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

NON-CALCULUS SOLUTION BY DIRTY TRICK (SUBSTITUTION)

T(x)= 5(36+x^2)^1/2 + 4(20-x)

let u = x/6

then T(u) = 5(36+36u^2)^1/2 + 4(20-6u)

= 30(1+u^2)^1/2 + 4(20 - 6u)

now let u = sinhv, then sqrt(1+u^2) = coshv (DIRTY TRICK)

Now we have

T(v) = 30coshv + 4(20 -6sinhv)

= 80 + 30coshv - 24sinhv

= 80 + 30(e^v + e^-v)/2 - 24(e^v-e^-v)/2

= 80 + 3e^v + 27e^-v

this is at a maximum when the second two terms are equal*, that is,

3e^v = 27e^-v

v + ln3 = ln27 - v

v = (3ln3 - ln3)/2 = ln3

then u = sinh(ln3) = 4/3

and x = 6u = 24/3 = 8

* this is a "conjecture" but it's true... look at a graph of ae^x + be^-x with various values for a and b. It's late so I'll leave the proof of that to someone else for now. Otherwise it was easy using this DIRTY TRICK!

Bad Wigins meant:

this is at a minimum when the second two terms are equal*

d/dx(ae^x + be^-x) = ae^x -be^-x and is zero at ae^x = be^-x. So that step is for sure valid. But there should be a simple non-calc proof for it.

Bad Wigins wrote:No high school student would be expected to solve the croc question without calculus. It's just straightforward derivation. There might be some kind of dirty trick you can pull with a trig substitution but to expect that would be cruel

you are "hypocrite" deriding trig & you know that

using high-powered trig in your hyperbolic solution later is still trig, no matter how you dress it up & way beyond me

( young relatives are not at age when i have to go learn advanced parabola to teach them )

or many other regulars here i suspect, but denigrating trig as non-essential to your solution is abject

totally lame.. (???!!!)

a hugely impressive performance though...

incomparable...

awesome. seriously.

I am asking, so please answer honestly. What is your full math background and is stuff like this easy for you?

Bad Wigins wrote:

NON-CALCULUS SOLUTION BY DIRTY TRICK (SUBSTITUTION)

T(x)= 5(36+x^2)^1/2 + 4(20-x)

let u = x/6

then T(u) = 5(36+36u^2)^1/2 + 4(20-6u)

= 30(1+u^2)^1/2 + 4(20 - 6u)

now let u = sinhv, then sqrt(1+u^2) = coshv (DIRTY TRICK)

Now we have

T(v) = 30coshv + 4(20 -6sinhv)

= 80 + 30coshv - 24sinhv

= 80 + 30(e^v + e^-v)/2 - 24(e^v-e^-v)/2

= 80 + 3e^v + 27e^-v

this is at a maximum when the second two terms are equal*, that is,

3e^v = 27e^-v

v + ln3 = ln27 - v

v = (3ln3 - ln3)/2 = ln3

then u = sinh(ln3) = 4/3

and x = 6u = 24/3 = 8

* this is a "conjecture" but it's true... look at a graph of ae^x + be^-x with various values for a and b. It's late so I'll leave the proof of that to someone else for now. Otherwise it was easy using this DIRTY TRICK!

Bad Wigins wrote:NON-CALCULUS SOLUTION BY DIRTY TRICK (SUBSTITUTION)

T(x)= 5(36+x^2)^1/2 + 4(20-x)

let u = x/6

then T(u) = 5(36+36u^2)^1/2 + 4(20-6u)

= 30(1+u^2)^1/2 + 4(20 - 6u)...

huge solution but i doubt a "very smart" 16/17y ole even with immense non-calculus ability was going to solve it, not even in asia

having ruminated over this in back of mind for coupla hours dealing with chumps who thought Ryun was laughably, only good for 3'30/3'31 ???

i kept coming thinking back to

Snell's Law

now, it is physics & not "pure maths", but i'm sure it will likely have "pure maths" origin, making it rock-solid for proof

it yields shortest path light can take in different media, which means shortest path in any 2 different media given relative speeds

look up if Snell's Law can be construed as proven as "pure maths

now, "basics" of problem took me helluva lot more figuring than "answer" :

namely, river is 6m wide & relative speeds on sea/land are 4 to 5 or actual 2.0 & 2.5m/s !!

