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 😉
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.
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
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.
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
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..
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.
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.
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.
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
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.
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
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!
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]
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)
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...
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
I see many went for the analytical solution,which solve the specific case easy but the the general case is pretty much impossible,this really shows you the difference between geometric and analytical solutions.
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
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.
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).
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. ^-^
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.
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.
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!
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 😉
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.
@@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.
@@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?
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
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.
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.
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.
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
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.
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.
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!
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.)
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 :)
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. 😄😊👍
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...)
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.
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
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
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 ?
for those confused by the area equivalence. try this. using original arithmetic problem by similar triangles xy=1. Thus if we define X=x+1 and Y=y+1 then L^2=16=X^2+Y^2 but XY= xy+x+y+1= X+Y , so 16=(X+Y)^2-2XY define T = X+Y , 16 =T^2 -2T note that T=w+2 in Presh’s notation. we can then solve for T as for w. T= 1+sqrt(17) = y+1/y +2 and use the quadratic formula again to solve for y. Note that use of pythagorean triples eg L=24, c=7 leads to simpler numbers at the end.
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
By similar triangles, y/1 = 1/x. By Pythagorean Theorem (x+1)^2 + (y+1)^ = 16. Substituting x = 1/y into this and rearranging gives (y^2+(1/y^2)) + 2(y+1/y) -14 = 0. This is a reciprocal equation which doesn't change when y is substituted for 1/y and can be converted to a quadratic by using the substitution z = y+1/y. This makes y^2+1/y^2 = z^2-2 and converts the equation into a quadratic. Each of the two solutions to this equation can be put back into z = y+1/y to give two more quadratics and get the 4 roots. Two roots are negative; the positive ones are 2.7609 and 0.36220.
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.
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
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?)
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* .
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.
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.
Presh, Solution you mention is a beautiful concept but IMHO, a tad complex unnecessarily. Here is a much simpler alternative. Break 4 into two parts a & b for smaller triangle and bigger triangle (where 4 length side is touched by squares right top vertex). So a + b = 4. We know sin² theta + cos² theta =1 => (1/a)² + (1/b)² =1 ==> a² +b² = a²b² ==> a² +b² +2ab = a²b² +2ab =>16 = (ab)² + 2ab ==> let’s take X = ab => X² +2X-16 = 0 ==> X = -1+ SQRT (17) = ab. We know a+b=4. So we now know b= 2+sqrt(5-sqrt(17)). So we know y = SQRT(b² -1) => y ~= 2.76.
