When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

I remember seeing some solutions of this one on the forum some time ago. I even remember posting one, but I'm sure it was quite different from this one: Let $B,C$ be ny two points in the plane, take $A$ s.t. $ABC$ is equilateral, and take $D,E$ on the segments $AB,AC$ respectively s.t. $AB=3AD=3AE$. Now let $U,V$ be the points where the perpendiculars through $B,C$ to $BC$ cut $DE$, and, finally, let $P=BU\cap AV$. Since the circle inscribed in $PAB$ is the same as that inscribed in $PUV$, we get $A+B=U+V$ (for a point $X$, the letter $X$ also denotes the value we assign to it). In the same way, we get $A+C=U+V$, from which we find $B=C$. Since $B,C$ were chosen arbitrarily, we're done.

Solution by mgao, Zhero, and myself: For a point $P$, let $f(P)$ be the number assigned to point $P$. Let $E, F$ be any two points in the plane; we want $f(E) = f(F)$. Construct regular hexagon $ABCDEF$ with center $O$. Then, $f(A) + f(C) + f(E) = 3 f(O) = f(B) + f(D) + f(F)$. Now, let $P$ be the intersection of lines $AB$ and $CD$. By reflection through the angle bisector of $APD$, it is easy to see that $APC$ and $DPB$ have the same incenter $I$, and we have $f(A) + f(P) + f(C) = 3 f(I) = f(D) + f(P) + f(B)$. Subtracting the equation $f(B) + f(D) + f(F) = f(A) + f(C) + f(E)$, we get the desired result $f(E) = f(F)$.

Take an equilateral triangle $ABC$ and construct triangles $XYZ$ and $DEF$ so that $XYZ$ is the medial triangle of $DEF$ which in turn is the medial triangle of $ABC$ ($AXD,BYE,CZF$ are lines and $\triangle{ABC},\triangle{DEF},\triangle{XYZ}$ share incenter $I$). Assume for the sake of contradiction that some two points in the plane don't have the same number; WLOG let $A$ and $X$ be these two points. Note that \begin{align} f(A)+f(B)+f(C)=f(D)+f(E)+f(F)=3f(I). \end{align}Also, because $\triangle{ABE}$ and $\triangle{ACF}$ share incircles and thus have a common incenter (this can be seen for example by reflecting about $AXID$ or noting that quadrilateral $AEIF$ is tangential), we have \[\frac{f(A)+f(B)+f(E)}{3}=\frac{f(A)+f(C)+f(F)}{3}\implies f(B)+f(E)=f(C)+f(F).\]Symmetry and (1) tell us that \[f(A)+f(D)=f(B)+f(E)=f(C)+f(F)=2f(I).\]But we clearly must also have \[f(D)+f(X)=f(E)+f(Y)=f(F)+f(Z)=2f(I)\]by applying the same argument to figure $DZEXFY$ as we did to $AFBDCE$, so $f(A)+f(D)=2f(I)=f(D)+f(X)$ and $f(A)=f(X)$, contradiction.

