Outline of solution:

Could you elaborate on this "simple" complex number application? I played with this idea a bit during the test, and the angle condition gave me cubes that made for complicated, hideous equations and were difficult to simplify.

Just curious: On the test I made a bit of leap in logic, that showed ACE is similar to the three triangles constructed on its edges, which implied that B is the reflection of E across AC, etc., and I concluded the lines came together at the orthocentre of ACE. How many points do you think that will garner? Perfectly correct conclusion, I just skipped the hardest step .

Bacteria wrote: Just curious: On the test I made a bit of leap in logic, that showed ACE is similar to the three triangles constructed on its edges, which implied that B is the reflection of E across AC, etc., and I concluded the lines came together at the orthocentre of ACE. How many points do you think that will garner? Perfectly correct conclusion, I just skipped the hardest step . What was the leap in logic Guessing that B is the reflection of E across AC, etc, alone will get you 1-2 points

Yay! I took the three triangles on the outside of ACE, and translated and rotated them to make a triangle, which I said could be done because the outside angles added to 180, and there were three distinct side lengths. It was convincing to me, but after a little thought it is obviously wrong.

I apologize to those who despise thoughtless bashes. But in my opinion it's kind of nice to know that it's feasible given the nature of the first two problems on the first day...

Edit: This is wrong (note it doesn't use, for instance, the fact that opposite sides are not parallel). I'll try to patch it, but no guarantees... Edit 2: OK, so fortunately it still works (and actually characterizes all valid hexagons, including those with some opposite sides parallel), but the bash is now messier and so I've left out the details. I might add them in later if I have time. Nevertheless, it's interesting that taking $(ABC)$ the unit circle is nicer than, say $e=0$ or $(ACE)$ the unit circle (which I'm not sure are even tractable, really).

math154 wrote: I apologize for people who despise thoughtless bashes. But in my opinion it's kind of nice to know that it's feasible given the nature of the first two problems on the first day...

To be honest though this problem was made for complex numbers, the first thing I tried was complex numbers though I decided I didn't want to get into the details.

Wait isn't what I did significantly different from your outline? (The only similarity is the phantoming?) In particular, I suspect that what you did during the test is probably nicer?

math154 wrote: Wait isn't what I did significantly different from your outline? (The only similarity is the phantoming?) In particular, I suspect that what you did during the test is probably nicer? It is significantly different. I'm not saying what I did during the test was that, but my first thought when looking at the problems was "complex numbers." Which turned out to be useful much later with the phantom hexagon lemma, though it's use there is much less bashy.

Can this problem not be proved by using Brianchon's theorem?If one proves that there exists a circle touching the sides of the hexagon.....

Note we can construct WLOG $\angle B$ is the smallest of $\angle B,\angle D\angle F$ so $\angle B<90$ $P,Q$ such that $FEP\cong PEQ\cong QED\cong ABC$ with $FP=OQ=QD=AC$ now we have $EP=AB=DE$ and $QE=BC=FE$ so $\angle QDE+\angle EFP+\angle ABC=180$ by the angle conditions $\angle B+\angle D+\angle F=180$ giving $\angle D+\angle F=\angle QDE+\angle EFP$ so $\angle QDC=\angle PFA$ and they have the same orientation. but $PF=QD$ and $AF=CD$ so $AFP\cong CDQ$ therefore $AP=CQ$ but sincfe $PQ=AC$ then $ACPQ$ is a paralelogram. If $P=A$ then clearly $Q=C$ and $AD,BE,CF$ meet at the orthocenter of $ACE$ but if not then $D$ is a translation of $F$ by the length $PQ=AC$ giving $FED\cong ABC$ so $\angle FED=\angle ABC$ so since $\angle E=3\angle B$ we have $\angle B=90$ but this is a contradiction.

This is identical to M2130 in a 2009 issue of Kvant: http://kvant.ras.ru/pdf/2009/2009-02.pdf (page 23) The proposer of that Kvant problem pointed this out to me. I wouldn't completely rule out the possibility of coincidence, but it certainly raises eyebrows.

