Let the pentagon be $ABCDE$ and tangent at opposite sides of $A$ is $A'$ etc. From Brainchon's on $ABCA'DE$ we get $AA',BD,CE$ concurrent. Let them meet at $A_1$. Similarly define $B_1$ to $E_1$. We prove the following fact (eqn. 1): $$\frac{\sin CAA'}{\sin DAA'}\frac{\sin BEE'}{\sin CEE'}\frac{\sin ADD'}{\sin BDD'}\frac{\sin ECC'}{\sin ACC'}\frac{\sin DBB'}{\sin EBB'}=1.$$Note that $\frac{\sin CAA'}{\sin DAA'}=\frac{CA'\cdot AD}{A'D\cdot AC}$, and $A'D=B'D$ etc., so if we convert all $5$ fractions into similar forms and multiply they cancel each other out. Part a) WLOG $AA',BB',CC',DD'$ concurrent at $X$. Ceva's on point $X$ and both triangle $AB_1C$ and $BC_1D$, combining eqn. 1, we get $$\frac{\sin BEE'}{\sin CEE'}\frac{\sin CB_1B}{\sin AB_1B}\frac{\sin CC_1D}{\sin BC_1C}=1$$$$\Rightarrow \frac{\sin BEE'}{\sin CEE'}\frac{\sin EB_1B}{\sin AB_1B}\frac{\sin CC_1D}{\sin EC_1C}=1\Rightarrow BB',CC',EE'\text{ concurrent}$$and we are done. Part b) We show that $AA',BB',CC'$ concurrent iff $BB',DD',EE'$ concurrent. This will imply the desired result. Backward Case: Suppose $BB',DD',EE'$ concurrent at $X$. Combining Ceva's on point $X$ for triangle $EB_1D$ and triangle $EBD$ we get $$\frac{\sin BEE'}{\sin CEE'}\frac{\sin ADD'}{\sin BDD'}\frac{\sin DBB'}{\sin EBB'}\frac{\sin CB_1B}{\sin AB_1B}=1$$and then using eqn. 1, we get $$\frac{\sin CAA'}{\sin DAA'}\frac{\sin AB_1B}{\sin CB_1B}\frac{\sin ECC'}{\sin ACC'}=1\Leftrightarrow AA',BB',CC'\text{ concurrent.}$$ Forward Case: Suppose $AA',BB',CC'$ concurrent, then reversing the steps as before we get $$\frac{\sin BEE'}{\sin CEE'}\frac{\sin ADD'}{\sin BDD'}\frac{\sin DBB'}{\sin EBB'}\frac{\sin CB_1B}{\sin AB_1B}=1.$$ Now let $BB'$ intersect $DD'$ at $X$, and using Ceva's on point $X$ for triangle $EB_1D$ and triangle $EBD$, we get $$\frac{\sin BEX}{\sin CEX}\frac{\sin ADD'}{\sin BDD'}\frac{\sin DBB'}{\sin EBB'}\frac{\sin CB_1B}{\sin AB_1B}=1,$$so $$\frac{\sin BEX}{\sin CEX}=\frac{\sin BEE'}{\sin CEE'}$$which implies $E,E',X$ concurrent, so we are done.

