"a" is how high the pole was cut, and "c" + "d" is the remainder of the pole. (image not to scale)
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given: a+c+d=100 feet
given: angle between lines "a" and "b" = 45 degrees (line "b" bisects a 90 degreee angle)

Part 1
the diagonal of the 10 x 10 squre is equivielent to the hypotenuse of a triangle with two sides equal to length 10:

therefore, using the Pythagorean Theorm we get 10² + 10² = b²
and after a couple of steps, we get b = 10 * sqrt(2)


Part 2

Angle P is the same as Angle Q, as they are corresponding interior angles, and are therefore the same:
angle P = angle Q

Part 3

Now I introduce the Law Of Cosines, which states that given a triangle with sides a,b,c and an angle X (that is opposite side c):
c²=a² + b² - 2*a*b*Cos(angle X)

in the above reference image the triangle is abc, where b= 10*sqrt(2) and angle X (opposite side c) = 45 degrees while a=a and c=c

plugging into the law of cosines gives: c²=a² + (10*sqrt(2))² - 2*a*(10*sqrt(2))*Cos(45 degrees)

this simplifies to: c²=a² + 200 - 2*a*10*sqrt(2)*(1/sqrt(2))

this further simplifies after a couple of steps to: c²=a² + 200 - 20*a

Part 4
The trigonomic function Sine gives the ratio between two sides of a triangle. This ratio is of a side opposite an angle and the hypotenuse. Sin(angle)= (side of triangle opposite angle)/(hypotenuse).

I write the equations from the reference image above:
Sin(angle P) = (a-10)/c
Sin(angle Q) = 10/d

since in Part 2 it was shown that angle P = angle Q, it can be said that Sin(angle P)= 10/d

I can then set the two equations togeather: 10/d=(a-10)/c

This can be solved for d, which looks like this: d=(10*c)/(a-10) after several algabraic steps

Part 5
Here are three of the main equations we have:
1) a + c + d = 100 ft
2) c²=a² + 200 - 20*a
3) d=(10*c)/(a-10)

we can now plug in equation 3 into equation 1: a + c + (10*c)/(a-10) = 100

for equation 2 we can solve for c, which then makes it: c = sqrt(a² + 200 - 20*a)

Before we plug this into a + c + (10*c)/(a-10) = 100 , I want to rewrite a + c + (10*c)/(a-10) = 100 so that there are no denominators. So, we rewrite it as follows:
a + c + (10*c)/(a-10) = 100
c + (10*c)/(a-10) = 100 - a
[c + (10*c)/(a - 10) = 100 - a] * (a - 10)
c*(a - 10) + (10 * c) = 100*(a - 10) - a*(a - 10)
c*a - c*10 + c*10 = 100*a - 1000 - a^2 + 10*a
c*a = a^2 + 110*a -1000

at this point we can now substitute for c, and we get:
a*sqrt(a² + 200 - 20*a) = a^2 + 110*a -1000


Now the sqare root needs to go, so we have to square both sides:
(a*sqrt(a² + 200 - 20*a) = a^2 + 110*a -1000)^2

after this there is some VERY nasty algabra. After FOILing and simplifying it becomes:
a^4 - 20*a^3 + 200*a^2 = a^4 - 220*a^3 + 14100*a^2 - 220000*a + 1000000

further simplifying gets to this cubic polynomial:
0 = -200*a^3+13900*a^2-220000*a+1000000

To solve a cubic like this would take many hours to do by hand if its even possible, so I used the aid of a computer to solve it:

1) 9.00544849
2) 11.2820903
3) 49.2124611

These three answers correspond to possible heights for cutting the pole. But as you can see, the first answer is about 9, which obviously doesn't work and is an extraneous solution. However the last two work as they are above 10. So, the two possible answers for cutting the pole are at

1) 11.28 ft
2) 42.21 ft

Bah! I give up. I probably don't know how to solve it anyway, but I'll kick myself when I find out because it will be easy.

The equation is cubic because there are three answers to the question which match the equation as derived. Both whimbrel and anthonytc22 got an equivalent cubic, from different derivations. This equation reflects the requirement that the total length of the bent pole is still 100 feet, but does not quite require that the bent pole actually touch the corner of the box.
So, given that hint, what does the diagram look like for the third case?

