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

I'm only trying to get my nephews to pass the "non-verbal reasoning" part of exam into top junior-high

it's damn hard !!!

I have to admit, once I learned Calculus, I continued to move forward for a couple more years, not backwards to the Renaissance.

I didn't have much occasion or interest to solve calculus problems without calculus. Therefore I found it odd, using quadratic equations where a,b, and c were not coefficients, but messy non-quadratic functions in x -- not because it's wrong, but because I've not come across this kind of solution. But it's still algebra, and not something like derivatives, where these kinds of substitutions would break.

Well finding a sign mistake explains one of the mysteries -- I was beginning to doubt I understood your solution.Re-arranging U and x, and attempting to solve for U using the same method fails for one basic reason. In our problem:- at x=8, we have U=18 (and T=98) is a local (and global) minimum- but at U=18, we have x=8, which is neither minimum nor maximum, even in our limited range of x in [0,20]This is why Pavel can use b^2=4ac, to look for a "unique root", but you cannot. You can only imagine a "unique root".In fact, applying the quadratic formula only brings us back to the original problem statement.I will be impressed if you find a real minimum x, even with calc.

ventolin^3 wrote:

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

I recall this can be solved with just algebra and geometry, without resorting to trig functions.The part that made me cry (figuratively), is the part skipped in the solution:"From there, square both sides, you'll have a quartic equation in x2 :yuck:"because solving that without errors is not obvious:https://en.wikipedia.org/wiki/Quartic_functiononce you have a,b,c,d, and e, you can find the four roots with:https://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg

I'm not sure you can approach it this way

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

You have to look at this as a stand-alone equation & solve for it not being aware we are looking at a problem where range of x is from 0 - 20

The question is whether this can be solved for a minimum value of U using only algebra, not that we know already answer is x = 8 to give minimum U =

18

That's back-tracking to pre-suppose the answer already

It does not appear this equation is amenable to any "easy" substitution for minimum value of U

That part is not more difficult than initial exam question so I assume you are referring to some elsewhere part of discussion

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

-> 2u * ( du/dx ) + 8u - 18x = 0

-> du/dx = (9x - 4u ) / u = (9x/u) - 4

for minimum, this = 0, -> u = 9/4x

subbing back :

-> (9/4x)^2 + 8(9/4x)x - 9x^2 - 900 = 0

-> 14 1/16 x^2 = 900

-> x^2 = 64

x= 8

It is attempting to solve the U equation algebraically which is requested part...

rekrunner wrote:I recall this can be solved with just algebra and geometry, without resorting to trig functions

Reducing to trig functions was an attempt to solve it, hoping it was easier

It wasn't

You ended up in same miasma

It maybe there is trig identity which fits bill, but sadly no one in their right mind goes hunting for trig identities

The only simplistic way to solve it is by similar triangles + basic algebra

The part that made me cry (figuratively), is the part skipped in the solution:

"From there, square both sides, you'll have a quartic equation in x2 :yuck:"

because solving that without errors is not obvious

You have to assume his quartic is correct because he got the correct answer in the end

The quartic route can never be requested puzzle solution albeit this guy was incredibly skilled & perseverant in doing so, because you are "supposed" to be able to do these on a piece of paper with basic calculator with no more than square root/log/trig on it

The take home lesson was simply for bright kid to realise the brute-force method was virtually impossible & that cunning was required

If they came to their own conclusion that a cunning method was required, then they should have been given full marks & highly commended as that is all that could have possibly been asked of them

I wasn't approaching it anyway, but just explaining why b^2=4ac fails for U, but not Pavel Karas.From the problem statement, we know we are looking to minimize U (over x between 0 and 20), and likewise we know that a minimum x (over U, either 0 or -oo, your choice) will not lead to the kind of "local parabolic extremum" that allows us to assume that b^2=4ac will give us the interesting result.You got off track with complex roots, because you thought that b^2=4ac was a valid step.Maybe I'm just getting confused with what you mean by which minimum, but you just used calc to find a minimum U, and not a minimum x.I'll take your word solving your U equation algebraically is not straightforward. I plugged it back in to the quadratic equation, and just got the original equation back. Maybe something clever will help.

that is not enough refutation

but calc solves u for x

therefore there is minimum & algebra shoud have shot at solving it

i asked for nice substitution method which may do it

i don't see why this is invalid

i made some error with signs but calc was easy method

calc shows u for x can be solved - it's simple

the algebraic substitution shoud make headway in solving it without calc

i thought that was obvious ?!

demanded is minimum u in x

it's requisite to solve back for x

this shoud be obvious to you ???

i'm not sure what problem you are evading

i ask 1 thing only :

can you ( or anyone ) solve :

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

for minimum value of x using algebra/substitution & no calculus

if you can, then offer it...

You also asked:

What's wrong is that b^2 =/= 4ac for your f(U), for any real x.

We are looking for which x gives minimum U, not which U gives minimum x.

At the interesting point we are looking for:

dU/dx = 0 (therefore, for f(x), b^2-4ac = 0)

dx/dU = 1/(dU/dx) = +/- oo (therefore, for f(U), b^2-4ac =/= 0)

Generally, for parabolas, b^2-4ac can be positive, negative or zero, depending on the nature of the roots.

