This was a tough brain teaser. In fact, I, the Big Giant Brain, did not have the answer to this one until a reader submitted it. (remember - BIG doesn't mean SMART...)
In an alley two ladders are placed cross-wise. The lengths of these ladders are respectively 2 and 3 meters. They cross one another at one meter above the ground.
What is the width of the alley?
Scroll down for the answer.
The Answer:
Many thanks to Bill Gleeson of Tucson Arizona for the following answer.
The alley is 1.231186 meters wide.
Since the ladders form a series of right triangles, we can use algebra to determine the width of the alley:
- Let W be the width of the alley
- Let a be the angle between the alley and the 3-meter ladder
- Let b be be the angle between the alley and the 2-meter ladder
- Let B be the horizontal distance between the wall the 3-meter ladder rests against and the point where the ladders cross.
- Let A be the horizontal distance between the wall the 2-meter ladder rests against and the point where the ladders cross.
We can also see that W = A+B.
From algebra, we learned that the cosine of the angle is equal to the adjacent side (the width of the alley) divided by the hypotenuse (the length of the ladders), and the tangent of the angle is equal to the opposite side divided by the hypotenuse.
Solving for the width (adjacent side):
W = 3cos(a) for the 3-meter ladder
W = 2cos(b) for the 2-meter ladder
Similarly, solving for A and B (for the smaller triangles):
A = 1/tan(a) and B = 1/tan(b)
Since the sum of the angles of any triangle equal 180 degrees, and we are dealing with right triangles, a (and b) can't be more than 90 degrees.
Pick any number between 0 and 90 for a and perform the following calculations.
Calculate W = 3cos(a)
Calculate b = acos(W/2)
Calculate A = 1/(tan(a))
Calculate B = 1/(tan(b))
Calculate X = A+B
Adjust a until W and X are equal.
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