I think the statement is correct

qwerty_ytrewq solved it in 30 minutes, but sadly I don't understand his solution Discord links: page 1 page 2 page 3

Let the pentagon be $P_1P_2P_3P_4P_5$ with circumcircle $\Omega$ centered at $O$ and incircle $\omega$ centered at $I$. Taking indices modulo $5$, let $I_i$ and $J_i$ be the incenters of $P_{i-1}P_iP_{i+1}$ and $P_{i-2}P_iP_{i+2}$. By Poncelet Theory, the pentagon $P_1P_3P_5P_2P_4$ has an incircle $\omega'$ that is coaxial with $\Omega$ and $\omega$, so it is centered at some point $J$ on $OI$. By the Pingpong Lemma, there exists a point $K$ on line $OI$ such that if $\tau_P:\Omega\rightarrow\Omega$ denotes projection through $P$, then $\tau_O\circ\tau_I\circ\tau_J\circ\tau_K$ is the identity. Let $Q_i=\tau_O\circ\tau_I(P_i)$ and $R_i=\tau_J(P_i)$, so $Q_iR_i$ passes through $K$. Note that $Q_i$ is the midpoint of arc $P_{i-1}P_{i+1}$ and $R_i$ is the midpoint of arc $P_{i-2}P_{i+2}$. It follows by Fact 5 that $Q_iR_i$ is the common perpendicular bisector of $I_{i-2}I_{i+2}$ and $J_{i-1}J_{i+1}$, so $KI_{i-2}=KI_{i+2}$ and $KJ_{i-1}=KJ_{i+1}$. Taking this over all $i$ gives the desired result.

CANBANKAN wrote: qwerty_ytrewq solved it in 30 minutes, but sadly I don't understand his solution Discord links: page 1 page 2 page 3 very nice ideas!

Feels too easy? Call $I_1$ the incenter of $A_1A_2A_5$ and $P_1$ the incenter of $A_1A_3A_4.$ Define $I_2, \dots I_5$ and $P_2, \dots P_5$ similarly. Note $$\angle A_1I_1A_2 = 90 + \angle A_1A_5A_2/2 = 90 + \angle A_1A_3A_2/2 = \angle A_1I_2A_2$$so $A_1I_1I_2A_2$ is cyclic and by symmetry, so are all its cyclic variants. Let $I$ be the incenter of $A_1A_2A_3A_4A_5.$ So $II_i \times IA_i$ is constant over all $i.$ An inversion with center $I$ exists mapping $(A_1A_2A_3A_4A_5)$ to $(I_1I_2I_3I_4I_5).$ So $(I_1I_2I_3I_4I_5)$ is cyclic. By a similar argument as before, $(A_1P_4I_2A_2), (A_2I_2I_3A_3), (A_3I_3P_1A_4), (A_4P_1P_4A_1)$ are all cyclic. Note $$\angle P_1P_4I_2 = 360^\circ - \angle P_1P_4A_1 - \angle A_1P_4I_2 = \angle P_1A_4A_1 + \angle I_2A_2A_1 = 90^\circ.$$By symmetry, $I_2P_4P_1I_3$ has all right angles and is in fact a rectangle. So $P_1P_4$'s perpendicular bisector coincides with $I_3I_2.$ By symmetry we find $(P_1P_2P_3P_4P_5)$ is cyclic and its center coincides with that of $(I_1I_2I_3I_4I_5)$ as desired. $\blacksquare$

Attachments:

Let $ABCDE$ be a convex pentagon such that the five vertices lie on a circle and the five sides are tangent to another circle inside the pentagon. There are ${5 \choose 3}= 10$ triangles which can be formed by choosing $3$ of the $5$ vertices. For each of these $10$ triangles, mark its incenter. Prove that these $10$ incenters lie on two concentric circles.

