The ladder and box problem - a classic challenge!
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4 жыл бұрынSpecial thanks this month to: Michael Anvari, Kyle. Thanks to all supporters on Patreon! A ladder leans against a wall, just touching the corner of a cubical box. If the ladder has a length of 4, and the box has a side of 1, what is the distance between the top of the box and the top of the ladder?
References
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@honeyraj 2 жыл бұрын
Moral of the story: Never put a ladder and a cube in the same room. Thank you for your understanding. 🙏
@Aman-bd7dv 2 жыл бұрын
Loooool 🤭🤭
@patrickjacquiot9073 2 жыл бұрын
Nice joke !
@sushilkumarlohani6709 2 жыл бұрын
YOUR NAME LOL
@Mathematician6124 2 жыл бұрын
👌👌🤣🤣
@pressanna1385 2 жыл бұрын
😂😂😂
@nuvamusic 3 жыл бұрын
You really chose the most complicated way of solving a relatively easy problem.
@badalkumar7185 3 жыл бұрын
No this is the only simplest way to solve it
@jakubedzior 3 жыл бұрын
@@badalkumar7185 wrong, there is an easier one, considering that y/c = c/x and (x+1)^2 + (y+1)^2 = l^2
@stef2499 3 жыл бұрын
@@jakubedzior thats what i used to solve it but i did a mistake somewhere 😅 oh well. Ratios all the way!
@mmattoso1 3 жыл бұрын
I called "x" the whole basis of larger triangle, by similarity we get x/(y+c) = c/y and then x = c.(1+c/y) By Pithagoras L^2 = (y+c)^2 + x^2 Replacing x and doing the calculations we get to the equation: (y^2 + c^4/y^2) + 2c(y + c^2/y) + 2c^2 - L^2 = 0 Then we should use the magic trick: let u = y + c^2/y we see that u^2 - 2c^2 = y^2 + c^4/y^2 Therefore we can replace u in previous equation and after simplifying we get the following u^2 + 2c.u - L^2 =0 from which we get u = sqrt(c^2+L^2) - c and from "u" we get y by solving: y^2 - u.y + c^2 = 0 By this way we get the general solution without the need for the magical geometric constructions 😉
@jakubedzior 3 жыл бұрын
I did it this way: y/c = c/x and (x+c)^2 + (y+c)^2 = l^2. You get y^4 + 2cy^3 + (2c^2 - l^2)y^2 + 2cy + 1 = 0, which in our example is y^4 + 2y^3 - 14y^2 + 2y+1=0. From this polynomial y=-0.2 or y=0.36 or y=2.76 or y=-4.92, so y=2.76 is the answer.
@btough4483 3 жыл бұрын
Me: Trying to do something with Pythagoras theorem😂
@yurakril3580 3 жыл бұрын
Me too, And my result was closed to correct y = 2.7957 :D
@shivgoyal8967 3 жыл бұрын
Me too and my answer was equal to 2.7
@AngelMorales-kd7cy 3 жыл бұрын
Mine was 2.76. I feel good.
@unknow11321 3 жыл бұрын
ngl same
@bijoychandraroy 3 жыл бұрын
how the hell did you all solve it with pythagoras theorem
@gloystar 3 жыл бұрын
As the others said in their comments, actually there is another solution which is approximately 0.3622 using pythagorean theorem and similar triangles to generate 2 equations and solve them. I didn't use X and kept the problem in one variable Y, so ended up with a quartic that has 2 negative roots and 2 positive roots. Disregard negatives and you get the solutions approximately 2.76 and 0.3622.
@matteovillella1482 Ай бұрын
I did the same. I think that since, based on the diagram, y>1, the 0.36 solution can be disregarded
@ritvikn9167 22 күн бұрын
The bi quadratic equation right? Yea, i did the same and gave up on factoring the roots. Used desmos
@jason_9276 6 күн бұрын
i came up with the same idea
@TheOfficialCzex 4 жыл бұрын
As others have pointed out, solving through similar triangles is another good way of looking at it.
@sukritgoyal6295 4 жыл бұрын
yup its easy to do with similar triangles
@SummitTidua 4 жыл бұрын
I did it that way as well. I think the reason he did it such a complicated way was to get the exact values as with similar triangles you only get a decimal answer
@cocgamer3511 4 жыл бұрын
Yes
@omjadhav9128 4 жыл бұрын
Yup
@InShadowsLinger 4 жыл бұрын
can you use similar triangles to solve for the general case of leader and cube side length? All the solutions I have seen involving similar triangles are for the specific version of 4 and 1.
@hrishikeshpurohit8676 4 жыл бұрын
Me: Trying to solve the problem by similar triangles. Presh: Creates a square using the diagram. Me: *surprised Pikachu face*
@SolathPrime 4 жыл бұрын
Wow that's mean that I wasn't alone When I thought that similarity of the triangle would solve it easily
@jayvillahermosa8550 4 жыл бұрын
Me too, I also use similar triangle principle, in solving the problem
@mhammamabid2275 3 жыл бұрын
tbh almost all of us, use similar triangles so, no surprise here lol.
@ashishgrm 3 жыл бұрын
Pi ka pi? 😳
@dasguptaarup8684 3 жыл бұрын
for the specific case of cube side 1 ..... it feels similar triangles is easier .... (bottom x+1 side 1+ 1/x) .... Or not 🤦
@erikblaas5826 2 жыл бұрын
I had just the same question as an end of term test at school.. but there the ladder had a length of 9, the cube 1 by 1 by 1 and the question was the distance of the base of the ladder from the wall. After finishing school, I forgot all about it and never figured it out again on my own, thanks for having this brought to me again.
@atharvasrivastava4711 3 жыл бұрын
Using similarity, simple algebra and Pythagoras theorem. I found the answer.
@rc210397 4 жыл бұрын
You should also consider the value of x as another possible solution as it is not given to us that y > x, and neither is the slope of the ladder given. So this problem has 2 solutions, 2.76 as well as 0.36
@lucianozaffaina9853 4 жыл бұрын
Rishi Chugh yes I Also found 2 solutions but I thought Y >X because of the diagram
@ANURAGGUPTA-fe4ek 4 жыл бұрын
Acc to figure the height y and the small base x=1/y ( which can be proved by using similiarity of triangle) so when you use y=2.76 , u get x=1/2.76(=0.362), Now if you put y=0.362 you get x=1/y(=2.76) , so its not new solution ,it can obtained when you solve question other way round taking height 1/y then base x=y ,its like you have solved the question when ladder has slipped to y=0.362 then x=2.76..
