anantmudgal09 wrote: Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$. Proposed by Pranjal Srivastava and Rohan Goyal Let $X$ be an immaculate point and $P, Q$ be the points where lines $AB, AC$ meet the perpendicular to line $MX$ at $X$. Since $XP=XQ$, we get $$-1 = A(PQ; X \infty) = (AB, AC; AX, \ell)$$where $\ell$ is the perpendicular to $MX$ from $A$. Let $D$ be the antipode of $A$ in $BC$. Look at lines perpendicular to lines of this pencil passing through $D$, we get $(DB, DC; DX, DK) = -1$ where $K$ lies on the line through $D$ parallel to line $XM$. Let tangents to $B$ and $C$ meet at $T$; then $T, K, X$ are collinear so we redefine $K$ as such and compute the conditions on $X$ with $XM \parallel DK$. We now use complex numbers with $(ABC)$ as the unit circle. Let $X$ be an immaculate point with coordinate $x^2$, with an appropriate sign to be chosen later. Suppose $M$ has coordinate $m \in \mathbb{R}$ by aligning the diameter line through $M$ with the real axis (using a rotation if necessary). Then $m = \overline{m}$. So $T$ has coordinate $\frac{1}{m}$. Now $$K \in \overline{TX} \iff \frac{k-x^2}{x^2-\frac{1}{m}} = \frac{\frac{1}{k}-\frac{1}{x^2}}{\frac{1}{x^2}-\frac{1}{m}} \iff k = \frac{x^2-\frac{1}{m}}{x^2\left(\frac{x^2-m}{mx^2}\right)} = \frac{mx^2-1}{x^2-m}$$and $$XM \parallel DK \iff \frac{k-d}{x^2-m} = \frac{\frac{1}{k}-\frac{1}{d}}{\frac{1}{x^2}-m} \iff kd = \frac{x^2(x^2-m)}{mx^2-1} \iff d = \frac{x^2(x^2-m)^2}{(mx^2-1)^2}.$$Fix a square root $\lambda$ of $d$. Choosing the value of $x$ such that $$\frac{x(x^2-m)}{(mx^2-1)} = \lambda \iff x^3-\lambda mx^2-mx+\lambda = 0.$$Thus, if $X, Y, Z$ are immaculate points, there exists complex numbers $x, y, z$ such that $X=x^2, Y=y^2, Z=z^2$, and $-m = xy+yz+zx$ by Vieta's formula, hence $m=-(xy+yz+zx)$. Thus, $M$ is either the incentre or one of the three excentres of $\triangle ABC$. Since $M$ lies in the interior of $(XYZ)$, we are done.

Let $U'$ denote the point on $(ABC)$ such that $(UU';BC)=-1$. Define $V'$ and $W'$ similarly. Projecting $(UU';BC)=-1$ from $A$ onto the line through $U$ perpendicular to $MU$, we get that $AU' \perp UM$. Note that we have the cyclic quadrilaterals $UU'OM$, $VV'OM$, $WW'OM$ where $O$ is the circumcentre of $ABC$. As $V'A \perp MV$ and $W'A \perp MW$, \[\measuredangle V'AW'=\measuredangle VMW\]\begin{align*} 2 \measuredangle V'AW' &=\measuredangle V'OW'\\ &= \measuredangle V'OM +\measuredangle MOW'\\ &= \measuredangle V'VM + \measuredangle MWW'\\ &= \measuredangle V'VU+\measuredangle UVM+ \measuredangle MWU+ \measuredangle UWW'\\ &= \measuredangle V'WU+\measuredangle UWW'+\measuredangle UVM+ \measuredangle MWU\\ &= \measuredangle V'AW'+ \measuredangle UVM+ \measuredangle MWU\\ \end{align*}\[\measuredangle VMW=\measuredangle UVM+ \measuredangle MWU\implies 2 \measuredangle VMW= \measuredangle VUW\]The centre $N_1$ of $(VMW)$ lies on $(UVW)$. Similarly, the centres $N_2,N_3$ of $(UMW)$ and $UMV$ lie on $(UVW)$. If any two of $N_1,N_2,N_3$ are the midpoints of the corresponding arcs not containing the opposite vertex, then $M$ is the incentre. If not, then $N_1,N_2,N_3$ are the arc midpoints of $\widehat{VUW},\widehat{UVW},\widehat{UWV}$. It follows that $M$ is the reflection of $U$ in $N_2N_3$, but then $UM$ internally bisects $\angle U$ and similarly $VM,WM$ are also internal angle bisectors which implies that $M$ must be the incentre.

