Can you solve the missing square puzzle?

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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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30 сен 2024

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Комментарии : 273

@MarsJenkar 5 месяцев назад

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

@emmettdja 5 месяцев назад

this is gold

@GamingDimiGD 5 месяцев назад

LOL

@JL-sm1gm 5 месяцев назад

Lmao

@LitoMike 5 месяцев назад

*screams of pain*

@stuchly1 4 месяца назад

And this one would be a perfect fit too lmao 🤣

@xpusostomos 5 месяцев назад

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 5 месяцев назад

Something something Banach Tarski

@yvessioui2716 5 месяцев назад

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.

@d1scocubes 5 месяцев назад

Don't be toxic. It's bad.

@penguincute3564 5 месяцев назад

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

@Arnikaaa 4 месяца назад

@@penguincute3564except for infinity

@trombonedavid1 5 месяцев назад

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 5 месяцев назад

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

@verkuilb 5 месяцев назад

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 5 месяцев назад

@@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 5 месяцев назад

@@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 5 месяцев назад

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.

@slmnchk 5 месяцев назад

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 5 месяцев назад

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

@arturovasquez5334 4 месяца назад

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

@Jonesy1701 4 месяца назад

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

@StephenMarkTurner 5 месяцев назад

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.

@STEAMerBear 5 месяцев назад

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 5 месяцев назад

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.

@WhiteGandalfs 5 месяцев назад

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.

@oldtimefarmboy617 5 месяцев назад

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

@trueriver1950 5 месяцев назад

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

@smeissner328 5 месяцев назад

@@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 5 месяцев назад

That's the key to *all* tricks.

@crinolynneendymion8755 5 месяцев назад

@@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 5 месяцев назад

@@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.

@hyperboloidofonesheet1036 5 месяцев назад

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

@deuce454 5 месяцев назад

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 4 месяца назад

Yep... we watched the video too lol.

@tedspens 4 месяца назад

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!

@SunriseLAW 3 месяца назад

Similar to how you make a "Football Cake", which can be made from any cake that is round. Cut a large piece out of the center equal to approximately 20% of the cake such that you have two equal-sized oval shapes left over. Push the remaining end pieces together and frost over the cut. Eat the remaining 20% of the cake.

@arthurvyater656 4 месяца назад

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.

@MichaelPuterbaugh 5 месяцев назад

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

@guilhermeottoni1367 5 месяцев назад

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

@olli1068 5 месяцев назад

... 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!! 😂

@verkuilb 5 месяцев назад

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 5 месяцев назад

sqrt(-1) tetris fr

@shininio 5 месяцев назад

best explanation to this popular trick. kudos Presh!

@paulromsky9527 5 месяцев назад

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 4 месяца назад

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.

@SoulLeister 3 месяца назад

The old adage applies "figures don't lie but but liars figure". Around 4:45 "assumption" is introduced? Inescapable facts provided: ...5x13 rectangle equals 65 divided by 2 equals the large right triangle we are given (32.5) 8 of which is not shaded (2x8 right triangle) 32.5 - 8 = 25.5 (shaded and further divided into three "supposedly" right triangles) ...BUT THE FOLLOWING "DISTRACTORS" result in: a shaded triangle 3x8 (12) and a shaded triangle 2x5 triangle (5) which trade places in relation to a 2x8 rectangle (16) equaling 33 not the expected 32.5!!! MAGICALLY (?) CHANGING THE 2X8 RECTANGLE INTO A 3X5 RECTANGLE, 16 VS 15 (where da one go? the triangles angles didn't change) ...SOME MUMBLE JUMBLE ABOUT A MINUTE CURVED LINE IS ALLEGED, AS A VIOLA REASON FOR A DISCREPANCY OF "ONE" WHEN TRIANGLES ARE MOVED? AND ANGLES BEING INCONSISTENT BLAH BLAH BLAH FACTS (rounded, even Einstein could only do six decimal places in head (Oppenheimer quipped): a) 5/13= the tangent of angle in question and converts to "21.03" *rounded 21 b) 3/8= the tangent converts to "20.56" c) 2/5= the tangent converts to "21.8" (why doesn't a and c match?, it is the same drawing, hmmm) ...the latter two (raw) avg ="21.18"; but as a ratio(s) over distance "20.92" *rounded 21 who knew .1 of a degree over 13cm (not even visually perceptible) would create a one square centimeter discrepancy, its magical!?!? we should apply this to surveying land and literally carve out hundreds of square miles for the homeless, creating something out of nothing, problem solved through simple math.

@erikaz1590 5 месяцев назад

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

@schwarzerritter5724 Месяц назад

The long side of one triangle is 5, the other 8. The smallest common divisor is 40. If you enlarge the triangles, so the long side of both is 40, the short side of the red triangle will be 15, the blue triangle 16. If the overall shape was a right triangle, both would be the same.

@lethalty6055 5 месяцев назад

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.

@Becky_Cooling 23 дня назад

that's exactly what I thought of.

@1a1u0g9t4s2u 5 месяцев назад

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.

