Solution to: Leaning Ladder
The picture below shows the situation.
Let a be the length of line segment AD, and let b be the length of line segment CF.
Because of the similarity of the triangles ADE and EFC, the following holds:
a : 1 = 1 : b
so
ab = 1.
According to the Pythagorean Theorem, the following holds:
(AB)2 + (BC)2 = (AC)2
so
(a + 1)2 + (1 + b)2 = 42
which can be rewritten to
a2 + 2 + b2 + 2(a + b) = 16.
Now we use the fact that ab=1, so 2=2ab, and we get:
a2 + 2ab + b2 + 2(a + b) = 16
which can be rewritten to
(a + b)2 + 2 × (a + b) - 16 = 0.
Because we know that a+b is greater than 0, using the quadratic formula we find that
a + b = √17 - 1.
Because of the similarity of the triangles ADE and EFC, the following holds:
a : 1 = 1 : b
so
b = 1 / a.
Now we know that
a + 1/a = √17 - 1
so
a2 + (1 - √17) × a + 1 = 0.
We know that a is greater than 0, and using the quadratic formula we find the following two values for a:
1⁄2 × (√17 - 1 + √( (1 - √17)2 - 4) ) ≈ 2.76
1⁄2 × (√17 - 1 - √( (1 - √17)2 - 4) ) ≈ 0.36
The ladder touches the wall 1 meter higher, which is at about 3.76 or 1.36 meters. In the figure, we can see that only the answer 3.76 can be correct.
Conclusion: the ladder touches the wall at about 3.76 meters.
Back to the puzzle