tmua.me Medium Set C Paper 2 · Question 17

The $2n$ distinct positive real numbers $a_1<a_2<\cdots<a_{2n}$ are divided into $n$ pairs. The sum of the numbers in each pair is found, and these $n$ sums are multiplied.

Consider this proof that the greatest product is obtained by pairing $a_1$ with $a_{2n}$ , $a_2$ with $a_{2n-1}$ , and so on.

Choose a pairing with greatest product. Write $a=a_1$ and $d=a_{2n}$ . Suppose that $a$ and $d$ are not paired together: their pairs are $(a,b)$ and $(c,d)$ . Then $a<b<d$ and $a<c<d$ .

[Missing argument]

This contradiction forces $a$ and $d$ to be paired together. The same argument applies to the remaining numbers, proving the result.

Which option correctly supplies the missing argument?

AReplace the two pairs by

(a,c)

and

(b,d)

. Their sum-product increases by

(c-b)(d-a)

, which is always positive.

BReplace the two pairs by

(a,d)

and

(b,c)

. Their sum-product increases by

(d-b)(c-a)>0

, while every other positive factor is unchanged.

CReplace the two pairs by

(a,d)

and

(b,c)

. Their sum-product increases by

(d-c)(b-a)>0

, while every other positive factor is unchanged.

DReplace the two pairs by

(a,d)

and

(b,c)

. Both new pair sums are greater than both original pair sums, so the product increases.

EReplace the two pairs by

(a,c)

and

(b,d)

. The total of the pair sums is unchanged, so their product increases.

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tmua.me Medium Set C Paper 2 · Q17