Problem of the Week

Problem 34 Solution

Congratulations to Matthew Whitten from The Cavendish School, Markus Kuan from Northstowe Secondary College, Aarna Singh from Wallington High School for Girls, Henry Boulton and Ishaan Agarwal from Cambourne Village College for successfully solving the problem!

Question 1 — The Common Chord

(a)

We have

$P(x)=x^2+2x+1,\qquad Q(x)=3x^2-4x+5.$

At the intersections,

$x^2+2x+1=3x^2-4x+5.$

Rearranging,

$2x^2-6x+4=0,$

$2(x-1)(x-2)=0.$

Hence $x=1$ or $x=2$ . Substituting into $P(x)$ , the intersection points are $(1,4)$ and $(2,9)$ .

The gradient of the line is

$m=\frac{9-4}{2-1}=5.$

Using $y=mx+c$ and the point $(1,4)$ ,

$4=5(1)+c\implies c=-1.$

Therefore, the line is

$\boxed{y=5x-1}.$

(b)

Method 1: Vieta’s Formulae (Not in GCSE):

We have

$P(x)=2x^2+3x-1,\qquad Q(x)=5x^2-4x+2.$

At the intersections,

$2x^2+3x-1=5x^2-4x+2,$

$3x^2-7x+3=0.$

Let the roots be $r$ and $s$ . By Vieta’s Formulae

$r+s=\frac73,\qquad rs=1.$

The corresponding $y$ -coordinates are $P(r)$ and $P(s)$ . The gradient of the line joining the points is

$m=\frac{P(s)-P(r)}{s-r}.$

Therefore we can substitute the values in to form:

$m=\frac{2s^2+3s-1-(2r^2+3r-1)}{s-r}$

$=\frac{2(s^2-r^2)+3(s-r)}{s-r}$

$=2(r+s)+3$

$=\frac{14}{3}+3$

$=\frac{23}{3}.$

Method 2: Elimination of Quadratic Method

At the intersections,

$y=2x^2+3x-1$

and

$y=5x^2-4x+2.$

Multiplying (1) by $5$ ,

$5y=10x^2+15x-5.$

Multiplying (2) by $2$ ,

$2y=10x^2-8x+4.$

Subtracting the second equation from the first eliminates $x^2$ :

$3y=23x-9.$

Rearranging,

$\boxed{y=\frac{23}{3}x-3}.$

Therefore, m = $\frac{23}{3}.$

(c)

The simplest solution is to set $P(x) = Q(x)$ .

$ax^2+bx+c=dx^2+ex+f.$

Rearranging provides the required structure

$\boxed{(a-d)x^2+(b-e)x+(c-f)=0}.$

(d)

$(a-d)(x^2-x(r+s)+rs)=(a-d)x^2 +(b-e)x+(c-f)$

$(a-d)(-x(r+s)+rs)=(b-e)x+(c-f)$

$(b-e)x +(a-d)(r+s)x=(a-d)rs-(c-f)$

$x(\frac{(b-e)}{a-d}+(r+s))=rs-\frac{(c-f)}{a-d}$

$(\frac{b-e}{a-d})x+\frac{c-f}{a-d}=(-r-s)x+rs$

$(\frac{e-b}{a-d})x+(\frac{f-c}{a-d})=(r+s)x+rs$

As both sides are equal and both sides contain a constant value and a variant value dependent on $x$ , the coefficients must be equal.

This leaves us with the result

$\frac{e-b}{a-d}=r+s$

(It is worthwhile to note that this can be done by inspection using sum of roots. However, the question asks us to figure this out with expansion instead.)

(e)

$m=\frac{P(s)-P(r)}{s-r}$

$P(s)= a(s^2) +b(s) +c$

$P(r)= a(r^2)+b(r)+c$

$P(s)-P(r)=a(s^2-r^2)+b(s-r)$

$\frac{P(s)-P(r)}{s-r}=\frac{a(s+r)(s-r)+b(s-r)}{s-r}$

$\frac{P(s)-P(r)}{s-r}=a(s+r)+b = m$

The second part follows:

$m=a(s+r)+b$

$s+r=\frac{e-b}{a-d}$

$m=a(\frac{e-b}{a-d})+b$

$m=a(\frac{e-b}{a-d})+\frac{b(a-d)}{a-d}$

$m=\frac{ae-ab}{a-d}+\frac{ba-bd}{a-d}$

$m=\frac{ae-ab+ab-bd}{a-d}$

$m=\frac{ae-bd}{a-d}$

(f)

