Geometric symmetry Resulting by dividing parallelogram in quarters gives four parts of octagon of equal area. Diagonally dividing the reduced parallelogram , that this 1/4 of octagon is in, reveals two small triangles in this area in question . The base of each is only half of a side the reduced parallelogram so if either were full height it would be 1/4 the area of the reduced parallelogram and together they would = 1/2 the area. But geometry shows the height of each to be only 1/3 (combined average ) of the width or height (left to right and bottom to top) so the area of each triangle is 1/12 and combined equal 1/6 of the area. The ratio is symmetrical so octagon is also 1/6.

[@brucehakami4489]

Ah Mr Talwalkar, this is the first time I am beaten by one of your problems, and I know why. It is because, in the words of Edward De Bono, this problem requires 'lateral thinking', which is not one of my strengths. Thank you for presenting the problem, which I enjoyed, despite failing it.

[@soup1322]

Assuming it works for all parallelograms, you can solve by using a square and solving by the integral. Something like: 2x[Integral from 1/4 to 1/3: 2x - (-2x+1) dx + Integral from 1/3 to 1/2: x/2 + 1/2 - (-x/2 +1/2) dx] This integral x 2 gives an area of 1/6 for the octagon with the square having an area of 1.

[@stevethecatcouch6532]

No details are given for the parallelogram so I will assume it is a square with sides of 4 units. The corners are W, X, Y and Z starting at the upper left and moving clockwise. The midpoint of WX is M. The midpoint of YZ is N. Call the center O. Label the vertices of the octagon A - H starting with the topmost one and moving clockwise. The distance AO is 1. By symmetry, angle AOB = 45 degrees. Draw segment HB. It is perpendicular to segment AO at point P. So angle PBO is 45 degrees. Meaning PB = OP.. Triangle MPB is similar to triangle MNY. MN = 4. NY = 2. PB/PM = NY/MN = 1/2. Therefore PO/PM = 1/2. PM = 2*PO. PO + PM = OM = 2. 3*PO = 2. So PB = PO = 2/3. The area of triangle OAB = (OA * PB) /2 = (1 * 2/3)/2 = 1/3. So the area of the octagon is 8/3. The area of the parallelogram is 16. The ratio is (8/3) / 16 = 1/6

[@seanocansey2956]

I like how you were honest that you couldn't figure out the problem 🙂

[@sporadicdrive5884]

Mhmm these geometry problems mainly appear in olympiads so they're hard for the average students

[@QuippersUnited]

I used the problem in square form. I then figured that cutting the diagram in half, the vertex of the octagon is .25 of the figure from each edge. That means the radius of the octagon is .25. Using trig, you can get an answer PRETTY close to the 1/6 mark.

[@bpark10001]

Distorted it to a square. Then wrote equations for 2 of the lines, to solve for the intersection. That gives dimensions of triangle which is 1/8th of octagon. As base & height known in units of square's side, solve for this area x 8 and divide by square's area.