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Source: 2024 MODS Geometry Contest, Problem 4 of 6

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387 posts

#1 h Mar 15, 2024, 2:18 AM • 3 Y

Y by ohiorizzler1434, MathLuis, Rounak_iitr

Let $ABCD$ be a parallelogram. Let line $\ell$ externally bisect $\angle DCA$ and let $\ell'$ be the line passing through $D$ which is parallel to line $AC$ . Suppose that $\ell'$ meets line $AB$ at point $E$ and $\ell$ at point $F$ , and that $\ell$ meets the internal bisector of $\angle BAC$ at point $X$ . Further let circle $EXF$ meet line $BX$ at point $Y \neq X$ and the internal bisector of $\angle DCA$ meet circle $AXC$ at point $Z \neq C$ .
Prove that points $D$ , $X$ , $Y$ , and $Z$ are concyclic.

Proposed by squarc_rs3v2m

This post has been edited 1 time. Last edited by Scilyse, Sep 26, 2024, 8:17 AM

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59 posts

#2 h Mar 16, 2024, 5:02 AM

Y by

Underrated problem

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387 posts

#3 h Mar 16, 2024, 6:39 AM • 1 Y

Y by ehuseyinyigit

I disagree. This is a barbarically boorish, awfully appalling, incredibly indecent and unacceptably unbecoming problem!

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129 posts

#4 h Mar 16, 2024, 8:30 AM

Y by

yooo! the trauma problem :skull:

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75 posts

invt

#5 h Oct 23, 2024, 3:58 PM

Y by

bump. does anyone have solution?

