The first step in solving is filtering out unneeded information

@GuyKeith And since none of the information is truly needed we can filter out the whole problem, right?

@WilliamFleming Well, yes, some of it is needed. I was not trying to snark, but unless I misunderstand the question, the ladder is irrelevant, no? Isn't it only about the box? A 4 x 4 x 4 box is all that is relevant to your problem, unless I am missing something. 4 ft. up the wall.

@GuyKeith No, the 12’ ladder is leaning at an angle, and is pushed up against the 4’ box. The ladder extends about 8 or 9 feet vertically. I’m not sure I can remember how to solve it. It’s harder than it looks.

Oh shoot! I had it wrong—fixed now.

@WilliamFleming That makes more sense. Before, it was just all about the box, an easy 4ft.

@WilliamFleming Is the answer in whole numbers?

@GuyKeith No.

@WilliamFleming OK, I'm heading out for a little trip now. When I get back I will dust off my trig and PT theorem and give it a shot

OH, and is the ladder propped up for maximum extension? There could be multiple solutions if not.

@GuyKeith It is just a 12’ long ladder. No trickiness here.

@WilliamFleming I understand. I am not talking about an extension ladder. The ladder could be laying against the cube at a low angle, or at an angle for maxium "extension" on the wall. I am assuming the latter for the ladder. [Walking out the door]

@WilliamFleming Dammit! See drawing.

@GuyKeith Either way it amounts to the same problem. For purposes of the problem assume the ladder is leaning high up on the wall.

@WilliamFleming Well, without googling for a solution, I think I've gone about as far with this as I'm willing to go. My Trig is pretty rusty, and my trying to solve as a quadric x^ 4 equation or as a f of x function runs into the phantom zone when I try to solve. I got as far as x^4 +8x^3-112x^2 +128x +256 = 0, which the calculator says is unsolvable. I'm confident I probably made a mistake, but I cannot find it.

I will gladly wave the white flag. It looks like Trig is the best way to go with this one.

@GuyKeith I don’t remember how to solve it myself. It does boil down to a quadratic equation. One solution is for the ladder leaning up high and the other for it low. I could never do it with trigonometry.

If you google “ladder problem” you’ll find the answer I think. Of course there’s no fun in that.

I ought not give out problems I can’t work myself, aye?

@WilliamFleming Wait a minute. I haven't looked, but I think I'm on the verge of solving. Hold on.

@WilliamFleming Hell Yes! I got it. 10.57 feet up the wall. I checked my work by plugging the results back into the PT c^2 = a^2 + b^2 and it checked out (approx). I made a silly, stupid mistake. After I solved the quadric it was just pretty easy to solve via the PT theorem.

@GuyKeith Great! Congratulations.

Maybe we should form a recreational math group, just for fun. What do you think?

@WilliamFleming I don't know, it sounds like too much work I went to college like a million years ago and it is pitiful how much I have forgotten. I had 26 hours of college math, up thru DE, but damn if I remember much of it. I guess if you know enough to understand the basic concepts, you can look up the rest. These online calculators can be really helpful with some of these gnarly equations. 3rd semester calculus had scholarship whizzes weeping into their pillows. No calculators back then, just slide rules. I got out of that course with a C and was happy to get it. My professor was an idiot.

Exactly, thank You Mr. Fleming. How did you get that? I probably solved it yesterday in bed when I went to sleep, but should check it now if I was good. It's amazing to know more ways of solving same problem.
Edit: No, for some reason I am not able to solve it the way I though. Anyway You did it.

@InSaNe_97 I solved for C^11-C^5.5=1 on my calculator using iteration. That’s because S12=S1C^11 and S6.5=S1C^5.5 where S1 is the first number and S12 is the last.

I’m very worried about that S6.5, the middle of the series. Is that legit? We are turning your list of numbers into a continuous function. Nothing wrong with that, but I feel unsure about how that function would be expressed.

f=f(1)C^(x-1) I suppose. Maybe it’s ok after all. f of x comes out distorted on my screen???

@WilliamFleming This is what threw me off. The original question has a 6,5 (comma) and I didn't understand what that meant.

@GuyKeith I think Europeans use commas as decimal points instead of periods. Gave me pause also.

It's okay, thanks for checking out.

-_-