$P(r)=mr+k$

$k=P(r)-mr$

$P(r)=ar^2+br+c$

$m=a(r+s)+b$

$ar^2+br+c-(a(r+s)+b)r=k$

$ar^2+br+c-(ar^2+asr+br)=k$

$c-asr=k$

$sr=\frac{c-f}{a-d}$

$c-\frac{ac-af}{a-d}=k$

$\frac{ac-cd}{a-d}-\frac{ac-af}{a-d}=k$

$\frac{af-cd}{a-d}=k$

The second part follows using information from (e):

$y=mx+k$

$m=\frac{ae-bd}{a-d}$

$k=\frac{af-cd}{a-d}$

$y=(\frac{ae-bd}{a-d})x+\frac{af-cd}{a-d}$

$y=\frac{(ae-bd)x+(af-cd)}{a-d}$

(g)

$y=\frac{(ae-bd)x+(af-cd)}{a-d}$

$a=3,b=-2,c=5$

$\frac{ae-bd}{a-d}=4$

$\frac{af-cd}{a-d}=7$

from this

$\frac{3e+2d}{3-d}=4$

$3e+2d=12-4d$

$2d=4-e$

and

$\frac{3f-5d}{3-d}=7$

$3f-5d=21-7d$

$2d=21-3f$

Substituting $d=6$ provides integer coefficients

$d=6, e=-8, f=3$

$P(x)=6x^2-8x+3$

General solution is

${Q(x)=kx^2+(4-2k)x+\frac{21-2k}{3},\qquad k\in\mathbb{R}, k\neq0}$

Question 2 — The Hidden Derivative

(a)

The coefficients are $2x$ , $3x^2$ , $8x^7$ respectively.

(Note: the coefficient of $h^1$ for the $n^{th}$ power of $(x+h)^n$ is $nx^{(n-1)}$ )

(b)

The easiest few ways is to realise that this is very similar to differentiation, and simply “undifferentiate” the given expression.

Applying backwards logic from (a) was expected and results in the same form

$P(x)=3x^4-4x^2+k,\qquad k\in\mathbb{R}$

(c)

Applying the same logic from previous part provides
$(27x^8-12x^5+20x^3-2x)$

Question 3 — Consecutive Squares

(a)

For every positive integer (n),

$n^2<n^2+n<(n+1)^2.$

Since $n^2$ and $(n+1)^2$ are consecutive integer squares, $n^2+n$ cannot be a perfect square.

(b)

Expanding the first expression,

$\boxed{(x^2+1)^2=x^4+2x^2+1}.$

Expanding the second,

$\boxed{(x^2+x)^2=x^4+2x^3+x^2}.$

(c*)

The idea for this part was to apply the same logic from (a) but with a bit more working.

An example solution is shown below, but any valid proofs are awarded full credit.

$N=n^4+n^3+n^2+n+1$

$Q=2n^2+n$

$Q^2=4n^4+4n^3+n^2$

$4N=4n^4+4n^3+4n^2+4n+4$

$\emph{if } N = p^2 \emph{ then } 4N=(2p)^2$

then

$4N-Q^2=3n^2+4n+4$

$\emph{as such } 4N>Q^2 \emph{ for all } n\ge0$

then the next consecutive perfect square from $(Q)^2$ would be $(Q+1)^2$

$(Q+1)^2=(2n^2+n+1)^2=4n^4+4n^3+5n^2+2n+1$

$4N-(Q+1)^2=-n^2+2n+3$

Test values of $n$ on $-n^2 +2n +3$ :

for $n=3$ we get $4N=(Q+1)^2$ . So, $N$ is a perfect square.

for $n\ge4$ , $4N<(Q+1)^2$

$Q^2<4N<(Q+1)^2 \emph{for }n\ge4$

Check $n=1,2,3$ :

$n=1, N=5$

$n=2, N=31$

$n=3, N=121$

as all other values leave $4N$ nested between 2 consecutive perfect squares.

If $4N$ is not a square then neither is $N$ a square, due to $4$ being a square.

the only value of $n$ giving perfect square of the form:

$N=n^4+n^3+n^2+n+1$

$\emph{is } n=3$

Problem of the Week

Problem 34 Solution

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