Consider two points $A$ and $B$, with values $a$ and $b$ respectively. Draw the perpendicular bisector, and pick an arbitrary point $X$ on it, with value $c$. Then, let $X'$ be the incenter of $ABX$, $X''$ of $ABX'$, and so on. The value of $X$ is $\frac{a+b+c}{3}$, $X'$ is $\frac{4a+4b+c}{9}$, $X''$ is $\frac{13a+13b+c}{27}$, and so on, with $X''...''$ where there are $n$ $'$s, having a value of $\frac{\frac{3^{n+1}-1}{2}(a+b)+c}{3^{n+1}}$. As $n \to \infty$, we see that the value of $X''...''$ goes to $\frac{a+b}{2}$. As our function is necessarily continuous (slight perturbations of vertices slightly perturb incenter location), the midpoint $M$ of $AB$ has a value equal to $\frac{a+b}{2}$ (the sequence $X, X', X'',...$ converges to $M$). Knowing this, we can divide a line segment $AB$ into successively smaller intervals, to find the value of a point dividing it in a certain ratio. The midpoint of $AM$ has value $\frac{3a+b}{4}$, the midpoint of that point and $A$ again is $\frac{7a+b}{8}$, and so on. Remembering that the function is continuous, and as all fractions can be written as an infinite sum/difference of fractions with denominators powers of $2$, it is clear that a point's value is the weighted average of the values of the endpoints of the segment (with the ratio of weights being equal to the ratio of the lengths, and the higher weight given to the endpoint it is nearest to). Now, consider a triangle with vertices at $A=(0,0), B=(2,0), C=(0,2)$. Let $D,E$ be the midpoints of $AB,AC$ respectively, and let $A,B,C$ have values $a,b,c$ respectively. $D$ then has a value of $\frac{a+b}{2}$ and $E$ has $\frac{a+c}{2}$. The midpoint $O$ of $DE$ is at $\left(\frac{1}{2},\frac{1}{2}\right)$, and thus has value $o=\frac{2a+b+c}{4}$ Also, the incenter $I$ of $ABC$ is located at $(2-\sqrt{2},2-\sqrt{2})$ and has a value of $i=\frac{a+b+c}{3}$. However, $O$ is also on the line connecting $A$ and $I$. The distance $AO$ is $\frac{\sqrt2}{2}$, and the distance $IO$ is $\sqrt2\left(\frac{3}{2}-\sqrt2\right)=\frac{3\sqrt2-4}{2}$, and $AI$ is $\sqrt2(2-\sqrt2)=2\sqrt2-2$ Then, $O$'s value of $\frac{2a+b+c}{4}$ is also equal to $\frac{a\cdot IO+i\cdot AO}{AI}=\frac{\frac{3\sqrt2-4}{2}a+\frac{\sqrt2}{2}\frac{a+b+c}{3}}{2\sqrt2-2}$ After some effort, this (should!) simplify to $a=\frac{b+c}{2}$. As the midpoint $M$ of $BC=\frac{b+c}{2}$ as well, all points between $M$ and $A$ have value $\frac{b+c}{2}$. Reflecting $A$ over $M$ to $A'$, we get the same result for segment $AA'$. We can keep doing this infinitely, so all point on the line $AM$ have the same value. As there is nothing special about $AM$, we have that all points on a given line have the same value. As any two points are collinear, we have that all points in the plane have an equal value.

My solution : Let $f(M)$ be the value at point $M$. Let $I$ be the center of $ABC$. Let $D$ be the center of $ABI$, $E$ be the center of $ACI$, and $F$ be the center of $BIC$, and let $O$ be the center of $DEF$. Then it is easy to see that $f(I) = f(O).$ Assume we can construct $ABC$ such that $I \neq O$. Now using rotation and scaling, any two points $P$ and $Q$ will be $I$ and $O$ hence have the same value. Thus our proof is complete provide that we can construct $ABC$ such that $I \neq O$. This is not too hard a geometry problem.

Let F(A) be the value assigned to point A. Let I be the incenter of triangle ABC. Let D, E, and F be the incenters of triangles AIB, BIC, and CIA respectively. One can easily verify that F(D)+F(E)+F(F) = F(A)+F(B)+F(C) = 3F(I) but also notice that if ABC is not equilateral then the incenter I' of DEF is distinct from I. Now if we consider any 2 points P and Q in the plane then we can show that F(P)=F(Q) because the configuration described above can be translated, rotated, and dilated to have P and Q correspond to I and I' respectively. Thus the only solution is F(X) = constant.

Take triangles $A_1B_1C_1$ and $A_2B_2C_2$. Let the two incenters be $I_1$ and $I_2$ respectively. By appropriate dilation and rotation of each triangle about its respective incenter, we can match up any two corresponding sides. It follows that for point $K_1\in\{A_1,B_1,C_1\}$, and any point $K_2\in\{A_2,B_2,C_2\}$, we have that $f(K_1)-f(K_2)=f(I_1)-f(I_2)$. It follows that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$, $I_1$ and $I_2$ all have the same weight, hence we can see that all other points must also have this weight since we can perform the same scaling and rotating on any other pair of triangles with at least one vertice from the original set that we worked with. $\square$

Mewto55555 wrote: As our function is necessarily continuous (slight perturbations of vertices slightly perturb incenter location) It's been more than seven years but I'd like to point out this is clearly false: by this logic if "incenter" was replaced by "centroid" the function $f$ would still be continuous, but there are noncontinuous solutions in this case: any "affine-linear" function satisfies the conditions.