Label the vertices of the pentagon $A,B,C,D,E$, and let $\omega$ be the inscribed circle of the pentagon. Suppose that $\omega$ is tangent to the side of the pentagon opposite vertex $A$ at $T_A$, and define the other $T's$ similarly. Let $X_A$ be the pole of $AT_A$ with respect to $\omega$, and define the other $X's$ similarly. Consider vertex $A$ for the moment. We know that the polar of $A$ with respect to $\omega$ is $T_C T_D$. We also know that the polar of $T_A$ is $DC$. Thus, by La Hire's Theorem, we know that $X_A = T_C T_D \cap DC$, and we get analogous things for the other $X$'s. Note that gergonnians being concurrent correspond to the poles of these gergonnians being collinear by La Hire's Theorem. Specifically, these points must all lie on the polar of the intersection point. Now, for part a, assume that $AT_A$, $BT_B$, $CT_C$, and $DT_C$ all concur at a point $X$, and let $\ell$ be the polar of $X$ with respect to $\omega$. To show that $ET_E$ also goes through $X$, it suffices to show that the polar of $ET_E$ lies on $\ell$. To show this, we use Pascal's Theorem. First, we apply Pascal's Theorem to degenerate hexagon $T_A T_A T_B T_C T_D T_D$ to get that $T_A T_D \cap T_B T_C \in \ell$. Now, by applying Pascal's Theorem to degenerate hexagon $T_A T_B T_C T_C T_E T_E$, we see that showing the polar of $ET_E$ lies on $\ell$ is equivalent to showing that $T_E T_C \cap T_A T_B \in \ell$. But this is true by applying Pascals Theorem to degenerate hexagon $T_A T_B T_B T_C T_D T_E$ and using the fact that $T_A T_D \cap T_B T_C \in \ell$, so we are done with part a. For part b, suppose that $AT_A$, $BT_B$, and $CT_C$ all concur at some point $X$, and once again let $\ell$ denote the polar of $X$ with respect to $\omega$. Once again, this means that $X_A$, $X_B$, and $X_C$ all lie on $\ell$. It suffices to prove now that $X_B$, $X_D$, and $X_E$ are collinear. To do this, once again we use Pascal's Theorem. Let $\ell^\prime$ be the line through $X_B$ and $X_E$. All we have to show is that $X_D$ lies on $\ell^\prime$. First, we apply Pascals' Theorem to degenerate hexagon $T_A T_A T_C T_C T_D T_E$ to get that $T_A T_C \cap T_D T_E$ lies on $\ell$. Because $T_D T_E$ only intersects $\ell$ once at $X_B$, we know that $X_B$ lies on $T_A T_C$. Now, we apply Pascal's Theorem to degenerate hexagon $T_B T_B T_C T_D T_E T_E$ to get that $T_E T_B \cap T_C T_D$ lies on $\ell^\prime$. Now, we apply Pascal's Theorem to degenerate hexagon $T_A T_B T_B T_C T_D T_E$ to get that $T_E T_A \cap T_D T_B$ lies on $\ell_\prime$, where we used that $T_B T _E \cap T_C T_D$ and $ T_A T_C \cap T_B T_B$ lie on $\ell_\prime$. But now we are done by applying Pascal's Theorem to degenerate hexagon $T_A T_B T_B T_D T_D T_E$ because this implies that $T_D T_D \cap T_A T_B$ lies on $\ell^\prime$, which implies that $X_D$ lies on $\ell^\prime$, so we are done with part b.

Solution to part (a): Denote that tangency points by $ A, B, C, D, E $ and the corresponding vertices of the pentagon by $ A', B', C', D', E'. $ Assume that $ BB' $ and $ CC' $ and $ DD' $ and $ EE' $ all concur at a point $ O. $ Now since this problem is purely projective and since there is a projective transformation that fixes the circumcircle of pentagon $ ABCDE $ and takes $ O $ to the center of this circle, we can assume WLOG that $ O $ is the center of the circle. Now we proceed with complex numbers. Let $ A, B, C, D, E, A', B', C', D', E', O $ have complex coordinates $ a, b, c, d, e, a', b', c', d', e', 0 $ respectively and WLOG let the circumcircle of pentagon $ ABCDE $ be the unit circle. Note that $ a' = \frac{2cd}{c + d} $ and $ b' = \frac{2ed}{e + d} $ and $ c' = \frac{2ea}{e + a} $ and $ d' = \frac{2ab}{a + b} $ and $ e' = \frac{2bc}{b + c}. $ Then since $ B, O, B' $ are collinear we have that $ \frac{b - 0}{\overline{b} - 0} = \frac{b' - 0}{\overline{b'} - 0} \Longrightarrow b^2 = de. $ Similarly we find that $ c^2 = ea $ and $ d^2 = ab $ and $ e^2 = bc. $ Letting $ a^2 = x $ and multiplying all of these equalities together yields that $ x = cd $ so $ A, O, A' $ are collinear as well, as desired. This same idea of projective transformation + complex numbers likely works for part (b) as well - I will edit this post to the include the solution of (b) later.

