International Mathematical Olympiad Shortlisted Problems

In all three solutions, we denote $\theta =\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume without loss of generality that $\theta ⩾0$.

Solution 1. Let $x=AB=DE$, $y=CD=FA$, $z=EF=BC$. Consider the points $P,Q$, and $R$ such that the quadrilaterals $CDEP$, $EFAQ$, and $ABCR$ are parallelograms. We compute $$\begin{array}{rl}\angle PEQ& =\angle FEQ+\angle DEP-\angle E=(180^{\circ }-\angle F)+(180^{\circ }-\angle D)-\angle E\\ & =360^{\circ }-\angle D-\angle E-\angle F=\frac{1}{2}(\angle A+\angle B+\angle C-\angle D-\angle E-\angle F)=\theta /2\end{array}$$ Similarly, $\angle QAR=\angle RCP=\theta /2$.



If $\theta =0$, since $△RCP$ is isosceles, $R=P$. Therefore $AB\parallel RC=PC\parallel ED$, so $ABDE$ is a parallelogram. Similarly, $BCEF$ and $CDFA$ are parallelograms. It follows that $AD$, $BE$ and $CF$ meet at their common midpoint.

Now assume $\theta >0$. Since $△PEQ$, $△QAR$, and $△RCP$ are isosceles and have the same angle at the apex, we have $△PEQ\sim △QAR\sim △RCP$ with ratios of similarity $y:z:x$. Thus $$\begin{array}{c}△PQR\text{ is similar to the triangle with sidelengths }y,z,\text{ and }x.\end{array}\text{\hspace{1em}(1)}$$ Next, notice that $$\frac{RQ}{QP}=\frac{z}{y}=\frac{RA}{AF}$$ and, using directed angles between rays, $$\begin{array}{rl}⋡(RQ,QP)& =\text{Varangle}(RQ,QE)+/(QE,QP)\\ & =\text{Varangle}(RQ,QE)+/(RA,RQ)=\text{Varangle}(RA,QE)=\text{Varangle}(RA,AF).\end{array}$$ Thus $△PQR\sim △FAR$. Since $FA=y$ and $AR=z$, (1) then implies that $FR=x$. Similarly $FP=x$. Therefore $CRFP$ is a rhombus.

We conclude that $CF$ is the perpendicular bisector of $PR$. Similarly, $BE$ is the perpendicular bisector of $PQ$ and $AD$ is the perpendicular bisector of $QR$. It follows that $AD$, $BE$, and $CF$ are concurrent at the circumcenter of $PQR$.

Solution 2. Let $X=CD\cap EF$, $Y=EF\cap AB$, $Z=AB\cap CD$, $X^{\prime }=FA\cap BC$, $Y^{\prime }=BC\cap DE$, and $Z^{\prime }=DE\cap FA$. From $\angle A+\angle B+\angle C=360^{\circ }+\theta /2$ we get $\angle A+\angle B>180^{\circ }$ and $\angle B+\angle C>180^{\circ }$, so $Z$ and $X^{\prime }$ are respectively on the opposite sides of $BC$ and $AB$ from the hexagon. Similar conclusions hold for $X,Y,Y^{\prime }$, and $Z^{\prime }$. Then $$\angle YZX=\angle B+\angle C-180^{\circ }=\angle E+\angle F-180^{\circ }=\angle Y^{\prime }{Z}^{\prime }{X}^{\prime }$$ and similarly $\angle ZXY=\angle Z^{\prime }{X}^{\prime }{Y}^{\prime }$ and $\angle XYZ=\angle X^{\prime }{Y}^{\prime }{Z}^{\prime }$, so $△XYZ\sim △X^{\prime }{Y}^{\prime }{Z}^{\prime }$. Thus there is a rotation $R$ which sends $△XYZ$ to a triangle with sides parallel to $△X^{\prime }{Y}^{\prime }{Z}^{\prime }$. Since $AB=DE$ we have $R(\overrightarrow{AB})=\overrightarrow{DE}$. Similarly, $R(\overrightarrow{CD})=\overrightarrow{FA}$ and $R(\overrightarrow{EF})=\overrightarrow{BC}$. Therefore $$\overrightarrow{0}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA}=(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EF})+R(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EF})$$ If $R$ is a rotation by $180^{\circ }$, then any two opposite sides of our hexagon are equal and parallel, so the three diagonals meet at their common midpoint. Otherwise, we must have $$\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EF}=\overrightarrow{0}$$ or else we would have two vectors with different directions whose sum is $\overrightarrow{0}$.