The shortest solution material (a.k.a. this can be reworded to three/four lines in a contest due to the sheer emptiness and simplicity of the argument) a hard P6 I've ever seen: probably the only problem which fits that category is in ($\bigstar$) of the Motivation. $\color{green} \rule{25cm}{2pt}$ $\color{green} \textbf{Combinatorial/Algebraic First Step.}$ Given two points $P$ and $P'$ of distance $1$. Then, if we have proven that \[ f(P) = f(P'); \]we are finished with the problem. $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \textbf{Should be obvious, but this part is interesting nonetheless.}$ There are two ways to finish here. 1. We will prove that $f(P_1) = f(P_2)$ for any two points of the plane. Let the distance between those two points be $d$. Then, apply the same argument on a plane magnified with homothety/dilation of scale $d$. 2. We'll first prove that $f(P_3) = f(P_4)$ for any two $P_3,P_4$ with distance at most $2$. To do so, form an isosceles with sides $(1,1,\text{distance of} \ P_3,P_4)$. To prove the general statement, let we wish to prove $f(P_5) = f(P_6).$ Then find a $P_7$ in the segment $\overline{P_5P_6}$ so that $d(P_5,P_7) = (\text{fractional part of their distance}) < 1$ and induct upwards to translate $P_7$ to $P_6$. Note that the first way puts into perspective that \[\begin{tabular}{p{12cm}} If we have enforced a way to force some two general points to be uniformly in order, then regardless of the size of the distance of any $P_1$ and $P_2$, the operation then can be magnified by $d$. \end{tabular} \\ \]Well, and the second way also puts it into perspective that \[\begin{tabular}{p{12cm}} No magnification is required: by translating a valid operation that forces uniform values on two points, we can actually match any two points to be equal! \end{tabular} \\ \]This ends this Section. $\blacksquare$ $\color{red} \rule{25cm}{2pt}$ $\color{red} \textbf{The versatility and wild nature of incenters.}$ This might be the most absurd-ly worded Claim I've made in my Solutions, but here goes: \[\begin{tabular}{p{12cm}} Given $P,P'$ with distance $1$. Draw the segment $\overline{PP'}$ and draw two perpendicular lines to that segment that passes through $P$ and $P'$. Call the strip bounded by those two lines to be the $\textit{parallel range}$ of $P$ and $P'$. Moreover, divide the $\textit{parallel range}$ equally, and call the region closer to $P$ (and $P'$, respectively) to be the $\textit{P-half-range}$ (and $\textit{P'-half-range}$, respectively.) \\ \\ For any $Q$ which lies inside $\textit{P-half-range}$, we can prove that \\ \\ $\quad \quad \quad \quad \quad \quad \quad \quad \quad f(Q) - f(Q') = \dfrac{f(P)-f(P')}{3}.$ \\ \\ where $Q'$ is the reflection of $Q$ towards the perpendicular bisector of $PP'$. \end{tabular} \\ \]$\color{red} \rule{25cm}{0.4pt}$ $\color{red} \textbf{Proof 2.}$ We will proof that we can find two points $A$ and $B$ so that $Q$ is the incenter of $\triangle PAB$ and $Q'$ is the incenter of $\triangle P'AB$. Let the point $\overline{PQ} \cap \overline{P'Q'} \cap \text{perp.bisector of} \ PP'$ be $X$. Then, draw a circle $\omega$ with center $X$ and radius $|XQ| = |XQ'|$. Pick two points $A$ and $B$ so that \[ (PP'X) \cap \omega = \{A,B\} \]Since $X \in (PAB), (P'AB)$ and $PX,P'X$ is the internal bisector of angles $\angle APB, AP'B$, this validates the reverse construction --- $Q,Q'$ is forced to be the respective incenters of the two triangle. $\blacksquare$ $\blacksquare$ $\blacksquare$ $\color{blue} \rule{25cm}{2pt}$ $\color{blue} \textbf{A Final Contradiction.}$ The reason I put three blacksquares above was that the conclusion should be in sight after the $\color{red} \textbf{versatility of incenters}$: pick $Q,Q'$ in the $\textit{parallel range}$ of $PP'$ --- and pick $RR'$ to be in the $\textit{parallel range}$ of $QQ'$. It should be immediately clear that $RR'$ must be in the parallel range of $PP'$. Since $R,R'$ lies in the parallel range of $PP'$, then \[ f(R)-f(R') = \dfrac{f(P)-f(P')}{3} \]but since $R,R'$ lies in the parallel range of $QQ'$ and $QQ'$ lies in the parallel range of $PP'$, \[ f(R)-f(R') = \dfrac{f(Q)-f(Q')}{3} = \dfrac{\frac{f(P)-f(P')}{3}}{3} \]rendering $f(P) - f(P') = 0$ We are truly done by here. $\color{blue} \blacksquare \ \blacksquare \ \blacksquare$

