Can you solve the missing square puzzle?
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- Опубликовано: 14 апр 2024
- 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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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
You are right, that is sneaky quadrilateral. Even for a quadrilateral, which are already pretty sneaky
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.
@@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.
@@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.
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.
Where does the missing chocolate go? That's right, it goes into the square hole.
this is gold
LOL
Lmao
*screams of pain*
And this one would be a perfect fit too lmao 🤣
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.
Something something Banach Tarski
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.
Don't be toxic.
It's bad.
It’s because infinite chocolate is not a thing (nothing in the world is infinite)
@@penguincute3564except for infinity
I would love to see a couple more steps of this process, so that the loss is clearly visible and grows with every iteration
it's not really repeatable, the two triangles are either in one configuration or the other
is repeatable if you reconfigure the colors and put a new color at the right bottom green area forming a new L shape
@@srkingdavy No it absolutely is repeatable. You don't use the same blocks, you re-shade them and repeat.
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.😊
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.
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.
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.
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
Yep... we watched the video too lol.
best explanation to this popular trick. kudos Presh!
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!
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.
and, as Pannenkoek explained, the slightly different angles of the "hypotenuse" allow the out-of-bounds area underneath the triangle to poke through...
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.
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.
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.
So the key to the trick is for the presenter to lie about the details.
No lies told: Presh never said the overall shape was a triangle...
@@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.
That's the key to *all* tricks.
@@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.
@@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.
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.
Using similar triangles, the problem is right if you change 2cm to 1.923 cm
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.
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
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.
Cool video! I really enjoyed watching it 😊❤
Selling a choco bar with the marks to split it in this way would be such a powerful marketing move
I knew it! Thank you for proving it.
In fact, the "hypotenuse" of the triangle is not a straight line.
... 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!!
😂
And if you take the limit you end up with the Banach-Tarski paradox. :P
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.
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.
The two triangles have different slopes, and thus give different areas when rearranged.
That finally answers that question.
Amazing! I love Mathematics, especially on your channel.❤
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).
Yes, the whole arctan thing was unneeded. If the slope is different, the angle has to be different.
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?
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
Yep... we watched the video too lol.
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
I've never been so early that I could only finish one piece of chocolate XD
.. anyway you are good, ans yr channel its GREAT
You can easily see by the different pitch of the blue (2/5) and red (3/8) triangles that their hypotenuses cannot be parallel.
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
I wish I had as much chocolate as Presh shows this "problem."
My diabetes consultant is happy that I don't😂
MYD where do you take suggestions for problems to solve? I have a good one.
One would, logically, assume that the person asking this question has a ruler and the ability to draw a straight line! LMAO.
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. 🤩
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? 😂
sqrt(-1) tetris fr
Working this on a slide rule, the problem is instantly made clear. Also, made me want chocolate.
First puzzle: The big triangle has a slope of 3/8, .375, the slope of the second is 2/5 .4.
0:10 NOM NOM
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.
It's really ultra amazing.
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.
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.
We should cut the chocolate bar into a finite number of non-measurable pieces and reassemble them into two copies of the original bar.
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.
This would be a fun trick to pull on some young kids; it would absolutely blow their minds!
Strahlensatz sagt mir: "5:13 2:5" - Das große Dreieck ist ein Viereck.
❤ Thank you.
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.
0:34 the chocolate was refilled right under the diagonal line. That's where the 1 square had gone.
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.
I’ve watched something similar to this! Can you solve the Alice in Wonderland riddle!
floor overlap... floor gap... this is a certified Cause #4 situation
(Context: Pannenkoek2012's video on Invisible walls in Mario 64)
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.
I suppose the god of the gaps took it.
Both triangles have slightly different angles, which is not very visible just looking at it.
GREAT! Thx
It's pretty clear the slope on the right angles is not the same. Just count the squares.
"we can do it all over again" how many times before the bump becomes obvious?
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 ☺️
COOL PUZZLE !
It works with any pair of numbers such that height and base are Nth and (N+2)th number from series of Fibonacci
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.
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.
0:06 take this square of charger and eat it......NOM NOM XD
The funny thing is that you actually are able to make area disappear using translations and rotations, like Banach and Tarski have shown.
Pls post videos more frequently
I deny these results so that I might delude myself into thinking I might create infinite chocolate.
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
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.
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.
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".
I figured out the slope wasn't straight before any calculation by looking to my cell phone screen from a side perspective.
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 🤦
The thing is this wouldn't work with squares of chocolate because they're scored at regular intervals and it would be obvious they don't line up.
So the little right angle symbol in the bottom left is just a lie?
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
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.
I've always hated this one, because it IS an illusion and isn't playing fair.
The other interesting observation is how big a role the Fibonacci numbers play in the measurements of the sides of the two figures.
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.
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
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.
You can see it if you draw it with a fine pencil.
Assumptions are what humans do best
"Nom, Nom."
- Mind your decisions, 2024.
7:44 before u put the outline over the bottom triangle could anyone else see the dip in the centre?
for the “diagram”, the two “right triangles” are not similar, meaning the larger right triangle isn’t actually a triangle, its a quadrilateral
by larger right triangle, i mean the whole diagram, not the unshaded triangle
It went to Hell.
So this is a very big mystery. Almost like with English pronunciation.
Tak toto je veľmi veľká záhada. Skoro ako s výslovnosťou angličtiny.
😄
The angle of the triangle got less steep
I can see the edge slightly protruded..
Very easy: the two triangles are NOT triangles, but quadrangles differing in area.
Hmm, it seem there another question, how much area that has been lost based only in picture 2?
The red herring of this problem is the “3 cm” dimension introduced at 1:19.
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.
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.
Height of the small triangle in the right corner should be 25/13.
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 ...".