Can you solve the missing square puzzle?

Рет қаралды 138,190

Ай бұрын

The infinite chocolate trick is one of my favorite illusions. How is this even possible to re-arrange areas and have 1 extra square? To fully To understand what is going on, it is useful to solve a homework question from Reddit AskMath.
Missing square illusion
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Пікірлер: 251

@trombonedavid1 Ай бұрын

I love the hint at 1:02 “we have this diagram…” The prompt never refers to the large shape as a triangle, due to the fact that it’s a sneaky quadrilateral

@jeff-jo6fs Ай бұрын

You are right, that is sneaky quadrilateral. Even for a quadrilateral, which are already pretty sneaky

@verkuilb Ай бұрын

Or, maybe it IS a triangle-and the incorrect assumption isn’t that the hypotenuse is straight, but that the corner of the unshaded triangle lies all by the large triangle’s hypotenuse. Neither is actually stated.

@jeff-jo6fs Ай бұрын

@@verkuilb its a lesson of, if you understand the parameters of the game, you can claim concise victories by manipulating the edges of what is barely perceivable.

@chrisg3030 Ай бұрын

@@jeff-jo6fs Yes, in that sense it's like a stage conjuring trick, except that what's barely perceivable is just spatially rather than also temporally tiny.

@hocules Ай бұрын

If a quad 4 sides must be specified. Else inderterministic. And intentionally make it lok like a triangle and not specified the 4 sides make this a inderterministic tricky riddle.

@MarsJenkar Ай бұрын

Where does the missing chocolate go? That's right, it goes into the square hole.

@emmettdja Ай бұрын

this is gold

@GamingDimiGD Ай бұрын

LOL

@JL-sm1gm Ай бұрын

Lmao

@LitoMike Ай бұрын

*screams of pain*

@stuchly1 25 күн бұрын

And this one would be a perfect fit too lmao 🤣

@slmnchk Ай бұрын

I would love to see a couple more steps of this process, so that the loss is clearly visible and grows with every iteration

@srkingdavy Ай бұрын

it's not really repeatable, the two triangles are either in one configuration or the other

@arturovasquez5334 24 күн бұрын

⁠is repeatable if you reconfigure the colors and put a new color at the right bottom green area forming a new L shape

@Jonesy1701 7 күн бұрын

@@srkingdavy No it absolutely is repeatable. You don't use the same blocks, you re-shade them and repeat.

@STEAMerBear Ай бұрын

This PERFECTLY illustrates the vulnerability of visual proofs. A numerical method, weighing the chocolate, will catch the theft. Comparing the result to the original gives an imperfect fit, but it’s hard to spot it. Visual proofs are pretty and often convincing, but they are not rigorous or precise.😊

@CorenusYT Ай бұрын

To go further into the subject, this case illustrate the absolute necessity to determine the accuracy/precision of the measure. From a far perspective, the accuracy would hide the actual bump in both quadrilaterals. From a close enough perspective, the accuracy of the measure will be noticably below the size of the bumps, making it quite clear in a visual fashion.

@xpusostomos Ай бұрын

If you're so damned smart, why couldn't you figure out how we can get the infinite chocolate? That would be more useful than debunking a perfectly good miracle.

@p111SC Ай бұрын

Something something Banach Tarski

@yvessioui2716 Ай бұрын

Because 'stating infinite chocolate' is a magician trick used to carry your mind away from a sound analysis, very helpful in designing a way to deceive people.

@user-fv5pz9ov7y Ай бұрын

Don't be toxic. It's bad.

@penguincute3564 Ай бұрын

It’s because infinite chocolate is not a thing (nothing in the world is infinite)

@Arnikaaa 25 күн бұрын

@@penguincute3564except for infinity

@StephenMarkTurner Ай бұрын

My friend had a wood version of this back in the early 70s. It was a baffler back then, although I did learn the trick a few years later.

@WhiteGandalfs Ай бұрын

Simple: 2/5 !== 5/13 !== 3/8, but if you draw the lines with a just so little distortion, naive bystanders will not notice the difference.

