Solution to:
Leaning Ladder
The picture on the right 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} = 4^{2}
which can be rewritten to
a^{2} + 2 + b^{2} + 2(a + b) = 16.
Now we use the fact that ab=1, so 2=2ab, and we get:
a^{2} + 2ab + b^{2} + 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 = sqrt(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 = sqrt(17)  1
so
a^{2} + ( 1  sqrt(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}×( sqrt(17)  1 + sqrt( ( 1  sqrt(17) )^{2}  4 ) ) ≈ 2.76
^{1}/_{2}×( sqrt(17)  1  sqrt( ( 1  sqrt(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.
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