Let $(O)$ be an arbitrary circle and let $\ell_1,\ell_2,\ell_3,\ell_4,\ell_5$ be arbitrary tangent lines to $(O)$ no two of which are parallel. Let the function mapping points to real numbers be $f$. Then we have \[f(\ell_i\cap\ell_j)+f(\ell_i\cap\ell_k)+f(\ell_j\cap\ell_k)=c\]for all distinct $i,j,k$ and some fixed $c$. Thus we have $10$ equations in $10$ variables. To solve these equations, note \[f(\ell_i\cap\ell_j)=c-f(\ell_1\cap\ell_i)-f(\ell_1\cap\ell_j)\forall i,j\ne 1,i\ne j.\]Let $f(\ell_1\cap\ell_i)=a_i$ so $c=c-a_i-a_j+c-a_j-a_k+c-a_k-a_i=3c-2[a_i+a_j+a_k]$. This implies $a_i+a_j+a_k$ is fixed at $c$, and so the desired follows: each of the $f(\ell_i\cap\ell_j)$ is equal to $c/3$. But this implies that any non-degenerate triangle has all its real numbers of vertices. To finish, note that for fixed $A$ and $B$, all points $C$ not on line $AB$ must have $f(A)=f(B)=f(C)$, so for particular $D\not\in\{A,B\}$ on line $AB$ and $C$ not on line $AB$, looking at $\triangle CDB$ finishes. Edit: This is wrong due to config issues, I'll try to find a correct approach. Edit: Here is a correct solution which turns out to basically be Evan Chen's. Let $f$ denote the map from the plane $\mathbb R^2$ to $\mathbb R$. Observation: For isosceles trapezoid $ABCD$ with $AB\parallel CD$ which is not a rectangle, $f(A)+f(C)=f(B)+f(D)$. Proof: Let $BC\cap DA=X,AC\cap BD=Y$. By Pitot's Theorem, whichever one of $XCYD$ or $XAYB$ is concyclic has an incircle. Thus $f(X)+f(A)+f(C)=f(X)+f(B)+f(D)$, finishing. Then considering regular pentagon $ABCDE$ finishes, as $f(A)+f(C)=f(B)+f(D)=f(C)+f(E)=f(D)+f(A)=f(E)+f(B)$ so $f(A)=f(E)=f(D)=f(C)=f(B)$.