@deuce454 Ай бұрын

the triangles aren't like-sided .. so the large "triangle" is actually a 4 sided quadrigon with either a convex or a concave angle on what appears to be the long side of the "triangle" that area accounts for then missing area

@Jonesy1701 7 күн бұрын

Yep... we watched the video too lol.

@paulromsky9527 Ай бұрын

Great illusion, but in mechanical drawings, if you have what appears to be large right triangle but actually has the "bow in" and you don't include "clear" dimesions, it can lead to interpetation errors (like what we see here). That is why if we have a line that looks straight but has a kink in it, we would show the angle differences and NOT some other odd linear dimension to be clear the line has a subtle kink. True, showing the drawing with linear dimensions only is "correct" as well, but there should be a detail at the kink that shows that there is a kink there. Inputting the deminsions into a CAD or CNC machine will yield the correct geometry but back in the days before that we would never describe a shape like that with just linear dimensions - as doing so indicates that all lines are linear. For example, if a machinest starts to frabicate the part, errors would show show up on the final part. I learned this is drafting class in high school - proper dimensions to prevent errors is most important... but this is a good trick! A nice way to win a drink at the bar if you could cut the chocolate bar ahead of time because cutting it with a straight edge in front of the "mark" would give it away.

@rickoffee 5 күн бұрын

The assumption that the big diagram including the "missing chocolate" square is a triangle is wrong: it is in reality a quadrilateral with two sides almost parallel making a seemingly straight line. It is a bit unfair because the human eye cannot detect/measure that with such precision.

@trueriver1950 Ай бұрын

The trick is in making it look like the diagonal line is straight: in fact it's got a kink in it where the triangles meet. The slope of the hypotenuse of the small triangle is rise/run = 2/5 = 0.400 The slope of the hypotenuse of the large triangle is 3/8 = 0.375 The difference in slope is 0.025, or one fortieth. In the configuration with the white square there is a "hump" in the diagonal, and in the other one it's a "dip". As up the difference in area between the hump and the dip, and i bet it comes to exactly one: and that's the answer: the "missing" square is smeared out along the diagonal.

@tedspens 26 күн бұрын

So basically, if it was all straight lines, the height at the 8cm mark would be less than 2cm. I always wondered about that puzzle. Thanks!

@MichaelPuterbaugh Ай бұрын

and, as Pannenkoek explained, the slightly different angles of the "hypotenuse" allow the out-of-bounds area underneath the triangle to poke through...

@shininio Ай бұрын

best explanation to this popular trick. kudos Presh!

@lethalty6055 Ай бұрын

There was a Ted-ED riddle about two different boards of a 64 and 65, but the total multiplication of all the involved pieces are 64, but rearranged in a way so it fits them both, but a unit of 1 was the missing slope. I think it was an Alice riddle. EDIT: Just rewatched the riddle.

@oldtimefarmboy617 Ай бұрын

So the key to the trick is for the presenter to lie about the details.

@trueriver1950 Ай бұрын

No lies told: Presh never said the overall shape was a triangle...

@smeissner328 Ай бұрын

@@trueriver1950 Technically true, but the intent was still to deceive the viewer into believing that the large shapes are both triangles. Edit: Not the intent of Presh, but the intent of most people who present this problem.

@eventhisidistaken Ай бұрын

That's the key to *all* tricks.

@crinolynneendymion8755 Ай бұрын

@@smeissner328 No, the intent was to show the effect of small variations in data leading to very significant consequences. There are very important lessons to be learned from this example, particularly for Engineers.

@smeissner328 Ай бұрын

@@crinolynneendymion8755 That's the intent of this video. I was talking about the intent of people who present a problem like this and pretend that it's unsolvable or a true duplication of matter.

@1a1u0g9t4s2u Ай бұрын

At first I thought Okay, we have seen this before. But something told me to give this a chance. Glad I did. The two methods of solving reinforced what was already known and through a different viewpoint explained why this illusion works. Thanks for sharing.