Nice Problem! Let $BB \cap CC = T$ and $TU,TV,TW \cap (ABC) = U^{*},V^{*},V^{*}$. Claim 1: $AU^{*} \perp MU$

Claim 2: If $O$ is the circumcenter of $ABC$, then $UU^{*}OM$ is cyclic.

Claim 3: The center of $(MUV)$ lies on $(ABC)$.

Claim 3 and cyclic variants imply $M$ is incenter of $UVW$.

Relabel the immaculate points on $(ABC)$ as $X_1,X_2,X_3$. Let $\omega=(ABC)$ with center $O$. Let $Y_i$ be the point on $\omega$ such that $(X_iY_i;BC)=-1$ for $i=1,2,3$. Claim 1 : $AY_i \perp MX_i$ and $X_iY_iOM$ is cyclic for $i=1,2,3$. Proof : Projecting the cross ratio $(X_iY_i;BC)=-1$, from $A$ onto the line through $X_i$ perpendicular to $MX_i$, we infer that $AY_i \perp X_iM$. If $BB\cap CC=T$, then $TX_i\cdot TY_i = TB^2 = TO\cdot TM$ which implies that $X_iY_iOM$ is cyclic. $\blacksquare$ Claim 2 : $M$ is orthocenter of $Y_1Y_2Y_3$. Proof : Angel chase $\measuredangle Y_1Y_2Y_3 = \measuredangle Y_1AY_3 = \measuredangle X_1MX_3 = \measuredangle Y_3MY_1$. The second equality follows since $AY_i \perp MX_i$; and the last equality follows by reflection over line $OM$, since it is angle bisector of $\angle X_iMY_i$. Similarly doing the angle chase for $\measuredangle Y_1Y_3Y_2$ implies the claim. $\blacksquare$ Claim 3 : $M$ lies on angle bisector $X_1X_3X_2$. Proof : Look at quadrilateral $X_3Y_2Y_3X_2$ with center $O$. Since $M = (X_3Y_3O) \cap (X_2Y_2O)$, $M$ is the spiral center of the quadrilateral. Hence $\measuredangle MX_3X_2 = \measuredangle MY_2Y_3 = 90 - \measuredangle Y_2Y_3Y_1$ by claim 2. Hence, $90 - \measuredangle MX_3X_2 = \measuredangle Y_2Y_3Y_1 = - \measuredangle Y_1Y_3Y_2 = -90 + \measuredangle MX_3X_1$, which implies the claim. $\blacksquare$ Hence $M$ is incentre or excentre of $X_1X_2X_3$. Since $M$ lies inside $(X_1X_2X_3)$, it must be the incentre and we are done!

Seems too simple, but I couldn't find a mistake Claim: $\measuredangle AWM+\measuredangle BMW=90^{\circ}$ Intersect the perpendicular to $MW$ through $W$ with $AB$ and $AC$ at $B_W$ and $C_W$. Let $W'$ be the point such that $AB_WW'C_W$ is a parallelogram. Then triangle $AB_W W'$ and $CWB$ are similar with opposite orientation so $\measuredangle WMB= \measuredangle C_W W A$ and the result follows.

We make the use of complex numbers. Toss $ABC$ onto the unit circle. Let $X$ denote an immaculate point and let $X_1,X_2$ be points of $AB,AC$ such that $X_1,X_2,X$ are collinear and that line is perpendicular to $MX$. First note that since $MX_1=MX_2$ and $MX\perp X_1X_2$, we must have $X$ to be mid-point of $\overline{X_1X_2}$. So, $2X-X_1$ must lie on line $AC$. We have, \[x_1+ab\overline{x_1}=a+b, \ \ \ x_2+ac\overline{x_2}=a+c, \ \ \ x_1+x_2=2x\]This systems gives \[x_1=\frac{2bc+ab+ac-\frac{2abc}{x}-2bx}{c-b},x_2=\frac{2bc+ab+ac-\frac{2abc}{x}-2cx}{b-c}\]Hence, $\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{ax\left(b+c-\frac{2bc}{x}\right)}{b+c-2x}$ Since $MX\perp X_1X_2$, we want \[ax\left(\frac{m-\frac{bc}{x}}{m-x}\right)=ax\left(\frac{b+c-\frac{2bc}{x}}{b+c-2x}\right)=\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{x-m}{\overline{m}-\frac1x}\]\[\Longrightarrow x(x-m)^2+a(mx-bc)(\overline{m}x-1)=0.\]The roots of the cubic are $x_1,x_2,x_3$, then we need to compute $-\left(\sqrt{x_1x_2}+\sqrt{x_2x_3}+\sqrt{x_3x_1}\right)$ The rest follows by Vieta's relations.