Incircle geo with angle conditions

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Incircle geo with angle conditions

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Source: 239 MO 2024 J4

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#1 h May 22, 2024, 8:04 PM • 2 Y

Y by Rounak_iitr, GeoKing

Let $I$ be the incenter of a triangle $ABC$ . The points $X, Y$ lie on the prolongations of the lines $IB, IC$ after $I$ so that $\angle IAX=\angle IBA$ and $\angle IAY=\angle ICA$ . Show that the line through the midpoints of $IA$ and $XY$ passes through the circumcenter of $ABC$ .

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#3 h May 23, 2024, 10:16 PM • 2 Y

Y by egxa, GeoKing

$\textbf{Claim:} \$ In triangle $ABC$ , denote $D,E,F$ as the altitudes from $A,B,C$ to $BC,CA,AB$ respectively. $AD,BE,CF$ intersect $(ABC)$ at $P,Q,R$ second time. $S$ is the antipode of $A$ on $(ABC)$ . Let the tangent at $P$ to $(ABC)$ and $QR$ meet at $T$ . If $O$ is the circumcenter of $(ABC)$ and $OD$ intersects $HS$ at $K$ , then $TK\parallel BC$ .

Let the tangent at $A$ intersect $BC$ at $U$ . $QR\cap AH=L$ Denote $M$ as the midpoint of $BC$ .
Pascal at $PPAQRC$ gives that $T,H,AQ\cap CP$ are collinear and pascal at $AAQBCP$ gives that $U,AQ\cap CP,H$ are collinear thus, $T,H,U$ are collinear.
$\frac{HL}{LA}=\frac{HT}{TU}\overset{?}{=}\frac{HK}{KM}=\frac{HD}{OM}=\frac{2.\cos{B}.\cos{C}}{\cos{A}}$ angle
$\frac{HL}{LA}=\frac{HR.\frac{\cos{C}}{\sin{90+B-C}}}{AR.\frac{\cos{A}}{\sin{90+B-C}}}=\frac{2HF}{AH}.\frac{\cos{C}}{\cos{A}}=\frac{2\cos{B}\cos{C}}{\cos{A}}$ Which gives the desired result.

Let $AI\cap (ABC)=\{A,M\}, BI\cap (ABC)=\{B,R\}, CI\cap (ABC)=\{C,S\}$ The tangent at $A$ to $(ABC)$ meets $BC$ at $T$ . The midpoint of the arc $BAC$ is $N$ . $OP\cap IN=K$ . Midpoints of $AI,XY$ are $P,G$ respectively.
By replacing we get that the perpendicular from $T$ to $AD$ (which is the perpendicular bisector of $AD$ ), $OP$ and $IN$ are concurrent.

$\angle XAD=\angle XBD$ yields $X=(ABD)\cap BI$ and $\angle DAY=\angle DCY$ yields $Y=(ACD)\cap CI$ . Since $XD=XA$ and $YD=YA,$ the line $XY$ is the perpendicular bisector of $AD$ which passes through $T$ . Also $IX.IB=IA.ID=IY.IC$ hence $(BCXY)$ is cyclic. And $IB.IR=IC.IS$ so $XY\parallel RS$ . $IG$ bisects $SR$ which gives that $I,G,N$ are collinear since $ISNR$ is parallelogram. By the claim, $IN,OP,XY$ are collinear on $G$ as desired. $\blacksquare$

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DottedCaculator

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DottedCaculator

#4 h May 24, 2024, 1:36 AM • 2 Y

Y by NO_SQUARES, GeoKing

Solved without v4913, CyclicISLscelesTrapezoid, ccarolyn4, ld414think, crazyeyemoody907, or kathy

Let $a=x^2$ , $b=y^2$ , $c=z^2$ . Then, $AX\cap(ABC)=xy$ , so if $d=X$ , then $\overline d=\frac{y^2-xz-x^2-xy}{-xy^2z-x^3y}=\frac{x^2+xy+xz-y^2}{xy(x^2+yz)}$ . Similarly, if $e=Y$ , then $\overline e=\frac{x^2+xy+xz-z^2}{xz(x^2+yz)}$ , so the conjugate of the midpoint of $XY$ is
$\frac{x^2z+xz^2+x^2y+xy^2-y^2z-yz^2+2xyz}{2xyz(x^2+yz)}=\frac{(y+z)(x^2-yz+xy+xz)}{2xyz(x^2+yz)}.$
The midpoint of $AI$ is $\frac{x^2-xy-yz-zx}2$ , and its conjugate is
$\frac{yz-xz-xy-x^2}{2x^2yz}.$ The quotient of these two is $-\frac{x(y+z)}{x^2+yz}$ , which is real. Therefore, the line through the midpoints passes through the circumcenter of $ABC$ .