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.
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!
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!
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
Nice problem! Unlike other 'solvers', as an old numerically-incompetent computer scientist, I try to solve things using iterated successive approximation methods. Not really all that hard. In this case, there are 2 simple observations (Pythagoras, similar triangles and their proportionality) № 1 - 𝒛² = 4² - (𝒙 ⊕ 1)² where 𝒛 is (𝒚 + 𝒄) in your diagram. № 2 - 𝒉 / 𝒙 = 4 / (𝒙 ⊕ 1 ) where '𝒉' is the hypotenuse of the ladder touching floor, and left side of unit cube. A little algebraic expansion becomes № 1a - 𝒛 = √( 15 - 2𝒙 - 𝒙²) № 2a - (𝒙 ⊕ 1) √( 𝒙² ⊕ 1 ) / 𝒙 = 4 Using '𝒅𝒙 = 0.01' and initial 𝒙 = 𝒅𝒙, iterating along, one finds that equation № 2a starts out high (above 4), and drops rapidly. Cool. As 𝒅𝒙 is halved, and halved again at the crossover, the FIRST solution is № S1 - 𝒙 → 0.362200 and 𝒛 → (𝒚 + 𝒄) → 3.760306 However, inspection shows that № 2a inflects below '4', and starts rising again. Solving for that section by a similar repeated halving of 𝒅𝒙 around '4' leads to № S2 - 𝒙 → 2.760906 and 𝒛 → (𝒚 + 𝒄) → 1.362200 Kind of makes sense: who says the ladder is going to be upright and tall? Could be almost lying down and would be 'a solution' too!!! There you are. I'ven't the math chops to do what you did in the video, but using computers to COMPUTE things like this to surprisingly easily achieved levels of precision (in my case, 16 digits internally in only 100 iterations!) is a piece of cake. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=- Here's the code: #!/usr/bin/perl -w # ---------------------------------------------------------------------------------------------------- # ---------------------------------------------------------------------------------------------------- use Math::Trig; use strict; my $x; # bit to the left of the box, touching the baseline; my $z; # the 'rise' up the wall; my $h; # the 'little hypotenuse' of 'x' and the cube's wall; my $m; # the test-value inequality. Determines when a solution has been found; my $box_side = 1.0; printf "box side is: %f ", $box_side; my $ladder = 3.0; printf "ladder is: %f ", $ladder; my $dx_reset = 0.1; my $dx = $dx_reset; my $phase = 0; # to 'switch' from finding the first, to the 2nd solution; my $loops = 0; # keep track of how many loops done; for( $x = $dx; $x < 5; $x += $dx ) { $loops++; $z = ($ladder ** 2 - $box_side ** 2) - $x ** 2 - 2 * $box_side * $x; next if $z < 0; $z = sqrt $z; $m = ($x + $box_side) * sqrt( $x ** 2 + $box_side ** 2 ) / $x; if( $phase == 0 ) { next if $m > $ladder; $x -= $dx; # back off the dx, try again; $dx /= 2.718281828; # see below; if( $dx < 1e-11 ) # 'done enough'; { $dx = $dx_reset; printf "loop %4d ... x = %f, z = %f, m = %f ", $loops, $x, $z, $m; $phase = 1; next; } } else { next if $m < $ladder; $x -= $dx; # see above; $dx /= 2.718281828; # surprisingly... 'e' is better than '2', on average!!!; if( $dx < 1e-11 ) # 'done enough'; { printf "loop %4d ... x = %f, z = %f, m = %f ", $loops, $x, $z, $m; last; } } } printf "if you see no solutions, then the value of 'box' an 'ladder' do not allow a solution. "; ⋅
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.
The geometric solution given above is quite elegant (although arcane). However, the algebraic solution is fairly straightforward. By similarity, c + x = c(y + c)/y. It's one side of the large right-angle triangle; so we have: (y + c)^2 + (c(y + c)/y)^2 = L^2 Expanding this would produce a quartic equation, which generally have complicated solutions, although this one is solvable because of its symmetry. However, the equivalent result can be obtained by factoring out (y + c)^2 in the above equation and expanding it to give: (y^2 + 2yc + c^2)(1 + c^2/y^2) = L^2 The trick here is to divide the first factor by y (since y != 0) and multiply the second by y to give: (y + c^2/y + 2c)(y + c^2/y) = L^2 This is a quadratic equation in (y + c^2/y), which can be solved in the usual way, which then gives another quadratic equation in y.
Two exact solutions for the angle formed between the wall and ladder for a given ladder "l" and box "c" can be calculated using this equation θ=sin⁻¹((2±2√1+(l/c)²)/(l/c)²)/2. This equation is derived from l=c/sinθ+c/cosθ using trig identities and the quadratic formula. The solutions for x and y are l·cosθ and l·sinθ respectively.
😵 ½·(√17 − 1 + √(14 − 2√17)) This solution kinda reminds me of: sin12° = ¼·√(7 − √5 − √(30 − 6√5)) Solving is not so intuitive, but shows the beauty of irrational expressions. Thanks, Presh!