MithsApprentice wrote: Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.7799430983503575, xmax = 10.220716282714264, ymin = -1.8198421720682916, ymax = 9.297129470225183; /* image dimensions */ pen qqwwzz = rgb(0,0.4,0.6); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqttzz = rgb(0,0.2,0.6); pen ffqqtt = rgb(1,0,0.2); draw((-2.48,0.34)--(1.56,0.36)--(3.562679491924312,3.8687426312891326)--(1.5253589838486228,7.357485262578265)--(-2.5146410161513773,7.337485262578268)--(-4.51732050807569,3.8287426312891357)--cycle, linewidth(0.4) + qqwwzz); /* draw figures */ draw((-2.48,0.34)--(1.56,0.36), linewidth(0.4) + qqwwzz); draw((1.56,0.36)--(3.562679491924312,3.8687426312891326), linewidth(0.4) + qqwwzz); draw((3.562679491924312,3.8687426312891326)--(1.5253589838486228,7.357485262578265), linewidth(0.4) + qqwwzz); draw((1.5253589838486228,7.357485262578265)--(-2.5146410161513773,7.337485262578268), linewidth(0.4) + qqwwzz); draw((-2.5146410161513773,7.337485262578268)--(-4.51732050807569,3.8287426312891357), linewidth(0.4) + qqwwzz); draw((-4.51732050807569,3.8287426312891357)--(-2.48,0.34), linewidth(0.4) + qqwwzz); draw((xmin, 0.004950495049504941*xmin + 0.35227722772277226)--(xmax, 0.004950495049504941*xmax + 0.35227722772277226), linewidth(0.4) + qqwwzz); /* line */ draw((xmin, -1.7124171761194096*xmin + 9.969536186268696)--(xmax, -1.7124171761194096*xmax + 9.969536186268696), linewidth(0.4) + qqwwzz); /* line */ draw(circle((3.031422123213444,1.846063139364821), 1.4787607512476095), linewidth(0.4) + yqqqyq); draw((1.56,0.36)--(1.5253589838486228,7.357485262578265), linewidth(0.4) + qqttzz); draw((-2.48,0.34)--(3.562679491924312,3.8687426312891326), linewidth(0.4) + qqttzz); draw((xmin, -0.5707684211618897*xmin + 3.5763031585065823)--(xmax, -0.5707684211618897*xmax + 3.5763031585065823), linewidth(0.4) + linetype("4 4") + ffqqtt); /* line */ /* dots and labels */ dot((-2.48,0.34),linewidth(4pt) + dotstyle); label("$A$", (-2.350296352528207,0.020716046854468963), NE * labelscalefactor); dot((1.56,0.36),linewidth(4pt) + dotstyle); label("$B$", (1.5700926537772786,-0.05290628190244146), NE * labelscalefactor); dot((3.562679491924312,3.8687426312891326),dotstyle); label("$C$", (3.6315178589707733,4.051538546295315), NE * labelscalefactor); dot((1.5253589838486228,7.357485262578265),dotstyle); label("$D$", (1.6069038181557338,7.54859916224856), NE * labelscalefactor); dot((-2.5146410161513773,7.337485262578268),dotstyle); label("$E$", (-2.4423242634743447,7.530193580059332), NE * labelscalefactor); dot((-4.51732050807569,3.8287426312891357),dotstyle); label("$F$", (-4.632588543992433,4.08834971067377), NE * labelscalefactor); dot((5.6,0.38),linewidth(4pt) + dotstyle); label("$K$", (5.67453748197504,0.5360723481528419), NE * labelscalefactor); dot((1.5484529946162078,2.692495087526089),linewidth(4pt) + dotstyle); label("$L$", (1.6621205647234167,2.9472036149416585), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider two arbitrary points $E$ and $F$ in the plane and let $n(X)$ denote the number written at point $X$. Let $ABCDEF$ be a regular hexagon and let $AB \cap DC = K, DB \cap AC = L$. We see that $\triangle ACE$ and $\triangle BDF$ have same incenter (center of hexagon) and so we have that $n(A) + n(C) + n(E) = n(B) + n(D) + n(F) \dots (1)$. Now, we see that if $\Gamma$ is the incircle of $\triangle KAC$, then the reflection of $\Gamma$ about perpendicular bisector of segment $\overline{EF}$ must be the incircle of $\triangle KBD$, but we see that $\Gamma$ is also the incircle of $\square KBLC$ and $\angle KBL = \angle KCL = 90^\circ$ implying that center of $\Gamma$ lies on line $\overline{KL}$ but this is exactly the perpendicular bisector of segment $\overline{EF}$, which means that $\Gamma$ is preserved under reflection through perpendicular bisector of $\overline{EF}$ due to symmetry, which means that $\Gamma$ is also the incircle of $\triangle KBD$, therefore $n(K) + n(B) + n(D) = n(K) + n(A) + n(C) \implies n(B) + n(D) = n(A) + n(C) \dots (2)$, subtracting $(2)$ from $(1)$ we get that $n(A) + n(C) + n(E) - n(A) - n(C) = n(B) + n(D) + n(F) - n(B) - n(D) \implies n(E) = n(F)$, but since $E$ and $F$ were arbitrarily chosen, we must have that all of the points in the plane must have same number written on them/assigned to them.