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khina

#5 h May 24, 2024, 2:05 AM • 1 Y

Y by GeoKing

Hmm, I feel the solutions in this thread are rather complicated.

Rephrasing in terms of the triangle formed by the midpoints of the minor arcs, it suffices to show the following.

Quote:

Let $ABC$ be a triangle and $H$ its orthocenter. Let $BB'$ , $CC'$ be diameters of $(ABC)$ , let $AH$ intersect $(ABC)$ at $D$ , and let $DB' \cap CH = F$ , $DC' \cap BH = E$ . Then show the line connecting the midpoints of $DH, EF$ passes through the circumcenter of $(ABC)$ .

To prove this, let $X = AD \cap BC$ , and $Y$ be the midpoint of $EF$ . By angle chasing one may show htat $DEHF$ and $DC'AB'$ are homothetic, so $EF$ and $BC$ are parallel. Hence $HY$ bisects $BC$ and $DY$ bisects $B'C'$ . Thus $XY$ passes through the midpoint of the midpoints of $BC$ and $B'C'$ , which is precisely the circumcenter of $(ABC)$ .

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#7 h Jun 11, 2024, 4:52 AM • 1 Y

Y by GeoKing

a_507_bc wrote:

Let $I$ be the incenter of a triangle $ABC$ . The points $X, Y$ lie on the prolongations of the lines $IB, IC$ after $I$ so that $\angle IAX=\angle IBA$ and $\angle IAY=\angle ICA$ . Show that the line through the midpoints of $IA$ and $XY$ passes through the circumcenter of $ABC$ .

Let $D$ , $T$ in order are insections of lines $XY$ , $BC$ with line $AI$
Cuz $\angle DBI = \angle XBA = \angle IAX$ so $D$ , $B$ , $A$ , $X$ are concyclic. Similar, $D$ , $A$ , $Y$ , $C$ are concyclic
Therefore $\overline{IX}. \overline{IB} = \overline{ID}. \overline{IA} = \overline{IY}. \overline{IC}$ . Lead to $B$ , $C$ , $X$ , $Y$ are concylic
Further, we have that $\angle YAD = \angle YCD = \angle YXB$ so $A$ , $T$ , $X$ , $Y$ are concyclic
Next, let $U$ , $R$ , $S$ in order are midpoints of arcs $BC$ , $CA$ , $AB$ which dont pass through $A$ , $B$ , $C$
$V$ is midpoint of arc $BAC$ and $N$ , $M$ is order are midpoints of segments $XY$ , $AI$
Follow Sawayama theorem, we have that $I$ , $N$ , $V$ are colinear
So, use Menelaus theorem for $\triangle IVU$ , we need to prove that $\frac{MI}{MU} = \frac{NI}{NV}$ $(*)$
(Since $X, Y$ lie on the prolongations of the lines $IB, IC$ , we see that $T$ lie on prolongation of the line $IU$ )
Cuz line $IT$ pass through $U$ ; $B$ , $C$ , $X$ , $Y$ are concylic and $U$ is circumcenter of $\triangle BIC$ , so $IT \perp XY$
Lead to $TN \| AV$ and $\frac{NI}{NV} = \frac{TI}{TA}$ (Follow Thales theorem ) $(1)$
Again, since $D$ , $A$ , $X$ , $B$ are concyclic and $B$ , $C$ , $X$ , $Y$ are concylic, we have :
$IM. ID = \frac{1}{2}. IA. ID = \frac{1}{2} IB. IX = IT. IU$ . Which means $\frac{IT}{IM} = \frac{ID}{IU}$ $(2)$
$AT = AX. \sin \angle AXT = BD. \frac{AI}{BI}. \sin \angle DIC$ (Cuz $AT \perp XY$ and $A$ , $T$ , $X$ , $Y$ are concyclic
From $(1)$ , $(2)$ , we have
$(*) \Leftrightarrow \frac{AT}{MU} = \frac{ID}{IU} \Leftrightarrow \frac{BD. AI. \sin \angle DIC}{MU. IB} = \frac{ID}{IU} \Leftrightarrow \frac{ \sin \angle DIC. \frac{\sin \angle IBA}{\sin \angle IAB}}{MU} = \frac{\sin \angle IBD}{\sin \angle DIB}. \frac{1}{IU}$ (Sin theorem for triangles $IAB$ , $IBD$ )
$\hspace{0.6cm} \Leftrightarrow \sin \angle DIC . \sin \angle DIB = \frac{MU}{UV}$ (Cuz $IU = UC = UV. \sin \angle UVC = UV . \sin \angle IAB$ )
$\hspace{0.6cm} \Leftrightarrow \sin \angle DIB = \frac{MU}{\sin \angle SVU . UV} = \frac{MU}{SU} = \sin \angle USM$ (Cuz $SR \perp AU$ at $M$ )
Therefore, cuz $\angle DIB = \angle USM$ , so equality $(*)$ obvious right, problem proved