@kooltyme 4 жыл бұрын
lol thank you i was so confused
@JDeWittDIY 4 жыл бұрын
It has 2 solutions because the latter could be touching the cube near its base, with most of the latter high on the wall, or the latter could be touching the cube near its top, with most of the latter near the floor. In both cases it's leaning against the wall and touching the cube.
@erenjeager7049 3 жыл бұрын
U hv 99 likes and I am gonna give u ur 100th like
@itaycohen7619 4 жыл бұрын
by using similar triangles we can show that the base of the small triangle in the bottom left corner is 1/y - therefore we can use the Pythagorean theorem to construct the equation (1+y)^2 + (1+1/y)^2 = 16 , 1 + 2y + y^2 + 1 + 2/y + 1/y^2 =16 , notice that we can write y^2 + 2 + 1/y^2 as (y+1/y)^2 therefor we get (y+1/y)^2 + 2(y+1/y) - 16 = 0 . we let t=y+1/y , and we get the quadratic t^2 + 2t - 16 =0 , from here on its pretty clear.
@nisargbhavsar25 4 жыл бұрын
I believe you want to say Gongu theorem!!
@nigelsw55 4 жыл бұрын
Essentially the same way I used to solve this problem.
@soorajsajikumar5580 3 жыл бұрын
but then exponential powers would all be different rite?
@Al.2 3 жыл бұрын
almost the same but maybe slightly easier: 16 = (1+y)^2 + (1+x)^2 = (x+y)^2 + 2(x+y) - 2xy +2 and the last two terms cancel each other out because xy = 1
@user-yp6eb5wj4w 3 жыл бұрын
Year, I solved like this way)) there is similar equation in Russian exam where you can solve with help of such methods
@bluehope42 3 жыл бұрын
I had found some equation with unknowns a and b using pythagoras, when I factored in the similar triangles equation I eliminated a and b and got a polynomial which wolfram alpha solved to this solution, I was very pleased! :D
@YamiSuzume 3 жыл бұрын
i look forward to the day, these videos are watchable again without stressing heartbeat
@tcfs 3 жыл бұрын
Just remember when you met similar questions written on a piece of paper... at 8:00 AM.... in high school.... :D :D
@JPL454 3 жыл бұрын
@@tcfs i don't have to remember, cuz i still meet that
@rajendrapathak7034 4 жыл бұрын
Pretty complicated solution!
@nickwilson3499 4 жыл бұрын
I always hope it’s something easy and simple you can just solve by thinking, but it’s usually something way out of my league
@prateekmourya9567 4 жыл бұрын
Use similar triangle for pretty easy solution
@rifnasriff166 4 жыл бұрын
drive.google.com/folderview?id=14F2jmF4E7AaCk9GJPgg_qx1itnwQ-_FQ
@fix5072 4 жыл бұрын
@@prateekmourya9567 no Calculator allowed
@polish3717 4 жыл бұрын
overcomplicated, actually
@andrewwoan 4 жыл бұрын
As someone who doesn't care about accuracy: 4 (length of ladder) - 1 (length of box) = 3 2.76 rounds up to 3 Close enough
@nycholasgr8112 4 жыл бұрын
typical engineering student
@albertomontori2863 3 жыл бұрын
true as long as “x+c” is enough small to allow the “overlapping” between “ l “and “y+c”:only in this case l (ladder) - c (box)=3
@lakshya4876 12 күн бұрын
Found the engineer!
@antoniopirescurto8138 3 жыл бұрын
I saw many solutions using similar triangles, but resulting in a 4th order equation, I just after reaching the conclusion that xy = 1 and pythagoras (x + 1) ² + (y + 1) ² = 16, x² + "2" + y² + 2 (x + y) = 16 replacing "2" with 2xy we get (x + y) ² + 2 (x + y) = 16 and now replacing (x + y) with k and solving the equation we find that k = (x + y) = -1 + √17 and with the initial information that xy = 1, x = 1/y so 1/y + y = -1 + √17 again solving this equation we get y = 1/2 ( -1 + √17 + √ (14-2√17)) or 2,76091...
@shrikrishnagokhale3557 2 жыл бұрын
I solved it in about 20 to30 minutes by using similarity of triangles and quadratic equations,and got the same answer.A very good problem.Thanks.
@ShreyRupani 4 жыл бұрын
I got the same answer(2.7609), but from a different(slightly easier) method. I used pytho and similar triangles. (x+c)^2 + (y+c)^2 = l^2 [from the initial big triangle where the ladder is the hypotenuse.] AND xy = c^2 [from similar triangles]
@subhajitsarkar2272 4 жыл бұрын
I have also solved it by this technique.. Hifi!!! 😊
@abhinavtiwari8481 11 ай бұрын
as any sane person should XD
@allenminch2253 4 жыл бұрын
This was an absolutely amazing problem! Thanks for doing it! The solution you present from Math Stack Exchange is very elegant in requiring only a quadratic equation. I solved it in the more obvious way where you use the similarity that you used in the solution you presented but then simply use the Pythagorean Theorem (or Gogou Theorem if you'd like to call it that) in the right triangle with the ladder length as the hypotenuse. Basically, (x+c)^2 + (y+c)^2 = l^2. The disadvantage of this method is that you end up with a quartic equation in y instead of a quadratic equation. This quartic equation has a constant term of c^4. However, recognizing that x and y must have a product of c^2 and that they are essentially interchangeable (you will get the same solutions regardless of which one you solve for), you can reason that two of the values you will get for y as solutions to the quartic equation are solutions to a quadratic equation with a constant term of c^2. It turns out that you can depress the quartic into two quadratics that each have a constant term of c^2; one of them yields the positive answers for y that you come up with and another one yields two negative values for y that don't make physical sense. One thing I think is interesting to do with this problem is to analyze the discriminant in the final quadratic equation for y. Of course, this discriminant has to be nonnegative in order to get values for y that make physical sense. I solved an inequality and found that for the discriminant to be nonnegative you must have l >= 2√2 * c. Interestingly what this means physically in this problem is that the length of the ladder must be at least two times the diagonal of a square face of the cube, since the diagonal's length is √2 * c. This actually makes intuitive sense: In the extreme case where l = 2√2 * c, it turns out you will have x = y and the two similar right triangles used in the problem are congruent. This means that the ladder will be at exactly a 45º angle with the ground. The diagonal of the square face then is exactly equal to half of the length of the ladder, and the ladder is exactly twice the length of the diagonal of the square face! It makes intuitive sense that this 45º case would be the smallest length the ladder could have for a given side length for the cube; therefore an algebraic result is exactly consistent with physical geometric intuition! It is so cool when you find out something like this!
