A wall rises from the ground at 90 degrees. In the vertex between wall and ground sits a unit cube. The window cleaner has a 10 unit ladder which must (obviously) have one end on the ground and must (equally obviously) have the other end resting on the wall. Less obviously, but the point of the problem, is that the ladder is required to touch the upper edge of the unit block. How high up the wall does the ladder reach?

I have been driving myself mad with this problem, but I am sure there is a simple solution that does not involve finding the real roots of a quartic equation. Is there a simple solution and if so what is it?

For a variation, replace the unit block by a cylinder of unit radius "tucked into" the vertex. This seems harder!

Maybe Dan's last statement can be used to solve the quartic without knowing the formula. If h is a root then sqrt(100-h² ) must be a root by Pythagoras. Can anyone now find a way of solving the quartic?

Michael

This question is a special case of 1991 BMO Q7, and one can obtain a quadratic by using the property stated above.
[BMO = British Mathematical Olympiad, a competition for school-aged mathematicians - The Editor]

Michael,

Try the Ferrari method of solving a quartic.

A quartic can always be reduced to

x⁴ +bx² +cx+d=0

Now rearraging to give

x⁴ +bx² +b² /4=-cx+(b² /4-d)=-cx+d'

The LHS factorises to (x² +b/2)²

Now the crucial step: If we add in a y inside the bracket, we have

(x² +b/2+y)² =2yx² -cx+(d'+2yb/2+y² )

We choose y such that the RHS is a perfect square, ie,

c² =8y(d+2yb/2+y² )

This is a cubic which can be solved, eg by using Cardan's formula after eliminating the y² term.

Thus, we have

(x² +b')² =(a'x+c')²

and we can solve the two resulting quadratic.

Thanks Kerwin. I was aware of the Ferrari formula but not how it was derived, so that should be useful. What I was wondering above was whether this particular quartic could be solved without knowing the Ferrari formula, using the fact that if h is a root then sqrt(100-h² ) is also a root.

Thanks,

Michael

PS You said that this was a special case. How did the BMO generalise it?

Hi, Michael

BMO generalise it to a ladder length l having a point equidistant d from the ground and the wall. So this is a special case in which l =10 and d =1.

Kerwin

Thanks Kerwin. I am very glad I didn't take the 1991 BMO otherwise I think I would have panicked as soon as I'd seen a quartic appear!

Michael

Well, I never claimed that the problem itself (or indeed its solution) had any intrinsic interest! What surprised my undeveloped mathematical intuition (JMB A level 1985) was that the solution was a quartic in the first place, given the way the problem is set up. Just a fact of life I guess.

You can see intuitively by the symetry of the problem that the two solutions are going to be closely related, but isn't that just because this happens to be one of a (preumably infinite) class of quartics two of whose roots are related in a particular way? Is there a more interesting number-theory type issue as to the way that the roots of quartics are related, given particular relationships between the coefficients? I am quite prepared to belive the answer is no!!!!!

I had a second look at the question. If we label the horizontal distance of contact from the wall to be f then we can find that

hf=d(h+f)
and (h+f)² =l² +2hf

Hence we can find a quadratic equation in the form

x² -px+q=0

where h and f are the solution. It turns out that

p=d+sqrt(d² +l² ),
q=d² +d*sqrt(d² +l² )

So we can solve, in this special case, the quadratic equation

x² -(1+sqrt(101))x+(1+sqrt(101))=0

Kerwin