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20090215—0220Here is a problem that can be solved many ways. My good friend C. Blanco sent it to me to solve using the methods of classical geometry, coordinate geometry, and trigonometry.
But being how I take the MacGyver approach to solving math problems, I have found an unconventional and improvised way to solve the problem.
Here is the problem:
Two buildings, I and II stand next to each other forming an alleyway between them. Two ladders, ladder A and ladder B in the alley cross each other touching a the point where they cross. The bottom of A sits against the base of building I, and leans over on building II. The bottom of ladder B sits against the base of building II, and leans over on building I. Ladder A is 3 meters long, ladder B is 4 meters long. The point where ladder A and ladder B cross is 1 meter above the ground. What is the width of the alleyway?
As shown in the following diagrams my approach is graphical. There is just about a graphical representation for anything in geometry. However a simple solution can be hidden among the lines drawn by the imagination.
I have tried to find the “circle inverted”. By this I mean taking a point and finding the circle that passes through it with the circle having a given radius. Not a perfect description, by any means, but if you can determine why the circles were drawn you, the reader, will understand.
This pictures are going to be confusing. In fact, it may not work at all. I am placing my early work here to get feedback and show the brainstorming process.
The geometric construction of the ladder problem uses only 3 to 4 values of circles to determine the angle of the ladder and the width of the alley. Those three values are the length of ladder A (which in the problem is 3 meters), the length of ladder B (4 meters), the height were the ladders cross (1 meter), and if needed the width of one ladder plus the distance where they cross (for example 1 + 3 or 41).
Construct 2 circles with a in proportion to a radius of 3 meters for one circle and a radius of 4 meters for the other respectively. From the very left, horizontal center of the 3 meter circle draw an arc of length 4 meters. (This is done because 3 + 1 = 4. The 1 is the length were 3 and 4 meet.) Where this arc is intersected by the horizontal of the center of the two beginning circles, is the end of a very important 3 meter circle. Where this 3 meter circle intersects the original 4 meter circle is the point where a new 3 meter circle and the original 4 meter circle intersect to for a length 1 meter high. The segment length from the center of the two original circles to this point of intersection is the length of the alley.
This is based on the theory that the alley way will always equal the horizontal length of one of the ladders. For a circle to intersect (given our measurements) at one it would have the arcs would have to move closer. And likewise they must be further apart to intersect at a higher height. (This description is not very strict mathematical description, but it helps to form a visual picture.)
And it is worth noting again that this is not proven. I have to review everything. It may also be impossible. Correct or incorrect the theory is presented. I will and more to the reasoning behind the drawing in future posts.
May the Creative Force be with You