, 2020

Solution:

Denote the angle bisector of $A$ by $a$ and similarly for the other bisectors. Thus, given that $a, c, e$ have a common point $M$, we need to prove that $b, d, f$ are concurrent. We write $\angle(x, y)$ for the value of the directed angle between the lines $x$ and $y$, i.e. the angle of the counterclockwise rotation from $x$ to $y$ (defined $\bmod 180^\circ$).

Since the sum of the angles of a convex hexagon is $720^\circ$, from the angle conditions we get that the sum of any two consecutive angles is equal to $240^\circ$. In particular, it now follows that $\angle(b, a)=\angle(c, b)=\angle(d, c)=\angle(e, d)=\angle(f, e)=\angle(a, f)=60^\circ$ (assuming the hexagon is clockwise oriented).

Let $X=AB \cap CD$, $Y=CD \cap EF$ and $Z=EF \cap AB$. Similarly, let $P=BC \cap DE$, $Q=DE \cap FA$ and $R=FA \cap BC$. From $\angle B+\angle C=240^\circ$ it follows that $\angle(ZX, XY)=\angle(BX, XC)=60^\circ$. Similarly we have $\angle(XY, YZ)=\angle(YZ, ZX)=60^\circ$, so triangle $XYZ$ (and similarly triangle $PQR$) is equilateral. We see that the hexagon $ABCDEF$ is obtained by intersecting the two equilateral triangles $XYZ$ and $PQR$.

We have $\angle(AM, MC)=\angle(a, c)=\angle(a, b)+\angle(b, c)=60^\circ$, and since
$$\angle(AM, MC)=\angle(AX, XC)=\angle(AR, RC)=60^\circ$$
$A, C, M, X, R$ are concyclic. Because $M$ lies on the bisector of angle $\angle XAR$, we must have $MR=MX$, so triangle $MRX$ is isosceles. Moreover, we have $\angle(MR, MX)=\angle(CR, CX)=\angle(BC, CD)$, which is angle $C$ of the hexagon. We now see that the triangles $MRX, MPY$ and $MQZ$ are isosceles and similar. This implies that there is a rotation centered at $M$ that sends $X, Y$ and $Z$ to $R, P$ and $Q$ respectively. In particular, the equilateral triangles $XYZ$ and $PQR$ are congruent.

It follows that there also exists a rotation sending $X, Y, Z$ to $P, Q, R$ respectively. Define $N$ as the center of this rotation. Triangles $NXZ$ and $NPR$ are congruent and equally oriented, hence $N$ is equidistant from $XZ$ and $PR$ and lies on the inner bisector $b$ of $\angle B$ (we know $N$ lies on the inner, not the outer bisector because the rotation centered at $N$ is clockwise). In the same way we can show that $N$ is on $d$ and on $f$, so $b, d, f$ are concurrent at $N$.

Remark. The key observation (a rotation centered at $M$ sends $\triangle XYZ$ to $\triangle RQP$) can be established in slightly different ways. E.g., since $A, M, R, X$ are concyclic and $A, M, Q, Z$ are concyclic, $M$ is the Miquel point of the lines $XZ, RQ, XR, ZQ$, hence it is the center of similitude $s$ sending $\overrightarrow{XZ}$ to $\overrightarrow{RQ}$. Repeating the same argument for the other pairs of vectors, we obtain that $s$ sends $\triangle XYZ$ to $\triangle RQP$. Moreover, $s$ is a rotation, since $M$ is equidistant from $XZ$ and $RQ$.

Remark. The reverse argument can be derived in a different way, e.g., defining $N$ as the common point of the circles $BXP D, D YQ F, FZR B$, and showing that $\triangle NXZ=\triangle NPR$, etc.

Solution:

As in Solution A, we prove that the hexagon $ABCDEF$ is the intersection of the equilateral triangles $PQR$ and $XYZ$.

Let $d(S, AB)$ denote the signed distance from the point $S$ to the line $AB$, where the negative sign is taken if $AB$ separates $S$ and the hexagon. We define similarly the other distances ($d(S, BC)$, etc). Since $M \in a$, we have $d(M, ZX)=d(M, QR)$. In the same way, we have $d(M, XY)=d(M, RP)$ and $d(M, YZ)=d(M, PQ)$. Therefore $d(M, ZX)+d(M, XY)+d(M, YZ)=d(M, QR)+d(M, RP)+d(M, PQ)$.

We now use the following well-known lemma (which can be easily proved using areas) to deduce that triangles $PQR$ and $XYZ$ are congruent.

Lemma. The sum of the signed distances from any point to the sidelines of an equilateral triangle (where the signs are taken such that all distances are positive inside the triangle) is constant and equals the length of the altitude.