@franciscoparedes1131 4 жыл бұрын
Very interseting for sure!!
@siralanturing9103 2 жыл бұрын
You wrote an entire mathematical paper
@bulbergaming9142 4 жыл бұрын
Am i the only one who dont know whats going on but still watches the full video
@winzkling 3 жыл бұрын
No, I fully understand everything without even following his explanations all too closely. Also I'm a big liar.
@ya33a 3 жыл бұрын
A triangle has 3 sides and a ham sandwich is lunch?
@koro-sensei9783 3 жыл бұрын
@@winzkling The last one got me😂
@marco7563 2 жыл бұрын
Like watching 5 dimensional beings
@profamitgupta 3 жыл бұрын
I derived using similar triangles and pythagoras theorem and the solution was obtained in much less number of steps. But i do appreciate your geometric solution for its physical interpretation.
@atharvius7522 4 жыл бұрын
Me: Watches the video thoroughly Also me: what just happened?
@fosskytheanswerer 4 жыл бұрын
1:51 "Now we are going to do a little trick" This is where things start to go wrong...
@projectmayhem6898 4 жыл бұрын
He did what's called a pro-gamer move.
@atharvius7522 4 жыл бұрын
Fossky The Answerer Ikr?!
@aatheeswarank7025 4 жыл бұрын
@@fosskytheanswerer bruh , he just pulled a sneaky on us
@AMa-us8de 4 жыл бұрын
The problem looks quite hard, but the moment you try solving it It's even harder
@acballer4232 3 жыл бұрын
And I thought I was good at math
@rayeittastay2458 3 жыл бұрын
I love these videos, they're so boring it helps me fall asleep faster.
@tagnetorare5401 3 жыл бұрын
You can solve y by those two equations directly: (x+c)^2+(y+c)^2 = l^2 and x=c^2/y Introducing w just make the representation of y more readable (otherwise you need to solve y from a quartic equation)
@IshanBanerjee 4 жыл бұрын
I have been watching your videos since a long time , you are an inspiration for me
@BambinaSaldana 4 жыл бұрын
*for a long time
@egillandersson1780 4 жыл бұрын
Wow ! Very difficult this one ! And how elegant the solution ! Thank you
@anildeshkar906 3 жыл бұрын
Your graphics during explaining is fantastic
@nobillismccaw7450 3 жыл бұрын
Thank you so much. I’ve struggled with this problem for years.
@Joschfrosch2 3 жыл бұрын
hey I'm surprised of myself, i went another way to find y ! at first, I used Pythagorean Theorem : 16 = (x+1)^2 + (y+1)^2 which leads us to 14 = x^2 + 2x + y^2 + 2y I used this as a circle-function and looked for the positive zeropoint on y-axis. The zeropoint (idk if native english speakers name it like that) guides us to y=2,873 which amazed me because it's quite close to Presh's solution! I hope i found something here and didn't just have luck with that. ^-^
@verheggn 2 жыл бұрын
By looking at the y-axis of the formula you found, you're basically saying x=0 and therefore you place the ladder at the bottom edge of the cube, at a distance of 1 from the wall. From thereon it's just Pythagoras, and the top of your ladder will be √(4^2 -1^2) = √15 from the floor. y will be √15 - 1 = 2,873. It will go straight through the cube however... Btw: zeropoint is root in english.
@mosesmuchina1308 2 жыл бұрын
Very interesting solution. Thanks Presh.
@jpico99 2 жыл бұрын
Since the triangles are similar, using Pythagoras you almost directly get the quadratic equation (y+1/y)^2+2(y+1/y)-16=0 where 1/y is your x. Then, define z=y+1/y and you get z=sqrt(17)-1. From that, you have y
@kozatas 2 жыл бұрын
Actually, it's (y+1)^2+((y+1)/y)^2-16=0 write this down to excel. Then "goal seek" the equation to zero. y=2,76
@bificommander 4 жыл бұрын
Tricky. I could see it was solvable from the number of unknowns and variables, I just couldn't do it without running into 4th order polynomials. Others found a way to turn those into 2nd order, but I quit before then. Also: Technically both the + and - answer of the quadratic formula are valid values of y, since you can put the ladder almost horizontally instead of almost vertically, and it still satisfies the requirements at the start of this problem. While the drawing put the cube edge below the middle of the ladder, you can't trust drawings on these problems.
@verheggn 2 жыл бұрын
@@KNYD Nope, all the in between solutions you suggest, won't touch the edge of the cube. For instance, if you put the ladder on 45°, the middle of the ladder will be √2 from both the floor and the wall (and not 1 like the cube). The fact that the cube is fixed, and the ladder can't extend beyond the floor or wall, really ties it down to only 2 solutions.
@verheggn 2 жыл бұрын
@@KNYD did it. Still only 2 solutions :-) can't get any more without changing the length of the ladder/pen or making a hole in the wall or floor. I know it doesn't have to touch the centre, just as in the original problem. But where else should it touch at my example of 45 degrees?
@ahmederfan3663 3 жыл бұрын
I got a relation between L and y and c in just three steps!! We will have the same triangle as in minute 1:54 From psyghorath equation, L^2 = (y+c)^2 + (x+c)^2 From triangles similarity, (x/c)=(c/y) Then x= (c^2)/y By substituting equ.2 in equ.1: L^2 = (y+c)^2 + ((c^2)/y + c)^2 Now we can substitute by numbers and get the same exact answer
@AntonBourbon 2 жыл бұрын
Your _elegant_ solution creates 6 additional squares (4 black, 1 blue, 1 green), a dozen additional triangles and additional variables: X, Y, Z and w. Really smart people could also add lots of Greek letters and a couple of *rhombic dodecahedrons* for the solution to be *lethally elegant* .