@ffggddss 13 күн бұрын

This is an old one. It appeared in Martin Gardner's _Mathematical Games_ column in Scientific American, some time around 1960. Not sure, but I believe it was attributed to one of two famous puzzlists of about a century-plus ago - American Sam Loyd or British Henry Ernest Dudeny. If you compute the slopes of the hypotenuses of the two triangular pieces, based on their "pivot" point being initially at (8,2), and later at (5,3), you'll find that they are different, so that the whole "right triangle" is really a quadrilateral, which is a tiny bit convex initially, and a tiny bit concave after removal of the little square & re-assembly. Thus, the area really is smaller after than before taking the piece out of it. Fred

@hyperboloidofonesheet1036 Ай бұрын

And if you take the limit you end up with the Banach-Tarski paradox. :P

@MarieAnne. 18 күн бұрын

The original chocolate triangle has height = 5 and base = 13 The orange triangle has height = 3 and base = 8 The blue triangle has height = 2 and base = 5 None of these triangles are similar, so the hypotenuse of the orange and blue triangles cannot lie along the hypotenuse of the original chocolate triangle. In fact, in the first arrangement, the hypotenuse of the orange and blue triangles lie slightly above the hypotenuse of the actual chocolate triangle, but in the rearrangement, they lie slightly below. This slight difference will make up the 1 square unit of the piece that was eaten. Perceived area of chocolate triangle = 1/2 × 5 × 13 = 32.5 Area of original shape (before piece of chocolate is taken away) = (1/2 × 3 × 8) + (1/2 × 2 × 5) + (8 × 2) = 12 + 5 + 16 = 33 Area of new shape (after pieces are rearranged) = (1/2 × 2 × 5) + (1/2 × 3 × 8) + (5 × 3) = 5 + 12 + 15 = 32

@Jonesy1701 7 күн бұрын

Yep... we watched the video too lol.

@jimlocke9320 Ай бұрын

At 7:30, red triangle has area (1/2)(8)(3) = 12 and blue triangle (1/2)(5)(2) = 5. In the top figure, there were a total of 16 green and yellow squares before a square was removed, so combined green and yellow area = 16 and total area = 33. In the bottom figure, there are a total of 15 green and yellow squares, total area = 32. So, the area has correctly been reduced by 1 after 1 unit of area was removed. Another method: in both figures, construct a line segment from the topmost vertex to the rightmost vertex. Its length is, by Pythagoras, √(5² + 13²) = √(25 + 169) = √(194). Now, compute the hypotenuse lengths for both the red and blue triangles. The red triangle's hypotenuse has length √(3² + 8²) = √(9 + 64) = √(73) and the blue triangle's hypotenuse has length √(2² + 5²) = √(4 + 25) = √(29). Clearly, these two hypotenuses do not add up to √(194). The three line segments do form a "sliver" triangle and Heron's formula, A = √(s(s - a)(s - b)(s - c)), may be used to compute its area. The side lengths are a, b and c, Let a = √(194), b = √(73) and c = √(29). The semi-perimeter s = (a + b + c)/2 = (√(194) + √(73) + √(29))/2. Using a calculator, I get approximately 0.5 for A. The vertical distance to the large triangle's hypotenuse 5 units from the right most vertex is (5/13)5 = 25/13, This is less than 2, so, in the top figure, the area of the sliver triangle must be added to the area of the large triangle to get the total area before the piece of chocolate was removed. So A = (1/2)(13)(5) + 0.5 = 32.5 + 0.5 = 33. In the bottom figure, the vertical distance to the large triangle's hypotenuse 8 units from the right most vertex is (8/13)5 = 40/13, which is more than 3. So, the area of the sliver triangle must be deducted and A = (1/2)(13)(5) - 0,5 = 32.5 - 0.5 = 32, matching the above calculations.

@arthurvyater656 24 күн бұрын

Less math heavy way to see it: Small triangle goes 5 across and 2 up. On the big triangle, when you go 5 across, you can see that it doesn't quite reach 2 units looking up.