For $N=b \cap d$ we now find $d(N, ZX)=d(N, RP)$ and $d(N, XY)=d(N, PQ)$. Using again the lemma for the point $N$, we get $d(N, ZX)+d(N, XY)+d(N, YZ)=d(N, QR)+d(N, RP)+d(N, PQ)$. Therefore $d(N, YZ)=d(N, QR)$, thus $N \in f$.

Remark. Instead of using the lemma, it is possible to use some equivalent observation in terms of signed areas.

Solution:

We use the same notations as in Solution A. We will show that $a, c$ and $e$ are concurrent if and only if
$$AB+CD+EF=BC+DE+FA,$$
which clearly implies the problem statement by symmetry.

Let $\vec{a}$ be the vector of unit length parallel to $a$ directed from $A$ towards the interior of the hexagon. We define analogously $\vec{b}$, etc. The angle conditions imply that opposite bisectors of the hexagon are parallel, so we have $\vec{a}\parallel\vec{d}$, $\vec{b}\parallel \vec{e}$ and $\vec{c} \parallel \vec{f}$. Moreover, as in the previous solutions, we know that $\vec{a}, \vec{c}$ and $\vec{e}$ make angles of $120^\circ$ with each other. Let $M_A=c \cap e, M_C=e \cap a$ and $M_E=a \cap c$. Then $M_A, M_C, M_E$ form an equilateral triangle with side length denoted by $s$. Note that the case $s=0$ is equivalent to $a, c$ and $e$ being concurrent.

Projecting $\overrightarrow{M_EA}+\overrightarrow{AB}=\overrightarrow{M_EB}=\overrightarrow{M_EC}+\overrightarrow{CB}$ onto $\vec{e}=-\vec{b}$, we obtain
$$\overrightarrow{AB} \cdot \vec{b}-\overrightarrow{CB} \cdot \vec{b}=\overrightarrow{M_EC} \cdot \vec{b}-\overrightarrow{M_EA} \cdot \vec{b}=\overrightarrow{M_EA} \cdot \vec{e}-\overrightarrow{M_EC} \cdot \vec{e}$$
Writing $\varphi=\frac{1}{2} \angle B=\frac{1}{2} \angle D=\frac{1}{2} \angle F$, we know that $\overrightarrow{AB} \cdot \vec{b}=-AB \cdot \cos (\varphi)$, and similarly $\overrightarrow{CB} \cdot \vec{b}=-CB \cdot \cos (\varphi)$. Because $M_EA$ and $M_EC$ intersect $e$ at $120^\circ$ angles, we have $\overrightarrow{M_EA} \cdot \vec{e}=\frac{1}{2} M_EA$ and $\overrightarrow{M_EC} \cdot \vec{e}=\frac{1}{2} M_EC$. We conclude that
$$2 \cos (\varphi)(AB-CB)=M_EC-M_EA$$
Adding the analogous equalities $2 \cos (\varphi)(CD-ED)=M_AE-M_AC$ and $2 \cos (\varphi)(EF-AF)=M_CA-M_CE$, we obtain
$$2 \cos (\varphi)(AB+CD+EF-CB-ED-AF)=M_EC-M_EA+M_AE-M_AC+M_CA-M_CE.$$
Because $M_A, M_C$ and $M_E$ form an equilateral triangle with side length $s$, we have $M_EC-M_AC=\pm s, M_CA-M_EA=\pm s$, and $M_AE-M_CE=\pm s$. Therefore, the right hand side $M_EC-M_EA+M_AE-M_AC+M_CA-M_CE$ equals $\pm s \pm s \pm s$, which (irrespective of the choices of the $\pm$-signs) is $0$ if and only if $s=0$. Because $\cos (\varphi) \neq 0$, we conclude that
$$AB+CD+EF=CB+ED+AF \Longleftrightarrow s=0 \Longleftrightarrow a, c, e \text{ concurrent, }$$
as desired.

Remark. Equalities used in the solution could appear in different forms, in particular, in terms of signed lengths.

Remark. Similar solutions could be obtained by projecting onto the line perpendicular to $b$ instead of $b$.

Solution:

We use the same notations as in previous solutions and the fact that $a \parallel d$, $b \parallel e$ and $c \parallel f$ make angles of $120^{\circ}$. Also, we may assume that $E$ and $C$ are not symmetric in $a$ (if they are, the entire figure is symmetric and the conclusion is immediate).

We consider two mappings: the first one $s: a \rightarrow BC \rightarrow d$ sending $A' \mapsto B' \mapsto S$ is defined such that $A'B' \parallel AB$ and $B'S \parallel b$, and the second one $t: a \rightarrow EF \rightarrow d$ sending $A' \mapsto F' \mapsto T$ is defined such that $A'F' \parallel AF$ and $F'T \parallel f$. Both maps are affine linear since they are compositions of affine transformations. We will prove that they coincide by finding two distinct points $A', A'' \in a$ for which $s(A')=t(A')$ and $s(A'')=t(A'')$. Then we will obtain that $s(A)=t(A)$, which by construction implies that the bisectors of $\angle B, \angle D$ and $\angle F$ are concurrent.