Solved with nukelauncher. The problem succumbs to the following: Claim: Let \(ABC\) be a scalene triangle with incenter \(I\). Let \(I_A\), \(I_B\), \(I_C\) be the incenters of \(\triangle IBC\), \(\triangle ICA\), \(\triangle IAB\), and let \(J\) be the incenter of \(\triangle I_AI_BI_C\). Then \(I\ne J\). Proof. Assume for contradiction \(I\) is the incenter of \(\triangle I_AI_BI_C\). Let \(X=\overline{AI}\cap\overline{I_BI_C}\), \(Y=\overline{BI}\cap\overline{I_CI_A}\), \(Z=\overline{CI}\cap\overline{I_AI_B}\), and let \((I_A)\), \((I_B)\), \((I_C)\) touch \(\overline{BC}\), \(\overline{CA}\), \(\overline{AB}\) at \(P\), \(Q\), \(R\). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,IA,IB,IC,X,Y,Z,P,Q,R; A=dir(130); B=dir(200); C=dir(340); I=incenter(A,B,C); IA=incenter(I,B,C); IB=incenter(I,C,A); IC=incenter(I,A,B); X=extension(A,I,IB,IC); Y=extension(B,I,IC,IA); Z=extension(C,I,IA,IB); P=foot(IA,B,C); Q=foot(IB,C,A); R=foot(IC,A,B); draw(incircle(A,B,C),linewidth(.4)); draw(A--I,gray); draw(B--I,gray); draw(C--I,gray); draw(IA--IB--IC--cycle,linewidth(.5)); draw(A--B--C--cycle,linewidth(.5)); draw(incircle(I,B,C),linewidth(.3)); draw(incircle(I,C,A),linewidth(.3)); draw(incircle(I,A,B),linewidth(.3)); dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(I\)",I,NE); dot("\(I_A\)",IA,S); dot("\(I_B\)",IB,NE); dot("\(I_C\)",IC,W); dot("\(X\)",X,dir(60)); dot("\(Y\)",Y,S); dot("\(Z\)",Z,dir(290)); dot("\(P\)",P,S); dot("\(Q\)",Q,NE); dot("\(R\)",R,NW); [/asy][/asy] Then \(\angle YII_A=\angle ZII_A\) and \(\angle YI_AI=\angle ZI_AI\), so \(\angle IYI_A=\angle IZI_A=:\theta\), Then \(\angle IXI_B=\angle IZI_B=180^\circ-\theta\) and \(\angle IXI_C=180^\circ-\theta\), so \(\theta=90^\circ\). This implies \((I_B)\) and \((I_C)\) are tangent at \(X\), so \(A\) lies on their radical axis, i.e.\ \(AQ=AR\). Analgoously \(BR=BP\) and \(CP=CQ\), so \(P\), \(Q\), \(R\) are the points where the incircle touches the sides of the triangle. Hence \(\overline{II_A}\perp\overline{BC}\), implying \(IB=IC\), and so \(\angle B=\angle C\), contradiction. \(\blacksquare\) Now in the above configuration, \begin{align*} f(J)&=\tfrac13\big(f(I_A)+f(I_B)+f(I_C)\big)\\ &=\tfrac19f(I)+\tfrac29\big(f(A)+f(B)+f(C)\big) =f(I). \end{align*}For any two points \(U\) and \(V\), it is possible to rotate and scale the above diagram so that \(I\mapsto U\) and \(J\mapsto V\). Hence \(f(U)=f(V)\), so \(f\) is constant.