@yassermachkour4291 Ай бұрын

Using similar triangles, the problem is right if you change 2cm to 1.923 cm

@erikaz1590 Ай бұрын

I've never been so early that I could only finish one piece of chocolate XD

@JavierSalcedoC Ай бұрын

Selling a choco bar with the marks to split it in this way would be such a powerful marketing move

@shawnmark3492 Ай бұрын

Take any 3 consecutive numbers in the Fibonacci Sequence (in this case: 5, 8, 13) and if you squared the middle number (8^2=64) then multiply the other two together (5*13=65). a*c=b^2(+/-)1 Meaning you can increase or decrease the scale of this illusion. From TED-ED's Can you solve Alice's Riddle?

@yvessioui2716 Ай бұрын

One helpful way for me to be sure of the different slopes without trig involvement: the blue hypotenuse slope is 2 over 5 (0,40) and the red one is 3 over 8 (0,375).

@keith6706 Ай бұрын

Yes, the whole arctan thing was unneeded. If the slope is different, the angle has to be different.

@verkuilb Ай бұрын

0:22 “…slide the yellow piece like a game of Tetris…” In what version of Tetris can you move your piece UP before going left or right? 😂

@reminderIknows Ай бұрын

sqrt(-1) tetris fr

@boggisthecat 14 күн бұрын

First shape has a larger area because of the two triangles being unequal in ratio. Transposing the triangles creates a shape with a smaller area - by one unit square. It may look like each shape encompasses the same triangular area, but that’s not so.

@rmcgraw7943 Ай бұрын

One would, logically, assume that the person asking this question has a ruler and the ability to draw a straight line! LMAO.

@3Cr15w311 Ай бұрын

I saw another version of this in 1987 where the width was 12 and where pieces were rearranged to fill a missing square or create a missing square but by growing or shrinking the height of the whole figure by 1 / 12.

@guilhermeottoni1367 Ай бұрын

In fact, the "hypotenuse" of the triangle is not a straight line.

@olli1068 Ай бұрын

... which he first said it was, but later said it's not. I tell a lie! I tell the truth! What I said changed! That's Illusion!! 😂

@smylesg Ай бұрын

I wish I had as much chocolate as Presh shows this "problem."

@trueriver1950 Ай бұрын

My diabetes consultant is happy that I don't😂

@jeremiahlyleseditor437 Ай бұрын

That finally answers that question.

@davidmehling4310 Ай бұрын

About fifty years ago, I had a child's magic show kit which had a very similar illusion. There was a plastic triangle frame with four triangular and two L ish shaped pieces with grid lines on them. A small two square piece would or would not fit depending on how you arranged the other pieces. Same illusion

@prufrock1977 Ай бұрын

I knew it! Thank you for proving it.

@kahvipaputyyppi 12 күн бұрын

I solved this once on a paper when someone told me about this problem, it was fun and I was pretty excited about it. When I showed the solution to that person they didn't care much. Geometry was one of my top favourite subjects in math. 🤩

@fabioberetz Ай бұрын

It works with any pair of numbers such that height and base are Nth and (N+2)th number from series of Fibonacci

@Ninja20704 Ай бұрын

Yes indeed. And the reason is because the ratio of consecutive terms in any fibonacci style sequence approaches the golden ratio, phi. So the ratio of Nth and (N+2)th will be very similar for different N’s, but not equal. (the slopes are very close to 1/phi^2 in fact) This is what makes the slopes so similar that they are hard to distinguish just by looking.

@williamperez-hernandez3968 Ай бұрын

Taking the Fibonacci numbers as F[5] = 5, F[6] = 8, F[7] = 13, the identity (F[n])(F[n+2]) = (F[n+1])^2 - (-1)^n, gives (5)(13) = 64+1. Thus taking away an area of 1 from the original shape creates the illusion upon rearranging the remaining area. But if we begin with lengths 8 and 21, then (8)(21) = 169 - 1. Then to create the illusion, an area of 1 must be added to the original shape. So if n=odd, we remove an area of 1, but for n=even, we must put in an area of 1.