This problem needs a nice solution. Let $\text{col}(X,Y,Z)$ represent the collinearity of points $X,Y,Z$ and $\text{conc}(x,y,z)$ represent the concurrency of lines $x,y,z$. Let $\mathcal{P}(\mathcal{H})$ denote that we used Pascal on cyclic hexagon $\mathcal{H}$ and let $\text{pp}$ represent the fact that 3 lines are concurrent iff their poles are collinear. Let $A,B,C,D,E$ denote the points of tangency of the cyclic pentagon (which form a convex pentagon), and $a,b,c,d,e$ be the sides that are opposite $A,B,C,D,E$ in the pentagon $ABCDE$. Let $A'B'C'D'E'$ be the circumscribed pentagon. I will prove $\text{conc}(AA',BB',CC') \Leftrightarrow \text{conc}(DD',BB',EE')$ which kills both parts of the problem. Note that $XX$ represents the tangent to a circle at point $X$. Indeed, $\text{conc}(AA',BB',CC') \Leftrightarrow \text{(pp) } \text{col}(AA \cap a, BB \cap b, CC \cap c) \Leftrightarrow $ $ (\mathcal{P}(AAEDCC)) \text{ col}(AA \cap a, BB \cap b, CC \cap c, CA \cap b) \Leftrightarrow \text{conc}(BA,CA,b) \Leftrightarrow $ $(\mathcal{P}(EBBDCA) \text{ conc}(BB,CA,b,\overline{(EB\cap a)(DB\cap c)}) \Leftrightarrow \text{ conc}(BB,b,\overline{(EB\cap a)(DB \cap c)}) \Leftrightarrow$ $ \text{col}(BB \cap b, EB \cap a, DB \cap c) \Leftrightarrow (\mathcal{ P}(EEBBCD,DDBBAE)) \text{ col}(EE \cap e, BB \cap b, DD \cap d) \Leftrightarrow$ $\text{(pp) } \text{conc}(EE',BB',DD') \text{ }\blacksquare$ It is worth noting that $BAB'C,BEB'D$ are harmonic (where $B'$ is such that $BB'$ is the polar of $AC \cap ED$, assuming $\text{conc}(AA',BB'CC')$. Therefore $\text{col}(EE\cap DD, B, AA \cap CC)$ and so by autopolarity of the triangle formed by the intersections of the diagonals of a cyclic quadrilateral, $\text{col}(EA \cap DC, EC \cap AD, B)$ and one can apply Desargues' Theorem to this to find that $ \text{col}(BB \cap b, EB \cap a, DB \cap c)$. Also, it is worth noting that one does not need the circle to be a circle, it could just as well be a conic. All pentagons have an inscribed conic.

This problem is insta-killed by taking a projective transformation sending the concurrency point (in either part) to the center of the circle while fixing the incircle.

Let the pentagon be $ABCDE$, and let the intouch pentagon be $A'B'C'D'E'$. We claim that the concurrency of $AA'$, $BB'$, and $EE'$ is equivalent to the concurrency of $AA'$, $CC'$, and $DD'$. This proves both part (a) and part (b). Let $BB'$ and $EE'$ intersect at $X$ and $CC'$ and $DD'$ intersect at $Y$. Let $BC$ and $DE$ intersect at $F$. Since quadrilateral $ABFE$ has an inscribed circle, $AF$, $BE$, $B'D'$, and $C' E'$ concur at a point $P$. We claim that $X$ and $Y$ are both on $AF$. Let $B'E'$, $BB'$, and $EE'$ intersect $AF$ at $Q$, $B_1$, and $D_1$. Projecting some harmonic pencils onto $AF$, we have that $B'(B',B;D',E')=(F,B_1;P,Q)=-1$ and $E'(E',E;C',B')=(F,E_1;P,Q)=-1$, so $B_1=E_1=X$ lies on $AF$. Hence, $AA'$, $BB'$, and $EE'$ concur if and only if $A'$ lies on $AF$. Now we claim that $A'$ lies on $AF$ if and only if $Y$ lies on $AF$, which is equivalent to the concurrency of $AA'$, $CC'$, and $DD'$. Suppose that $A'$ lies on $AF$. Let $CC'$ and $DD'$ intersect $AF$ at $C_1$ and $D_1$. Projecting some harmonic pencils onto $AF$, we have that $C'(C',C;E',A')=(A,C_1;P,A')=-1$ and $D'(D',D;B',A')=(A,D_1;P,A')=-1$, so $C_1=D_1=Y$ lies on $AF$. Now suppose that $Y$ lies on $AF$. Let $CA'$ intersect $AF$ at $C_2$ and $DA'$ intersect $AF$ at $D_2$. Projecting the same harmonic pencils onto $AF$, we have that $C'(C',C;E',A')=(A,Y;P,C_2)=-1$ and $D'(D',D,B',A')=(A,Y;P,D_2)=-1$, so $C_2=D_2=A'$ lies on $AF$. Hence, the concurrency of $AA'$, $BB'$, and $EE'$ is equivalent to the concurrency of $AA'$, $CC'$, and $DD'$, so we are done.