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! 😊
I did ....i did... MAN, put screenshoot, or some prove that you solve. Easy to say but not put solution. If you are subscriber here, you should, theoretical, be seriously, man
For similar triangles, we would get: x = 1/y From Pythagoras Theorem, we would get: (1+x)^2 + (1+y)^2 = (1+1/y)^2 + (1+y)^2 = 4^2 = 16 => (1+2/y+1/y^2) + (1+2y+y^2) = 16 Rearranging terms we get: => 2 + 2(y + 1/y) + (y^2 + 1/y^2) = 16 Assuming z = y+1/y, squaring both side, z^2 = y^2 + 1/y^2 + 2. Substituting in above equation, we get: 2+ 2z + z^2-2 = 16 Solving for z, we get: z = -1+sqrt(17) = y+1/y Above quadratic equation can be solved for 'y'
Same result using Pythagoras and the original drawing (without the helping central square): From the two small similar triangles, we have xy = c^2. Now, from the ladder big triangle, we have (x+c)^2 + (y+c)^2 = L^2, (let me use capital L, so that it won't be confused as 1 (one), or capital I). this means x^2 + 2cx + c^2 + y^2 + 2cy + c^2 = L^2 => x^2 + 2cx + c^2 + y^2 + 2cy = L^2 - c^2 => x^2 + 2cx + c^2 + y^2 + 2cy + 2xy = L^2 - c^2 + 2xy => (x+y+c)^2 = L^2 - c^2 + 2c^2 .........(since xy = c^2) => (x+y+c)^2 = L^2 + c^2 => x + y + c = +- sqrt(L^2 + c^2) .........(take positive only, since x+y+c > 0) Let w = x + y, w = -c + sqrt(L^2 + c^2). This is exactly the same result as that in the video. Starting here, the rest is the same as the 2nd part (i.e. w = x + y = c^2/y + y) of the video, since this 2nd formula requires only the similar triangle relationship, no need for the helping central square.
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).
Moral of the story: Never put a ladder and a cube in the same room. Thank you for your understanding. 🙏
Loooool 🤭🤭
Nice joke !
YOUR NAME LOL
👌👌🤣🤣
😂😂😂
You really chose the most complicated way of solving a relatively easy problem.
No this is the only simplest way to solve it
@@badalkumar7185 wrong, there is an easier one, considering that y/c = c/x and (x+1)^2 + (y+1)^2 = l^2
@@jakubedzior thats what i used to solve it but i did a mistake somewhere 😅 oh well. Ratios all the way!
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 😉
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.
As others have pointed out, solving through similar triangles is another good way of looking at it.
yup its easy to do with similar triangles
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
Yes
Yup
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.
Me: Trying to do something with Pythagoras theorem😂
Me too, And my result was closed to correct y = 2.7957 :D
Me too and my answer was equal to 2.7
Mine was 2.76. I feel good.
ngl same
how the hell did you all solve it with pythagoras theorem
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
Rishi Chugh yes I Also found 2 solutions but I thought Y >X because of the diagram
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..
lol thank you i was so confused
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.
U hv 99 likes and I am gonna give u ur 100th like
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.
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.
I believe you want to say Gongu theorem!!
Essentially the same way I used to solve this problem.
but then exponential powers would all be different rite?
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
Year, I solved like this way)) there is similar equation in Russian exam where you can solve with help of such methods
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.
I did the same. I think that since, based on the diagram, y>1, the 0.36 solution can be disregarded
The bi quadratic equation right? Yea, i did the same and gave up on factoring the roots. Used desmos
i came up with the same idea
Me: Trying to solve the problem by similar triangles.
Presh: Creates a square using the diagram.
Me: *surprised Pikachu face*
Wow that's mean that I wasn't alone
When I thought that similarity of the triangle would solve it easily
Me too,
I also use similar triangle principle, in solving the problem
tbh almost all of us, use similar triangles so, no surprise here lol.
Pi ka pi? 😳
for the specific case of cube side 1 ..... it feels similar triangles is easier .... (bottom x+1 side 1+ 1/x) .... Or not 🤦
i look forward to the day, these videos are watchable again without stressing heartbeat
Just remember when you met similar questions written on a piece of paper... at 8:00 AM.... in high school.... :D :D
@@tcfs i don't have to remember, cuz i still meet that
Pretty complicated solution!
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
Use similar triangle for pretty easy solution
drive.google.com/folderview?id=14F2jmF4E7AaCk9GJPgg_qx1itnwQ-_FQ
@@prateekmourya9567 no Calculator allowed
overcomplicated, actually
I love these videos, they're so boring it helps me fall asleep faster.
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
typical engineering student
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
Found the engineer!
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
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!
Very interseting for sure!!
You wrote an entire mathematical paper
Using similarity, simple algebra and Pythagoras theorem. I found the answer.
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]
I have also solved it by this technique.. Hifi!!! 😊
as any sane person should XD
Your graphics during explaining is fantastic
The problem looks quite hard, but the moment you try solving it
It's even harder
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)
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...
