I found this question on YouTube on mindyourdecisions. I got easy method to solve this puzzle. Click here to get alternative method to solve this puzzle. Give this puzzle a try and watch the solution when you are ready.
Lets begin to this solution. In this puzzle, a square shaped box of side 1 unit is kept on a corner of a wall. Beside it, a ladder is kept which have length of 4 units. But the condition is ladder touching the box. We have to find distance between between top of box and top and top of ladder as shown in figure.
To solve this question lets convert it into a Geometric Problem and draw it's diagram in Cartesian coordinate system. We represent ladder as a straight line, floor and wall as X-axis and Y-axis respectively. as shown in figure.
Lets mark the details given. Height and width of bow be 1 units, length of ladder be 4 units and ladder touches box at (-1,1).
Let us consider slope of line be 'm' then equation of line will be:
y = mx+c
But the line is passing through (-1,1)
∴ 1 = -m+c
⇒ c = 1+m Put this value in 'y=mx+c'
⇒ y = mx+1+m
Ladder touches wall which is represented by Y-axis (x-coordinate is 0)
⇒ y = 1+m
Ladder touches floor which is represented by X-axis (y-coordinate is 0)
⇒ 0 = mx+1+m
⇒ x = -(1+m)/m
Let's mark coordinate points as shown in figure.
By observing above figure we can notice, we have to find value of 'm' as distance between between Origin and A is '1+m' ⇒ Distance between top of box and top of ladder is 'm'.
By the given condition, length of ladder is 4 units ⇒ Distance between A and B is 4 units.
By using distance formula:
\[\sqrt{\left(\frac{1+m}{m}\right)^{2}+\left(1+m\right)^{2}}=4\]
\[\left(\frac{1+m}{m}\right)^{2}+\left(1+m\right)^{2}=16\]
\[\left(\frac{1+2m+m^{2}}{m^{2}}\right)+\left(1+2m+m^{2}\right)=16\]
\[1+2m+m^{2}+m^{2}+2m^{3}+m^{4}=16m^{2}\]
\[m^{4}+2m^{3}-14m^{2}+2m+1=0\]
Divide both sides by m^2, we get
\[m^{2}+2m-14+\frac{2}{m}+\frac{1}{m^{2}}=0\]
\[\left(m^{2}+\frac{1}{m^{2}}\right)+2\left(m+\frac{1}{m}\right)-14=0\]
\[\left(m^{2}+\frac{1}{m^{2}}+2\right)+2\left(m+\frac{1}{m}\right)-16=0\]
As we know,
\[\left(m+\frac{1}{m}\right)^{2}=\left(m^{2}+\frac{1}{m^{2}}+2\right)\]
Apply this in our equation
\[\left(m+\frac{1}{m}\right)^{2}+2\left(m+\frac{1}{m}\right)-16=0\]
It is a quadratic equation
By using Brahmagupta's Quadratic Formula
\[m+\frac{1}{m}=\frac{-2\pm\sqrt{4-4(-16)}}{2}\]
\[m+\frac{1}{m}=\frac{-2\pm2\sqrt{1+16}}{2}\]
\[m+\frac{1}{m}=-1\pm\sqrt{17}\]
\[m+\frac{1}{m}=-1+\sqrt{17}\]
or
\[m+\frac{1}{m}=-1-\sqrt{17}\]
Bur m is a positive value ⇒ (m+1/m) is a positive value
\[\Rightarrow m+\frac{1}{m}\neq-1-\sqrt{17}\]
\[\Rightarrow m+\frac{1}{m}=-1+\sqrt{17}\]
Now multiply both sides by 'm'
We get,
\[m^{2}+1=\left(-1+\sqrt{17}\right)m\]
\[m^{2}+\left(1-\sqrt{17}\right)m+1=0\]
By using Brahmagupta's Quadratic Formula
\[m=\frac{-\left(1-\sqrt{17}\right)\pm\sqrt{\left(1-\sqrt{17}\right)^{2}-4}}{2}\]
\[m=\frac{\left(\sqrt{17}-1\right)\pm\sqrt{1-2\sqrt{17}+17-4}}{2}\]
\[m=\frac{\left(\sqrt{17}-1\right)\pm\sqrt{14-2\sqrt{17}}}{2}\]
\[m_{1}=\frac{\sqrt{17}-1+\sqrt{14-2\sqrt{17}}}{2}=2.7609\]
or
\[m_{2}=\frac{\sqrt{17}-1-\sqrt{14-2\sqrt{17}}}{2}=0.3622\]
Here we can observe we got two values this is because ladder can be arranged in two different positions. One is in which we solved this puzzle and other is as shown in figure below.
If distance from top of box to top of ladder be m1 then distance between left corner of box and foot of ladder will be m2.