DDIT says hi! Let the pentagon be $ABCDE$ and $A_1, B_1, C_1, D_1$ be the tangency points opposite to $A,B,C,D,E$ resp. We claim that $AA_1, CC_1, DD_1$ are concurrent $\iff AA_1, BB_1, EE_1$ are concurrent. Claim : $AA_1, CC_1, DD_1$ are concurrent $\iff BC, DE, AA_1$ are concurrent. Proof : I will prove only the forward direction. The argument is reversible. Let $CD$ intersects $C_1D_1$ and $BE$ at $K, K'$ and $AA_1\cap CC_1\cap DD_1 = P$. By DDIT on $CD_1DC_1$, we get an involution swapping $(AB, AC), (AD, AE), (AP, AK)$. By Brianchon's theorem, we get $BD, CE, AA_1$ are concurrent hence DDIT on $BCDE$ gives involution swapping $(AB, AC), (AD, AE), (AP, AK')$ hence $K=K'$. Let $T=BC\cap DE$ By Newton's theorem on $ABTE$, $K$ also pass through $B_1E_1$. Thus $AA_1$ is polar of $K$, which obviously pass through $T=B_1B_1\cap E_1E_1$. Hence we are done. Claim : $AA_1, BB_1, EE_1$ are concurrent $\iff BC, DE, AA_1$ are concurrent. Proof : I will prove only the forward direction. The argument is reversible. Let $T=BC\cap DE$ and $P=AA_1\cap BB_1\cap EE_1$. By DDIT on $BB_1EE_1$, we get an involution swapping $(AB, AE), (AB_1, AE_1)$ and $(AP, AT)$. Hence by degenerate DDIT on $B_1, E_1$ gives an involution swapping $(AB, AE)$ (both tangents), $(AB_1, AE_1)$ and $(AT, AT)$. Hence $A, P, T$ are colinear, implying the conclusion.

Unless I'm mistaken the above posters have all severely overcomplicated the problem. We in fact only need Brianchon's Theorem. Denote the vertices of the pentagon as $A, B, C, D, E$ and the opposite touch points as $A', B', C', D', E'$, respectively. It is clear that it suffices to show that: $AA', BB', EE'$ concurrent $\iff$ $AA', CC', DD'$ concurrent. Let the extensions of $AB$ and $CD$, $BC$ and $DE$, and $CD$ and $EA$ intersect at $E_1$, $A_1$, and $B_1$ respectively. Brianchon on $AEDA'ED'$ gives that $AA'$, $DD'$, and $EE_1$ concur, and similarly Brianchon on $ABCA'BC'$ gives that $AA'$, $CC'$, and $BB_1$ concur, so $AA', CC', DD'$ concurrent $\iff$ $AA', BB_1, EE_1$ concurrent. Now Brianchon on $A_1BE_1A'BE$ gives that $A'A_1$, $BB_1$, and $EE_1$ concur, so $AA', BB_1, EE_1$ concurrent $\iff$ $A, A', A_1$ collinear. But finally Brianchon on $ABE'A_1B'E$ gives that $AA_1$, $BB'$, and $EE'$ concur so $A, A', A_1$ collinear $\iff$ $AA', BB', EE'$ concurrent. Stringing together these results gives the desired proposition so we are done with 4 applications of Brianchon's Theorem. $\blacksquare$