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
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
Me: Watches the video thoroughly Also me: what just happened?
1:51
"Now we are going to do a little trick"
This is where things start to go wrong...
He did what's called a pro-gamer move.
Fossky The Answerer Ikr?!
@@fosskytheanswerer bruh , he just pulled a sneaky on us
I see many went for the analytical solution,which solve the specific case easy but the the general case is pretty much impossible,this really shows you the difference between geometric and analytical solutions.
Can't we do it with similar triangles
Yes.. similarity can be used
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
Yes. Check my comments in the video
@@rabindranathghosh31 i got that too, but how to get exact value of y??
I tried to do that I ran into complex numbers. So it's harder
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.
I have been watching your videos since a long time , you are an inspiration for me
*for a long time
Thank you so much. I’ve struggled with this problem for years.
its too complex; simply use the concept of similar triangle, and Pythagoras theorem. Its simple....Any way your solution is interesting as well
noob!!
Jakob Fredin I did for general equation also and got equation of fourth power which on solution got the required answer
It's true. I use similar triangle and have the same result ~2.76
yes, but did you have to use a computer to solve a 4th degree polynomial?
Solving a quartic equation is hardly "simple," at least for the kind of audience these videos are for.
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).
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. ^-^
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.
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.
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!
sir which software do you use?
pleaese respond
@@gardenmenuuu I actually use PPT for effects in graphics
MathFlix yours are amazing as well
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 😉
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.
@@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.
@@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?
I used the tap measure from my moms sewing kit. And she recognised her son solved it in the fastest way.
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
Yes I did in in the same way
i did same dude
Same here.
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.
Doesn't that give you a quartic?
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.
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.
What? Have 2.7M subscribers and do something straightforward? Please be reasonable in what you expect :)
“How many variables do you want?”
Him: yes
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
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.
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?
I'm disappointed that nobody answered this question...
I'd very much like to know this answer too
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.
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!
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
Absolutely!
Then I had a degree 4 equation :
y^4+2y^3-14y^2+2y+1=0
Maurice Vassilitch 😂 ppl need to actually solve before commenting how “easy” their “solution” is.
@@prateekmourya9567 I really liked how you turned the 4th degree equation to a 2nd
Same
first equation ok but for the second why not use x=1/y and get a symetric equation of 4th degree?
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.)
Sorry, y used twice but you get the drift.
i thought pythagoras theorem(gougu theorem ) will be used in this video!!!
lol
Actually you need nothing more than pythagoras' theorem and similar triangles to solve this.
@@DrHades is it??
how!!!
@@sumathip3687 drive.google.com/folderview?id=14F2jmF4E7AaCk9GJPgg_qx1itnwQ-_FQ
@@rifnasriff166 hope presh talwalker realises your work!!
and thank you for your answer!!!
Constructing these squares is the same thing as using p. Theorem and similar triangles, just a longer path
Wow ! Very difficult this one ! And how elegant the solution ! Thank you
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.
Omg same out of curiosity did you also get 2.82
@@therion8469 yes
@@therion8469 though i used calculator
@@therion8469 how did you get 2.82 ?
If possible please write the solution
@@codingkrtehum.. Considering the upper triangle,
cos( theta)=1/3 --> theta = 70.52
then, tan (theta) = y/3
y=2.8
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 :)
y^4+2y^3-14y^2+2y+1=0 ??
(1+1/y)^2＋(1+y)^2-16=0
Am i the only one who dont know whats going on but still watches the full video
No, I fully understand everything without even following his explanations all too closely. Also I'm a big liar.
A triangle has 3 sides and a ham sandwich is lunch?
@@winzkling The last one got me😂
Like watching 5 dimensional beings
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.
😄😊👍
And I thought I was good at math
Very interesting solution. Thanks Presh.
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...)
U R in which class?
@@ashmitmehta5088 At that time, I was in the 11th class, but that's some 25 years ago.
@@jbglaw oh!!
Actually i am just promoted few days back to 10th...
So everything went out of my head
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.
After seeing the problem, I was like aight bet...until I saw the rest of the video.
