# Thread: Very difficult A-level problem from 1971 (UK)

1. ## Very difficult A-level problem from 1971 (UK)

Hi.

I'm having real trouble answering a question posted online as appearing in a 1971 A level exam (as part of a discussion of whether exams have got easier over the years - which this example would seem to confirm!). It can be found at bit.ly/mZoyP8, and goes like this:

"If a + b + c=a2+b2+c2=a3+b3+c3=2, find by considering values of (a+b+c)2 and (a+b+c)3, or otherwise, the values of (i) ab+bc+ca, (ii) abc.
Hence find the equation whose roots are a, b and c."

According to my calculations, ab+bc+ca = 1 & abc = -2/3. But I've been struggling for hours to complete the final step. You end up with a set of three simultaneous (?) equations (the two implied by these answers, plus the original "a+b+c = 2") from which it seems impossible to derive a (presumably?) cubic equation that will give have the three roots.

Anyone got any ideas?

Thanks...

2. ## Re: Very difficult A-level problem from 1971 (UK)

Incidentally, where the question comes out here as "a2+b2+c2" etc., it means a squared plus b squared, etc. Not sure if a superscript facility is available here...

3. ## Re: Very difficult A-level problem from 1971 (UK)

a + b + c=aa+bb+cc=aaa+bbb+ccc=2
(a+b+c)(a+b+c) = (aa+bb+cc) + 2(ab+bc+ac) = 4 = 2 + 2(ab+bc+ac)
ab+bc+ac = 1
(aa+bb+cc)(a+b+c) = (aaa+bbb+ccc)+(aab+abb+bbc+bcc+aac+acc) =
= 2*2 = 2+(aab+abb+bbc+bcc+aac+acc)
(aab+abb+bbc+bcc+aac+acc) = 2
(ab+bc+ac)(a+b+c) = 3abc +(aab+abb+bbc+bcc+aac+acc) = 1*2 = 3abc+2
abc = 0
At least, one among a, b, c is nul.
Suppose c=0 then
a+b+c = 2 = a+b
ab+bc+ac = 1 = ab

Finally, the solutions are :
( a=1 , b=1 , c=0 )
or ( a=0 , b=1 , c=1 )
or ( a=1 , b=0 , c=1 )

4. ## Re: Very difficult A-level problem from 1971 (UK)

off the top of my head, the first thing to do is consider

$4 = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$

$= a + b + c + 2ab + 2ac + 2bc = 2(1 + ab + ac + bc)$

dividing by 2, and subtracting 1 from each side:

$1 = ab + ac + bc$

the next thing i would do is consider:

$8 = (a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2c + 3ab^2 + 3ac^2 + 3bc^2 + 6abc$

$= 2 + 3a(ab + ac + bc) + 3b(ab + ac + bc) + 3c(ab + ac + bc) - 3abc$

$= 2 + 3(a + b + c)(ab + ac + bc) - 3abc$

$= 2 + 6 - 3abc$, so abc = 0.

now the polynomial with roots a,b and c is:

$(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab + ac + bc)x - abc$

$= x^3 - 2x^2 + x = x(x-1)^2$

so 2 of the numbers are 1, and the third is 0.

5. ## Re: Very difficult A-level problem from 1971 (UK)

Originally Posted by Montaillou
Hi.

I'm having real trouble answering a question posted online as appearing in a 1971 A level exam (as part of a discussion of whether exams have got easier over the years - which this example would seem to confirm!). It can be found at bit.ly/mZoyP8, and goes like this:

"If a + b + c=a2+b2+c2=a3+b3+c3=2, find by considering values of (a+b+c)2 and (a+b+c)3, or otherwise, the values of (i) ab+bc+ca, (ii) abc.
Hence find the equation whose roots are a, b and c."

According to my calculations, ab+bc+ca = 1 & abc = -2/3. But I've been struggling for hours to complete the final step. You end up with a set of three simultaneous (?) equations (the two implied by these answers, plus the original "a+b+c = 2") from which it seems impossible to derive a (presumably?) cubic equation that will give have the three roots.

Anyone got any ideas?

Thanks...
It may be a change in the sylabus. This for the time this is tediously routine, but otherwise easy (or rather the method is obvious the real difficulty is keeping algebraic errors at bay).

(I took taken A-Level maths ca 1970)

CB

6. ## Re: Very difficult A-level problem from 1971 (UK)

well this problem has one first-rate idea, cloaked in a hopeless mire of applying the distributive law repeatedly.

and the good idea is this: the coefficients of a polynomial are themselves symmetric polynomials of the roots.

however, even in a degree as low as 3, you have a lot of bookkeeping to do (quartics are even worse).

if i may illustrate: the thread starter deduced (albiet incorrectly for abc) that ab + ac + bc and abc could be deduced from the equation:

$a+b+c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3 = 2$

and this is because both $a^2+b^2+c^2$ and $a^3+b^3+c^3$

can be expressed in terms of symmetric polynomials of lesser degree (something useful to know, if tedious at times to apply).

but seeing that symmetric polynomials occur as the coefficents of:

$(x-a)(x-b)(x-c)$

did not occur to the original poster, nor apparently to the first respondent. which is a shame, because (in my opinion) that's the whole point of this problem (finding a,b and c is just a bonus, and isn't even specifically asked for).

so, CB, if you had seen this problem in 1970, would you have come up with the polynomial in short order?

7. ## Re: Very difficult A-level problem from 1971 (UK)

Originally Posted by Deveno
so, CB, if you had seen this problem in 1970, would you have come up with the polynomial in short order?
The time is not relevant, length of argument is. Then the answer is probably not, but then the time constraint was not that tight. I don't recall exactly but something like 15 minutes per question seems about what I recall and that is plenty.

Also, I'm pretty sure the candidate is supposed to view this in the light of Viète's formulas

Looking at this problem again, I'm inclined to ask: are we sure this is from an A-level paper and not Additional Maths (an O-level)? Maybe I will look at some text books when I get home this evening.

CB

8. ## Re: Very difficult A-level problem from 1971 (UK)

Ah, of course. Thanks for the responses - that's what comes of attempting relatively simple algebra problems at 3.30am!

9. ## Re: Very difficult A-level problem from 1971 (UK)

Originally Posted by CaptainBlack
Also, I'm pretty sure the candidate is supposed to view this in the light of Viète's formulas

CB
yes! that is the one first-rate idea. the rest is just....*ugh*....turning the crank