Subfield: Geometry


Question: Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent.

Solution:
we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume without loss of generality that $\theta \geqslant 0$.

Let $x=A B=D E, y=C D=F A, z=E F=B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P, E F A Q$, and $A B C R$ are parallelograms. We compute

$$
\begin{aligned}
\angle P E Q & =\angle F E Q+\angle D E P-\angle E=\left(180^{\circ}-\angle F\right)+\left(180^{\circ}-\angle D\right)-\angle E \\
& =360^{\circ}-\angle D-\angle E-\angle F=\frac{1}{2}(\angle A+\angle B+\angle C-\angle D-\angle E-\angle F)=\theta / 2
\end{aligned}
$$

Similarly, $\angle Q A R=\angle R C P=\theta / 2$.

<img_3816>

If $\theta=0$, since $\triangle R C P$ is isosceles, $R=P$. Therefore $A B\|R C=P C\| E D$, so $A B D E$ is a parallelogram. Similarly, $B C E F$ and $C D F A$ are parallelograms. It follows that $A D, B E$ and $C F$ meet at their common midpoint.

Now assume $\theta>0$. Since $\triangle P E Q, \triangle Q A R$, and $\triangle R C P$ are isosceles and have the same angle at the apex, we have $\triangle P E Q \sim \triangle Q A R \sim \triangle R C P$ with ratios of similarity $y: z: x$. Thus

$$
\triangle P Q R \text { is similar to the triangle with sidelengths } y, z, \text { and } x \text {. }
\tag{1}
$$

Next, notice that

$$
\frac{R Q}{Q P}=\frac{z}{y}=\frac{R A}{A F}
$$

and, using directed angles between rays,

$$
\begin{aligned}
\sphericalangle(R Q, Q P) & =\sphericalangle(R Q, Q E)+\sphericalangle(Q E, Q P) \\
& =\sphericalangle(R Q, Q E)+\sphericalangle(R A, R Q)=\sphericalangle(R A, Q E)=\sphericalangle(R A, A F) .
\end{aligned}
$$

Thus $\triangle P Q R \sim \triangle F A R$. Since $F A=y$ and $A R=z,(1)$ then implies that $F R=x$. Similarly $F P=x$. Therefore $C R F P$ is a rhombus.

We conclude that $C F$ is the perpendicular bisector of $P R$. Similarly, $B E$ is the perpendicular bisector of $P Q$ and $A D$ is the perpendicular bisector of $Q R$. It follows that $A D, B E$, and $C F$ are concurrent at the circumcenter of $P Q R$.
we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume without loss of generality that $\theta \geqslant 0$.

Let $X=C D \cap E F, Y=E F \cap A B, Z=A B \cap C D, X^{\prime}=F A \cap B C, Y^{\prime}=$ $B C \cap D E$, and $Z^{\prime}=D E \cap F A$. 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 $B C$ and $A B$ from the hexagon. Similar conclusions hold for $X, Y, Y^{\prime}$, and $Z^{\prime}$. Then

$$
\angle Y Z X=\angle B+\angle C-180^{\circ}=\angle E+\angle F-180^{\circ}=\angle Y^{\prime} Z^{\prime} X^{\prime},
$$

and similarly $\angle Z X Y=\angle Z^{\prime} X^{\prime} Y^{\prime}$ and $\angle X Y Z=\angle X^{\prime} Y^{\prime} Z^{\prime}$, so $\triangle X Y Z \sim \triangle X^{\prime} Y^{\prime} Z^{\prime}$. Thus there is a rotation $R$ which sends $\triangle X Y Z$ to a triangle with sides parallel to $\triangle X^{\prime} Y^{\prime} Z^{\prime}$. Since $A B=D E$ we have $R(\overrightarrow{A B})=\overrightarrow{D E}$. Similarly, $R(\overrightarrow{C D})=\overrightarrow{F A}$ and $R(\overrightarrow{E F})=\overrightarrow{B C}$. Therefore

$$
\overrightarrow{0}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D E}+\overrightarrow{E F}+\overrightarrow{F A}=(\overrightarrow{A B}+\overrightarrow{C D}+\overrightarrow{E F})+R(\overrightarrow{A B}+\overrightarrow{C D}+\overrightarrow{E F})
$$

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

$$
\overrightarrow{A B}+\overrightarrow{C D}+\overrightarrow{E F}=\overrightarrow{0}
$$

or else we would have two vectors with different directions whose sum is $\overrightarrow{0}$.