Here is another synthetic solution which uses vector. Hope that there is no flaw in logic. WLOG $A,B,C,D,E,F$ are aligned clockwise. Let $ e= CD = FA$, $a = BC = EF$ and $c = AB = DE$. The key idea is we will rearrange vectors $\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD}, \overrightarrow{DE}, \overrightarrow{EF}, \overrightarrow{FA}$ into new hexagon which is easier to work with. [asy][asy] unitsize(0.6cm); defaultpen(fontsize(10pt)); pair A = (1.5,4); pair C = (0,0); pair E = (5,0); pair B = 2*foot(E,C,A)-E; pair D = 2*foot(A,C,E)-A; pair F = 2*foot(C,A,E)-C; pair T =extension(B,C,F,E); draw(B--A,red); draw(D--E,red); draw(B--C,green); draw(E--F,green); draw(C--D,blue); draw(F--A,blue); draw(C--T--E,dashed+green); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(T); label("$F$",F,NE); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$T$",T,S); pair A1 = (12,5); pair B1 = A1 + A - F; pair C1 = B1 + D - C; pair D1 = C1 + C - B; pair E1 = D1 + F - E; pair F1 = E1 + E- D; draw(A1--B1--C1,blue); draw(C1--D1--E1,green); draw(E1--F1--A1,red); draw(A1--C1--E1--cycle,dashed); dot(A1);dot(B1);dot(C1);dot(D1);dot(E1);dot(F1); label("$A_1$",A1,N); label("$B_1$",B1,NW); label("$C_1$",C1,SW); label("$D_1$",D1,S); label("$E_1$",E1,SE); label("$F_1$",F1,NW);[/asy][/asy] As shown in the figure above, vectors $\overrightarrow{A_1B_1}$, $\overrightarrow{B_1C_1}$, $\overrightarrow{C_1D_1}$, $\overrightarrow{D_1E_1}$, $\overrightarrow{E_1F_1}$, $\overrightarrow{F_1A_1}$ are equal to vectors $\overrightarrow{FA}$, $\overrightarrow{CD}$, $\overrightarrow{BC}$, $\overrightarrow{EF}$, $\overrightarrow{DE}$, $\overrightarrow{AB}$ respectively. First, let $T=BC\cap EF$, then $\angle CTE = 360^{\circ} - 3\angle D-\angle B - \angle F = 180^{\circ} - 2\angle D$. Thus $\triangle C_1D_1E_1$ are isosceles with base angle $\angle D$. Similarly $\triangle A_1B_1C_1, \triangle E_1F_1A_1$ are isosceles with base angle $\angle B$ and $\angle E$ respectively. Observe that $\angle B_1A_1F_1 = 360^{\circ} - \angle A$ thus $\angle C_1A_1E_1 = 360^{\circ} - 3\angle D - \angle B - \angle E = 180^{\circ} - 2\angle D$. But by Law of Sine on $\triangle C_1D_1E_1$, we get $C_1E_1 = a\cdot\frac{\sin 2\angle D}{\sin \angle D}$. Thus Law of Sine on $\triangle A_1C_1E_1$ gives $$a\cdot\frac{\sin 2\angle D}{\sin\angle D} : c\cdot\frac{\sin 2\angle F}{\sin\angle F} : e\cdot\frac{\sin 2\angle B}{\sin\angle B} = \sin 2\angle D : \sin 2\angle F : \sin 2\angle B$$or $a : c : e = \sin\angle D : \sin\angle F : \sin\angle B$. Hence $\triangle CDE\sim\triangle FAE \sim\triangle CAB$ which is enough to conclude that $B,D,F$ are reflections of $E,A,C$ across $AC, CE, EA$ respectively. Thus $AD, BE, CF$ are concurrent at orthocenter of $\triangle ACE$.

I think the following is a purely synthetic solution.

