Mother of all Questions (1 Viewer)

[OLDMAN]

Sorry guys, it wasn't much of a Mother after all. Will try to make up for it later towards the exams.

Couldn't see Affinity's answer (nor Xayma's), but imagine it follows from the expansion of (i-i)^10 and collect the 1's. Truth is I had this solution before I went looking for a problem to match the solution.

Wogboy's and CM_tutor's approaches, although had more explaining, give the way to handle more general n-gon thus not tied only to a square.

Just wondering whether examiners will accept mod arithmetic in a solution. My guess is yes, considering that hundreds of talented maths students went through the Enrichment series of the AMT(Australian Maths Trust) in which the topic is introduced, plus the fact that it really is quite simple to define. What does CM_tutor think?

 

[McLake]

Originally posted by OLDMAN
Just wondering whether examiners will accept mod arithmetic in a solution. My guess is yes, considering that hundreds of talented maths students went through the Enrichment series of the AMT(Australian Maths Trust) in which the topic is introduced, plus the fact that it really is quite simple to define. What does CM_tutor think?

Couldn't a student quickly (in about 5 lines) define "mod" and then use it in their solution?

 

[OLDMAN]

I agree, McLake.

 

[CM_Tutor]

Originally posted by OLDMAN
Couldn't see Affinity's answer (nor Xayma's), but imagine it follows from the expansion of (i-i)^10 and collect the 1's. Truth is I had this solution before I went looking for a problem to match the solution.

What is the problem with Xayma's? I think it's a nice approach to the square, albeit not easily generalised to an
n-gon. Can you just not download it, or ...?

Just wondering whether examiners will accept mod arithmetic in a solution. My guess is yes, considering that hundreds of talented maths students went through the Enrichment series of the AMT(Australian Maths Trust) in which the topic is introduced, plus the fact that it really is quite simple to define. What does CM_tutor think?

I think McLake has the solution - I have actually done something like this in a Uni exam (not for modulo arithmetic, though). I wasn't at all sure that the marker would understand, so I sketched out the relevant theory before applying it.

On the more general approach, applying without first explaining, that is a risky approach - esp. in the trials. I had a student a couple of years ago who was way better at Maths than his teacher. He lost marks in the trials for answers that didn't have enough working, mostly because the teacher wasn't capable of some of the leaps he was making. His ability to see algebraic simplifications and more generally to see solutions was awesome. I have serious doubts that that marker would have known about any theory beyond the syllabus...

Also, just assuming a piece of theory not in the syllabus and then applying it can raise more general marking problems. How, for example, would you mark someon who did the last part of the 1991 4u HSC in two lines by starting from the AM-GM-HM inequality? What about in four lines from two applications of AM-GM?

 

[Affinity]

Learn the art of expressing advanced ideas in terms of those in the syllabus...

and I heard some Uni lecturers and others mark the papers.. so they should understand anyway.

 

[OLDMAN]

Sorry for not being clear. Couldn't seem to download Xayma's. Though from the discussion it sounded correct.

 

[Merforga]

The thing I don't get is why you can't just put down "it is because it is and works" ^___^

 

[Grey Council]

Originally posted by Merforga
The thing I don't get is why you can't just put down "it is because it is and works" ^___^

that, my friend, is the bane of my existence. ^__^

some things are so obvious, yet you have to prove them. I mean, its true . . . BECAUSE. It has to be. lol

my year 10 teacher made a comment in my report that I had to learn how to prove something, not just say its true because it has to be.

btw, does that mean that we ARE allowed to use the AM-GM thing in an exam?

ooh, on a related note, i recommend you check out Pender's harder 3u inservice thingo, he makes free use of the AM-GM result thingo. hrm, i think we ARE allowed to use it.
*thinking*

 

[CM_Tutor]

Grey Council, IMO you should memorise a proof of AM-GM for the cases n = 2, 3 and 4 - then you can prove it if you need it. Unless a question mandates the method of proof to use, you can then prove what you need an move on. Many inequalities are just re-arrangements of this AM-GM result.

Example: Prove that, for x, y, and z > 0, x³ + y³ + z³ => 3xyz.