@tamajongmichaelnkeh1978 4 жыл бұрын
This solution is great! Lots of beautiful geometry. This is a good problem which shows how algebra is translated to geometry and vice versa
@luhdooce 4 жыл бұрын
Before watching, my solution was that: If you have 2 similar triangles, the upper larger one has a side to base ratio of y/1 The lower, smaller one has a ratio of 1/a, which must be equal to y/1 If y = 1/a than a must = 1/y, which is the space between the bottom of the ladder and the cube. If we use pythagorean, (1 + 1/y)^2 + (y + 1)^2 = 4^2 We can than expand this out to solve for y values
@sayamchakraborty5873 4 жыл бұрын
Yes I did in in the same way
@bhabanishankarrath3973 4 жыл бұрын
i did same dude
@x_abyss 4 жыл бұрын
Same here.
@twwc960 4 жыл бұрын
Yeah, Presh's solution is way, way more complicated than it needs to be. What I did was to set up a Cartesian coordinate system, and noted that the equation of the line representing the ladder has a slope equal to y (which I renamed as m since y is used in the equation of the line), and then I found the x and y intercepts of the line and noted that the length of the ladder squared is the sum of the squares of the x and y intercepts, so setting that equal to 4^2, I ended up with the same equation as you. The tricky part is that you end up with a quartic equation. I first tried to find a linear factor, but it doesn't factor nicely that way. It does turn out you can factor it into two quadratics though. (I noted the leading coefficient and constant term in the quartic were one, so I assumed a factorization into the form (x^2+ax+1)(x^2+bx+1) and was able to solve for a and b.) I then solved the resulting quadratics (both have the same roots, even though a and b are different) and obtained the same solution as Presh in terms of l and c. Plugging in the given values, I got the right answer.
@SimonClarkstone 4 жыл бұрын
Doesn't that give you a quartic?
@iqbalchowdhury11 4 жыл бұрын
I used the tap measure from my moms sewing kit. And she recognised her son solved it in the fastest way.
@jax6648 2 жыл бұрын
“How many variables do you want?” Him: yes
@scarletevans4474 Жыл бұрын
Very clever! I did it the hard way, here's how to do it for c=1, l=4, can be similarly done in general case 👍 Denote α as an angle between the wall and the ladder. We have 1/y=tan(α), then from Intercept Theorem and Pythagorean Theorem we quickly arrive at: y^4+2y^3-14y^2+2y+1=0 This is SYMMETRICAL quartic equation, so we can easily solve it! 🙂 It gives us the solution you show us above, plus three extra ones that we have to cast away. Much more work in general, but still works for people who didn't found the clever solution 😉
@IceMetalPunk 3 жыл бұрын
Wait, but... around 3:40, how do we know that exactly half the area is Z and the other half is X+Y? What's stopping it from being 2/3 X+Y and 1/3 Z?
@Halfmania 2 жыл бұрын
I'm disappointed that nobody answered this question...
@chrisjeaaalbertos3802 2 жыл бұрын
I'd very much like to know this answer too
@Halfmania Жыл бұрын
I don't know why, but today, after 8 month, i randomly came back to this video and understood immediatly how he found out this result. The orange area (A) is equal to X+Y+Z. If you look closely, you can see that Y can be written Y=Z-X (conversely X=Z-Y and Z=X+Y) by slicing Z area in two with a vertical line passing by its right angle. Then in A=X+Y+Z, you can substitute Y by Z-X. You find A=X+(Z-X)+Z. Once you simplify you get A=Z+Z=2Z. Therefore A/2=2Z/2=Z. I know it's funny to post an answer 8 month later and maybe you already find the explanation, but i felt i had to post what i found out for those who still didn't understand. Have a nice day.
@parispapadopoulos8892 3 жыл бұрын
Alternatively, we have three similar right angled triangles (large, medium and small) and it is relatively easy to find their ratios, then use the pythagorean theorem to form an equation and finally solve for y.
@phugoid Жыл бұрын
What? Have 2.7M subscribers and do something straightforward? Please be reasonable in what you expect :)
@gouravtada 3 жыл бұрын
That was really clever. You can also equate the area of big triangle and sum of 2 triangles and sqare, and then apply Pythagoras theorem
@smittymcjob2582 2 жыл бұрын
The general solution offered in this video is quite elegant. People offering solutions using Pythagorean don't realize that if the little square side wasn't 1 then they couldn't solve the equations they get. The Pythagorean gives you the relation between (x-squared + y-squared). the solution in the video gives you the relation between (x+y) which then gives you a closed form solution for y given any value of c.
@sergten 4 жыл бұрын
If you just try solving it algebraically, you end up with a 4th degree polynomial. Good news is that the polynomial is symmetric, and easy to solve with the standard "t = x + 1/x" substitution trick.
@MindYourDecisions 4 жыл бұрын
Thanks. Though I didn't present it, your method is definitely worth mentioning! The second answer on Math StackExchange used that substitution for those that want a few more details: math.stackexchange.com/a/1345007/84271
@zomzombos8257 4 жыл бұрын
Ah yes, I stuck on the 4th degree polynomial and forgot to use t=x+1/x thanks sir now I can solve this! (Also why it says that this was from 4 days ago but the video just came 10 minute ago?)
@sergten 4 жыл бұрын
@@zomzombos8257 The video was unlisted for a while, available to Patreon supporters though - so I got an early access.
@lucianozaffaina9853 4 жыл бұрын
I used a simpler solution just by solving 2 equations (x+1)^2+(y+1)^2=16; y^2+1=(4-√(x^2+1))^2
@mauricevassilitch9507 4 жыл бұрын
Absolutely! Then I had a degree 4 equation : y^4+2y^3-14y^2+2y+1=0
@peptobismolveins 4 жыл бұрын
Maurice Vassilitch 😂 ppl need to actually solve before commenting how “easy” their “solution” is.
