BWM Runde 1 Mock Paper 4: Questions, Hints and Solutions

Because the nonzero numbers sum to zero, two have one sign and the third has the opposite sign. Permuting the variables and changing all three signs do not change the expression. We may therefore suppose $a,b>0$ and $c=-(a+b)$.

Set $u=a/b+b/a$. Since $(a-b)^2\ge0$, we have $u\ge2$, with equality exactly when $a=b$. Also

$$\begin{gathered}a^2+b^2+c^2=2ab(u+1),\\ \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=\frac1{ab}\left(u+\frac1{u+2}\right).\end{gathered}$$

Here $(a+b)^2=ab(u+2)$, which explains the last denominator. Multiplication gives $2(u+1)^3/(u+2)$. Its difference from the proposed bound is

$$\begin{aligned}\frac{2(u+1)^3}{u+2}-\frac{27}{2}&=\frac{4(u+1)^3-27(u+2)}{2(u+2)}\\&=\frac{(u-2)(4u^2+20u+25)}{2(u+2)}\\&\ge0.\end{aligned}$$

All factors apart from $u-2$ are positive for $u\ge2$. Equality is therefore equivalent to $a=b$, giving $(a,b,c)=(t,t,-2t)$ after undoing the sign change and permutation, where $t\ne0$. Every such triple meets the condition and attains the bound.

Conclusion: Equality holds exactly for permutations of $(t,t,-2t)$, with $t\ne0$.

Review the idea: Algebraic identities · Inequality rules and signs

If $n=p$ is prime. For $1\le k<p$,

$$k!\binom pk=p(p-1)\cdots(p-k+1).$$

The right side is divisible by $p$, whereas none of $1,\ldots,k$ is. Euclid’s prime-divisor rule therefore implies $p\mid\binom pk$.

If $n$ is composite. Choose a prime divisor $p$ of $n$. Then $p<n$, so the alleged property would apply at $k=p$. Write $\binom np=nt$ if it holds. The factorial identity

$$p\binom np=n\binom{n-1}{p-1}$$

then gives $\binom{n-1}{p-1}=pt$. But

$$(p-1)!\binom{n-1}{p-1}=(n-1)(n-2)\cdots(n-p+1)\equiv(-1)^{p-1}(p-1)!\pmod p.$$

The integer $(p-1)!$ is coprime to $p$, so it can be cancelled modulo $p$. We get $\binom{n-1}{p-1}\equiv(-1)^{p-1}\not\equiv0\pmod p$, a contradiction. This also handles $p=2$, where $-1$ is nonzero modulo 2. Thus no composite $n$ has the property.

Conclusion: The divisibility condition holds exactly for prime $n$.

Review the idea: Combinations and binomial coefficients · Prime numbers · Divisibility

We first build one closed route through all cells. Number rows 1 to 6 from top to bottom and columns 1 to 8 from left to right. Starting at $(1,1)$, go right along row 1 to column 8. Snake through rows 2 to 6 using columns 2 to 8: travel left along row 2, right along row 3, left along row 4, right along row 5, and left along row 6. Move from $(6,2)$ to $(6,1)$, then go up column 1 to the starting cell.

Every step joins cells sharing a side, and each cell occurs exactly once before the route closes. Along this cycle the colours alternate. If two cells of opposite colours are removed, the number of steps between them along either direction of the cycle is odd. Hence the number of cells strictly between them is even.

The remaining route therefore consists of two paths, each containing an even number of cells; one path may be empty if the removed cells were neighbours on the route. On each nonempty path, pair its first cell with its second, its third with its fourth, and so on. Each pair shares a side and forms a domino. The pairs are disjoint and cover precisely all remaining cells, proving the claim.

Conclusion: Every pair of opposite-coloured removed cells admits a domino tiling.

Review the idea: Parity · Counting with bijections · Proof methods

The diagonals of a convex quadrilateral meet inside it. Thus the parallel line crosses the two legs $AD$ and $BC$, with $E,O,F$ in that order.

Triangles $AOB$ and $COD$ have equal vertical angles at $O$ and corresponding angles from $AB\parallel CD$, so they are similar. Therefore

$$\begin{gathered}\frac{BO}{OD}=\frac{AO}{OC}=\frac ab,\\ \frac{DO}{DB}=\frac{CO}{CA}=\frac b{a+b}.\end{gathered}$$

Because $EO\parallel AB$, triangles $DEO$ and $DAB$ are similar. It follows that

$$\begin{gathered}\frac{EO}{AB}=\frac{DO}{DB}=\frac b{a+b},\\\text{so } EO=\frac{ab}{a+b}.\end{gathered}$$

Likewise $FO\parallel BA$, so triangles $CFO$ and $CBA$ are similar. Hence $OF/BA=CO/CA=b/(a+b)$, giving the same value for $OF$. In particular, $O$ is the midpoint of the parallel section and $EF=2ab/(a+b)$.

Conclusion: $EO=OF=\dfrac{ab}{a+b}$.

Review the idea: Similar triangles · Quadrilaterals and their diagonals