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Source: The 1st India-Iran Friendly Competition Problem 6

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1979 posts

#1 h Jun 13, 2024, 3:42 PM • 4 Y

Y by GeoKing, Supercali, Rg230403, kiyoras_2001

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

This post has been edited 1 time. Last edited by anantmudgal09, Jun 13, 2024, 3:42 PM

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1979 posts

#2 h Jun 13, 2024, 3:49 PM • 3 Y

Y by GeoKing, Rijul saini, Supercali

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.

Anecdote

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#3 h Jun 13, 2024, 5:04 PM • 2 Y

Y by GeoKing, anantmudgal09

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.

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659 posts

#4 h Jun 13, 2024, 5:26 PM • 1 Y

Y by GeoKing

Nice Problem!
Let $BB \cap CC = T$ and $TU,TV,TW \cap (ABC) = U^{*},V^{*},V^{*}$ .
Claim 1: $AU^{*} \perp MU$
Proof
Claim 2: If $O$ is the circumcenter of $ABC$ , then $UU^{*}OM$ is cyclic.
Proof
Claim 3: The center of $(MUV)$ lies on $(ABC)$ .
Proof
Claim 3 and cyclic variants imply $M$ is incenter of $UVW$ .

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#5 h Jun 14, 2024, 7:13 AM • 2 Y

Y by GeoKing, Om245

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!

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871 posts

#6 h Aug 18, 2024, 9:24 PM • 1 Y

Y by GeoKing

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.

Cyclic variants also hold. Thus $\angle WMV=-\measuredangle BMW-\measuredangle VMC=\measuredangle AWM+\measuredangle MVA=\measuredangle WAV-\measuredangle WMV\iff 2\measuredangle WMV=\measuredangle WUV$ Cyclic variants imply $M$ is the incenter of $UVW$ as desired. remark

Attachments:

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#7 h Oct 1, 2024, 2:24 PM

Y by

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.

This post has been edited 1 time. Last edited by Pyramix, Oct 1, 2024, 2:24 PM

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