Fml 🤦🏽♂️🤦🏽♂️🤦🏽♂️
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
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
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 ?
for those confused by the area equivalence. try this.
using original arithmetic problem
by similar triangles xy=1. Thus if we define X=x+1 and Y=y+1 then L^2=16=X^2+Y^2
but XY= xy+x+y+1= X+Y , so 16=(X+Y)^2-2XY define T = X+Y , 16 =T^2 -2T
note that T=w+2 in Presh’s notation.
we can then solve for T as for w. T= 1+sqrt(17) = y+1/y +2
and use the quadratic formula again to solve for y.
Note that use of pythagorean triples eg L=24, c=7 leads to simpler numbers at the end.
Use similarity and quadratic directly. (y+1) ^2 + (1/y +1)^2 = 4^2
It's incorrect bruh...How can you take horizontal distance (base) as 1/y +1 ....Moreover ans isn't coming through this
Also need similar triangle
This is a 4th order equation and hard to solve. I think there is a way to do it but is very complex.
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
By similar triangles, y/1 = 1/x. By Pythagorean Theorem (x+1)^2 + (y+1)^ = 16. Substituting x = 1/y into this and rearranging gives (y^2+(1/y^2)) + 2(y+1/y) -14 = 0. This is a reciprocal equation which doesn't change when y is substituted for 1/y and can be converted to a quadratic by using the substitution z = y+1/y. This makes y^2+1/y^2 = z^2-2 and converts the equation into a quadratic. Each of the two solutions to this equation can be put back into z = y+1/y to give two more quadratics and get the 4 roots. Two roots are negative; the positive ones are 2.7609 and 0.36220.
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.
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
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?)
@@zomzombos8257 The video was unlisted for a while, available to Patreon supporters though - so I got an early access.
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* .
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.
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.
Your visuals are amazing .... But keep calm and use similarity ... 😉😉
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.
Presh,
Solution you mention is a beautiful concept but IMHO, a tad complex unnecessarily. Here is a much simpler alternative. Break 4 into two parts a & b for smaller triangle and bigger triangle (where 4 length side is touched by squares right top vertex). So a + b = 4. We know sin² theta + cos² theta =1 => (1/a)² + (1/b)² =1 ==> a² +b² = a²b² ==> a² +b² +2ab = a²b² +2ab =>16 = (ab)² + 2ab ==> let’s take X = ab => X² +2X-16 = 0 ==> X = -1+ SQRT (17) = ab. We know a+b=4. So we now know b= 2+sqrt(5-sqrt(17)). So we know y = SQRT(b² -1) => y ~= 2.76.
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
Lol Same
I rolled some dice and got a guess of 2.76090563295442
@@mohamedhusam8189 How do you get this from rolling dice?
Walter r/woooosh
Walter I have a dice that have the number from 1 to 6 with 73 decimals
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.
This level of intuition😳😳 that's why I'm poor at maths....I would use lenght tape
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!
I found out that y= 0.36 if x= 2.76 or y=2.76 if x= 0.36 (approximate answers)
I was using just 9th standard concept........ But..... The solution is heart touching.... ♥️♥️Love that visualisation........
he:
waits three seconds for me
but at the same time he:
solves at five minutes
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!
When you think you are the only person to solve it but then sees comment section
RIP English
These people are math nerds . What did you expect lol
@@aatheeswarank7025 this is what i want to say
@@rishit_gandash we are math and physics (ok,ok i am only fan of physics not other people here isn't) nerds bro .
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
Presh: an elegant solution
Me: huh "elegant"?
I mean, there are far easier ways to solve it xd
Elegant is beautiful
Actually, the other way we will end up with a 4th degree polynomial xd
Nice problem!
Unlike other 'solvers', as an old numerically-incompetent computer scientist, I try to solve things using iterated successive approximation methods. Not really all that hard.
In this case, there are 2 simple observations (Pythagoras, similar triangles and their proportionality)
№ 1 - 𝒛² = 4² - (𝒙 ⊕ 1)² where 𝒛 is (𝒚 + 𝒄) in your diagram.
№ 2 - 𝒉 / 𝒙 = 4 / (𝒙 ⊕ 1 ) where '𝒉' is the hypotenuse of the ladder touching floor, and left side of unit cube.