@dirkhaar2243 5 күн бұрын

Strahlensatz sagt mir: "5:13 2:5" - Das große Dreieck ist ein Viereck.

@user-fp9kz6xv6l Ай бұрын

Cool video! I really enjoyed watching it 😊❤

@michaelbarnard8529 11 күн бұрын

The two triangles have different slopes, and thus give different areas when rearranged.

@rafaelallenblock 21 күн бұрын

I solved it a third way: I calculated the area of the far right triangle, then mentally split the left one into two right triangles, then added: 5+12+8=25.

@Alex-gi7sm 6 күн бұрын

You can easily see by the different pitch of the blue (2/5) and red (3/8) triangles that their hypotenuses cannot be parallel.

@theplasmawolf Ай бұрын

This may have been the first mathematicalprovlems I ever saw. Must have been no later than 1998. My dad showed it, and I didn't understand as I was too young. Pretty fun puzzle and good to see it still shows up 26 years later

@engineboy_1449 2 күн бұрын

0:06 take this square of charger and eat it......NOM NOM XD

@lorentzinvariant7348 26 күн бұрын

Working this on a slide rule, the problem is instantly made clear. Also, made me want chocolate.

@THall-vi8cp Ай бұрын

Before 7:59 I could already see the bow in the "hypotenuse" of the lower figure. In the upper figure it wasn't so noticeable. Cool problem. It highlights the tendency to make assumptions rather than observations.

@Fernandez218 Ай бұрын

MYD where do you take suggestions for problems to solve? I have a good one.

@matthewgraham2619 Ай бұрын

I remember seeing a problem in a magazine and thinking my high school geometry made easy work. The issue was, the diagram wasn't lined up with the information given. If you solved the triangle as given, it came out to a 180 degree straight line. Might have been an april fools joke.

@ShawnF6FHellcat 7 күн бұрын

This would be a fun trick to pull on some young kids; it would absolutely blow their minds!

@eventhisidistaken Ай бұрын

If you get different answers by different valid methods, then all you know is that the information is inconsistent. You don't know *what* is inconsistent. The problem setup only tells us that the 'figure' on the right is a triangle. It didn't say the figure on left is a triangle, nor that any of the other lines (except the triangle on the right) are straight. To set it up correctly, the two blue areas and the nonshaded area need to be stated to all be triangles.

@raffimolero64 12 күн бұрын

floor overlap... floor gap... this is a certified Cause #4 situation (Context: Pannenkoek2012's video on Invisible walls in Mario 64)

@timwestlund3072 Ай бұрын

We should cut the chocolate bar into a finite number of non-measurable pieces and reassemble them into two copies of the original bar.

@InformationEngineer59 Күн бұрын

First puzzle: The big triangle has a slope of 3/8, .375, the slope of the second is 2/5 .4.

@rogerkearns8094 Ай бұрын

I suppose the god of the gaps took it.

@abdulmateen1250 Ай бұрын

Amazing! I love Mathematics, especially on your channel.❤

@egillandersson1780 Ай бұрын

I read this paradox first in a Sam Loyd's book, many years ago. I don't know if he created it or just collected from a previous author.

@claudiamanta1943 Ай бұрын

0:48 The total coloured area in both cases does not represent half of the chocolate tablet (true half being 32.5 if a small square side is 1). In the beginning the coloured area is 33 = true half the tablet + 0.5. In the second instance the coloured area is 32 = true half the tablet area - 0.5. There, two wrongs make a right sometimes 😄 (0.5+0.5=1). I don’t know why it made me think of bargaining (ask for more and give the impression you lose in order to get the price you want). You just almost imperceptibly to the eye reconfigured your piece of chocolate- you always had the same area/ quantity but you presented it in two different ways. The smaller the chocolate tablet the more difficult it would be to play this trick, I guess. I suspect that in both cases you were messing with the coloured areas in the squares through which the false diagonal passed (to my eye is more visible in the second case- it looks like a curve). The (x6, y4) was a giveaway if you compare the two. So, you had more than half the chocolate tablet to start with, ate a square, then melted it again and reshaped it cutting the diagonal with a shaky hand (probably you felt guilty about it 😁). Please be kind if you comment. I’m not very bright and I hated maths with a vengeance, but this was good fun. And I love chocolate 🍫😋 PS after watching to the end. OMG, I was right 😃 PS2- You just made me think you had the same amount of chocolate. QED I’m not THAT bright after all that’s why I insist on not being lied to 😄 This was truly delightful, thanks again ☺️