once figured out

( please help me with eloquent solution to these )

then using Snell of :

sin A / sin B = relative speeds

arc sin ( 4 / 5 ) = 53.13 degrees

-> for adjacent triangle, "sneaky trig" not "dressed up trig in a skirt - it's still all mega-python-in-his-pants- euclid" :

x / 6 = tan 53.13 degrees

-> x / 6 = 1.333333*

->

x = 8

Bad Wigins wrote:

d/dx(ae^x + be^-x) = ae^x -be^-x and is zero at ae^x = be^-x. So that step is for sure valid. But there should be a simple non-calc proof for it.

let a, b, and c be real constants and consider the expression

ac^x + bc^-x

this is equivalent to

c^log(a)c x c^x + c^log(b)c x c^-x

= c^(x+log(a)c) + c^(-x + log(b)c)

looking at the first term, c^(x + a constant) has the same graph as c^x shifted to the left by that constant, for example it equals 1 at -log(a)c instead of zero. Likewise the second term is a horizontally shifted c^-x and is symmetric to c^x about a vertical line.

The graph of their sum forms a bowl shape. c^-x has negative slope increasing toward 0 on (-inf,inf), but is itself decreasing on that same interval, which means that if c^-y < c^-x then the absolute value of the slope at x is greater. And if c^y < c^x then that expression has greater slope at x than at y. This is to say, anywhere "higher up" on the graph of c^-x is changing height faster than anywhere "lower down" on either graph, and vice versa.

This means that if ac^x =/= bc^-x, then in either the positive or negative direction, one of those terms is decreasing faster than the other is increasing. So there exists y sufficiently close to x such that

ac^y + bc^-y < ac^x + bc^-x

Therefore the minimum cannot occur anywhere ac^x =/= bc^-x. Since the graph has one minimum this must be where ac^x = bc^-x.

Now my solution is 100% calculus-free.

Why is attempting to solve this problem without calculus a thing? The Scottish maths Higher syllabus covers the relevant calculus. If the question was "find the derivative of T and the value when T'=0, the kids would not have had an issue with it.

The question caused a stir because the syllabus did not spend time on interpretation of 'real world' examples and so the kids didn't recognise the tools required.

There's your answer.

Ah those kids will be fine. Im good at math but that doesn't do much for me in terms of life success because im also sorely lacking in some other areas, namely social skills. If these kids grow up and learn only good social skills (if that is possible to isolate) they will be fine.

ventolin^3 wrote:

look up if Snell's Law can be construed as proven as "pure maths

it can, as was expected & not just thru trillions of repeat experiments as "physics" "proof"

the mathematical proof of Snell's Law requires calculus, so offered solution of Snell's Law is a calculus based one & fails quest

can anyone come up with a non-calculus, algebraic &/or trigonometric solution not involving hyperbolics ?

finally !!!a beautiful simple non-calculus solution in comments : http://www.iflscience.com/editors-blog/math-question-baffled-scottish-teensfrom "Pavel Karas" :

Nice problem but still quite easy. The challenge is to solve part b with simple algebra, without using derivations or graphic solution. Here's how to do that:

http://imgur.com/bLtaEMd

or

beautiful solution

( i must admit, i tried similar but going for

"U = T - 80 + 4x"

which got me nowhere

( but i don't do lot of these type of attacks as basic calculus just crushes it & is method demanded )

instead of elegance personified of :

"U = T - 80" )

appreciate it

very few of us will be 1'43 or 3'30 guys, but you can work easily to this kinda beautiful logic...

But calculus is pretty much based on algebra, dividing changes in y by changes in x.

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

http://www.letsrun.com/forum/flat_read.php?thread=3913652

Even better with a proof.

A non-exhaustive brute force method has found an arrangement with 4647.

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 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'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 ?!

A 17 year old cried over a math problem?

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 ?!

bump

no yoyos please

question is, why can't the alternative route be taken ?!

"smart guys" ?