This allows us to consider a triangle $LMN$ with $\overrightarrow{LM}=\overrightarrow{EF}$, $\overrightarrow{MN}=\overrightarrow{AB}$, and $\overrightarrow{NL}=\overrightarrow{CD}$. Let $O$ be the circumcenter of $△LMN$ and consider the points $O_1,O_2,O_3$ such that $△A{O}_{1}B$, $△C{O}_{2}D$, and $△E{O}_{3}F$ are translations of $△MON$, $△NOL$, and $△LOM$, respectively. Since $F{O}_3$ and $A{O}_1$ are translations of $MO$, quadrilateral $AF{O}_3{O}_1$ is a parallelogram and $O_3{O}_1=FA=CD=NL$. Similarly, $O_1{O}_2=LM$ and $O_2{O}_3=MN$. Therefore $△O_1{O}_2{O}_3\cong △LMN$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.

Let $T$ be the circumcenter of $△O_1{O}_2{O}_3$. We claim that $AD$, $BE$, and $CF$ meet at $T$. Let us show that $C$, $T$, and $F$ are collinear. Notice that $C{O}_2=O_{2}T=T{O}_3=O_{3}F$ since they are all equal to the circumradius of $△LMN$. Therefore $△T{O}_{3}F$ and $△C{O}_{2}T$ are isosceles. Using directed angles between rays again, we get $$\begin{array}{c}⋡\left(TF,T{O}_3\right)=\text{Varangle}\left(F{O}_3,FT\right)\text{ and }/\left(T{O}_2,TC\right)=\text{Varangle}\left(CT,C{O}_2\right).\end{array}\text{\hspace{1em}(2)}$$ Also, $T$ and $O$ are the circumcenters of the congruent triangles $△O_1{O}_2{O}_3$ and $△LMN$ so we have $\text{Varangle}\left(T{O}_3,T{O}_2\right)=\text{Varangle}(ON,OM)$. Since $C{O}_2$ and $F{O}_3$ are translations of $NO$ and $MO$ respectively, this implies $$\begin{array}{c}⋡\left(T{O}_3,T{O}_2\right)=\text{Varangle}\left(C{O}_2,F{O}_3\right).\end{array}\text{\hspace{1em}(3)}$$ Adding the three equations in (2) and (3) gives $$⋡(TF,TC)=\text{Varangle}(CT,FT)=-\text{Varangle}(TF,TC)$$ which implies that $T$ is on $CF$. Analogous arguments show that it is on $AD$ and $BE$ also. The desired result follows.

Solution 3. Place the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A,B,C,D,E,F$ represent the corresponding complex numbers. Also consider the complex numbers $a,b,c,a^{\prime },b^{\prime },c^{\prime }$ given by $B-A=a$, $D-C=b$, $F-E=c$, $E-D=a^{\prime }$, $A-F=b^{\prime }$, and $C-B=c^{\prime }$. Let $k=|a|/|b|$. From $a/b^{\prime }=-k{e}^{i\angle A}$ and $a^{\prime }/b=-k{e}^{i\angle D}$ we get that $\left(a^{\prime }/a\right)\left(b^{\prime }/b\right)=e^{-i\theta }$ and similarly $\left(b^{\prime }/b\right)\left(c^{\prime }/c\right)=e^{-i\theta }$ and $\left(c^{\prime }/c\right)\left(a^{\prime }/a\right)=e^{-i\theta }$. It follows that $a^{\prime }=ar$, $b^{\prime }=br$, and $c^{\prime }=cr$ for a complex number $r$ with $|r|=1$, as shown below.



We have $$0=a+cr+b+ar+c+br=(a+b+c)(1+r).$$ If $r=-1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise $$a+b+c=0.$$ Therefore $$A=0,B=a,C=a+cr,D=c(r-1),E=-br-c,F=-br.$$ Now consider a point $W$ on $AD$ given by the complex number $c(r-1)\lambda$, where $\lambda$ is a real number with $0<\lambda <1$. Since $D\ne A$, we have $r\ne 1$, so we can define $s=1/(r-1)$. From $r\bar{r}=|r{|}^2=1$ we get $$1+s=\frac{r}{r-1}=\frac{r}{r-r\bar{r}}=\frac{1}{1-\bar{r}}=-\bar{s}.$$ Now, $$\begin{array}{rl}W\text{ is on }BE& ⟺c(r-1)\lambda -a\Vert a-(-br-c)=b(r-1)⟺c\lambda -as\Vert b\\ & ⟺-a\lambda -b\lambda -as\Vert b⟺a(\lambda +s)\Vert b.\end{array}$$ One easily checks that $r\ne \pm 1$ implies that $\lambda +s\ne 0$ since $s$ is not real. On the other hand, $$\begin{array}{rl}W\text{ on }CF& ⟺c(r-1)\lambda +br\Vert -br-(a+cr)=a(r-1)⟺c\lambda +b(1+s)\Vert a\\ & ⟺-a\lambda -b\lambda -b\bar{s}\Vert a⟺b(\lambda +\bar{s})\Vert a⟺b\Vert a(\lambda +s),\end{array}$$ where in the last step we use that $(\lambda +s)(\lambda +\bar{s})=|\lambda +s{|}^2\in {\mathbb{R}}_{>0}$. We conclude that $AD\cap BE=CF\cap BE$, and the desired result follows.