@michaelroditis1952 4 жыл бұрын
@@prateekmourya9567 I really liked how you turned the 4th degree equation to a 2nd
@azzanporter4377 4 жыл бұрын
Same
@MrKA1961 4 жыл бұрын
first equation ok but for the second why not use x=1/y and get a symetric equation of 4th degree?
@nirupmalkani788 3 жыл бұрын
By observation, it appears that there can be a range of solutions and not just one exact number.
@ABc-es4nv 3 жыл бұрын
Used similar triangles to get the relations of y and the horizontal side which I named x. Got a 4th degree polynomial, tried to factor it using different methods, didn't really work out, opted for the calculator and after plugging it in got the right solution. If you have a calculator on you this really is not a very difficult problem, but if your doing everything manualy so to say, you'll be doing a lot of work.
@pkmath12345 4 жыл бұрын
I like the graphics you used! Geometry def requires nice visuals, and your video is a good example of it. I also use graphics for conceptual videos for math topics. Glad you do the same! Nice work!
@gardenmenuuu 4 жыл бұрын
sir which software do you use?
@gardenmenuuu 4 жыл бұрын
pleaese respond
@pkmath12345 4 жыл бұрын
@@gardenmenuuu I actually use PPT for effects in graphics
@skyblue4558 4 жыл бұрын
MathFlix yours are amazing as well
@ravirajshelar250 4 жыл бұрын
Me :-becomes happy after easily solving by trignometry and I expected the same solution. The solution:I'm about to end this mans whole career.
@therion8469 3 жыл бұрын
Omg same out of curiosity did you also get 2.82
@hououinkyouma4388 3 жыл бұрын
@@therion8469 yes
@hououinkyouma4388 3 жыл бұрын
@@therion8469 though i used calculator
@codingkrtehum.. 3 жыл бұрын
@@therion8469 how did you get 2.82 ? If possible please write the solution
@braindamage9083 2 жыл бұрын
@@codingkrtehum.. Considering the upper triangle, cos( theta)=1/3 --> theta = 70.52 then, tan (theta) = y/3 y=2.8
@Javiemperator 3 жыл бұрын
Another way of looking at it could be: The condition of the ladder resting on the wall and floor: (y+c)^2+(x+c)^2=l^2 The condition of the ladder touching the corner of the cube: x*(y+c)/(x+c)=c You have a system of two equations with two unknowns x and y, so the problem is defined and we can solve it (by substitution for example). This will provide four solutions, from which we can easily determine the only two possible ones (x and y have to be real and positive).
@kevinstewart2572 2 жыл бұрын
Amazing how a problem so simple to state becomes surprisingly harder to solve than expected. Not sure if someone has already shown this, but I have this twist for readers: Some commenters have used (x+1)/4 = x/√(1+x^2) to arrive at the quartic x^4+2x^3-14x^2+2x+1=0 and then resorted to their calculators. But one technique here is to recognize that, because of its symmetry, the expression can be written as product of two quadratics: (x^2+ax+1)(x^2+bx+1) giving a+b=2 and ab=-16 and hence values for a and b of 1+/-√17. Then the quadratics are easily solved and extraneous solutions eliminated.
@it8755 4 жыл бұрын
Can't we do it with similar triangles
@Akash-ul2vd 4 жыл бұрын
Yes.. similarity can be used
@rabindranathghosh31 4 жыл бұрын
In the top right triangle, height:base = y:1 In the Big right triangle, height= y+1 Base= √[4^2-(y+1)^2] Using similarity, y/1=y+1/√(16-(y+1)^2) => y√(15-y^2-2y)=y+1 Squaring both sides and simplifying, we get a quartic in y: y^4+2y^3-15y^2+y+1=0 I think that explains why the answer is so complicated
@prashantkrishnan5856 4 жыл бұрын
Yes. Check my comments in the video
@muhammadalfaatihsalsabila1266 4 жыл бұрын
@@rabindranathghosh31 i got that too, but how to get exact value of y??
@arbs-5164 4 жыл бұрын
I tried to do that I ran into complex numbers. So it's harder
@sumathip3687 4 жыл бұрын
i thought pythagoras theorem(gougu theorem ) will be used in this video!!! lol
@DrHades 4 жыл бұрын
Actually you need nothing more than pythagoras' theorem and similar triangles to solve this.
@sumathip3687 4 жыл бұрын
@@DrHades is it?? how!!!
@rifnasriff166 4 жыл бұрын
@@sumathip3687 drive.google.com/folderview?id=14F2jmF4E7AaCk9GJPgg_qx1itnwQ-_FQ
@sumathip3687 4 жыл бұрын
@@rifnasriff166 hope presh talwalker realises your work!! and thank you for your answer!!!
@pedrohenriqueribeiroabraha5593 4 жыл бұрын
Constructing these squares is the same thing as using p. Theorem and similar triangles, just a longer path
@georiashang1120 5 ай бұрын
the shorter length does exit and it is one of the answers.Take the wall as the ground in the illustration and the ground the wall,and you will get that. So the geometry and algebra correspond to each other!
@viniciusfernandes2303 3 жыл бұрын
Thanks for the video!!
@ptviwatcher 4 жыл бұрын
It would be way faster and easier to divide the ladder into x and 4-x and then do 1/x=y/(4-x) together with the Pythagoras theorem (4-x)^2=1+y^2. These could also be generally solved for any length of the ladder and cube side.
@LogicalMath 4 жыл бұрын
there are two ways of leaning a ladder against a wall symmetrically to the diagonal of the square (by swapping and rotating the two triangles), so Presh why you did not examine the second one?
@lucbrink-morrison5593 3 жыл бұрын
Subtract from l the inverse of: Sin(Arccos(c/l)) And then use Pythagorean theorem. Calculator gives y~2.79 so a little less precision but I’m still glad I found this myself
@bradleyjames3709 2 жыл бұрын
I was introduced to this problem many years ago when working as a draughtsman on an Aerospace design team. Although the numbers were different the problem was the same. There are 2 possible answers to this, the ladder could touch the wall, box and floor in the steep presentation shown or a shallow presentation. This lead me initially to a quadratic solution in many ways similar to the one shown but I’m most pleased with the following trig solution. Let the apex angle be a, the length of the ladder above the box intersection x and below the box intersection y. The vertical height floor to the top of the ladder will be H which equals y + 1. ccs a + sec a = x/1 + y/1 = 4 interpolating from Godfrey and Sidon four figure tables angle a = 19.9109deg ( or 70.0891deg) expressed as a decimal because I used a calculator for the next bit. cos 19.9109 = H/4, H=3.761, so y=2.761. (Ladder at steepest angle.)