Clean solution but hard to motivate. For any $P \in \mathbb{R}^2$ define $f(P)$ to be the number it was assigned. We would like to show $f(X)=f(Y)$ for all $X,Y \in \mathbb{R}^2$. Indeed, consider points $A, B, C, D$ where $ABCDXY$ is a regular hexagon. Triangles $ACX$ and $BDY$ share an incenter so \[ f(X)+f(A)+f(C) = f(Y)+f(B)+f(D) \]So, it is sufficient to show that $f(A)+f(C) = f(B)+f(D)$. Indeed, if we let $AB$ and $CD$ intersect at $T$ then we notice that triangles $ACT$ and $DBT$ share an incenter. So, \[ f(A) + f(C) + f(T) = f(B) + f(D) + f(T) \Leftrightarrow f(A) + f(C) = f(B) + f(D) \]as desired.

This might just be the most cursed solution to grace the entire AoPS forum to date. Let $f(P)$ be the number assigned to a point $P$. Consider the following configuration. Let $A,B,C$ be (fixed) points with $C$ on segment $\overline{AB}$. and $\ell$ be a line perpendicular to and passing through segment $\overline{AC}$. Let $I$ be a variable point on $\ell$. Let $D$ be the point such that $I$ is the incenter of $\triangle ABD$, and let $E$ be the point such that $I$ is the incenter of $\triangle CDE$. The latter two points don't always exist, but they will when $I$ lies within a certain distance of $\overline{AB}$. Observe that we have $$f(A)+f(B)+f(D)=3f(I)=f(C)+f(D)+f(E) \implies f(E)=f(A)+f(B)-f(C),$$which is fixed. I claim that $E$ varies along some projective algebraic plane curve as $I$ varies. That is, there exist real rational functions $f,g$ such that $E$ (on some subset of where it exists) can be parameterized as $(f(t),g(t))$ where $t$ varies in some interval. This is due to the theory of moving points: if we move $I$ linearly, we can obtain $\overline{AD}$ and $\overline{BD}$ by reflecting $\overline{AB}$ over $\overline{AI}$ and $\overline{BI}$, so $D$ has a well-defined degree. Then we can obtain $\overline{DE}$ and $\overline{CE}$ by reflecting $\overline{CD}$ over $\overline{DI}$ and $\overline{CI}$, so $E$ has a well-defined degree too. $E$ thus varies along some projective algebraic plane curve as desired. By implicit differentiation, the value of the second derivative $\tfrac{d^2y}{dx^2}$ at $(f(t),g(t))$ is now a real rational function in $t$, so there exist finitely many points where its sign changes. Take some finite contiguous section $\mathcal{S}$ of the locus of $E$ where the curve is always concave or always convex (WLOG concave) and where no two points on this curve have the same $y$-coordinate. Fix a tiny open disc $\mathcal{D}$ lying close to but entirely above $\mathcal{S}$ (so there exists a point on $\mathcal{S}$ lying directly below any point in $\mathcal{D}$, and $\mathcal{D}$ does not intersect $\mathcal{S}$) and let $Q$ be a point directly above the center of $\mathcal{D}$ which lies very far above $Q$. Let $P$ be any point in $\mathcal{D}$: I claim that there exist points $X,Y \in \mathcal{S}$ such that $P$ is the incenter of $\triangle QXY$. To do this, we vary a circle $\Omega$ of radius $r$ centered at $B$ and let $X',Y'$ be the points where the tangents from $Q$ to $\Omega$ first intersect $\mathcal{S}$, which exist by our definition of $\mathcal{D}$. Consider the expression $d(P,\overline{X'Y'})-r$, where $d$ denotes signed distance and is positive if $P$ lies above line $\overline{X'Y'}$ and negative otherwise ($\overline{X'Y'}$ being vertical never happens). If this expression is positive, then $\Omega$ lies entirely in the interior of $\triangle QX'Y'$, and if it's negative, then $\Omega$ lies partially outside $\triangle QX'Y'$. The expression is clearly continuous in $r$ and positive as $r \to 0^+$. Moreover, when $r$ is chosen minimally so that $\Omega$ intersects $\mathcal{S}$ (so they're tangent), because $\mathcal{S}$ is convex, the intersection point lies nonstrictly $\overline{X'Y'}$ and thus outside $\triangle QX'Y'$, so the expression is nonpositive. Thus this expression is $0$ for some value of $r$ by the intermediate value theorem, at which point $\Omega$ will be the incenter of $\triangle QX'Y'$ as desired (by choosing $(X,Y)=(X',Y')$ for this value of $r$). Then, since $f(\mathcal{S})$ is fixed and $f(Q)$ is fixed, $f(P)$ should also be fixed, so we find that $f$ is constant on $\mathcal{B}$. Now, consider two points $A,B \in \mathcal{B}$ such that the circle with diameter $\overline{AB}$ is contained entirely in $\mathcal{B}$. Varying a point $I$ along the perpendicular bisector of $\overline{AB}$ lying inside this circle with diameter $\overline{AB}$ and letting $C$ be the point such that $\triangle ABC$ has incircle $I$, we find that the locus of $C$ is the entire perpendicular bisector minus its intersection with $\overline{AB}$, so $f$ is also constant (and equal to its constant value on $\mathcal{B}$) along the entire perpendicular bisector. Now rotating $\overline{AB}$ about its midpoint and continually applying this fact implies $f$ is constant everywhere, as desired. $\blacksquare$