One approach would be to prove the AM-GM for n = 3, ie (a + b + c) / 3 => cubert(abc), for a, b, and c > 0, and then take a = x³, b = y³ and c = z³ and re-arrange.

My question to Oldman was more difficult, as the last part of q8 in 1991 was never meant to be done with the general case of AM-GM, and certainly not with AM-GM-HM - it makes the question trivial. The problem is, what do you do if it is not forbidden by the question and someone does use it?

 

[maniacguy]

Relying on the marker's understanding is dangerous.

In the case of q8 in 1991, the fact that the writer left a 'or otherwise' in there implies that they were happy for the AM-GM relationship to be used, however trivial it makes the question. The AM-GM is after all mentioned in the syllabus, and it is reasonable to expect 4u students to know of the general case, even if they've never proved/used it.

The use of the AM-HM is somewhat more problematic, since the HM is not actually mentioned anywhere in the syllabus. I would expect the student to at the least write down a two-line proof of the AM-HM (trivial after the AM-GM).

Bear in mind that the HSC does aim to instil some basic level of knowledge into the student. It doesn't matter if a student can write down a proof of the Riemann hypothesis if they can't do any basic arithmetic (before the esteemed buchanan shoots me, an example would be if they couldn't work out that $1 million was greater than $1, and traded their prize for a $1 coin... or even trading it for nothing!).

Ultimately, therefore, things that diverge too much from the syllabus are going to be penalized at some stage, and you need to work out where that will be. (Hint: If you've had to prove the Riemann hypothesis to answer the question, you're probably diverging a bit too far!!*)

AM-GM-HM is a simple progression to make, but other such progressions might not be (use of Ceva's Theorem in a triangle geometry, for instance, might raise eyebrow)

Incidentally, I'm interested to find out how many people can:
i) Prove the AM-GM for n variables
ii) Prove it without calculus
iii) Prove it by induction (assuming you didn't use induction in part ii) )

* Or you can predict the identity of one of your examiners

 

[Grey Council]

Originally posted by CM_Tutor
Grey Council, IMO you should memorise a proof of AM-GM for the cases n = 2, 3 and 4 - then you can prove it if you need it. Unless a question mandates the method of proof to use, you can then prove what you need an move on. Many inequalities are just re-arrangements of this AM-GM result.

Would someone be so kind as to post up / refer me to a proof of these results? Or should I try proving them myself . . .
Although I remember Turtle_2468 saying somehwere that proving the AM-GM is a question 8 in itself. :-\

btw, maybe its just the way Bill Pender set out that inservice thing, but it seems to me that all of the harder inequalities in question 8 are related to this result. Would this be correct?

And the AM-GM is this, is it not:
(a+b)/2 >= (ab)^1/2
not sure I know what you mean by n = 2, 3, 4
hrm, sorry, Don't understand, could you please explain?

And Maniacguy? Thanks for your post. ^_^ As always, smooth. heheh. What is the HM though? AM-GM-HM? wtf
EDIT:
Did a bit of research. Hrm, i wrote the case of n=2, yup? heheh
this is the AM-GM inequality:
(a₁ +...+ a_n) / n >= (a₁... a_n)^1/n

gawd, hope I wrote that correctly. So many subscripts and superscripts. hmph

 

[Grey Council]

hokay, for n=2:

Suppose by way of contradiction that
(a+b)/2 < (ab)^1/2
a^2 + 2ab + b^2 < 4ab
a^2 - 2ab + b^2 < 0
(a - b)² < 0
Contradiction!
Therefore (a + b) / 2 => sqrt(ab), and equality occurs when a = b

 

[turtle_2468]

Cool. Now do the rest

 

[CM_Tutor]

Proving AM-GM for the general case could be a question 8, but proving specific cases, n = 2, 3 and 4 aren't that hard - n = 3 is the hardest of these three, IMO.