Let $O$ be the incenter of the pentagon. Let $I_A$ be the incenter of $\triangle EAB$, etc. Let $M_{XY}$ be the midpoint of minor arc $XY$ of $(ABCDE)$ where $X,Y$ are arbitrary points Notice that $I_A$ and $I_B$ lie on $OA$ and $OB$ respectively, while $OI_A\times OA=OI_B\times OB$ as $I_A,I_B,A,B$ lie on the circle centered at the $M_{AB}$. Hence by symmetry we have $OI_A\times OA=OI_B\times OB=OI_C\times OC=OI_D\times OD=OI_E\times OE$, hence an inversion Now let $J$ be the center of $(I_AI_BI_CI_DI_E)$. Let $J_A$ be the incenter of $\triangle ADC$, etc. Notice that $M_{AE}J$ is the perpendicular bisector of $I_EI_A$, hence $$\angle JM_{AE}I_A=\frac{1}{2}\angle I_EM_{AE}I_A=\frac{1}{2}\angle DM_{AE}B=\angle M_{BD}M_{AE}B=\angle M_{BD}M_{AE}I_A$$Hence $J,M_{AE},M_{BD}$ are collinear. Meanwhile we notice that $$\angle J_BM_{BD}M_{AE}=\angle EM_{BD}M_{AE}=\frac{1}{2}EM_{BD}A=\frac{1}{2}J_BM_{BD}J_D$$Hence $M_{AE}M_{BD}$ is the perpendicular bisector of $J_BJ_D$, which implies $JJ_B=JJ_D$, by symmetry $J_A,J_B,J_C,J_D,J_E$ all lie on a circle centered at $J$ so we are done.

Let $I_A,I_B,I_C,I_D,I_E$ be the incenters of triangles $EAB,ABC,BCD,CDE,DEA,$ respectively. Let $J_A,J_B,J_C,J_D,J_E$ be the incenters of triangles $ACD,BDE,CAE,DAB,EBC.$ Claim: $AI_AJ_DI_BB$ is cyclic. Proof. Notice \begin{align*}\angle AI_AB&=90+\tfrac{1}{2}\angle AEB\\&=90+\tfrac{1}{2}\angle ACB=\angle AI_BB\\&=90+\tfrac{1}{2}\angle ADB=\angle AJ_DB.\end{align*}$\blacksquare$ Similarly, $BI_BJ_EI_CC,$ etc, are cyclic. Claim: $AJ_CJ_AC$ is cyclic. Proof. We know $$\angle AJ_CC=90+\tfrac{1}{2}\angle DEA=90+\tfrac{1}{2}\angle CDA=\angle AJ_AC.$$$\blacksquare$ Similarly, $AJ_DJ_AD,$ etc, are cyclic. Claim: $I_AI_BI_CI_DI_E$ is cyclic. Proof. We see $$II_A\cdot IA=II_B\cdot IB=II_C\cdot IC=II_D\cdot ID=II_E\cdot IE$$so there is an inversion at $I$ that maps $A$ to $I_A$ and so on. Since $ABCDE$ is cyclic, $I_AI_BI_CI_DI_E$ is also cyclic. $\blacksquare$ Claim: A circle concentric with $(I_AI_BI_CI_DI_E)$ circumscribes $J_AJ_BJ_CJ_DJ_E.$ Proof. By symmetry, it suffices to show the perpendicular bisector of $\overline{I_BI_C}$ is congruent to the perpendicular bisector of $\overline{J_DJ_A}.$ Note $$\angle I_BI_CJ_A=360-(180-\angle I_BBC)-(180-\angle CDJ_A)=\tfrac{1}{2}\angle ABC+\tfrac{1}{2}\angle ADC=90.$$Similarly, all other angles of $I_BI_CJ_AJ_D$ are right, so it is a rectangle. $\blacksquare$ $\square$

Is this possible to do just by angle chasing and showing that for any four points among Ia,Ib,Ic,Id,Ie it satsfies inscribed angle theroem ? And similarily for Ja,Jb... ?