Solved with davidkong, TheBeast5520, RubiksCube3.1415. Let $p$ be the real number assigned to $P$. The key idea is the following picture: We will be using the triangles $\triangle ABC, \triangle DEF, \triangle ABG, \triangle DFG$. Clearly, $\triangle ABC$ and $\triangle DEF$ share an incenter, so $a+b+c=d+e+f$. Notice that $AXDG$ is a kite, so it has an incircle. This incircle is the incircle of $\triangle ABG$ and $\triangle DFG$, so they also share an incenter: $a+b+g=d+f+g\implies a+b=d+f$. Therefore, $c=e$. Rotating and translating the picture to replace $C, E$ with any points $P, Q$, we get the desired result.

Let $f(X)$ be the corresponding number for some point $X$ in the plane. Let $ABCDEF$ be a regular hexagon with the incenter of both $\triangle ACE$ and $\triangle BDF$ being $I$. Then let $AB \cap CD = G$. Then notice that $\triangle GAC$ is a reflection of $\triangle GDB$ over the angle bisector of $\angle AGD$ so the angle bisectors of $\angle GAC$, $\angle BDG$ and $\angle AGD$ all concur. This implies that $\triangle AGC$ and $\triangle BDG$ have a shared incenter, so $f(A) + f(G) + f(C) = f(B) + f(D) + f(G)$. However from earlier we know that $\triangle ACE$ and $\triangle BDF$ share an incenter so $f(A) + f(C) + f(E) = f(B) + f(D) + f(F)$, so combining the two equations gives $f(E) = f(F)$. Varying hexagon $ABCDEF$ gives us that all points are equal, done.

Hmm here is an interesting way to state the main lemma of this problem. Also this is not 35 MOHS. LEMMA: There exists a hexagon $ABCDEF$ such that $ACE$ and $BDF$ share an incenter $I$, such that if the incenters of $ABC$ and $DEF$ are $I_1$ and $I_2$ respectively, then $\angle{I_1II_2}$ is obtuse but not straight. Proof: Pick $ACE, BDF$ equilateral and cocentric such that $\angle{ACB}$ is small. Details are left to the reader. $\blacksquare$. Now in the diagram above, note that since $\angle{I_1II_2}$ is obtuse but not straight, there exists a point $X$ such that $I$ is the incenter of $I_1I_2X$. Note however that the arithmetic mean of $f(I_1)$ and $f(I_2)$ is the arithmetic mean of $f(A), f(B), f(C), f(D), f(E), f(F)$, which is $f(I)$. But this is also the arithmetic mean of $f(I_1), f(I_2), f(X)$, so we get $f(X) = f(I)$. Now a composition of a rotation, dilation, and translation maps the pair $(X, I)$ onto any pair of points in the plane, as desired.