Is this wrong? Compared to the above solutions, it seems way too simple. Let the pentagon be $ABCDE$ and the tangency points be $A'B'C'D'E'$. a) Take a projective transformation fixing the circle and sending the concurrency point to the center of the circle. WLOG assume $BB', CC', DD', EE'$ concur at the center. It's easy to see (because symmedian passing through circumcenter implies isosceles) that we must have $B'D'=B'E', C'E'=C'A', D'A'=D'B', E'B' = E'C'$. But this implies that $A'B'C'D'E'$ is regular, so $AA'$ obviously passes through the center of the circle. b) Take the same projective transformation as before. We have 2 cases: Case 1: The three concurrent gergonians come from 3 consecutive vertices. WLOG assume $AA', BB', CC'$ concur. Similar to part a), we note that $A'C'=A'D', B'D'=B'E', C'E'=C'A'$. This obviously forces $A'B'C'D'E'$ to be symmetric about the perpendicular bisector of $A'C'$, so $DD'$ and $EE'$ obviously meet on line $BB'$. Case 2: The three concurrent gergonians come from 3 nonconsecutive vertices. WLOG assume $AA', CC', DD'$ concur. As before, note that $A'C'=A'D', C'E'=C'A', D'A'=D'B'$. This obviously forces $A'B'C'D'E'$ to be symmetric about the perpendicular bisector of $C'D'$, so $BB'$ and $EE'$ obviously meet on line $AA'$ as desired.

@above v_Enhance wrote: This problem is insta-killed by taking a projective transformation sending the concurrency point (in either part) to the center of the circle while fixing the incircle. i think your solution is fine

This may be the worst approach yet . Let the points of tangency be $A_1,A_2,A_3,A_4,A_5$ in that order (with indices taken modulo $5$). Project the concurrency point to the center of the inscribed circle. Let the center of the inscribed circle be $O$. Observe that if the Gergonnian from $A_i$ passes through $O$, then $A_i$ lies on the perpendicular bisector of $A_{i+2}A_{i+3}$. Throw the diagram on the complex unit circle and let $\arg A_i = \theta_i$ for each $i$. Hence, the Gergonnian from $A_i$ passes through $O$ iff \[2\theta_i\equiv \theta_{i+2}+\theta_{i+3}\pmod{2\pi}.\]Note this reformulation immediately implies part a; if this result holds for four of the $i$ and not for the fifth, then summing yields a contradiction. Now, after noting symmetry, we have two cases: either the Gergonnians through $A_1,A_2,A_3$ concur, or the Gergonnians through $A_1,A_2,A_4$ concur. Case 1: The Gergonnians through $A_1,A_2,A_3$ concur. We have \begin{align*} 2\theta_1&\equiv \theta_3+\theta_4\pmod{2\pi}\\ 2\theta_2&\equiv \theta_4+\theta_5\pmod{2\pi}\\ 2\theta_3&\equiv \theta_5+\theta_1\pmod{2\pi}. \end{align*}Sum the first and last equation to yield $2(\theta_1+\theta_3)\equiv \theta_3+\theta_4+\theta_5+\theta_1\pmod{2\pi}$, hence $\theta_1+\theta_3\equiv\theta_4+\theta_5\equiv2\theta_2\pmod{2\pi}$. This implies $A_1A_3\parallel A_4A_5\perp A_2O$. That is, the pentagon $A_1A_2A_3A_4A_5$ maps back to itself under reflection about line $A_2O$. Hence, the Gergonnians through $A_4$ and $A_5$ must concur on line $A_2O$ by symmetry, but because $A_2O$ is the Gergonnian through $A_2$, we are done. Case 2: The Gergonnians through $A_1,A_2,A_4$ concur. We have \begin{align*} 2\theta_1&\equiv \theta_3+\theta_4\pmod{2\pi}\\ 2\theta_2&\equiv \theta_4+\theta_5\pmod{2\pi}\\ 2\theta_4&\equiv \theta_1+\theta_2\pmod{2\pi}. \end{align*}Sum the first and last equation to yield $\theta_1+\theta_4\equiv\theta_2+\theta_3\pmod{2\pi}$ as before. Sum the second and last equation to yield $\theta_2+\theta_4\equiv\theta_1+\theta_5$. Now, rewrite $\theta_3\equiv\theta_1+\theta_4-\theta_2,\theta_5\equiv\theta_2+\theta_4-\theta_1$. Sum to yield $\theta_3+\theta_5\equiv 2\theta_4$. Hence, we have $\theta_3+\theta_5\equiv\theta_1+\theta_2\equiv\theta_4$, and can finish similarly to our argument in the first case.