Edit: wow, this and ISL 2013 G5 are pretty similar (both can be solved easily with this trick)

Here is another way to show the hexagon is unique up to scaling, for the lazy. Let $\vec u = (\overrightarrow{AB} + \overrightarrow{DE})/\Vert\overrightarrow{AB}\Vert$, $\vec v = (\overrightarrow{BC} + \overrightarrow{EF})\Vert\overrightarrow{BC}\Vert$, and $\vec w = (\overrightarrow{CD} + \overrightarrow{FA})/\Vert\overrightarrow{CD}\Vert$. WLOG assume $A$ is the origin and $B$ lies on the positive $x$-axis, then these are fixed. Then, if $x = \Vert\overrightarrow{AB}\Vert$, $y = \Vert\overrightarrow{BC}\Vert$, and $z = \Vert\overrightarrow{CD}\Vert$, then we see that $(x,y,z) \stackrel{T}{\longrightarrow} x\vec u + y\vec v + z \vec w$ is a linear map from $\mathbb R^3 \to \mathbb R^2$, and the possible values of $(x,y,z)$ is precisely $\ker T$, so if the hexagon is not unique up to scaling we would have $\operatorname{im} T \neq \mathbb R^2$ by Rank-Nullity. In particular, $\dim \operatorname{im} T = 1$, so $\vec u, \vec v, \vec w$ are parallel. We will show that this cannot happen by complex numbers. Let the unit complex numbers in the directions of $\overrightarrow{AB}$, $\overrightarrow{BC}$, etc. be $a, b, c, \ldots, f$. Then, if $x = e^{iB}$, $y = e^{iD}$, $z = e^{iF}$, we have: \[ a = 1, b = x, c = xz^3, d = xyz^3, e = x^4yz^3, f = x^4yz^4. \]Suppose now by contradiction that $a+d\parallel b+e$, so that $(a+d)/(b+e)\in \mathbb R$, then \[ \frac{a+d}{b+e} = \frac{\frac 1a + \frac 1d}{\frac 1b + \frac 1e} = \frac{be}{ad} \frac{a+d}{b+e}, \]so that we force $ad = be = cf$, or \[ xyz^3 = x^5yz^3 = x^5yz^7. \]Then, $x^4 = 1 = y^4$, but this causes us to conclude $\angle B, \angle D \in \left\{ 0, \frac \pi2, \pi, \frac{3\pi}{2}\right\}$, which is absurd since $\angle B, \angle D \in (0, \pi)$ and $\angle B + \angle D < \pi$. $\blacksquare$

Here's my writeup of the OTIS walkthrough. All hexagons here are up to scaling. We will call a good hexagon any hexagon $ABCDEF$ such that $A$ and $D$ are reflections over $\overline{CE}$ and cyclic permutations. In particular, the lines $\overline{AD}, \overline{BC}, \overline{DF}$ concur at the orthocenter of triangle $ACE$ in any good hexagon. We will prove that any hexagon that satisfies the given conditions is good. Pick any such triangle $ABCDEF$, and note that as $\alpha = \angle B$, $\beta = \angle D$, and $\gamma = \angle F$ satisfy $\alpha+\beta+\gamma=\pi$, we may construct a good hexagon $A'B'C'D'E'F'$ that has identical angles to $ABCDEF$. To do so, let $\mathbf x$, $\mathbf y$, $\mathbf z$ be the sums of unit vectors in the directions $\overrightarrow{AB}$ and $\overrightarrow{DE}$. Notice that we have $$AB \cdot \mathbf x + BC \cdot \mathbf y + CD \cdot \mathbf z = 0 = A'B' \cdot \mathbf x + B'C' \cdot \mathbf y + C'D' \cdot \mathbf z.$$Now, the non-parallel condition implies that none of $\mathbf x, \mathbf y, \mathbf z$ are nonzero, and an argument computation yields that no two are scalar multiples of each other (say, just set $\overrightarrow{AB}$ horizontal and compute all the angles in terms of $\alpha, \beta, \gamma$.) So $\mathbf x$, $\mathbf y$, $\mathbf z$ are linearly independent, implying that $AB, BC, CD$ are in the same ratio as $A'B', B'C', C'D'$. This implies that $ABCDEF \sim A'B'C'D'E'F'$, as needed.