@bradleyjames3709 2 жыл бұрын
Sorry, y used twice but you get the drift.
@ThePowerfulOne07 4 жыл бұрын
After seeing the problem, I was like aight bet...until I saw the rest of the video. Fml 🤦🏽♂️🤦🏽♂️🤦🏽♂️
@jbglaw 3 жыл бұрын
That's actually the very exact problem my great math teacher gave me during a break to give it a try. (Though in those times, I stranded with a 4th order polynominal which I couldn't solve back then...)
@ashmitmehta5088 3 жыл бұрын
U R in which class?
@jbglaw 3 жыл бұрын
@@ashmitmehta5088 At that time, I was in the 11th class, but that's some 25 years ago.
@ashmitmehta5088 3 жыл бұрын
@@jbglaw oh!! Actually i am just promoted few days back to 10th... So everything went out of my head
@jixpuzzle Жыл бұрын
I solved it by transforming it into coordinates space. And then the ladder would be represented by the line x/a+y/b=1. Where a and b are it's x and y intercepts. Now the line passed through (1,1), that is it touches the box. It's also has a length of 4. So, 1/a + 1/b = 1, √(a^2+b^2)=4 Solving these two equations we get (1.36,3.76) or (3.76,1.36) as positive solutions. Now the required length is 1.36-1=0.36 or 3.76-1=2.76. Since in the given figure the required length is longer it would be 2.76 as the answer!
@Mr.Kim.T Жыл бұрын
If you focus on lines or angles you can easily end up with a 4th order polynomial which is difficult to solve. By focusing on areas instead of lines you essentially break the problem down into two quadratics. The two quadratics in question are H^2 - PH + CP = 0 and P^2 - 2CP - L^2 = 0. Where H is the distance up the wall you want to find (from the ground). Solve for H in the first and solve for P in the second. As a matter of interest, P is actually the side of the whole diagram (W + 2C). You don’t actually need to derive the second quadratic because its solution “P = C + SQR(L^2 + C^2)” can be deduced from the diagram in a similar way that Presh did; i.e. that the area of the square (P - C)^2 equals the area of the square L^2 plus the area C^2.
@rick57hart 2 жыл бұрын
I calculated: 1/sin alpha + 1/cos alpha = 4 Then I tried some numbers, and found out, that the angle is 19.91 °. Than I calculated 1/tan 19.91 and got 2.76. I'm fine with that, despite trying out numbers is not a pure mathematical method.
@SanjayaNeupane1987 4 жыл бұрын
its too complex; simply use the concept of similar triangle, and Pythagoras theorem. Its simple....Any way your solution is interesting as well
@Champiansh 4 жыл бұрын
noob!!
@SanjayaNeupane1987 4 жыл бұрын
Jakob Fredin I did for general equation also and got equation of fourth power which on solution got the required answer
@toanbui5339 3 жыл бұрын
It's true. I use similar triangle and have the same result ~2.76
@ribbonsofnight 2 жыл бұрын
yes, but did you have to use a computer to solve a 4th degree polynomial?
@Choinkus 2 жыл бұрын
Solving a quartic equation is hardly "simple," at least for the kind of audience these videos are for.
@alastairbateman6365 4 жыл бұрын
@Rajesh Jain @Maurice Vassalitch @Prateek Mourya Brilliant video and equally brilliant method of solution. Well done the guys who came up with it. I particularly like the diagram that they end up with as it is a composite of the Pythagoras theorem diagram and the Quarter Squares diagram plus 4 corner squares and is 'The Simpletons' mathematical coat of arms. I solved the same problem for a 9 foot ladder and box of 3 foot square cross section way back around 1980. I mentally moved the ladder around the inside of a 9 foot square box and decided that the point of interest on the ladder described an ellipse within the box. So adding the data to the equation for an ellipse and simplifying I also ended up with a 4th degree equation and I solved for the roots on a programmable Texas SR-56 calculator that had a root finding program. The Bolzano method of bisecting the interval that contained the root.
@SimplyLogical 3 жыл бұрын
complicated solution , thoroughly enjoyed the video.
@amitsinghrajawat4080 4 жыл бұрын
Amazing puzzles keep it up.😊♥️
@prashantkrishnan5856 4 жыл бұрын
Use similarity and quadratic directly. (y+1) ^2 + (1/y +1)^2 = 4^2
@omdave1008 4 жыл бұрын
It's incorrect bruh...How can you take horizontal distance (base) as 1/y +1 ....Moreover ans isn't coming through this
@SanjayaNeupane1987 4 жыл бұрын
Also need similar triangle
@deanshulze931 4 жыл бұрын
This is a 4th order equation and hard to solve. I think there is a way to do it but is very complex.
@prashantkrishnan5856 4 жыл бұрын
If you take the 2 white triangles that are similar you get the equation y/1 =1/x so xy is 1 and x = 1/y. Substitute 2.76 in place of y in the equation it will fit
@peterribers 3 жыл бұрын
Forget the complicated figure. Algebra on paper is enough! With c=1 and l=4 similar triangles gives you: y=1/x xy=1. Pythagoras yields 16 = (x+1)^2 + (y+1)^2 = x^2+y^2+2x+2y+2 Rewrite the last term 2 = 2xy, add 1 on both sides to get: 17=x^2+y^2+2x+2y+2xy+1 Clever rewriting of the righthand side produces: 17=(x+y+1)^2 x+y=sqrt(17)-1. Rewrite x as 1/y, multiply by y and solve the resulting second degree equation for y.
@johnjoyce2508 3 жыл бұрын
Ahh when you put it that way it makes so much sense!!!