VulcanForge wrote: Is this wrong? Compared to the above solutions, it seems way too simple. Let the pentagon be $ABCDE$ and the tangency points be $A'B'C'D'E'$. a) Take a projective transformation fixing the circle and sending the concurrency point to the center of the circle. WLOG assume $BB', CC', DD', EE'$ concur at the center. It's easy to see (because symmedian passing through circumcenter implies isosceles) that we must have $B'D'=B'E', C'E'=C'A', D'A'=D'B', E'B' = E'C'$. But this implies that $A'B'C'D'E'$ is regular, so $AA'$ obviously passes through the center of the circle. b) Take the same projective transformation as before. We have 2 cases: Case 1: The three concurrent gergonians come from 3 consecutive vertices. WLOG assume $AA', BB', CC'$ concur. Similar to part a), we note that $A'C'=A'D', B'D'=B'E', C'E'=C'A'$. This obviously forces $A'B'C'D'E'$ to be symmetric about the perpendicular bisector of $A'C'$, so $DD'$ and $EE'$ obviously meet on line $BB'$. Case 2: The three concurrent gergonians come from 3 nonconsecutive vertices. WLOG assume $AA', CC', DD'$ concur. As before, note that $A'C'=A'D', C'E'=C'A', D'A'=D'B'$. This obviously forces $A'B'C'D'E'$ to be symmetric about the perpendicular bisector of $C'D'$, so $BB'$ and $EE'$ obviously meet on line $AA'$ as desired. Nice i did the same . Sincerely, Aayam

Let $ABCDE$ be the pentagon, and let their opposite points of tangency be $A_1,B_1,\dots,E_1$. Lemma. $AA_1,CC_1,DD_1$ concur if and only if $AA_1,BB_1,EE_1$ concur. Proof. By Brianchion on $ABCA_1DE$, lines $CE$ and $BD$ intersect at some point $A_2$ on $AA_1$. Define $B_2,\dots,E_2$ similarly. By Desargues on $\triangle AD_2C_2$ and $\triangle A_2DC$, lines $AA_2,DD_2,EE_2$ concur if and only if lines $BE,B_2E_2,CD$ concur. By Desargues on $\triangle AB_2E_2$ and $\triangle A_2BE$, lines $AA_2,BB_2,EE_2$ concur if and only if lines $BE,B_2E_2,CD$ concur, so we are done. $\square$ For part (a), suppose WLOG that $AA_1,\dots,DD_1$ concur, then by the lemma $EE_1$ passes through the same concurrence point. Part (b) is essentially the lemma restated.

Darn I didn't see the proj transform at all. Here's a non-overpowered solution. Let $ABCDE$ be the pentagon, and let $A',B',C',D',E'$ be the points on $\omega$ such that $\overline{AA'},\overline{BB'},$etc are the gergonnians. Let $P=\overline{BC}\cap\overline{DE}$, I will show that the conditions: $\overline{AA'},\overline{BB'},\overline{EE'}$ concur $\overline{AA'},\overline{CC'},\overline{DD'}$ concur are both equivalent to $A-A'-P$. By Brianchon's on $ABE'PB'E$, $\overline{AP},\overline{BB'},\overline{EE'}$ concur, proving the first claim. Let $Q=\overline{AB}\cap\overline{CD}$, let $R=\overline{AE}\cap\overline{CD}$. Let $S=\overline{D'C'}\cap\overline{CD}$, and let $T=\overline{D'C'}\cap\overline{AA'}$. Then in $\triangle AQR$, we see that $(D',C';S,T)=-1$. If $\overline{AA'},\overline{CC'},\overline{DD'}$ concur, then $-1=(D',C';S,T)=(D,C;A',T)$, and the reverse is true as well. By Pole-Polar Transfromation, $(D,C;A',T)=A'(E',B';D,A)=(E',B';A',\overline{AA'}\cap\omega)$. This is harmonic iff $A-A'-P$, as desired.