@Vienticus Ай бұрын

I deny these results so that I might delude myself into thinking I might create infinite chocolate.

@stefanschneider3681 Ай бұрын

GREAT! Thx

@VineetJangra-wu8ou Ай бұрын

It's really ultra amazing.

@henp99 Ай бұрын

❤ Thank you.

@lapisanyta Ай бұрын

0:34 the chocolate was refilled right under the diagonal line. That's where the 1 square had gone.

@val-e6968 Ай бұрын

The funny thing is that you actually are able to make area disappear using translations and rotations, like Banach and Tarski have shown.

@Michael-sb8jf Ай бұрын

after you see the difference in slope you can't unsee it and that was 20 years ago when my calc teacher pointer it out

@RedRad1990 13 күн бұрын

Dude, I will always be fooled by visual tricks and magicians. When he did the zoom in on the "hypotenuse" at 6:01 and said "it's not straight", it still looks straight to me 🤦

@jaybling6687 Ай бұрын

I remember this. This big shape is ultimately not a triangle. Because the blue triangle is 2:5 and the red triangle is 3:8. 2:5 =/= 3:8. That's what makes it an illusion.

@JoshRendall Ай бұрын

I’ve watched something similar to this! Can you solve the Alice in Wonderland riddle!

@zecuse Ай бұрын

Rather than taking 3 to be a correct value in the problem, if the shapes are in fact triangles sharing a coincident bottom edge, hypotenuse, and bottom right angle, you can find the missing length that is the height of the smaller shaded triangle by similar triangles. You'll find it's height isn't 2 (it'll be 25/13) which means either the 3 in the problem is wrong (unlikely) or the triangles aren't actually similar and therefore expose this "kink".

@ofofofff Ай бұрын

.. anyway you are good, ans yr channel its GREAT

@MegaUpstairs Ай бұрын

So the main trick is that even the original long side is not a straight line, but contains 2 segments in an angle. One can validate that quickly by dividing the 2 legs of the 2 smaller triangles. The ratio is different.

@crushermach3263 Ай бұрын

The way I figured this out is that it's a 13 by 5 right triangle. There are no common divisors for 13 and 5 (never mind that they're both primes anyway) so the length of the hypotenuse should never intersect with a corner. Yet it clearly does, so the only solution must be that the angle subtly changes to accommodate making it not a true triangle.

@trueriver1950 Ай бұрын

Interestingly, this is obvious if you look at the chocolate bar in Presh's graphic: the line is drawn so it intersects at 8,2 but then clearly doesn't intersect at 3,4 where you'd expect it to on the basis of the trick question

@marwynnsworld9390 Ай бұрын

0:10 NOM NOM

@Tiqerboy Ай бұрын

Simple. The diagonal is NOT a straight line before and after. I saw that immediately by observing the ratios of the sides of red and blue triangles with respect to the large 'triangle'. If it was a straight line, you'd expect similar triangles. They aren't. If you methodically calculate the areas of the colored components at the start, you find you do NOT have half the chocolate bar's area (8 + 8 + 12 + 5 = 33 vs the actual half which is 32.5). The new area of the colored components is 32. In both cases it looks close enough to 32.5, but it's not. The drawing is an illusion. Before the unit square of chocolate is removed, the 'straight' line is slightly convex, after it is removed and the pieces rearranged, it is slightly concave. ** EDIT ** After watching Presh's video, I see that's exactly it, though I'm surprised he didn't use the argument of similar triangles as I did. BUT it was nicely explained.