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teoira

#8 h Jul 10, 2024, 11:22 AM • 1 Y

Y by GeoKing

Synthetic solution
https://dgrozev.wordpress.com/2024/07/10/step-by-step-solution-of-a-geometric-problem-from-239-open-mathematical-olympiad-2024/

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#9 h Oct 17, 2024, 12:33 PM • 1 Y

Y by p.lazarov06

I completely loved this problem, so many properties!
Here goes my solution, completely synthetic:

Firstly, let $D,E,F$ be the touchpoints of the incircle of $ABC$ with $BC, AC, AB$ respectively. Also let $L=AI\cap BC$ . As stated in some of the solutions above, $L$ is symmetric to $A$ w.r.t. $XY$ . Also $AIXY$ and $BCXY$ are concyclic by angle-chase (this was also in some of the above solutions, so I won't reprove it). Since $XY\perp AI$ and $EF \perp AI$ , $EF\parallel XY$ . By abgle-chase again, $LY\parallel DF$ and $LX\parallel DE$ . Now $XE, YF, LD$ are concurrent at the center of homothety that sends one triangle to the other. Let this point be $G$ , so $G\in LD\equiv BC$ .

Define by $O_1$ the cirucmecnter of $AXYI$ . By angle-chase again $(AXYI)$ and $(ABC)$ are tangent at $A$ , so $O, O_1, A$ are collinear. Now if $M$ is the midpoint of $AI$ and $N$ is the midpoint of $XY$ , if we prove the symmetric point to $O_1$ wrt to $N$ and the symmetric point to $A$ wrt $M$ also lie on a line with $O$ , clearly $O, M, N$ would be collinear. Now the symmetric of $O_1$ wrt $N$ is $O_2$ - the circumcenter of $XYL$ , and the symmetric of $A$ wrt $M$ is $I$ . So it's enough to prove that $O, I, O_2$ are collinear. However, $I$ is the circumcenter of $DEF$ and $O_2$ is the circumcenter of $XYL$ , so $O_2, I, G$ are collinear. (reminder that $G$ was the center of homothety sending $XYL$ to $EFD$ ). Furthermore, let $R$ be the orthocenter of $DEF$ . Since $I$ is the orthocenter of $XYL$ , $G, I, R$ are collinear. Finally, inversion at $I$ (claim proved below) shows us the Euler line of the touchpoint triangle coincides with $OI$ , so $O, R, I$ are collinear. So $G, O_2, R, I$ are collinear and we are done. $\square$

For completeness, here is the proof $O, I, R$ are collinear (which is apparently also in the Wikipedia article for Inversive geometry) -- the center of the nine-point circle for $DEF$ lies on it Euler line, so on $RI$ . Also inversion at $I$ preserving the circumcircle of $DEF$ sends its midpoint triangle to $ABC$ , so the centers of $(ABC)$ , the nine-point circle of $(DEF)$ and $I$ are collinear, hence we're done.

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