This is probably similar and possibly the same as some of the above Brianchon solutions. We will prove (b) which clearly implies (a). Let the pentagon at hand be $ABCDE$ and $A', B', C', D', E'$ be the corresponding points of tangency. We will prove the following claim which clearly suffices. $\textbf{Claim:}$ $AA', BB', CC'$ concur if and only if $AA', DD', EE'$ concur $\textbf{Proof)}$ Let $X,Y,Z$ be the intersections of $AE$ and $CD$, $AB$ and $CD$, and $ED$ and $BC$, respectively. By Brianchon's Theorem on $AEDA'CYD'$ we get that $DD'$ passes through $AA' \cap EY$ and similarly by Brianchon on $ABA'DXC'$ we get that $CC'$ passes through $AA' \cap BX$ and we therefore have to prove that $AA', BX, EY$ concur iff $AA', BB', EE'$ are concurrent. Notice that $Z$ lies on $AA'$ by Brianchon on $ABE'ZB'E$ and $Z$ iff $AA', BB', EE'$ are concurrent also by Brianchon on $ZBYA'XE$ we have that $ZA', BX, EY$ concur, yet $ZA'$ is $AA'$ iff $AA', BB', EE'$ concur which must then also contain the intersection of $BX$ and $EY$ as desired. $\blacksquare$

Take a homography keeping the circle the same and mapping the point of concurrency to the center of the circle. Let the points of tangency be labeled $A$, $B$, $C$, $D$, and $E$. (a) Assume without loss of generality that the gergonnians through $A$, $B$, $C$, and $D$ are concurrent at the center of the circle. Then, using complex numbers where the circle is the unit circle, we must have $a^2=cd$, $b^2=de$, $c^2=ea$, and $d^2=ab$. Therefore, multiplying gives $a^2b^2c^2d^2=a^2bcd^2e^2$, so $e^2=bc$. Therefore, the gergonnian through $E$ also passes through the center of the circle, so all five gergonnians are concurrent. (b) There are two possible configurations of three gergonnians being concurrent. Case 1: The gergonnians through $A$, $B$, and $C$ are concurrent. Then, $a^2=cd$, $de=b^2$, and $c^2=ea$, so multiplying gives $a^2c^2de=ab^2cde$, so $ac=b^2$. Therefore, since $de=b^2$, this means that $ABCDE$ is symmetric about the altitude from $B$ to $DE$. Therefore, the gergonnians through $D$ and $E$ intersect on the altitude from $B$ to $DE$, which is the gergonnian through $B$. Case 2: The gergonnians through $A$, $C$, and $D$ are concurrent.\newline Then, $a^2=cd$, $c^2=ea$, and $d^2=ab$, so multiplying gives $a^2c^2d^2=a^2bcde$, so $be=cd=a^2$. Therefore, since $a^2=cd$, this means that $ABCDE$ is symmetric about the altitude from $A$ to $CD$. Therefore, the gergonnians through $B$ and $E$ intersect on the altitude from $A$ to $CD$, which is the gergonnian through $A$. Therefore, if one triple of three gergonnians concur, then another triple of three gergonnians concur.

Let $P$ , $Q$ , $R$ , $S$ and $T$ denote the tangency points opposite vertices $A$ , $B$ , $C$ , $D$ and $E$ respectively. (a) Say gergonnians $AP$ , $BQ$ , $CR$ and $ET$ concur. Take a homography which sends $PQRT$ to a rectangle while preserving its circumcircle. Note that by symmetry, the concurrence point must lie on the perpendicular bisector of segment $RT$. But by symmetry again, this perpendicular bisector is simply line $\overline{DS}$, implying that the five gergonnians concur. (b) Say gergonnians $AP$, $BQ$ and $ET$ concur. Then, after taking a homography which sends $PQRT$ to a rectangle, the finish is similar to the above case. Say gergonnians $AP$ , $BQ$ and $DS$ concur. Then, take a homography which sends $PQRT$ to a rectangle while preserving its circumcircle. By symmetry again, $CT$ and $ER$ must concur on the perpendicular bisector of $RT$, so gergonnians $CT$ , $ER$ and $DS$ also concur.