We will choose $A'$ to be the reflection of $C$ in $e$ and $A''$ to be the reflection of $E$ in $c$. They are distinct since otherwise $C$ and $E$ would be symmetric in $a$. Applying the above maps $a \rightarrow BC$ and $a \rightarrow EF$ to $A'$, we get points $B'$ and $F'$ such that $A'B' C D E F'$ satisfies the problem statement. However, this hexagon is symmetric in $e$, hence the bisectors of $\angle B', \angle D, \angle F'$ are concurrent and $s(A')=t(A')$. The same reasoning yields $s(A'')=t(A'')$, which finishes the solution.

Remark. This solution is based on the fact that two specific affine linear maps coincide. Here it was proved by exhibiting two points where they coincide. One could prove it in another way, exhibiting one such point and proving that the 'slopes' are equal.

Remark. There are similar solutions where claims and proofs could be presented in more 'elementary' terms. For example, an elementary reformulation of the 'slopes' being equal is: if $b'$ passes through $B'$ parallel to $b$, and $f'$ passes through $F'$ parallel to $f$, then the line through $b \cap f$ and $b' \cap f'$ is parallel to $a$ (which is parallel to $d$).

Solution:

We use the same notations as in previous solutions.

Since the sum of the angles of a convex hexagon is $720^{\circ}$, from the angle conditions we get $\angle B+\angle C=720^{\circ} / 3=240^{\circ}$. From $\angle B+\angle C=240^{\circ}$ it follows that the angle between $c$ and $b$ equals $60^{\circ}$. The same is analogously true for other pairs of bisectors of neighboring angles.

Consider the points $O_a \in a, O_c \in c, O_e \in e$, each at the same distance $d'$ from $M$, where $d'>\max \{MA, MC, ME\}$, and such that the rays $AO_a, CO_c, EO_e$ point out of the hexagon. By construction, $O_a$ and $O_c$ are symmetrical in $e$, hence $O_aO_c \perp b$. Similarly, $O_cO_e \perp d, O_eO_a \perp f$. Thus it suffices to prove that perpendiculars from $B, D, F$ to the sidelines of $\triangle O_aO_cO_e$ are concurrent. By a well-known criterion, this condition is equivalent to equality
$$O_aB^2-O_cB^2+O_cD^2-O_eD^2+O_eF^2-O_aF^2=0$$
To prove $(* )$ consider a circle $\omega_a$ centered at $O_a$ and tangent to $AB$ and $AF$ and define circles $\omega_c$ and $\omega_e$ in the same way. Rewrite $O_aB^2$ as $r_a^2+B_aB^2$, where $r_a$ is the radius of $\omega_a$, and $B_a$ is the touch point of $\omega_a$ with $AB$. Using similar notation for the other tangent points, transform $(* )$ into
$$B_aB^2-B_cB^2+D_cD^2-D_eD^2+F_eF^2-F_aF^2=0.$$
Furthermore, $\angle O_cO_aB_a=\angle MO_aB_a+\angle O_cO_aM=(90^{\circ}-\varphi)+30^{\circ}=120^{\circ}-\varphi$, where $\varphi=\frac{1}{2} \angle A$. (Note that $\varphi>30^{\circ}$, since $ABCDEF$ is convex.) By analogous arguments, $\angle O_aO_cB_c=\angle O_eO_cD_c=\angle O_cO_eD_e=\angle O_aO_eF_e=\angle O_eO_aF_a=120^{\circ}-\varphi$. It follows that rays $O_aB_a$ and $O_cB_c$ (being symmetrical in $e$) intersect at $U_e \in e$ forming an isosceles triangle $\triangle O_aU_eO_c$. Similarly define $\triangle O_cU_aO_e$ and $\triangle O_eU_cO_a$. These triangles are congruent (equal bases and corresponding angles). Therefore we have $O_aU_c=U_cO_e=O_eU_a=U_aO_c=O_cU_e=U_eO_a$. Moreover, we also have $B_aU_e=O_aU_e-r_a=O_aU_c-r_a=F_aU_c=x$, and thus similarly $D_cU_a=B_cU_e=y$, $F_eU_c=D_eU_a=z$.

Now from quadrilateral $BB_aU_eB_c$ with two opposite right angles $B_aB^2-B_cB^2=B_cU_e^2-B_aU_e^2=y^2-x^2$. Similarly $D_cD^2-D_eD^2=D_eU_a^2-D_cU_a^2=z^2-y^2$ and $F_eF^2-F_aF^2=F_aU_c^2-F_eU_c^2=x^2-z^2$. Finally, we substitute this into $(**)$, and the claim is proved.

Remark. Circles $\omega_a, \omega_c$ and $\omega_e$ could be helpful in some other solutions. In particular, the movement of $A$ along $a$ in Solution D is equivalent to varying $r_a$.