A little algebraic expansion becomes
№ 1a - 𝒛 = √( 15 - 2𝒙 - 𝒙²)
№ 2a - (𝒙 ⊕ 1) √( 𝒙² ⊕ 1 ) / 𝒙 = 4
Using '𝒅𝒙 = 0.01' and initial 𝒙 = 𝒅𝒙, iterating along, one finds that equation № 2a starts out high (above 4), and drops rapidly. Cool. As 𝒅𝒙 is halved, and halved again at the crossover, the FIRST solution is
№ S1 - 𝒙 → 0.362200 and 𝒛 → (𝒚 + 𝒄) → 3.760306
However, inspection shows that № 2a inflects below '4', and starts rising again. Solving for that section by a similar repeated halving of 𝒅𝒙 around '4' leads to
№ S2 - 𝒙 → 2.760906 and 𝒛 → (𝒚 + 𝒄) → 1.362200
Kind of makes sense: who says the ladder is going to be upright and tall? Could be almost lying down and would be 'a solution' too!!!
There you are. I'ven't the math chops to do what you did in the video, but using computers to COMPUTE things like this to surprisingly easily achieved levels of precision (in my case, 16 digits internally in only 100 iterations!) is a piece of cake.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-
Here's the code:
#!/usr/bin/perl -w
# ----------------------------------------------------------------------------------------------------
# ----------------------------------------------------------------------------------------------------
use Math::Trig;
use strict;
my $x; # bit to the left of the box, touching the baseline;
my $z; # the 'rise' up the wall;
my $h; # the 'little hypotenuse' of 'x' and the cube's wall;
my $m; # the test-value inequality. Determines when a solution has been found;
my $box_side = 1.0; printf "box side is: %f
", $box_side;
my $ladder = 3.0; printf "ladder is: %f
", $ladder;
my $dx_reset = 0.1;
my $dx = $dx_reset;
my $phase = 0; # to 'switch' from finding the first, to the 2nd solution;
my $loops = 0; # keep track of how many loops done;
for( $x = $dx; $x < 5; $x += $dx )
{
$loops++;
$z = ($ladder ** 2 - $box_side ** 2) - $x ** 2 - 2 * $box_side * $x;
next if $z < 0;
$z = sqrt $z;
$m = ($x + $box_side) * sqrt( $x ** 2 + $box_side ** 2 ) / $x;
if( $phase == 0 )
{
next if $m > $ladder;
$x -= $dx; # back off the dx, try again;
$dx /= 2.718281828; # see below;
if( $dx < 1e-11 ) # 'done enough';
{
$dx = $dx_reset;
printf "loop %4d ... x = %f, z = %f, m = %f
", $loops, $x, $z, $m;
$phase = 1;
next;
}
}
else
{
next if $m < $ladder;
$x -= $dx; # see above;
$dx /= 2.718281828; # surprisingly... 'e' is better than '2', on average!!!;
if( $dx < 1e-11 ) # 'done enough';
{
printf "loop %4d ... x = %f, z = %f, m = %f
", $loops, $x, $z, $m;
last;
}
}
}
printf "if you see no solutions, then the value of 'box' an 'ladder' do not allow a solution.
";
⋅
Search for x*y = c*c on the circle (x+c)**2 + (y+c)**2 = l*l . Wouldn't be easier to parameterize the circle?
3:32 how do you know that?
I've the same question
It's because the area of the triangle is equal to half the total area of a rectangle that circumscribes it.
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.
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 🙂
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
"We get the exact answer for Y which is approximately ..." I laughed a little bit here.
Someone just commented 5 days ago before the video is uploaded
Maybe patreon?
Joan González Peña yep
Paradox confirmed~
Ahh when you put it that way it makes so much sense!!!
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 🏃🏃🏃
Bruh
Nerd
The geometric solution given above is quite elegant (although arcane). However, the algebraic solution is fairly straightforward.
By similarity, c + x = c(y + c)/y. It's one side of the large right-angle triangle; so we have:
(y + c)^2 + (c(y + c)/y)^2 = L^2
Expanding this would produce a quartic equation, which generally have complicated solutions, although this one is solvable because of its symmetry.