Throughout the solution all indicates are taken modulo $5.$ Denote the pentagon as $A_1A_2A_3A_4A_5,$ as $I_i,J_i$ incenters of $A_{i-1}A_iA_{i+1},$ $A_{i-2}A_iA_{i+2}$ respectively. It's well-known that $A_iA_{i+1}I_iI_{i+1}$ are concyclic, moreover lines $A_iI_i$ pass through incenter of pentagon, so by inversion $I_1I_2I_3I_4I_5$ are concyclic with center $O.$ But well-known that $I_iI_{i+1}J_{i+2}J_{i-1}$ is rectangle, so perpendicular bisectors of $J_{i+2}J_{i-1}$ pass through $O,$ which clearly complete our proof.

Solved with mueller.25, starchan, Siddharth03, AdhityaMV Lemma: In a cyclic quadrilateral $ABCD$, the incenters of the four possible triangles form a rectangle. Proof: Let $I_1, I_2, I_3, I_4$ be the incenters of $BCD, CDA, DAB, ABC$. First note that $DI_2I_1C$ is cyclic since $\angle DI_2C = 90 + \frac{\angle DAC}{2} = 90 + \frac{\angle DBC}{2} = \angle DI_1C$. Similarly $BI_4I_1C$ is cyclic as well. So, $\angle I_4I_1I_2 = 360 - \angle I_4I_1C - \angle I_2I_1C = 360 - (180 - \angle I_4BC) - (180 - \angle I_2DC) = \frac{\angle ADC + \angle ABC}{2} = 90^\circ$. Similarly all angles of the quadrilateral are right, so it is a rectangle. $\square$ Now, let $ABCDE$ be the bicentric pentagon and $I_1, I_2, I_3, I_4, I_5$ the incenters of $EAB, ABC, BCD, CDE, DEA$ and let $I$ be the incenter of the quadrilateral. Note that due to the cyclic quadrilaterals mentioned above, we get by power of point, $II_1 \cdot IA = II_2 \cdot IB = \cdots = II_5 \cdot IE$. Considering an inversion with radius, square root of this length we see that $(ABCDE)$ swaps with $(I_1I_2I_3I_4I_5)$, so these five incenters lie on a circle. Suppose this circle has center $O$. Let $I_1', I_2', I_3', I_4', I_5'$ be the incenters of $ACD, BDE, CEA, DAB, EBC$. Then note that since $I_1I_2I_5'I_3'$ is a rectangle (by the Lemma), we have that $OI_3' = OI_5'$. Similarly we get $OI_k' = OI_{k+2}'$ for all $k$ with indices taken mod $5$. Since $5$ is odd, this implies that $OI_k'$ is fixed across all $k$, so the points $I_1', I_2', I_3', I_4', I_5'$ also lie on a circle with center $O$. So the ten incenters do indeed lie on two concentric circles, as desired. $\blacksquare$

