BWM Runde 2 Mock Paper 2: Questions, Hints and Solutions

At $x=0$ the expression is 1. For $x\ne0$, division by the positive number $x^2$ changes it into

$$\begin{gathered}t^2+at+b-2,\\ t=x+\frac1x.\end{gathered}$$

The range of $t$ is exactly $( -\infty,-2]\cup[2,\infty)$: the equation $x^2-tx+1=0$ has a nonzero real solution exactly when its discriminant $t^2-4$ is nonnegative.

Put $u=|a|$. For each $s=|t|\ge2$, the smaller value between the two signs of $t$ is $s^2-us+b-2$. Both signs are allowed, so this smaller value must be nonnegative for every $s\ge2$.

If $0\le u\le4$, write $s=2+h$, where $h\ge0$. Then

$$s^2-us+b-2=h^2+(4-u)h+b+2-2u.$$

The first two terms are nonnegative and vanish together at $h=0$. Thus the condition is $b\ge2u-2$.

If $u\ge4$, completing the square gives

$$s^2-us+b-2=(s-u/2)^2+b-2-u^2/4.$$

Its vertex $s=u/2$ is allowed, so the condition is $b\ge2+u^2/4$. The formulas agree at $u=4$.

Boundary equality. When $a=0,b=-2$, the polynomial is $(x^2-1)^2$, with zeros $x=\pm1$. When $0<|a|\le4$ and $b=2|a|-2$, the required $t$ is $-2\operatorname{sgn}(a)$, so $x=-\operatorname{sgn}(a)$. Here $\operatorname{sgn}(a)$ is 1 for positive $a$ and $-1$ for negative $a$. When $|a|>4$ and $b=2+a^2/4$, the required $t$ is $-a/2$, giving the two roots of $x^2+(a/2)x+1=0$. The discriminant is positive in this case. These are all equality cases, because the minimisations above account for all allowed $t$.

Conclusion: For $|a|\le4$, $b\ge2|a|-2$; for $|a|\ge4$, $b\ge2+a^2/4$. Boundary zeros are described in the solution.

Review the idea: Equations and quadratics · Absolute value inequalities

The answer consists of the following families, with all displayed letters denoting primes:

$$1;\quad p,p^2,p^3;\quad pq,pq^2\ (p<q);\quad p^2q\ (p<q<p^2);\quad pqr\ (p<q<r<pq).$$

To prove completeness, a composite $N$ with smallest composite divisor $m$ has the required property exactly when $N<m^2$. Equality would put the forbidden composite divisor at $\sqrt N$.

One distinct prime. For $N=p^a$, the cases $a=1$ work immediately. If $a\ge2$, the smallest composite divisor is $p^2$, so $p^a<p^4$ is equivalent to $a\le3$.

Two distinct primes. Write $N=p^a q^b$, with $p<q$ and $a,b\ge1$. If $a\ge2$, the smallest composite divisor is $p^2$, requiring $p^a q^b<p^4$. If $a\ge3$, then $p^a q^b\ge p^3q>p^4$; if $b\ge2$, then $p^a q^b\ge p^2q^2>p^4$. Thus only $a=2,b=1$ remains, and its condition is $q<p^2$. If $a=1$, the smallest composite divisor is $pq$. The condition becomes $pq^b<p^2q^2$, or $q^{b-2}<p$. It holds for $b=1,2$, and fails for $b\ge3$ because $q>p$.

Three distinct primes. Write the primes as $p<q<r$. If $p$ occurs twice, then $N\ge p^2qr>p^4$, while $p^2$ is a composite divisor; this fails. If $q$ occurs twice, $N\ge pq^2r>p^2q^2$; if $r$ occurs twice, $N\ge pqr^2>p^2q^2$. In each case the composite divisor $pq$ is at most $\sqrt N$. Thus all exponents must be 1. Now the smallest composite divisor is $pq$, so precisely $r<pq$ is required.

Four or more distinct primes. For the four smallest $p<q<r<s$, we have $N\ge pqrs>(pq)^2$. The composite divisor $pq$ violates the condition.

Finally, 1 works because its only divisor is 1. In every family listed, either there is no composite divisor below the required threshold, or the smallest composite divisor satisfies the strict inequality just proved. This verifies sufficiency as well as necessity.

