Puzzle Corner
R. Ramanujam, The Institute of Mathematical Sciences, Chennai

The problem:
There are two walls, one 5 units tall and the other 3 units tall.  Two ladders reach from the bottom of each wall exactly to the top of the other.  At what height do the ladders cross?  (See Figure 1 for how the situation looks.)

Solution:
Clearly, we have two overlapping right anged triangles. Let us name them as in Figure 2, as ABC and DBC.
E is the point at which the ladders cross, so we are looking for the height h of the side EF.
Now let us pull out triangles ABC and EFC. Can you observe that they are similar triangles? We therefore get the equation:
5/(x+y)=h/y.
Hence h/5=y/(x+y)
It is also easy to see that triangles DBC and EBF are similar, and so we get:
3/(x+y)=h/x
Hence h/3=x/(x+y)
Adding the left hand sides as well as the right hand sides of equations (2) and (4) gives us:
h/5+h/3=1.
Therefore, we get h=3.5/(3+5)=1.875 units.
More generally, if the walls were r and s units tall respectively, the height at which the ladders cross is given by:
h=(r.s)/(r+s).
Equivalently:
1/r + 1/s = 1/h.
Martin Gardner presents and discusses this problem in his famous 1979 book of mathematical puzzles.
The problem is far more challenging and interesting when the heights of the two ladders are known, as also the height at which they cross, and we are asked to find how far apart the walls are. This is the puzzle discussed by Gardner and there is a lot of very interesting algebra to get through. If you solve the problem, write to Jantar Mantar and share it with all of us!