Incenter and concurrency

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Source: 2025 Nepal ptst p3 of 4

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5 posts

#1 h Mar 15, 2025, 2:40 PM

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Let the incircle of $\triangle ABC$ touch sides $BC$ , $CA$ , and $AB$ at points $D$ , $E$ , and $F$ , respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$ , respectively. Prove that the lines $DX$ , $D'Y$ , and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)

This post has been edited 2 times. Last edited by jenishmalla, Mar 15, 2025, 3:00 PM
Reason: formatting

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#2 h Mar 15, 2025, 2:45 PM

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Main idea for nuke

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#3 h Mar 15, 2025, 4:03 PM • 1 Y

Y by ABYSSGYAT

My problem. We present two solutions :love:

:love:

Solution 1: Let lines $D'Y$ and $DX$ intersect at $T$ . Define $I$ as the incenter of $\triangle{ABC}$ and let $Z = DD' \cap AT$ . First, observe that $D'$ is the orthocenter of $\triangle{TAD}.$ Since $IZ \perp AT$ , it follows that $Z$ lies in a circle with diameter $AI$ . Consequently, $AZFIE$ is a cyclic pentagon.

Now consider the radical axes of the circles $(AZFE), (EFXD)$ , and $(AZXD)$ . This implies that $AZ$ , $EF$ , and $DX$ are concurrent at $T$ , meaning $EF$ and $DX$ intersect at $T$ . However, since $YD'$ and $DX$ also intersect at $T$ , it follows that $DX$ , $D'Y$ , and $EF$ are concurrent at $T$ , as desired.

Solution 2: Observe that $FYED$ is a harmonic quadrilateral. Brianchon's Theorem on hexagon $YD'FXDE$ , we see that $YX$ and $D'D$ concur on line $FE$ . Lastly, by Pascal’s Theorem on hexagon $D'YYXDD$ , the desired result follows.

This post has been edited 1 time. Last edited by Thapakazi, Mar 15, 2025, 5:02 PM

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#4 h Apr 7, 2025, 7:25 AM

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Let $G=D'Y\cap XD$
For the complete quadrilateral $D'XDY$ Let $J$ be the center of spiral similarity that sends $D'Y\rightarrow XD$
$\implies$ that $AG\parallel BC$ as $D'XDY$ is cylcic
Let $GE$ meet the incircle again at $F'$
$\implies GE\cdot GF'=GX\cdot GD=GJ\cdot GA$
$\therefore F\equiv F'$

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#5 h Apr 7, 2025, 12:33 PM

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This can also be easily bashed with complex numbers:

Take the incircle of $\Delta ABC$ to be the unit circle.

Then $a=\frac{2ef}{e+f}, b=\frac{2df}{d+f}, c=\frac{2de}{d+e}, d'=-d$ .

$A, D', X$ are collinear $\implies -d'x=\frac{a-d'}{\overline{a}-\overline{d'}}\implies dx=\frac{\frac{2ef}{e+f}+d}{\frac{2}{e+f}+\frac{1}{d}}=d\cdot\frac{2ef+de+df}{2d+e+f}\implies x=\frac{2ef+de+df}{2d+e+f}$ .

$A, D, Y$ are collinear $\implies -dy=\frac{a-d}{\overline{a}-\overline{d}}\implies -dy=d\cdot\frac{2ef-de-df}{2d-e-f}\implies y=\frac{2ef-de-df}{e+f-2d}$ .

Let $P=EF\cap DX\implies p=\frac{ef\cdot\frac{2d^2+2de+2df+2ef}{2d+e+f}-\frac{d(2ef+de+df)}{2d+e+f}(e+f)}{ef-\frac{d(2ef+de+df)}{2d+e+f}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}$ .

Let $Q=EF\cap D'Y\implies q=\frac{ef\cdot\frac{2d^2-2de-2df+2ef}{e+f-2d}+\frac{d(2ef-de-df)}{e+f-2d}(e+f)}{ef+\frac{d(2ef-de-df)}{e+f-2d}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}=p$ , so $P=Q$ and we're done.

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34 posts

#6 h Apr 7, 2025, 2:12 PM

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Trivial by Six-point-line!

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238 posts

#7 h Apr 7, 2025, 11:34 PM

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All pole-polars are w.r.t incircle. It is well known midpoint of $\overline{BC}$ say $M$ is pole of $\overline{DX}$ (and $A$ is pole of $\overline{EF}$ ); hence we just need to prove pole of $\overline{YD'}$ lies on $\overline{AM}$ by La Hire. Now $\overline{AM}$ , $\overline{DD'}$ , $\overline{EF}$ concurrence is well known and since $(YD;EF)=(D'X;EF)=-1$ , we have the concurrency point also lies on $\overline{YX}$ . Now Pascal at $YYDD'D'X$ finishes.

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#9 h Apr 20, 2025, 4:06 PM

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So, like one little solution which i noticed during exam time :blush:

:blush:

Define:
$P = EF \cap AD, \quad Q = EF \cap AD'.$
We claim that lines $DX$ , $D'Y$ , and $EF$ are concurrent.

Note that the cross-ratios satisfy:
$(A, Q; D', X) = (A, P; Y, D) = -1.$
This follows because $P$ and $Q$ are defined by the intersections of $AD$ and $AD'$ with $EF$ , and $(A, P; Y, D)$ and $(A, Q; D', X)$ are harmonic divisions.

Moreover, since $D'$ is the reflection of $D$ across the incenter $I$ of triangle $ABC$ , it follows that $D'$ and $Y$ lie on the same side of $EF$ , and likewise for $D$ and $X$ .

Now, by the **Prism Lemma**, which states that if $(A, P; Y, D) = -1$ and $(A, Q; D', X) = -1$ and if $D'$ and $X$ lie on the same side of $EF$ , then the lines $DX$ , $D'Y$ , and $EF$ concur, the result follows.

REMARK: : we haven't used the property that D is the intouch point of the circle with side BC, and D' is the antipode of D, It holds for general point on the incircle, this property of D opens the path for the nice synthetic solution by radical axis, only required property was that EDFD' harmonic!

This post has been edited 1 time. Last edited by brute12, Apr 20, 2025, 4:13 PM
Reason: just for adding nice remark

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