Local Mathematical Competitions

Let us build $E^{\prime }$, the symmetrical of $E$ with respect to $P$. If $E$ is latticeal, then $E^{\prime }$ is latticeal, and this implies $E^{\prime }\notin \text{int}(ABC)$. We deduce that $E^{\prime }=A$, otherwise $A\in (E^{\prime }P)$ and through a translation we get that there exists a latticeal point lying on $PE$, contradiction. So in this case $\frac{AP}{PE}=1$. If $E$ is not a latticeal point, and on ($BC$) lies some other latticeal point $Q$, since $Q\ne E$ we can restrict the problem to either triangle $ABQ$ or $ACQ$, with the same initial hypothesis. Thus we can assume that the only latticeal points on the sides of the triangle are on $[AB]$ and $[AC]$. We shall prove that, over all configurations, $$\max \frac{AP}{PE}=5,$$ value which can actually be reached by a proper configuration. Rather than looking for $\max \frac{AP}{PE}$, we will establish $\min \frac{PE}{AE}=\min \frac{[BPC]}{[ABC]}$ (where by $[XYZ]$ we mean the area of $△XYZ$). Let us notice that for the triangle $BPC$, the only latticeal points that are lying either on its sides, or in its interior, are just its vertices. We deduce from Pick's Theorem that $[BPC]=\frac{1}{2}$. Let us denote by $\beta$ and $\gamma$ the number of latticeal points on the sides ($AB$), respectively ($AC$). Availing ourselves again of Pick's Theorem, we find that $[ABC]=\frac{\beta +\gamma +3}{2}$. We get that $\frac{PE}{AE}=\frac{1}{\beta +\gamma +3}$. We want to prove that $\frac{PE}{AE}\ge \frac{1}{6}$. We contrariwise assume $\frac{PE}{AE}<\frac{1}{6}$, and thus $\beta +\gamma \ge 4$.

Denote by $B_0,B_1,\dots ,B_{\beta +1}$ the lattice points on $[AB]$, with $B_0=A$, $B_{\beta +1}=B$, and by $C_0,C_1,\dots ,C_{\gamma +1}$ the lattice points on $[AC]$, with $C_0=A$, $C_{\gamma +1}=C$. Let us consider the triangles of the type $P{C}_i{C}_{i+1}$ with $i=0,\dots ,\gamma$. They share the property that no lattice points lie inside or on the sides, other than the vertices, so $[P{C}_i{C}_{i+1}]=\frac{1}{2}$. It follows that $A{C}_1=C_1{C}_2=\cdots =C_{\gamma }C$. With a similar reasoning we have also $A{B}_1=B_1{B}_2=\cdots =B_{\beta }B$. Thus $\frac{A{C}_j}{C_j{C}_{j+1}}=\frac{A{B}_j}{B_j{B}_{j+1}}$, and so $C_i{B}_i\parallel C_{i+1}{B}_{i+1}$, for all $i\in \{1,2,\dots ,\min \{\beta ,\gamma \}\}$. We also deduce that on the segment $(B_i{C}_i)$ lie $i-1$ lattice points. This is because we can construct $i-1$ parallelograms with vertices of the type $B_l{C}_l{B}_{i}D$ with $l<i$, or of the type $B_l{C}_l{C}_{i}D$ with $D\in (B_i{C}_i)$. From effectively writing the relationships between coordinates in a parallelogram, we get that $D$ is a lattice point. Thus if $\min \{\beta ,\gamma \}\ge 2$, the segments $(B_i,C_i)$ would contain at least three lattice points (the ones from $(B_2{C}_2)$ and $(B_3{C}_3)$), which is obviously false since $P$ is on at most one of the segments $(B_i{C}_i)$. This means $\min \{\beta ,\gamma \}\le 1$, with say $\beta \ge \gamma$.

First case. $\gamma =1$. From $\beta +\gamma \ge 4$ follows that $\beta \ge 3$ and that $C_1$ is the midpoint of $AC$. Now since $B_1$ is the midpoint of $A{B}_2$ we have that if we take $M$ the midpoint of $C{B}_2$, we get the parallelogram $B_2{B}_1{C}_{1}M$, and since all the points $B_2,B_1,C_1$ are lattice, we get $M$ lattice. Since $M$ is inside the triangle $ABC$, it must be $P$. The quadrilateral $B{B}_{\beta }PC$ is convex since $B_{\beta }\in (B_{2}B)$. We note that the triangles $B{B}_{\beta }C$ and $B{B}_{\beta }P$ have the property that there are no lattice points inside or on their sides, except their vertices, so $[B{B}_{\beta }P]=[B{B}_{\beta }C]=\frac{1}{2}$, hence $B{B}_{\beta }\parallel PC$, which is false.

Second case. $\gamma =0$. It follows that $\beta \ge 4$.

LEMMA. Let $XYZT$ be a convex quadrilateral with lattice vertices, with the property that no lattice points lie in its interior, nor on its sides, except for its vertices. Then $XYZT$ is a parallelogram. Proof. From Pick's theorem we know that $[XYZ]=[XYT]=\frac{1}{2}$, and so $XY\parallel ZT$. Similarly $XT\parallel YZ$, which ends the proof. $\square$

Let us consider the quadrilateral $A{B}_{1}PC$, with $A{B}_1\cap PC\ne \varnothing$. If it is convex, the previous LEMMA implies a contradiction. Thus $[A{B}_1]\cap PC\ne \varnothing$. Also consider the quadrilateral $B{B}_{\beta }PC$. By the same reasoning, we must have $[B{B}_{\beta }]\cap PC\ne \varnothing$. This is the desired contradiction and ends the solution of the problem.