BWM Runde 1 Mock Paper 5: Questions, Hints and Solutions

Since $P$ is monic of degree $n$, expanding its leading term gives $(x+1)^n-x^n=nx^{n-1}+\cdots$. Terms of lower degree in $P$ contribute differences of degree at most $n-2$. Thus $Q$ has degree exactly $n-1$.

Fix $i$. The gap assumption places both $a_i+1$ and $a_{i+1}-1$ strictly between $a_i$ and $a_{i+1}$. In that interval, the factorisation

$$P(x)=\prod_{j=1}^{n}(x-a_j)$$

has exactly $n-i$ negative factors. Its sign is therefore constant and nonzero there. Consequently $Q(a_i)=P(a_i+1)$ and $Q(a_{i+1}-1)=-P(a_{i+1}-1)$ have opposite signs.

A polynomial is continuous: as its argument changes without a jump, so does its value. The intermediate value theorem therefore gives at least one zero between these two endpoints. The $n-1$ intervals are disjoint, so we have $n-1$ distinct real roots of $Q$.

A nonzero polynomial of degree $n-1$ cannot have more than $n-1$ roots counted with multiplicity: each root supplies a linear factor, lowering the remaining degree by 1. These already account for its entire degree. Hence there is exactly one in each interval, each is simple, and there are no others.

Conclusion: Exactly one simple root lies in each specified interval; these are all the roots of $Q$.

Review the idea: Polynomial roots and multiplicity · Polynomial functions

The equation is symmetric, so use the sum $s=x+y$ and product $p=xy$. Both are positive integers, $s\ge2$, and the equation becomes

$$\begin{gathered}s^3-3sp=2(s^2-p),\\ p(3s-2)=s^2(s-2).\end{gathered}$$

The left side is positive, so $s>2$. The inequality $(x-y)^2\ge0$ gives $4p\le s^2$. Substituting the required value of $p$ and cancelling positive quantities gives

$$\begin{gathered}\frac{s^2(s-2)}{3s-2}\le\frac{s^2}{4},\\4(s-2)\le3s-2,\\ s\le6.\end{gathered}$$

Only $s=3,4,5,6$ remain. They give $p=9/7,16/5,75/13,9$, respectively. The first three are not integers. For $s=6,p=9$,

$$(x-y)^2=s^2-4p=0,$$

so $x=y=3$. Direct substitution gives $27+27=2(9+9+9)$, proving that this pair works and that there are no other positive integer solutions.

Conclusion: $(x,y)=(3,3)$ only.

Review the idea: Factorisation integer solutions · Algebraic identities

Number rows and columns from 0 to $n-1$. In cell $(i,j)$, place the unique remainder of $i+2j$ on division by $n$.

We use a basic fact: if $\gcd(c,n)=1$, then the remainders of $ct+d$, for $t=0,\ldots,n-1$, are all different. Indeed, equal remainders at $t_1,t_2$ imply $n\mid c(t_1-t_2)$, hence $n\mid t_1-t_2$. The difference has absolute value less than $n$, so it must be zero. Since there are $n$ different remainders, all possible remainders occur.

• In a fixed row $i$, the entries are $2j+i$; the coefficient 2 is coprime to $n$.
• In a fixed column $j$, they are $i+2j$; the coefficient 1 is coprime to $n$.
• On the main diagonal, $j=i$, so the entries are $3i$; the coefficient 3 is coprime to $n$.
• On the other diagonal, $j=n-1-i$, so the entries are congruent to $-i-2$; the coefficient $-1$ is coprime to $n$.

The assumed $\gcd(n,6)=1$ supplies both needed coprimality conditions. Thus every required line contains each symbol exactly once. For $n=1$, the construction is the single entry 0 and also works.

Conclusion: Place $i+2j\pmod n$ in cell $(i,j)$, using representatives $0,\ldots,n-1$.

Review the idea: Complete and reduced residue systems · Counting with bijections

Let $BC=a$, $AC=b$, and $AB=c>b$. The internal angle-bisector theorem gives $BD/DC=c/b$, hence

$$BD=\frac{ac}{b+c}.$$

For completeness, the external ratio is $BE/CE=c/b$. One way to see it is to compare the areas of triangles $ABE$ and $ACE$. Using bases on line $BC$, the ratio is $BE/CE$, since they share an altitude from $A$. Using the common side $AE$, the ratio is $AB\sin\angle BAE/(AC\sin\angle CAE)=c/b$: the two indicated angles are supplementary for an external bisector and therefore have equal sines. As $E$ is beyond $C$, $CE=BE-a$, giving

$$\begin{gathered}\frac{BE}{BE-a}=\frac cb,\\ BE=\frac{ac}{c-b}.\end{gathered}$$

Since $c>b$, we have $BD>a/2$, so the order is $B,M,D,C,E$. It follows that

$$\begin{gathered}MD=\frac{ac}{b+c}-\frac a2=\frac{a(c-b)}{2(b+c)},\\ ME=\frac{ac}{c-b}-\frac a2=\frac{a(c+b)}{2(c-b)}.\end{gathered}$$

Multiplying cancels the side-length ratios and leaves $MD\cdot ME=a^2/4=MB^2$, as required.

Conclusion: $MD\cdot ME=BC^2/4=MB^2$.

Review the idea: Parallel lines and angle bisectors · Triangle area ratios · Trigonometry in geometry