@johngonon1507 16 күн бұрын

Both triangles have slightly different angles, which is not very visible just looking at it.

@donwald3436 12 күн бұрын

"we can do it all over again" how many times before the bump becomes obvious?

@gordonweir5474 Ай бұрын

The other interesting observation is how big a role the Fibonacci numbers play in the measurements of the sides of the two figures.

@hhgygy Ай бұрын

I knew this problem and it is easier to see that the chocolate table is not right: the big triangle and the smaller ones should be similar but it would imply that 5/13 is equal to 3/8 as well as to 2/5 as manipulated with the chocolate pieces. You do not even need to calculate this.

@N.i.c.k.H Ай бұрын

I think the thing that sells it is the apex of the triangles being at/near the corner of the square. If they weren't people would probably be more likely to see the trick.

@keith6706 Ай бұрын

You need a large enough grid to make it work. The smaller the number of cells, the more obvious the kink in the "hypotenuse" has to be in order to have the round numbers in the problem.

@googa8 12 күн бұрын

I figured out the slope wasn't straight before any calculation by looking to my cell phone screen from a side perspective.

@justsaadunoyeah1234 Ай бұрын

"Nom, Nom." - Mind your decisions, 2024.

@opendstudio7141 Ай бұрын

And now we know, GEOMETRY is the reason chocolate bars keep getting smaller and costing more. 😜

@toastyburger Ай бұрын

It's pretty clear the slope on the right angles is not the same. Just count the squares.

@biabtwo 11 күн бұрын

Assumptions are what humans do best

@JablkacMatus Ай бұрын

So this is a very big mystery. Almost like with English pronunciation. Tak toto je veľmi veľká záhada. Skoro ako s výslovnosťou angličtiny. 😄

@T0MT0Mmmmy Ай бұрын

Damm, for one second I thought "how do you know that the hight of the overall triangle is 2cm at 5cm" (which would have lead me to "this is not a true triangle, therefore the first calculation must be wrong"). But I then assumed "oh, it is probably a true triangle, because he would not tell us so if it isn't ...".

@SwanandChawathe Ай бұрын

Pls post videos more frequently

@SlamSector 12 күн бұрын

You can see it if you draw it with a fine pencil.

@dan2800 Ай бұрын

I need a video where it is shown removing a square one by one till there's nothing left or it's impossible to make a triangle from remaining square's

@nichodemus10 Ай бұрын

It seems you have the technology, please make a short of 5-10 iterations so we can watch the deformation of the "triangle". The initial assumption is that the figure would get more concave, but because you arent changing that part of the two triangles it shouldnt do that, and when you move them for the second time you would have a convex shape again. I just dont get what could happen next.

@HauntedKnight-cj8kv Ай бұрын

It went to Hell.

@ItsWhatever2517 Ай бұрын

The red herring of this problem is the “3 cm” dimension introduced at 1:19.

@FellowInterneter Ай бұрын

If I have understood correctly, It is actually a hint to a correct answer. It is used in the formula for the right answer but specifically ignored in the wrong answer.

@ItsWhatever2517 Ай бұрын

Actually, neither answer is correct. I used AutoCAD for this and the “3 cm” dimension used in BOTH methods is supposed to be 3.077 cm, yielding an area of 24.8 sq. cm. As I said earlier, the “3 cm” dimension is the red herring of this riddle.

@selfdeveloment4084 Ай бұрын

This was in my school for some time.

@janami-dharmam Ай бұрын

I too remember this puzzle from our school book; there were other similar cases under "the importance of drawing accurately

@Tahgtahv Ай бұрын

@@janami-dharmam Drawing accurately has nothing to do with it, and it's bizarre if that was the point your school book was trying to make from this. The diagrams were drawn accurately in the video, but it's still close enough to be hard to tell (that's the whole point of the puzzle really.) You should never rely on a diagram, but use any stated information about the shapes, (lengths, angles, hash/arc marks) instead.