ABCDE wrote: Let the pentagon be $P_1P_2P_3P_4P_5$ with circumcircle $\Omega$ centered at $O$ and incircle $\omega$ centered at $I$. Taking indices modulo $5$, let $I_i$ and $J_i$ be the incenters of $P_{i-1}P_iP_{i+1}$ and $P_{i-2}P_iP_{i+2}$. By Poncelet Theory, the pentagon $P_1P_3P_5P_2P_4$ has an incircle $\omega'$ that is coaxial with $\Omega$ and $\omega$, so it is centered at some point $J$ on $OI$. By the Pingpong Lemma, there exists a point $K$ on line $OI$ such that if $\tau_P:\Omega\rightarrow\Omega$ denotes projection through $P$, then $\tau_O\circ\tau_I\circ\tau_J\circ\tau_K$ is the identity. Let $Q_i=\tau_O\circ\tau_I(P_i)$ and $R_i=\tau_J(P_i)$, so $Q_iR_i$ passes through $K$. Note that $Q_i$ is the midpoint of arc $P_{i-1}P_{i+1}$ and $R_i$ is the midpoint of arc $P_{i-2}P_{i+2}$. It follows by Fact 5 that $Q_iR_i$ is the common perpendicular bisector of $I_{i-2}I_{i+2}$ and $J_{i-1}J_{i+1}$, so $KI_{i-2}=KI_{i+2}$ and $KJ_{i-1}=KJ_{i+1}$. Taking this over all $i$ gives the desired result. Diagram for ABCDE solution. [asy][asy] import geometry; size(12cm); point p1, p2, p3, p4, p5, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, i, o, j, k, q1, r1; o = (0, 0); p1 = dir(140); real r = 0.673; i = (0.3, 0); circle in = circle(i, r); circle ci = circle(o, 1); line tangent = tangents(in, p1)[1]; point[] pipi1 = intersectionpoints(tangent, ci); p2 = pipi1[0] + pipi1[1] - p1; point nextTangent(point pi, point pin, circle in, circle ci) { line tangent0 = tangents(in, pi)[0]; point[] pk = intersectionpoints(tangent0, ci); line tangent1 = tangents(in, pi)[1]; point[] pl = intersectionpoints(tangent1, ci); return pk[0] + pl[0] + pk[1] + pl[1] - 2 * pi - pin; } p3 = nextTangent(p2, p1, in, ci); p4 = nextTangent(p3, p2, in, ci); p5 = nextTangent(p4, p3, in, ci); i1 = incenter(p5, p1, p2); i2 = incenter(p1, p2, p3); i3 = incenter(p2, p3, p4); i4 = incenter(p3, p4, p5); i5 = incenter(p4, p5, p1); j1 = incenter(p4, p1, p3); j2 = incenter(p5, p2, p4); j3 = incenter(p1, p3, p5); j4 = incenter(p2, p4, p1); j5 = incenter(p3, p5, p2); j = intersectionpoint(bisector(line(p1, p3), line(p2, p4), 90), bisector(line(p2, p4), line(p3, p5), 90)); k = circumcenter(i1, i2, i3); point[] qmp1 = intersectionpoints(line(p1, i), ci); point qm = qmp1[0]+qmp1[1] - p1; point[] q1qm = intersectionpoints(line(qm, o), ci); q1 = q1qm[0]+q1qm[1] - qm; point[] r1p1 = intersectionpoints(line(j, p1), ci); r1 = r1p1[0]+r1p1[1]-p1; draw(p1--p2--p3--p4--p5--cycle, springgreen); draw(p1--p3--p5--p2--p4--cycle, green); draw(line(o,i), purple); draw(ci, springgreen); draw(in, springgreen); draw(circle(2*j-foot(j, p1, p3), foot(j, p1, p3)), green); draw(q1--r1, dashed+red); draw(circle(i1,i2,i3), dashed+purple); draw(circle(j1,j2,j3), dashed+purple); draw(i3--i4, red); draw(j2--j5, red); draw(p2--r1--p5, red); draw(p3--q1--p4, red); dot("$P_1$", p1, dir(140)); dot("$P_2$", p2, dir(60)); dot("$P_3$", p3, dir(10)); dot("$P_4$", p4, dir(-30)); dot("$P_5$", p5, dir(280)); dot("$I_1$", i1, dir(10)); dot("$I_2$", i2, dir(80)); dot("$I_3$", i3, dir(230)); dot("$I_4$", i4, 1.2*dir(90)); dot("$I_5$", i5, dir(240)); dot("$J_1$", j1, dir(340)); dot("$J_2$", j2, 1.2*dir(180)); dot("$J_3$", j3, dir(210)); dot("$J_4$", j4, dir(100)); dot("$J_5$", j5, dir(210)); dot("$I$", i, dir(220)); dot("$O$", o, dir(50)); dot("$J$", j, dir(90)); dot("$K$", k, dir(90)); dot("$Q_1$", q1, dir(140)); dot("$R_1$", r1, dir(350)); [/asy][/asy]