However, the equivalent result can be obtained by factoring out (y + c)^2 in the above equation and expanding it to give:
(y^2 + 2yc + c^2)(1 + c^2/y^2) = L^2
The trick here is to divide the first factor by y (since y != 0) and multiply the second by y to give:
(y + c^2/y + 2c)(y + c^2/y) = L^2
This is a quadratic equation in (y + c^2/y), which can be solved in the usual way, which then gives another quadratic equation in y.
This is literally the same question asked on the exam to get into 7th grade in my country when I was in 6th grade
which country are you from?
WHICH COUNTRY?!☠️☠️
Two exact solutions for the angle formed between the wall and ladder for a given ladder "l" and box "c" can be calculated using this equation θ=sin⁻¹((2±2√1+(l/c)²)/(l/c)²)/2. This equation is derived from l=c/sinθ+c/cosθ using trig identities and the quadratic formula. The solutions for x and y are l·cosθ and l·sinθ respectively.
I've solved this same problem using coordinate geometry when I was in 11th.. Pretty complicated one though!
😵 ½·(√17 − 1 + √(14 − 2√17))
This solution kinda reminds me of:
sin12° = ¼·√(7 − √5 − √(30 − 6√5))
Solving is not so intuitive, but shows the beauty of irrational expressions.
Thanks, Presh!
this can be easily proved by assuming the lower angle Theta
and then tan(theta)=1/x
and also tan(theta)=y+1/1+x
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! 😊
I’m so lost 😂
Why to lost....it was not difficult?
@@rifatzehra6546 it's definitely not simple. It took me 3 hours.
How much did it take you?
@@xXDarQXx i was lost from the very beginning xD
complicated solution , thoroughly enjoyed the video.
I solved it with coordinate geometry it's easier than this method
I did ....i did... MAN, put screenshoot, or some prove that you solve. Easy to say but not put solution. If you are subscriber here, you should, theoretical, be seriously, man
@@cristinicca And if you have any doubt you can ask
@@light.236 ok, pls, add your solution. Tks
@@cristinicca prntscr.com/sui0mu
Bro this is the solution
@@light.236 tks, bro! (Or sis?)
For similar triangles, we would get: x = 1/y
From Pythagoras Theorem, we would get: (1+x)^2 + (1+y)^2 = (1+1/y)^2 + (1+y)^2 = 4^2 = 16
=> (1+2/y+1/y^2) + (1+2y+y^2) = 16
Rearranging terms we get:
=> 2 + 2(y + 1/y) + (y^2 + 1/y^2) = 16
Assuming z = y+1/y, squaring both side, z^2 = y^2 + 1/y^2 + 2. Substituting in above equation, we get:
2+ 2z + z^2-2 = 16
Solving for z, we get: z = -1+sqrt(17) = y+1/y
Above quadratic equation can be solved for 'y'
No "Gou Gu" theorem, very disappointed. 😉😀
Same result using Pythagoras and the original drawing (without the helping central square):
From the two small similar triangles, we have xy = c^2.
Now, from the ladder big triangle, we have (x+c)^2 + (y+c)^2 = L^2, (let me use capital L, so that it won't be confused as 1 (one), or capital I).
this means x^2 + 2cx + c^2 + y^2 + 2cy + c^2 = L^2
=> x^2 + 2cx + c^2 + y^2 + 2cy = L^2 - c^2
=> x^2 + 2cx + c^2 + y^2 + 2cy + 2xy = L^2 - c^2 + 2xy
=> (x+y+c)^2 = L^2 - c^2 + 2c^2 .........(since xy = c^2)
=> (x+y+c)^2 = L^2 + c^2
=> x + y + c = +- sqrt(L^2 + c^2) .........(take positive only, since x+y+c > 0)
Let w = x + y, w = -c + sqrt(L^2 + c^2). This is exactly the same result as that in the video.
Starting here, the rest is the same as the 2nd part (i.e. w = x + y = c^2/y + y) of the video, since this 2nd formula requires only the similar triangle relationship, no need for the helping central square.
Hi presh sir you are awesome...
I watched all of your videos ,can you plz solve jee papers also...A small request 😊 my like =100
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).