@rogerkearns8094 5 месяцев назад

I suppose the god of the gaps took it.

@yassermachkour4291 5 месяцев назад

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

@rmcgraw7943 5 месяцев назад

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

@JablkacMatus 5 месяцев назад

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

@MarieAnne. 4 месяца назад

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 4 месяца назад

Yep... we watched the video too lol.

@mitchellclark4377 5 месяцев назад

That white outline you add at 0:31 is doing a lot of heavy lifting to disguise the switch back.

@ffggddss 4 месяца назад

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

@smylesg 5 месяцев назад

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

@trueriver1950 5 месяцев назад

My diabetes consultant is happy that I don't😂

@Vienticus 5 месяцев назад

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

@fabioberetz 5 месяцев назад

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

@Ninja20704 5 месяцев назад

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 5 месяцев назад

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.

@strictlyeducationalmagick 5 месяцев назад

You don't get it. Better look at the chocolate again

@JavierSalcedoC 5 месяцев назад

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

@verkuilb 5 месяцев назад

You claim that the calculation of the shaded area as 24.5 is “wrong”. I don’t think that’s the case. Both answers require you to make assumptions which are not stated in the original problem. For the answer Presh claims is “right” to be true, you need to assume that the unshaded triangle actually intersects what seems to be the hypotenuse of the larger “triangle”. But that is not stated! If you make that assumption, Presh’s answer is correct. But there’s another, just as legitimate approach. Assume that the larger triangle is indeed a triangle-but instead, don’t assume that the upper right tip of the unshaded triangle actually intersects the hypotenuse! If you do that, the first method of calculation is indeed correct!! Neither is “right” or “wrong”-the two methods just make different assumptions, both of which are equally “right” and “wrong”.

@dashgandhi5829 5 месяцев назад

There is no question of any assumption. Both answers are wrong. Because both assume that inner smaller triangle hypotenuse would be 2. The actual value would be 25/13. Hence the right answer is 20 + 25/13 * 5/2 ie 24.86.

@verkuilb 5 месяцев назад

@@dashgandhi5829 incorrect. The height of the main “triangle” is stated as 5. The height of the internal triangle is stated to be 3 less than that, so it is correctly calculated (not “assumed”) to be 2.

@matthewgraham2619 5 месяцев назад

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.

@jimlocke9320 5 месяцев назад

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.

@shawnmark3492 5 месяцев назад

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?

@craigdeg1 2 месяца назад

I started to notice something funny, because I would have worked out the blue shaded triangles individually, but as 3 right angle triangles, the one at the bottom was easy as it was clearly 2 x 5 cm, and then I was going to split the top one in half, but the point was 2cm and the halfway mark was2.5cm, making irregular shapes. I ended up just watching your video for the answer. It's a good trick. Please keep making more of these puzzles, I've binged watch them for the last hour and am enjoying them.

@claudiamanta1943 5 месяцев назад

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 ☺️

@Tiqerboy 5 месяцев назад

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.

@raffimolero64 4 месяца назад

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

@TwinSimian 3 месяца назад

0:42 The "illusion" comes from the illustration being inaccurate. The blue triangle does not exist, because the diagonal that is drawn across a 13:5 grid would NOT cross exactly on a vertex of the grid, nor would it line up on any vertex of the grid seeing as how 13 and 5 have no common denominator. Its not a math trick, it's just a deliberately misleading drawing. Smh

@dashgandhi5829 5 месяцев назад

@genmihryzhkov2636 5 месяцев назад

Оба метода вычисления заштрихованной площади не ВЕРНЫ! ОБА метода! Доказательство: в треугольнике со сторонами 5 и 13 см нельзя построить треугольник со сторонами 2 и 5 см, потому что такие треугольники не подобны: 2/5 НЕ РАВНО 5/13 !!! Вычисление площади большого заштрихованного треугольника 5х8/2 - это правильно, а вот треугольника 2х5/2 - не правильно, потому что такой треугольник с высотой 2 см - нельзя выделить в треугольнике 5х13 см. Both methods of calculating the shaded area are not CORRECT! BOTH methods! Proof: in a triangle with sides 5 and 13 cm, it is impossible to build a triangle with sides 2 and 5 cm, because such triangles are not similar: 2/5 IS NOT EQUAL TO 5/13 !!! Calculating the area of a large shaded triangle of 5x8/2 is correct, but a triangle of 2x5/2 is incorrect, because such a triangle with a height of 2 cm cannot be distinguished in a triangle of 5x13 cm.

@Yezpahr 5 месяцев назад

Scheming tax officials: **reverse polarity** Instead of a perpetual chocolate printer it's now perpetually one piece short **after** doing this gimmick. Set a fine/tax for that missing piece which is counted as overreporting your taxes which you unfairly deducted.

@joeschmo622 5 месяцев назад

It's true. I took a chocolate bar and kept sneaking out and eating that 1 block over and over and over, probably hundreds of times. I almost ate myself into a diabetic coma and *still* had my original chocolate bar, seemingly untouched. Now I know how Jesus did it with the loaves and fishes.

