There is a 6 metres wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 metres long, the other 12 metres, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape.

What is the distance from the point of intersection from the ground?

By Pythagoras, the ladders touch the wall respectively 8 metres and 12 sqrt(3) metres above the ground.

Drop a perpendicular of length h from the point of intersection of the ladders to the ground.  Let x be the distance from the point where the perpendicular meets the ground to the base of the 10 metre ladder.

Then, by similar triangles, we have:

x/h = 6/8 = 3/4, and
(6 - x)/h = 6/(6 sqrt(3)) = sqrt(3)/3.

Adding: 6/h = 3/4 + sqrt(3)/3 = (9 + 4 sqrt(3))/12.

Hence h = 72/(9 + 4 sqrt(3)) = (216 - 96 sqrt(3))/11 metres.

(Note that if a and b are the heights above the ground where the ladders meet the wall, we have 1/h = 1/a + 1/b.)

Edited on June 13, 2004, 12:20 pm

Posted by Nick Hobson on 2004-06-13 12:08:42