@erotic_diaries 3 жыл бұрын
I found out that y= 0.36 if x= 2.76 or y=2.76 if x= 0.36 (approximate answers)
@walter7822 4 жыл бұрын
I have no idea how to solve this and I just guessed and got 2.75 4-1=3 1/4=0.25 3-0.25=2.75
@masterchieftheconqueror2631 4 жыл бұрын
Lol Same
@mohamedhusam8189 4 жыл бұрын
I rolled some dice and got a guess of 2.76090563295442
@walter7822 4 жыл бұрын
@@mohamedhusam8189 How do you get this from rolling dice?
@nemesismatrix9603 4 жыл бұрын
Walter r/woooosh
@lucianozaffaina9853 4 жыл бұрын
Walter I have a dice that have the number from 1 to 6 with 73 decimals
@tomtheultimatepro 3 жыл бұрын
I solved it using Pythagoras theorem and similar triangles. The resulting 4th degree equation can actually be rearranged nicely so one can use substitution and solve two quadratic equations. This method yields the exact solution of y = 1/2 (-1 + sqrt(17) + sqrt(14 - 2 sqrt(17))) which is approximately 2.76. No calculator needed :)
@myasiatech9849 3 жыл бұрын
y^4+2y^3-14y^2+2y+1=0 ??
@myasiatech9849 3 жыл бұрын
(1+1/y)^2＋(1+y)^2-16=0
@piyushpriyadarshandas5630 3 жыл бұрын
I was using just 9th standard concept........ But..... The solution is heart touching.... ♥️♥️Love that visualisation........
@manav821 4 жыл бұрын
Your visuals are amazing .... But keep calm and use similarity ... 😉😉
@Choinkus 2 жыл бұрын
I feel like looking at what the answer is, that shows something along the way is clearly a bit more complicated. And it turns out that if you use similarity and Pythagoras, you end up with a quartic equation.
@joeljolly5856 2 жыл бұрын
4:57 x÷c = c÷y c=1 (side of cube) So , x÷1 = 1÷y ie x = 1÷y so, xy = 1 (1×1=1) So, x = 1 y = 1 c = 1 Everything is 1 . Ok ,THANK YOU 🏃🏃🏃
@theunknownguy265 2 жыл бұрын
Bruh
@theunknownguy265 2 жыл бұрын
Nerd
@monroeclewis1973 7 ай бұрын
I solved this problem with geometry, algebra, and a graphing calculator. The later just saved time instead of graphing a 4th degree equation by hand. There are two solutions: .362 and 2.761. Here goes: by the Pythagorean equation, 4^2 = (y + 1)^2 + (z + 1)^2. (Call the small leftover space adjacent square, “z.”) By similar triangles, 1/y = z/1. Therefore, “z” = 1/y. Now substitute 1/y for z in the Pythagorean equation above. After simple algebra we get the following equation: y^4 + 2y^3 - 14y^2 +2y + 1 = 0. This equation doesn’t factor by synthetic division, grouping or any other purely algebraic method. So I graphed using a popular app. The line crossed the y axis where x = 0 in two places as discussed above at 3.62 and 2.761. And that’s the answer! 😊
@anotherpromotor 29 күн бұрын
I wanted to try a problem on my own for once and I actually got pretty close. I got the sqrt(13) - 1, which is about 2.6, which is only about .1 off from the correct answer. I'm very proud of myself for this one.
@rtfacts5317 4 жыл бұрын
When you think you are the only person to solve it but then sees comment section
@splitzerr6194 4 жыл бұрын
RIP English
@aatheeswarank7025 4 жыл бұрын
These people are math nerds . What did you expect lol
@bhabanishankarrath3973 4 жыл бұрын
@@aatheeswarank7025 this is what i want to say
@bhabanishankarrath3973 4 жыл бұрын
@@splitzerr6194 we are math and physics (ok,ok i am only fan of physics not other people here isn't) nerds bro .
@kvamshi4561 3 жыл бұрын
I've solved this same problem using coordinate geometry when I was in 11th.. Pretty complicated one though!
@tdranger6888 4 жыл бұрын
Two solutions of y are 0.36 & 2.76 found by solving a 4th degree polynomial with coefficients 1, 2, -14, 2, 1 from Pythagoris' Theorem, somewhat more difficult than your elegant construction.
@parkerpurciful7676 3 жыл бұрын
For my solution, I worked in the first quadrant. I realized that our three constraints were that (1) the solution must be a line, (2) that the point (1,1) must fall on this line, and (3) that the length of the line in the first quadrant must be 4. We satisfy (1) by taking y=mx+b. Using (2), we find that m=1-b and that the x-intercept is b/(b-1). To satisfy (3), we use the Pythagorean theorem, finding that 4^2=b^2+(b/(b-1))^2. This can be simplified to b^4-2b^3-14b^2+32b-16=0. Solve this equation and discard the solutions with b
@satyampriyadarshi5349 3 жыл бұрын
This level of intuition😳😳 that's why I'm poor at maths....I would use lenght tape
@RobertShield62 4 жыл бұрын
3:32 how do you know that?
@Ra.Sallam 4 жыл бұрын
I've the same question
@devnakz 4 жыл бұрын
It's because the area of the triangle is equal to half the total area of a rectangle that circumscribes it.
@NordicMe 4 жыл бұрын
Z has a base of x+y and height c If you take triangle X and paste it to the left of Y, you create a triangle (X&Y) with base x+y and height c. Since area of triangle=base*height/2, these two triangles have the same area. In general, if you cut up a rectangle starting from two corners, making 3 triangles in total, the area of the big one will equal the area of the two smaller ones.
@treesarecool12345678 4 жыл бұрын
Draw a line from the right angle in triangle Z perpendicular to the side of the big square... you should be able to see it then 🙂
@arjundevchoudhary2585 4 жыл бұрын
Because the first and second triangle ( second triangle is merge of two X and Y triangle) have same height and same base so area is also same
@chigeryelam4061 8 ай бұрын
"We get the exact answer for Y which is approximately ..." I laughed a little bit here.
@Joniruchi 9 күн бұрын
Nice and visual solution! Personally I prefer the more direct algebraic solution, which is also mathematically intuitive in its own way, but this was a pleasing video for sure!