If you can't see the invalidity, maybe you can defend the validity of b^2 = 4ac, for your f(U),

particularly since b^2 is >= 0 for all real x, and 3600 > 4ac, for all real x.

Maybe it can be solved by algebra, but not by assuming b^2-4ac=0.

yes

but the trivial calc shows it has a minimum

no

you got question wrong

it is a quadratic in U with x as variable

i requested which x solves for minimum U using only only algebra

it is a simple stand-alone equation needing no obfuscation

i already posted the trivial calc solution

it was not asked to invert derivatives

the implicit differentiation was used at your request to refute this was not an easy calc problem

it is a trivial calc problem, even for 16y oles, that's why i'm not interested in it

( but as i said, i have some teaching to do for younger relatives over next few years to save on tutoring expenses & the "dummies" books on algebra/pre-calc/calc 1 & 2 are fantastic !!! )

i'm interested in a basic substitution method to solve :

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

for minimum value of U

calc shows it has minimum value

i'm interested in a solution with algebra not obfuscation

it clearly means you haven't found a "correct" substitution, which is nothing to be ashamed of

all very good

it is damn obvious the U^2 parabola doesn't cut x-axis so we can't use discriminant = 0 ?!, therefore a complex root

perhaps complex algebra will obtain a root ?!, but it will likely involve calc

however, we know U has a minimum & ask was to find it algebraically

i stated that long ago

it needs a fiendish substitution

i asked for a fiendish substitution as they are are so, so impressive

if you can't provide it, it is no slur, so i don't know why you prevaricate ?!

it woud take a very very smart guy to do it, looking in awe at Tartaglia's work for years, somehow using some fiendish substitution to depress the equation into a more basic quadratic where there is a workable discriminant...?!

ventolin^3 wrote:

http://www.telegraph.co.uk/education/educationnews/11921671/Can-you-answer-the-tricky-crocodile-maths-question-that-left-students-in-tears.html

now, i woud like to hear american 17y/18y oles on how "hard" this problem is & how long it took them to solve 8b

( note : no "grown-ups" disparage efforts of youngsters, as it shoudn't take to solve 8b, more than the 1'39.9 that Ryun '67 was capable of on a '70s synthetic off perfect drafting to bell

i managed 8b in about 1'00 but i'm doing some maths teaching to young relatives & have been read "dummies" on precalculus / calculus 1 / calculus 2, so am "trained up", so that 1'00 is unfair to ask for most "grown-ups" )

Your use of the English language brings me to tears.

chucky

you are my very good friend & i like you a lot, but this is not your thread

ventolin^3 wrote:

It maybe there is trig identity which fits bill, but sadly no one in their right mind goes hunting for trig identities

cos^2x + sin^2x = 1

cosh^2x - sinh^2x = 1

are endlessly useful for getting rid of nasty radicals. Others are used commonly too but these are indispensable and very easy to apply.

Bad Wigins wrote:

cos^2x + sin^2x = 1

cosh^2x - sinh^2x = 1

are endlessly useful for getting rid of nasty radicals. Others are used commonly too but these are indispensable and very easy to apply

you are smartest guy who posts on the board regularly

& by long way

such that whoever is 2nd last from non-dumbest didn't even realise how out-stoopidest he was ?!

it's still a basic-algebra-substitution like Pavel

if substitution can't be done for U, then proof required

U

has no minimum from basic methods of algebra...

I think half the problem is that most of us (myself included) are having difficulty understanding what ventolin is trying to say.

When thinking on the problem, I envisioned the function

f(x, y) = 9x^2 - 8yx + 900 - y^2

This is a hyperbolic parabaloid, though it's not obvious what it looks like. Of course it's intersect with a plane (f(x,y)=0) should naively be some 2-D curve. So you get your function,

9x^2 - 8yx+900-y^2 = 0

defining some curve in 2-dimensions. Well this curve is a hyperbola, but it's a little odd looking. From inspection you notice that it is an odd function, with a discontinuity at x=0, it is not injective or surjective, is well-defined for domain X, but not well defined for domain Y, i.e. (x1, y) and (x2, y) ordered pairs, x1 = x2 does not hold in general.

This hints at why ventolin's attempts to frame the problem using U (what I called y above), does not yield the same success as using x.

Just for to relate to the given algebraic solution, we find y=y' such that the contour

f(x) = 9x^2 - 8y'x+900-y'^2

has a unique solution. Since the contours of f(x,y) of constant y are parabolas, this is the same as placing this parabola such that the extrema point intersects the f(x,y)=0 plane.

Purely conjecture, and moving more towards mathematical rigor, I suspect the failure of ventolin's method is related to the fact that the hyperbolic paraboloid is not invariant under rotations about the z-axis by 90 degrees. It is symmetric about reflections across the y-z plane and the x-z plane, but not the x-y plane, which is actually equivalent to rotation by 90 degrees.

Needless to say, a change of coordinates involves a 90 degree rotation and symmetric reflection about the original y-z plane, so the paraboloid is not invariant under change of x and y coordinates (which is what ventolin's attempt amounts to).

thanks

these very

"interestings"

can you solve for U

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

for minimum U in x without calc, with only fiendish substitution(s) ?

if not, if you proved above it can't be done, then i apologise