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523 posts

#6 h Feb 22, 2025, 3:54 PM • 1 Y

Y by MS_asdfgzxcvb

I agree that this problem is little underrated since it has a rather clean sol using lin pop hahaha. Let $G$ be the point on $AB$ that $CG \perp AB$ and $M$ be the midpoint of $AC$ . We can easily see that $AX \perp XC$ and $Z$ is the reflection of $X$ wrt $M$ . We fix rectangle $AXCZ$ then $B$ lies on the reflection of $AC$ wrt $AX$ , and $D$ is the reflection of $B$ wrt $M$ . Also, it is easy to see that $DE = AC$ . Since $EF \parallel AC$ , we can easily see from reim, that $EFXG$ are concyclic. Now, it is enough to prove that $B$ lies on the radical axis of $(DXZ)$ and $(EGXF)$ . We define a function $f, g : \mathbb{R}^2 \to \mathbb{R}$ such that for any point $P$ on the plane, where $(M)$ is considered as a degenerate circle.
$f(P) = \text{Pow}_{(EGXF)}(P) - \text{Pow}_{(DXZ)}(P) \quad \text{ and } \quad g(P) = \text{Pow}_{(EGXF)}(P) - \text{Pow}_{(M)}(P)$ It is known that $f$ is linear.
$[asy] import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.987454545454577, xmax = 10, ymin = -5.2052727272727175, ymax = 11.558363636363643; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-5.7972189915724694,0.6186385583434268)--(-5.611826580758478,0.9568544429365202)--(-5.950042465351571,1.1422468537505122)--(-6.135434876165563,0.8040309691574188)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw(circle((-2.131,1.154), 4.019698620543585), linewidth(1) + blue); draw((-3.99,4.718)--(-0.272,-2.41), linewidth(1)); draw((-4.0631546184616365,-2.3708766688151477)--(-0.1988453815383635,4.678876668815148), linewidth(1)); draw((-0.1988453815383635,4.678876668815148)--(-3.99,4.718), linewidth(1)); draw((-3.99,4.718)--(-4.0631546184616365,-2.3708766688151477), linewidth(1)); draw((-0.272,-2.41)--(-0.1988453815383635,4.678876668815148), linewidth(1)); draw((xmin, 1.8243243243243246*xmin + 11.997054054054054)--(xmax, 1.8243243243243246*xmax + 11.997054054054054), linewidth(1)); /* line */ draw((xmin, -0.010319634813714826*xmin-2.412806940669331)--(xmax, -0.010319634813714826*xmax-2.412806940669331), linewidth(1)); /* line */ draw((-6.826455825524731,-0.45661535737619996)--(2.564455825524731,2.7646153573762), linewidth(1)); draw((-6.135434876165563,0.8040309691574182)--(-0.272,-2.41), linewidth(1)); draw(circle((0.6709285911209192,2.982834714254523), 7.146591453865348), linewidth(1) + qqwuqq); draw((-6.826455825524731,-0.45661535737619996)--(-0.272,-2.41), linewidth(1)); draw((2.564455825524731,2.7646153573762)--(-3.99,4.718), linewidth(1)); draw((-0.272,-2.41)--(2.564455825524731,2.7646153573762), linewidth(1)); draw((-1.1535441744752688,9.8926153573762)--(5.293518670323751,-2.4674341202266534), linewidth(1)); /* dots and labels */ dot((-3.99,4.718),linewidth(4pt) + dotstyle); label("$A$", (-4.605636363636391,5.19472727272728), NE * labelscalefactor); dot((-0.272,-2.41),linewidth(4pt) + dotstyle); label("$C$", (-0.2056363636363876,-3.2052727272727175), NE * labelscalefactor); dot((-2.131,1.154),linewidth(4pt) + dotstyle); label("$M$", (-2.9147272727272986,1.0856363636363723), NE * labelscalefactor); dot((-0.1988453815383635,4.678876668815148),linewidth(4pt) + dotstyle); label("$Z$", (0.06709090909088533,5.067454545454553), NE * labelscalefactor); dot((-4.0631546184616365,-2.3708766688151477),linewidth(4pt) + dotstyle); label("$X$", (-4.532909090909118,-3.1143636363636267), NE * labelscalefactor); dot((-6.135434876165563,0.8040309691574182),linewidth(4pt) + dotstyle); label("$G$", (-6.82381818181821,0.7947272727272815), NE * labelscalefactor); dot((-6.826455825524731,-0.45661535737619996),linewidth(4pt) + dotstyle); label("$B$", (-7.5692727272727565,-0.478), NE * labelscalefactor); dot((2.564455825524731,2.7646153573762),linewidth(4pt) + dotstyle); label("$D$", (2.9034545454545237,2.90381818181819), NE * labelscalefactor); dot((-1.1535441744752688,9.8926153573762),linewidth(4pt) + dotstyle); label("$E$", (-1.5329090909091159,10.322), NE * labelscalefactor); dot((5.293518670323751,-2.4674341202266534),linewidth(4pt) + dotstyle); label("$F$", (5.358,-3.2234545454545356), NE * labelscalefactor); dot((-7.854309236923274,-2.3317533376302957),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Now, from $f$ linear, we have that $f(B) + f(D) = 2f(M)$ and $2g(M) = g(A) + g(C)$ . From $\text{Pow}_{(DXZ)}(M) = - MX^2$ and $\text{Pow}_{(M)}(A) = \text{Pow}_{(M)}(C) = MA^2 = MC^2 = MX^2$ , we have that $\begin{aligned} 2g(M) + f(B) + f(D) &= g(A) + g(C) + 2f(M) \\ &= 2\text{Pow}_{(EGXF)}(M) + \text{Pow}_{(EGXF)}(A) + \text{Pow}_{(EGFX)}(C) \\ &= 2g(M) + AG \cdot AE + CX \cdot CF \end{aligned}$ Therefore, we obtain that $f(B) + DE \cdot DF = AG \cdot AE + CX \cdot CF$ . Since $DE = CA$ is a constant and $CX, AG$ are constant, and when $D$ moves on the ray $CD$ linearly, the lenghts $DF, CF, AE$ are proportional with $CD$ , by checking one case of $D$ , we can prove that $f(B) = 0$ holds for any time. Since when $D$ is the reflection of $A$ wrt $Z$ , the problem is very straightforward, we proved that $f(B) = 0$ holds for any choice of $D$ one ray $CD$ . Hence, the desired conclusion follows.

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#7 h Apr 2, 2025, 12:35 PM • 1 Y

Y by alexanderhamilton124

Let $AC\cap BD=M$ (center of $(AXC)$ ) and let $T,R$ be the foot of perpendiculars from $C$ to $AB$ and $Z$ to $BC$ .

Note that $R$ is actually the midpoit of $BY$ because $\frac{TB.BE}{TB.BA}=\frac{XB.BY}{XB.BR}=\frac{1}{2}$ .

It is easy to see that $AX\perp CX$ so we get that $(ARXCZ)$ is cyclic and since $\angle REF=\angle RAC= 180^{\circ}-\angle RXF$ we get that $(REXF)$ is also cyclic.

Since $(ARXCZ)$ has diameter $AC$ , $XCZA$ is a rectangle and because $M$ is the midpoint of $XZ$ we have $M$ as the midpoint of $BD$ and we get that $\angle XDZ=\angle XBZ$ and $\angle XBZ=\angle XYZ$ .

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