<img_3570>

This allows us to consider a triangle $L M N$ with $\overrightarrow{L M}=\overrightarrow{E F}, \overrightarrow{M N}=\overrightarrow{A B}$, and $\overrightarrow{N L}=\overrightarrow{C D}$. Let $O$ be the circumcenter of $\triangle L M N$ and consider the points $O_{1}, O_{2}, O_{3}$ such that $\triangle A O_{1} B, \triangle C O_{2} D$, and $\triangle E O_{3} F$ are translations of $\triangle M O N, \triangle N O L$, and $\triangle L O M$, respectively. Since $F O_{3}$ and $A O_{1}$ are translations of $M O$, quadrilateral $A F O_{3} O_{1}$ is a parallelogram and $O_{3} O_{1}=F A=C D=N L$. Similarly, $O_{1} O_{2}=L M$ and $O_{2} O_{3}=M N$. Therefore $\triangle O_{1} O_{2} O_{3} \cong \triangle L M N$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.

Let $T$ be the circumcenter of $\triangle O_{1} O_{2} O_{3}$. We claim that $A D, B E$, and $C F$ 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 $\triangle L M N$. Therefore $\triangle T O_{3} F$ and $\triangle C_{2} T$ are isosceles. Using directed angles between rays again, we get

$$
\sphericalangle\left(T F, T O_{3}\right)=\sphericalangle\left(F O_{3}, F T\right) \quad \text { and } \quad \sphericalangle\left(T O_{2}, T C\right)=\sphericalangle\left(C T, C O_{2}\right) \text {. }
\tag{2}
$$

Also, $T$ and $O$ are the circumcenters of the congruent triangles $\triangle O_{1} O_{2} O_{3}$ and $\triangle L M N$ so we have $\sphericalangle\left(T O_{3}, T O_{2}\right)=\sphericalangle(O N, O M)$. Since $C O_{2}$ and $F_{3}$ are translations of $N O$ and $M O$ respectively, this implies

$$
\sphericalangle\left(T O_{3}, T O_{2}\right)=\sphericalangle\left(C O_{2}, F O_{3}\right) .
\tag{3}
$$



Adding the three equations in (2) and (3) gives

$$
\sphericalangle(T F, T C)=\sphericalangle(C T, F T)=-\sphericalangle(T F, T C)
$$

which implies that $T$ is on $C F$. Analogous arguments show that it is on $A D$ and $B E$ also. The desired result follows.
we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume without loss of generality that $\theta \geqslant 0$.

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 $a^{\prime}=a r$, $b^{\prime}=b r$, and $c^{\prime}=c r$ for a complex number $r$ with $|r|=1$, as shown below.

<img_3725>

We have

$$
0=a+c r+b+a r+c+b r=(a+b+c)(1+r) \text {. }
$$

If $r=-1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise

$$
a+b+c=0
$$

Therefore

$$
A=0, \quad B=a, \quad C=a+c r, \quad D=c(r-1), \quad E=-b r-c, \quad F=-b r
$$

Now consider a point $W$ on $A D$ given by the complex number $c(r-1) \lambda$, where $\lambda$ is a real number with $0<\lambda<1$. Since $D \neq A$, we have $r \neq 1$, 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,

$$
\begin{aligned}
W \text { is on } B E & \Longleftrightarrow c(r-1) \lambda-a\|a-(-b r-c)=b(r-1) \Longleftrightarrow c \lambda-a s\| b \\
& \Longleftrightarrow-a \lambda-b \lambda-a s\|b \Longleftrightarrow a(\lambda+s)\| b .
\end{aligned}
$$

One easily checks that $r \neq \pm 1$ implies that $\lambda+s \neq 0$ since $s$ is not real. On the other hand,

$$
\begin{aligned}
W \text { on } C F & \Longleftrightarrow c(r-1) \lambda+b r\|-b r-(a+c r)=a(r-1) \Longleftrightarrow c \lambda+b(1+s)\| a \\
& \Longleftrightarrow-a \lambda-b \lambda-b \bar{s}\|a \Longleftrightarrow b(\lambda+\bar{s})\| a \Longleftrightarrow b \| a(\lambda+s),
\end{aligned}
$$

where in the last step we use that $(\lambda+s)(\lambda+\bar{s})=|\lambda+s|^{2} \in \mathbb{R}_{>0}$. We conclude that $A D \cap B E=$ $C F \cap B E$, and the desired result follows.