"Impossible" Maths Question reduces Scottish 17y/18y oles to tears !!!

I should say the link looked at a graph of f over x while you're looking at g over u. I'm not going to worry about whether they're inverse, the important thing is they are not the same graph.

Bad Wigins wrote:to put it simply, just because a function has a unique minimum doesn't mean the inverse function does too

yes

but then it means the solution involved possibility of going up dead-end initially, then tackling other route which solves it

you have still not offered interest for converse route

The linked solution was looking at a graph of U over x. You are looking at a graph of x over U

yes

that was my approach

why does that approach look a dead-end ?

However we already know the unique minimum of x over U, it's 0 as given in the problem

then why don't we look for unique maximum(s) of U over x

why doesn't it solve ?

looking at the function as f(x) allows you to search for values of u such that the vertex is on the x axis.

looking at it as g(u) doesn't. It doesn't cross the u axis. h(u) = g(u) - c does for c > 900, but results in different values of x for each constant so it won't be useful in finding the value of x where T is minimum.

As an aside, speaking as a native English speaker, I often struggle to understand your English. Not in an "Antonio or Renato" kind of way, but as if you are speaking some kind of hip kid-speak that is lost on my generation.Nevertheless, I knew what you wanted but was less clear why.

ventolin^3 wrote:

I posted (got deleted by mods ) very nice post about

"Chaldeans/Medianites/Asian.etc" likely solving this algebraically 7000+y ago on clay

+

"Churlish to throw stacking problems"

It followed with :

~ "Unstipulated, from my comment on 1st page, it was asking compressing mulch into compact quadratic to solve, expanding it up again to shoveable mulch..."

You knew what you were "asked" from instant you saw problem without my prompt

You will try better next time, as you are level beyond that !

I would say 5 is an overestimation.

Bad Wigins wrote:

rekrunner wrote:

If someone really wants to impress, they can try to find a maximal answer to "mcguirkthejerk"'s ping pong ball question:

In the unlikely event that you succeed, all 5 people in the world obsessed with that problem will be very impressed.

I hope it's not a false lead, but it's not clear to me yet that your "elegance personified" solution from Pavel Karas is mathematically valid. Maybe he got lucky.When finding the discriminant of his quadratic equation in x, 9x^2 - 8Ux + 900 - U^2 = 0, he treats U, as if it were a constant (he called it a parameter) value, independent of x, rather than some messy function of x. The "b" and "c" in his formula are not constant coefficients.My next step would be to work out why "D=(8U)^2 - 4*9*(900 - U^2)=0" should be considered a valid next step, and also why it produced the right answer despite what looks to me like a mathematical mistake.But maybe this is a likely reason why attempting to use the same approach for the function U fails.

ventolin^3 wrote:

i'd appreciate some analysis of the algebraic solution :

http://imgur.com/bLtaEMd

now, using substitution of :

U = T - 80

an equation evolved :

9x^2 - 8Ux + 900 - U^2 = 0

the guy proceeded to solve a quadratic for x with the discriminant = to 0 as that woud provide a unique solution which yields minimum time, T

now how about re-arranging above, so focus more on U :

U^2 + 8ux - 9x^2 - 900 = 0

woud not the unique solution for U also lead to the minimum time, T ?

however, applying same method to discriminant only yields complex roots

what is wrong with premise

U = T - 80

having a unique root, which it shoud do at minimum T but not solvable with quadratic formula ?!

Double-check your math -- you shouldn't find complex roots.

Assuming U and x are independent (which only yields extra "false" solutions), you should find these solutions for U and X:

U = +/- 30 when x = 0

U = -82, or 18, when x = 8

U = -18, or 82, when x = -8

Of course you can reject many of these solutions, but it you found a complex root along the way, then I suspect a math error.

While this is a fairly simple question, there are no crocodiles in Scotland. Do the Scottish math authorities expect students to travel to Egypt to put this knowledge into practice? Colonialism at its best.This is the most blatant flag waving I've seen since the commonwealth games.

I don't find it very difficult. The (a) question simply requires you to substitute two values in the given formula. The (b) question is a simple derivative, nothing that could be considered impossible for a 17/18y student.

Sorry too many posts. I pretty much concur with Bad Wiggins.

It boils down to whether it makes sense to set the discriminate = 0, in order to find the minumum. T (and U) have a "parabolic" minimum at the point of interest, when x=8.

The same cannot be said looking at x, as a function of U, for the interesting point, U=18. While the roots are not complex, the discriminant is positive, and not 0.

rekrunner wrote:When finding the discriminant of his quadratic equation in x, 9x^2 - 8Ux + 900 - U^2 = 0, he treats U, as if it were a constant (he called it a parameter) value, independent of x, rather than some messy function of x. The "b" and "c" in his formula are not constant coefficients.

My next step would be to work out why "D=(8U)^2 - 4*9*(900 - U^2)=0" should be considered a valid next step, and also why it produced the right answer despite what looks to me like a mathematical mistake

I am not sure if there has been mistake in your discriminant

The equation is :

U^2 + 8ux - 9x^2 - 900 = 0

a = 1

b = 8x

c = -9x^2 - 900

for a unique root : b^2 = 4ac

(8x)^2 = 4 ( -9x^2 - 900 ) = -36x^2 - 3600

-> 64x^2 = -36x^s - 3600

-> 100x^2 = -3600

-> x^2 = -36

That gives a complex root.

I believe I have calculated discriminant correctly.

The question was why we have a complex root if looking for a unique solution to U ?