In retrospect, I could ask for an equation that states P, I, and Q are co-linear. I would write P = I + x*(Q-I) for some x. When x = 0, P = I; when x = 1, P = Q; otherwise, P is elsewhere on the line. This translates to two equations so I can eliminate x.

(0,1+b) = (1,1) + x*((1+e,0)-(1,1))
(0,1+b) = (1,1) + x*(e, -1)
(0,1+b) = (1,1) + (x*e, -x)
(0,1+b) = (1 + x*e, 1 - x)
0 = 1 + x*e, and 1 + b = 1 - x
-x = 1/e, and -x = b
b = 1/e.

Actually, I wrote

(b+1)^2 + (e+1)^2 = (c+d)^2, expanded, and subtracted b^2+1 = c^2 and e^2+1 = d^2. This simplifes to b+e = cd. I squared this (which allows c or d to be negative without requiring they both be negative) and substituted for c^2 and d^2. This gives a perfect square, (be - 1)^2 = 0, or be = 1, b = 1/e.

Interesting. That means if you look at the X and Y intercepts of any line through (1,1), so that Y intercept is b+1 (f(0) = b+1) and X intercept is e (f(e) = 0), then b = 1/e.

A general line is y = m x + (b+1). When x = 0, y = (b+1). y intercept is b+1.

Now add the requirement that the line goes through (1,1). 1 = m + (b+1). m = -b. So a general line through (1,1) with y intercept (b+1) is y = -b x + (b+1).

Solve for x. x = (-1/b)(y - (b+1)) = -y/b + (b+1)/b = -y/b + (1 + 1/b). When y = 0, x is (1 + 1/b).

The length of the pole extending from O to P is (1+b). The length from P to I is c, and the length from I to Q is d. So, the equation is

10 = 1 + b + c + d.

I also have that the two small triangles are right triangles,

b^2 + 1 = c^2, and e^2 + 1 = d^2,

So I get

b + sqrt(b^2+1) + sqrt(1/b^2+1) = 9.

Now this equation is a perfectly good definition of b, and it has three roots. Two are shared with the cubic found before, and the third is different.

The third root of the cubic arising from similar triangles is also a root of

b - sqrt(b^2+1) + sqrt(1/b^2+1) = 9.

The 6th degree polynomial turns out equivalent to:

b ± sqrt(b^2+1) ± sqrt(1/b^2+1) = 9.

The equation with plus or minus square roots is just as solvable with numerical methods as the 6th degree polynomial, and the bisection method yields 6 roots.

To get the polynomial square c = 9 - (b+d). This permits c to be negative. Substitute for c^2 and d^2 (but not d). b^2 cancels. Move d(2b - 18) to the right hand side and square both sides. This permits d to be negative as well. Substitute (1/b^2 + 1) for d^2. Multiply through by b^2. Don't expand (avoid useless work). The result is

(1 + 81b^2 - 18b^3)^2 - b^2(b^2 + 1)(2b - 18)^2 = 0.

This equation obviously expands to a 6th degree polynomial. As written, it solves just fine numerically and gives the same roots as the simpler equation with plus or minus square roots.

The first root satisfies b + sqrt(b^2+1) - sqrt(1/b^2+1) = 9, but the pole doesn't even reach point I, let alone Q. This root comes from allowing d to be negative while requiring
1 + b + c + d = 10.

The second root is one of the expected solutions.

The third root is the other expected solution.

The fourth root satisfies b - sqrt(b^2+1) + sqrt(1/b^2+1) = 9. The pole does not reach the ground. It comes from allowing c to be negative while requiring 1 + b + c + d = 10.

The fifth root also satisfies b - sqrt(b^2+1) + sqrt(1/b^2+1) = 9. The pole is bent to the left at point P and touches the ground at point Q. It does not touch point I but point I is on the line determined by P and Q. In this case allowing c to be negative makes perfect sense, since it reflects the fact that the pole does not actually go over to point I and back.

The last root satisfies b + sqrt(b^2+1) + sqrt(1/b^2+1) = 9, with b negative. The pole bends at point P to the right and extends to I. Then it reverses and extends back through P to Q.