@ploypraplusboontanaakrapat6365 3 жыл бұрын
This is literally the same question asked on the exam to get into 7th grade in my country when I was in 6th grade
@somethingelse9228 3 жыл бұрын
which country are you from?
@sudhajaiswal1029 3 жыл бұрын
WHICH COUNTRY?!☠️☠️
@user-ed7oz3ee2u 4 жыл бұрын
Presh: an elegant solution Me: huh "elegant"? I mean, there are far easier ways to solve it xd
@Astitva 3 жыл бұрын
Elegant is beautiful
@boranchen3246 3 жыл бұрын
Actually, the other way we will end up with a 4th degree polynomial xd
@MACIEJ454545 2 жыл бұрын
I ended with a 4th degree polynomial and after not finding any intiger fractions I gave in and watched the rest I did not expect such a complex answer
@kavithavelu8282 5 ай бұрын
If you have continued,you would have got the answer
@titan1235813 3 жыл бұрын
I reached to the equation y^8 - 32 y^6 + 190y^4 - 32y^2 + 1 = 0, and one of the solutions was indeed y = 2.7069 (to 4 decimal places).
@muhittinylar697 3 жыл бұрын
he: waits three seconds for me but at the same time he: solves at five minutes
@morez39400able 4 жыл бұрын
you must have 2 solutions ; 2.76 and 0.36 , ( y > x , y < x) , for(x,y) ={(2.76,0.36),(0.36,2.76)} (y+1)² + (x+1)² = 4² and y=1/x replace y by 1/x and multi by x² x⁴ +2x³ -14x² + 2x + 1 = 0
@szehoonglew1863 4 жыл бұрын
Yes, i think so. If we consider the mirror image of the diagram and rotate the mirror image anticlockwise 90 degree, then the y' can be x. Is there any other conditions to make y > x ?
@BCSEbadulIslam 4 жыл бұрын
There should be two answers to this question. One when ladder makes an angle greater than 45° with floor, and one for when ladder makes an angle less than 45° with floor (that corresponds with x).
@patrickjacquiot9073 2 жыл бұрын
Obvious.
@rcnayak_58 4 жыл бұрын
The present solution is obviously very complex. you cannot imagine to start like this in order to get the answer. It is simple one. No construction required. From similar triangles, x /1 = 1/y. Or x = 1/y. Again, (1+x)^2 + (1+y)^2 = 16. Put the value of x in this equation and just rearrange the terms. It will be (y+1/y)^2 + 2 (y+1/y) -16 = 0. Assuming y+1/y = m, you have a quadratic equation involving m, solve for m which is -2+sqrt(17) which is 3.1231, taking positive value as both x and y is positive. Again solve y from m value, i.e., y+1/y = 3.1231, leading y = 2.7609. You have a general solution for y having 4 values of y (considering negative values). Presh made it only complicated.
@lucaschai5788 4 жыл бұрын
Someone just commented 5 days ago before the video is uploaded
@joan.gonzalez 4 жыл бұрын
Maybe patreon?
@xfire3778 4 жыл бұрын
Joan González Peña yep
@qqq1234x 4 жыл бұрын
Paradox confirmed~
@taylaarbon5611 4 жыл бұрын
I’m so lost 😂
@rifatzehra6546 4 жыл бұрын
Why to lost....it was not difficult?
@xXDarQXx 4 жыл бұрын
@@rifatzehra6546 it's definitely not simple. It took me 3 hours. How much did it take you?
@qqq1234x 4 жыл бұрын
@@xXDarQXx i was lost from the very beginning xD
@fredericmazoit1441 3 жыл бұрын
Also this solution is indeed elegant, it make this problem look needlessly complicated. This can be solved only by quite basic calculus. Let us stop the figure before the completion with the 3 copies of the triangle. First, let us note that we have 4 letters (x, y, c and l). If we scaled everything by a factor 1/c, we obtain a relation between x/c, y/c, 1 and l/c. Let thus X=x/c, Y=Y/c and L=l/c. If we can find Y (and X), then y=cY and we are done. What is nice is that the square now has size 1. We remark that we have 2 triangles that are similar. Thus 1/X=Y (eq1). Now, by Pythagoras, we have (X+1)^2+(1+Y)^2=L^2 (eq2). If we develop eq2, we obtain (X^2+2X+1)+(1+2Y+Y^2)=L^2. (eq2') This equation is completely symmetric in X and Y. It thus can be expressed in term of (X+Y) and XY. But note that XY=1. By playing a little bit, we easily obtain(eq2') (X+Y)^2+2(X+Y)=L^2 (eq2'). And now everything is straightforward. We have an equation of degree 2 in X+Y which can easily be solved. We thus obtain X+Y=foo which is also an equation of degree 2 in Y (X=1/Y).
@anandk9220 4 жыл бұрын
I ORALLY constructed the simplified equation to this. Here's what I did orally :- The lower (smaller) and upper (larger) right triangle are similar. So their ratio of corresponding sides is 1/y and hence the length of base of lower right triangle is 1/y (because base length of upper right triangle is 1). Now, applying Pythagoras Theorem to lower and upper right triangles respectively, I obtained their respective hypotenuses to be, Upper hypo = √[1 + (y^2)] Lower hypo = √[1 + 1/(y^2)] Adding both the above, √[1 + (y^2)] + √[1 + 1/(y^2)] = 4 Simplifying the equation step by step orally, √[1 + 1/(y^2)] = 4 - √[1 + (y^2)] Squaring both sides and simplifying, [1 + 1/(y^2)] = 17 + (y^2) - 8√[1 + (y^2)] Taking LCM and shifting terms to left, { [(y^2) + 1]/(y^2) } -17-(y^2) = - 8√[1 + (y^2)] Simplifying, 1 - 16(y^2) - (y^4) = - 8(y^2) × √[1 + (y^2)] Now, I think there's no alternative but to square both sides and simplify, and I guess even that won't suffice. I tried till here orally (in 1 successful attempt) and since further oral solving was out of bounds confusion, I chose peace by using Wolfram Alpha and obtained 6 different answers, of which 3 were negative hence invalid, one was 0.36 (too less as per diagram hence invalid) and the other was greater than 4, hence invalid. So the only leftover was the correct answer i.e. 2.7609. I hope I've done fine work if not remarkable. 😄😊👍
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