I was commenting on the "elegance personified" solution from Pavel Karas.The discriminate I quoted was his, and not yours, as I didn't fully understand how he thought it was quadratic.It looked odd as his a, b, and c were functions in x, and not static coefficients, but I guess algebraically, it still works.I get it now.Similarly, the answer to your question is that there is no unique solution for your equation.Your b^2-4ac is quite positive. This should be obvious by looking at your coefficients:b=8x4ac = 4 * 1 * (-9x^2 -900) = -(36x^2 + 3600)b^2 is surely positive for all real x, especially, between 0 and 20, as in the case of the crocodile/zebra problem, except at x=0, where b^2 is 0.4ac is surely negative for all x, even at x=0.So there is no "real" x for which we can expect b^2-4ac to be anything but positive. Attempting to set it to 0 cannot work for any real x, as you've discovered.The implication here is, not that the solutions are complex, but that there is no unique solution.Since b^2-4ac is a real positive number, the quadratic formula still gives two real solutions, rather than the unique assumption you wanted to assume (like Pavel Karas assumed).See here for an explanation of what positive, zero, and negative discriminates says about the nature of the roots: http://www.mathwords.com/d/discriminant_quadratic.htmSo for Pavel Karas, since there is a local minimum at his "parabolic" equation, there is a unique root with multiplicity of 2. (His equation isn't parabolic, but still seems to have a unique solution and a local minimum).Your equation doesn't have any local minimum or maximum, like a parabola, because you've turned it sideways.

I thought you would have been quite familiar with these substitution tricks

I alluded to it on 1st page asking for non-calc solution involving an ellipse ( should have said quadratic as in a parabola )

These substution tricks are usual non-calc way of doing these

That is not anything to worry about as these usually work out finding the unique solution to quadratic with the substitution

I've seen quite a few in the past on various puzzles ( but can't quote 1 at present )

You should be familiar with these

How else do you think those fiendishly clever Renaissance guys solved such problems without use of calculus ?

Look at Tartaglia's substitution method for finding roots of a cubic which is far more complicated

I'll think about it some more

I can't agree with this

This still comes out as

64x^2 = -36x^2 - 3600

-> 100x^2 = -3600

-> x^2 = -36

meaning complex roots

That's not true

Try x = 5 or 10 & you get complex roots

You only get real roots when x > 11.33

Not true

You only get real roots when x > 11.33

nor for x= 5 or 10 either, but only when x > 11.33

Not sure about that

U = T - 80

At minimum T of 98 from other methods, above shoud have unique value of 18 which shoudn't appear again in our range of x between 0 - 20 ?!

Above indicates unique solution at T = 98

but this needs more thought

Pavel was correct : i have seen his method used a few times in past & see Tartaglia for far more complicated use of substitutions

It is usual expectation from past experience

That's not true

That quadratic in U is a standard quadratic/parabola with a minimum not sideways

This does all need some more thought from everyone...

( albeit looking at x > 11.33 maybe somewhere to start )

I was like you and thought the same thing friend. But it's different now. My daughter is at a very high level of math as a sophomore in high school, a special program. She has breakdowns, physical breakdowns and will be overwhelmed and start crying incessantly. We've had long, long talks about not being able to solve everything or that sometimes it takes a bit longer, sometimes a lot longer. Kids are different now and have enormous pressures, pressures that many actually put on themselves, not always the parent. It's really gotten out of hand. Laugh at my post, that's okay, but I'm real and everything that I posted is real. If I put the name of the program in my post, it's possible that she will be identified as 2 of her teachers in the program run and come on here from time to time and they know that I run. This is the type of thread that they'd look at. So yes, although my daughter is 15, she has cried more than once over a math problem and although I do get it, I'm also perplexed that it's gotten to that point.

The Donger wrote:

A 17 year old cried over a math problem?

I believe that there are kids who put so much pressure on themselves that they cry over academics. I believe that there are kids in a calc course who can't solve the problem under discussion. But I would be surprised if there was any overlap between the two groups. I just can't see how anyone could put that kind of pressure on themselves and still be that bad at calculus.

ventolin^3 wrote:That's not true

Try x = 5 or 10 & you get complex roots

You only get real roots when x > 11.33

correction

put in wrong sign in part of discriminant

the discriminant is +ve for all values of x between 0 - 20

i've had a good look now & it is very hard to solve for minimum U in x without calc

it probably needs another "reducing down" substitution to get anywhere - a fiendish one & then work back 2 substitutions !

the "Pavel" method in solving minimum x in U looks only viable non-calc route using basic algebraic substitutions

this may help

ask her to solve this very seemingly simple problem

( only let her spend 15 minutes on it ) :

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

after 15 minutes tell her to stop & show her the algebraic solution here ( last post ) :

tell her that not all problems can be solved unless you are a ~ genius & to solve something like this seemingly trivial problem woud need a top-level degree graduate ( tell her Professor of maths at Ivy League college ) with great patience

( this problem can only realistically be solved with trigonometry using similar triangles - that's how it was solved in the puzzle book

what she needs to take away from this problem is to draw her own conclusions that this problem is "unsolvable" algebraically & needs some sort of trig of similar triangles

if she comes to this conclusion, tell her that she is super-bright & most definitely a top, top maths student

sometimes with the hard sciences you need to realise that the "standard" "grinding" methods don't work & that some alternative methods will have to be used

if she can come up with this insight, tell her once more she is super-smart & to be highly commended

sometimes solution isn't required but showing insight into the problem is asked for )

Thanks a lot! That was extremely nice of you. I have to know, are you a math or physics teacher and where at? Honestly, I have no clue who you are but you're obviously big on physics and math. I will share it with her tomorrow (she's studying Java at the moment). Again, I sincerely thank you for taking the time to post what you did.

I'm just trying to get nephews to pass entrance exams to top school in locale, which means no fees !