## 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}= 4^{2}

which can be rewritten to

a^{2}+ 2 +b^{2}+ 2(a+b) = 16.

Now we use the fact that *a**b*=1, so 2=2*a**b*, 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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