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

The first and third expressions balance at a useful candidate. More importantly, their average gives a lower bound valid for every positive $x$:

$$\begin{aligned}M(x)&\ge\frac12\left(x+\frac1x+4x+\frac1{4x}\right)\\&=\frac52x+\frac5{8x}\\&\ge 2\sqrt{\frac52x\cdot\frac5{8x}}\\&=\frac52.\end{aligned}$$

The last inequality is AM–GM; it follows from $(\sqrt u-\sqrt v)^2\ge0$ for positive $u,v$. Equality there requires $\frac52x=\frac5{8x}$, hence $x=\frac12$. At this value the three original expressions are $\frac52,2,\frac52$. Thus their maximum really is the lower bound, and the equality condition also proves uniqueness.

Conclusion: $\min M=\frac52$, attained only at $x=\frac12$.

Review the idea: Arithmetic geometric and harmonic means · Inequality rules and signs

Separate the common factor from the coprime parts: let $a=gx$, $b=gy$, with $x\le y$ and $\gcd(x,y)=1$. Then the lcm is $gxy$, so the given equation is equivalent to

$$\begin{gathered}g(xy-x-y+1)=6,\\ g(x-1)(y-1)=6.\end{gathered}$$

Neither $x$ nor $y$ can equal 1. Thus $x-1,y-1$ are positive and $g$ is one of $1,2,3,6$.

• For $g=1$, the ordered factor pairs of 6 with first entry no larger are $(1,6)$ and $(2,3)$. They give coprime pairs $(x,y)=(2,7),(3,4)$.
• For $g=2$, the only factor pair of 3 is $(1,3)$, giving $(2,4)$, which is not coprime.
• For $g=3$, the factor pair $(1,2)$ gives $(x,y)=(2,3)$, hence $(a,b)=(6,9)$.
• For $g=6$, the factor pair $(1,1)$ gives $(2,2)$, again not coprime.

The surviving original pairs are $(2,7),(3,4),(6,9)$. Their gcd/lcm sums are respectively $1+14=15$, $1+12=13$, and $3+18=21$, which equal $a+b+6$ in each case. The factor list proves that none are missing.

Conclusion: $(a,b)=(2,7),(3,4),(6,9)$.

Review the idea: Greatest common divisor · Factorisation integer solutions

Why the condition is necessary. Each move reverses two cells in each affected row. Its black-cell count changes by $-2$, 0, or 2, so its parity stays even. The same reasoning applies to each column.

Why it is sufficient. Take any colouring with the stated parity property. We shall return it to all white. Process rows 1 to 4 in order. Within a row, process columns 1 to 6 from left to right. If the cell being processed is black, flip the adjacent $2\times2$ block having that cell as its upper-left cell. The cell becomes white. The other changed cells lie either to its right or in the next row, so no cell already processed is changed.

After these 24 decisions, all cells in the first four rows and first six columns are white. In each of the first four rows only its last cell could still be black. Row parity is even throughout, so that last cell is also white. In each of the first six columns only the bottom cell could remain black; column parity forces it white. Finally the bottom-right cell is white by the parity of its row.

Every move is its own inverse. Reversing this finite clearing sequence takes the all-white board to the desired colouring. This proves both directions.

Conclusion: Exactly the colourings with an even number of black cells in every row and every column are reachable.

Review the idea: Parity · Proof methods

The intersections lie inside the diagonal: the segment from a vertex of a convex parallelogram to a point inside the opposite side crosses the diagonal separating them.

Since $BE\parallel DC$, while $B,P,D$ and $E,P,C$ are collinear, triangles $BEP$ and $DCP$ are similar. Therefore

$$\frac{BP}{PD}=\frac{BE}{DC}=\frac{AB/2}{AB}=\frac12.$$

So $BP=BD/3$. Likewise, $DF\parallel BC$ makes triangles $DFQ$ and $BCQ$ similar, and

$$\frac{DQ}{QB}=\frac{DF}{BC}=\frac{AD/2}{AD}=\frac12.$$

Thus $DQ=BD/3$ and $BQ=2BD/3$. In particular $BP<BQ$, giving the required order. Subtracting yields $PQ=BQ-BP=BD/3$, equal to both end pieces.

Conclusion: $BP=PQ=QD=BD/3$.

Review the idea: Similar triangles · Quadrilaterals and their diagonals