Conclusion: $1$; $p,p^2,p^3$; $pq,pq^2$ for $p<q$; $p^2q$ for $p<q<p^2$; $pqr$ for $p<q<r<pq$.

Review the idea: Unique prime factorisation · Counting and summing divisors

Start with any division into two groups; an empty group is permitted. Count the edges whose endpoints lie in different groups, calling them crossing edges.

If a vertex has at least two neighbours in its own group, move it to the other group. Suppose it has $i$ neighbours inside its old group and $j$ outside. The assumption gives $i\ge2$ and $i+j\le3$, so $j\le1$. Moving the vertex changes its $i$ internal edges into crossing edges and its $j$ crossing edges into internal ones. No other edge changes status. Thus the number of crossing edges increases by $i-j\ge1$.

This count is an integer between 0 and $e$. Consequently there can be at most $e$ such moves, regardless of the choices made. On termination, no vertex has two or more neighbours in its own group. That is exactly the required property. In particular, within either group each connected piece consists of either a single vertex or a single edge: a longer connected piece would contain a vertex with two internal neighbours.

Conclusion: Repeatedly move any vertex with at least two same-group neighbours; at most $e$ moves suffice.

Review the idea: Proof methods · Counting

Straighten the route. Write $A_0=\angle BAC$. Reflect $X$ in $AB$ and $AC$, calling the images $X_1,X_2$. Reflection fixes $Z$ on $AB$ and $Y$ on $AC$, so $XZ=X_1Z$ and $XY=X_2Y$. Repeated use of the triangle inequality gives

$$XY+YZ+ZX=X_1Z+ZY+YX_2\ge X_1X_2.$$

Both reflected points are distance $AX$ from $A$. Reflecting a ray in the two sides of an angle places its images at an angular separation of $2A_0$. Since $A_0<90^\circ$, this is the smaller angle between them. Bisecting the isosceles triangle $AX_1X_2$ gives $X_1X_2=2AX\sin A_0$. The right triangle $ADX$ gives $AX^2=AD^2+DX^2$, so

$$XY+YZ+ZX\ge2AX\sin A_0\ge2AD\sin A_0.$$

The latter equality requires $X=D$.

Show that the altitude feet attain the bound. Reflect $D$ in $AB,AC$, obtaining $D_1,D_2$. We prove that $D_1,F,E,D_2$ lie on one straight line in that order. Let $B_0=\angle ABC$ and $C_0=\angle BCA$.

The points $A,C,D,F$ are cyclic because the angles at $D,F$ subtending $AC$ are right angles. Their opposite angles give $\angle DFA=180^\circ-C_0$, hence $\angle DFB=C_0$. Also $B,C,E,F$ are cyclic, since $\angle BEC=\angle BFC=90^\circ$. They give $\angle BFE=180^\circ-C_0$. Reflection in $AB$ takes ray $FD$ to $FD_1$, on the opposite side of $AB$. These angle relations make $FD_1$ and $FE$ opposite rays.

At the other side, $A,B,D,E$ are cyclic because $\angle ADB=\angle AEB=90^\circ$. Thus $\angle DEC=B_0$. The circle through $B,C,E,F$ gives $\angle CEF=180^\circ-B_0$. Reflection of $ED$ in $AC$ is $ED_2$, so these relations make $ED_2$ and $EF$ opposite rays. Hence the claimed order is $D_1,F,E,D_2$.

All feet are inside the sides because the triangle is acute. Reflection now gives

$$DF+FE+ED=D_1F+FE+ED_2=D_1D_2=2AD\sin A_0.$$

Finally, equality in the broken-route triangle inequality requires $Z,Y$ to lie, in that order, on the straight segment $X_1X_2$. We already know $X=D$; that line meets $AB$ at $F$ and $AC$ at $E$, by the established construction. These intersections are unique, so $Z=F,Y=E$. This proves the equality classification.

Conclusion: The unique minimum is $2AD\sin\angle BAC$, attained by the triangle of altitude feet $DEF$.

Review the idea: Triangle inequalities · Cyclic and tangential quadrilaterals · Trigonometry in geometry