Grey Council asked about the HM part of AM-GM-HM. The full result is that, for any group of positive values
a₁, a₂, ..., a_n, the arithmetic mean is greater than or equal to the geometric mean, which is greater than or equal to the harmonic mean. ie:

(a₁ + a₂ + ... + a_n) / n => (a₁a₂...a_n)^1/n => n / [(1 / a₁) + (1 / a₂ + ... + (1 / a_n)]

In each case, equality occurs when a₁ = a₂ = ... = a_n - it is easy to see that this must lead to equality, but it's harder to show that it is the only time equality occurs.

Grey Council, your proof for n = 2 is correct, except that the last line should say that "(a + b) / 2 => sqrt(ab), and equality occurs when a = b". A shorter proof exists, by starting from (sqrt(a) - sqrt(b))² => 0 - it is this proof that I would memorise. (PS: Hint - In proving the n = 4 case, I would make use of the n = 2 case.)

Another useful exercise would be to find the proof for AM-HM making use of AM-GM for the general case - As maniacguy said, there is an easy proof.

 

[Grey Council]

argh! you guys are supposed to encourage me!

allright, will try to do n=3 and n=4, make no promises though. I can't even see a way to start. hrm

 

[CM_Tutor]

Originally posted by Grey Council
allright, will try to do n=3 and n=4, make no promises though. I can't even see a way to start. hrm

If you haven't made any progress by tonight, I'll post these up in an exam question style. I have them on my desk at home, but I'm at Uni at the moment.

 

[Grey Council]

make that tommorow night. I don't do maths unless its before 11:59AM. After that, other subjects call.

Start at 7ish though.

and thanks CM_Tutor.

 

[CM_Tutor]

OK, exam-style as promised:

1. n = 2 case (a) If a and b are positive reals, prove that (a + b) / 2 => sqrt(ab)

(b) Prove that equality occurs if and only if a = b

2. n = 3 case (a) If x and y are positive reals, prove that x² + y² => 2xy

(b) If x, y and z are positive reals, prove that x² + y² + z² - xy - yz - zx => 0

(c) Hence, show that if a, b and c are positive reals, (a + b + c) / 3 => cubert(abc)

3. n = 4 case (a) show that, for any positive reals x and y, (x + y) / 2 => sqrt(xy)

(b) Taking x = (a + b) / 2 and y = (c + d) / 2, where a, b, c and d are all positive reals, prove that
(a + b + c + d) / 4 => 4thrt(abcd)

4. AM-HM If x₁, x₂, ..., x_n are all positive reals, such that (x₁ + x₂ + ... + x_n) / n => nthrt(x₁x₂...x_n), prove that (x₁ + x₂ + ... + x_n) / n => n / [(1 / x₁) + (1 / x₂) + ... + (1 / x_n)]

5. AM-HM, modified from 1991 HSC Suppose that x and y are positive real numbers, and that s = x + y.

(a) By considering the function f(x) = (1 / x) + 1 / (s - x), prove that (1 / x) + (1 / y) => 4 / s

(b) Suppose that x_i > 0 for i = 1, 2, ..., n, where n > 1. Use mathematical induction to prove that
(1 / x₁) + (1 / x₂) + ... + (1 / x_n) => n² / s, where s = x₁ + x₂ + ... + x_n

(c) Hence, show that (x₁ + x₂ + ... + x_n) / n => n / [(1 / x₁) + (1 / x₂) + ... + (1 / x_n)]

6. n = 2 case, proven geometrically Consider two unequal circles with centres at M and N which touch externally at a single point C. MCN is a straight line with MC = r and NC = s, where r > s. A common tangent to the two circles touches the circle with centre M at A and the circle with centre N at B, and P is a point on AM such that AB and PN are parallel. By considering triangle PMN, find an expression for AB, and hence show that sqrt(rs) < (r + s) / 2

7. Application question, n = 2 case, from Q8 of 2000 Independent Trial (a) If p > 0 and q > 0 are positive real numbers, show that p + q => 2 * sqrt(pq)