@nichodemus10 5 месяцев назад

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.

@RedRad1990 4 месяца назад

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 🤦

@IamExeller 5 месяцев назад

at least it is not KitKat or I would've been caught red-handed in any way (They have some kind of superpower)

@kahvipaputyyppi 4 месяца назад

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. 🤩

@mayorb3366 17 дней назад

If you see this EXACTLY to scale. the differing angles are obvious.

@Alex-gi7sm 4 месяца назад

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

@opendstudio7141 5 месяцев назад

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

@InformationEngineer59 4 месяца назад

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

@eventhisidistaken 5 месяцев назад

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.

@timwestlund3072 5 месяцев назад

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

@googa8 4 месяца назад

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

@marcokostadinov73 20 дней назад

In such cases it helps to change the perspective. Looking from one corner to the other along the "hypotenuse" makes it easy to catch the trick.

@yvessioui2716 5 месяцев назад

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 4 месяца назад

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

@the_andrewest_andrew 5 месяцев назад

the fact that you put a square on the bottom left would lead anyone to assume the bigger triangle is a right triangle... if the actual exercise was presented as such, it would be misleading as the very least

@buddyreed2623 3 месяца назад

Why do you have two angles marked as 90-degree angles at 0:58? If that is true, then the hypotenuse is straight.

@ofofofff 5 месяцев назад

if u eat 1 and its the same then repeat 75 times if you still have it a full choclate, I'm going walmart and eat everyday 1 4free😅

@VineetJangra-wu8ou 5 месяцев назад

It's really ultra amazing.

@selfdeveloment4084 5 месяцев назад

This was in my school for some time.

@janami-dharmam 5 месяцев назад

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

@Tahgtahv 5 месяцев назад

@@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.

@lorentzinvariant7348 4 месяца назад

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

@JoshRendall 4 месяца назад

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

@aresorum 5 месяцев назад

0:26 That’s not a triangle; it has four edges! Then you make it a triangle. Who is fooled by this?

@johngonon1507 4 месяца назад

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

@ShawnF6FHellcat 4 месяца назад

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

@rickochet48 4 месяца назад

If you go to 0:30 and fast forward 5 seconds, you'll notice something interesting...

@pouzzler 3 месяца назад

It wasn't new for my baby boomer dad during his childhood, and even less for me forty years ago

@cryptocoinkiwi8272 5 месяцев назад

So the little right angle symbol in the bottom left is just a lie?

@koputai 5 месяцев назад

Eddie Woo did this one 9 years ago, and you’ve even used the same colours as he did! ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-7iSZ4rPycS0.htmlsi=1-Ft5siNclsBzzFc

@st-ex8506 5 месяцев назад

Very easy: the two triangles are NOT triangles, but quadrangles differing in area.

@DCSWCCkingpin1 5 месяцев назад

For the love of God, DO NOT START EVERY SENTENCE WITH THE WORD "so"

@chuckpeasley212 2 месяца назад

4:13 "...one of them has to be correct, and the other one has to be wrong..." Couldn't they both be wrong?

@Frie_Jemi 5 месяцев назад

Hypotenuse is not a straight line. SORRY. Cheap trick busted

@macsnafu 5 месяцев назад

I've always hated this one, because it IS an illusion and isn't playing fair.

@donwald3436 4 месяца назад

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

@sandybruce9092 5 месяцев назад

I figured this out just looking at the picture!!!! Is this supposed to be hard???

@sundareshvenugopal6575 4 месяца назад

This is not accurate. The slope of the hypoteneuse is not the same. The two triangles are not identical.

@HauntedKnight-cj8kv 5 месяцев назад

It went to Hell.

@Secret64462 5 месяцев назад

So the entire puzzle is that we're lied to Amazing message about the governmen-

@makeislameasy4206 5 месяцев назад

Height of the small triangle in the right corner should be 25/13.

@stevesmodelbuilds5473 4 месяца назад

The square was always there in the same place -- it's the color that went missing... 😏

@GraveUypo 5 месяцев назад

this is so old and frequently shown everywhere that i'm surprsied there's even anyone who's not seen it before

@buffuniballer 5 месяцев назад

If you look at these as similar triangles, you can see they don't work.

@hocules 5 месяцев назад

3/15 != 2/4. Blue & red don't have the same aspect ratio

@_baconality_ 5 месяцев назад

if it has a secret bend it is not a triangle and therefore there is no right answer to the question because they say its a triangle

@toastyburger 5 месяцев назад

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

@krispyking2450 5 месяцев назад

7:44 before u put the outline over the bottom triangle could anyone else see the dip in the centre?

@huetang 4 месяца назад

fantastic...another non-problem created for click baiting.

@engineboy_1449 4 месяца назад

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

@fredashay 4 месяца назад

The the whole premise of the problem is a lie because the large shape is not a triangle.

@carolinian2009 Месяц назад

This is the mathematical equivalent of a dirty rotten trick.

@Preview43 4 месяца назад

Banks do this kind of sh!t with our money every single day.

@crushermach3263 5 месяцев назад

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 5 месяцев назад

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