(b) Hence, show that

(i) [cubert(p) + cubert(q²)] * [cubert(q⁴) + cubert(p⁵)] => 4pq

(ii) cubert(p / q) + cubert(q² / p²) + cubert(q⁴ / p⁴) + cubert(p⁵ / q⁵) => 4

8. Application question, n = 3 case, from Q8 of 2001 Cherrybrook Technology High Trial Given any three positive real numbers a, b and c, it is known that (a + b + c) / 3 => cubert(abc). If a > 0, b > 0 and c > 0 are real numbers such that a + b + c = 1, use the given result to show that:

(a) 1 / abc => 27

(b) (1 / a) + (1 / b) + (1 / c) => 9

(c) (1 - a)(1 - b)(1 - c) => 8abc

9. Application question, general case It is know that the arithmetic mean of n positive real numbers is always greater than or equal to their geometric mean. That is, if a₁, a₂, ..., a_n are positive real numbers, then
(a₁ + a₂ + ... + a_n) / n => nthrt(a₁a₂...a_n). Prove that n! <= [(n + 1) / 2]ⁿ for any positive integer n => 1, either by using this result or by induction.

10. A question suitable for q 7 / 8, which includes these inequalities

(a) Consider a circle, centre at I, with radius r, which touches each of the sides of a triangle ABC. The distances AB, BC and CA are c, a and b, respectively. Let angle BAC be @, let s be the semi-perimeter of triangle ABC,
ie s = (a + b + c) / 2, and let DELTA be the area of triangle ABC.

(i) By considering the areas of the triangles AIB, BIC and CIA, or otherwise, show that DELTA = rs

(ii) By uising the formula DELTA = (1 / 2) * bc * sin@, show that DELTA² = [4b²c² - (2bc * cos@)²] / 16

(iii) Use the cosine rule to show that DELTA² = (1 / 16) * [a² - (b - c)²][(b + c)² - a²]

(iv) Hence, prove Heron's formula for the area of a triangle, which states that DELTA = sqrt[s(s - a)(s - b)(s - c)]

(b)(i) If x > 0, y > 0, show that x + y => 2 * sqrt(xy)

(ii) Hence, show that if x > 0, y > 0 and z > 0, then (x + y)(y + z)(z + x) => 8xyz

(iii) We know (from (a)(iv)) that the area of a triangle with sides a, b and c is given by
DELTA = sqrt[s(s - a)(s - b)(s - c)], where s is the semi perimeter. By choosing suitable values for x, y and z, show that DELTA² <= (a + b + c)abc / 16

 

[George W. Bush]

1. a) (sqrta - sqrtb)^2 >= 0
a - 2 sqrt ab + b >= 0.
a + b >= 2sqrt ab
(a+b)/2 >= sqrt ab
b) For equality, (sqrta - sqrtb)^2 = 0
i.e. sqrt a = sqrtb
Therefore equality iff a=b

2. (x-y)^2 >= 0
x^2 - 2xy + y^2 >= 0
x^2 + y^2 >= 2xy
b) From a, x^2 + y^2 - 2xy >= 0 (1) , y^2 + z^2 - 2yz >= 0 (2) , z^2 + x^2 -2zx>= 0 (3).
(1) + (2) + (3)
2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx >= 0
x^2 + y^2 + z^2 - xy - yz - zx >= 0

Rest later (although someone will beat me to it), I seem to recall q2 from a 4u q8 paper, which is really quite easy compared to other q8s

 

[Affinity]

Cool. Now do the rest

you guys are suppose to encourage me

alright alright.. here's a small encouragement:
assuming you have the 2 case,

let A= (x+y+z)/3

then
A
= [A + x + y + z]/4
= [(A+x)/2 + (y+z)/2]/2
>= [sqrt(Ax) + sqrt(yz)]/2
>= (Axyz)^(1/4)

so A^(3/4) >= (xyz)^(1/4)
A > (xyz)^(1/3)

and the bad way to prove the 4 case.

assume 2 and 3 cases:

let A = (w+x+y+z)/4

A
= [ A + A +w + x + y + z]/6
= [(A + A + w) /3 + (x + y + z)/3]/2
>= [cbroot(wA^2) + cbroot(xyz)]/2
>= (wxyzA^2)^(1/6)

so A^(4/6) >= (wxyz)^(1/6)
A >= (wxyz)^(1/4)

now your turn to prove the general case