Parallelogram and a simple cyclic quadrilateral

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Parallelogram and a simple cyclic quadrilateral

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Source: DGO 2023 Individual P1

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Noob_at_math_69_level

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Noob_at_math_69_level

#1 h Dec 18, 2023, 5:23 PM • 2 Y

Y by MathLuis, Rounak_iitr

Let $ABC$ be an acute triangle with point $D$ lie on the plane such that $ABDC$ is a parallelogram. $H$ is the orthocenter of $\triangle{ABC}.$ $BH$ intersects $CD$ at $Y$ and $CH$ intersects $BD$ at $X.$ The circle with diameter $AH$ intersects the circumcircle of $\triangle{ABC}$ again at $Q.$ Prove that: The circumcircle of $\triangle{XQY}$ passes through the reflection point of $D$ over $BC.$

Proposed by MathLuis

This post has been edited 1 time. Last edited by Noob_at_math_69_level, Dec 18, 2023, 7:40 PM

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#2 h Dec 18, 2023, 8:07 PM • 1 Y

Y by MathLuis

If $Z = BC \cap XY$ then $Z$ has equal power with respect to $(XBCY), (BQAD'C)$ , so it suffices to show that $Z, Q, D'$ are collinear, where $D'$ is the reflection of $D$ over $BC$ . If $J = DH \cap XY$ then by Ceva-Menelaus, $(Z, J; B, C) = -1$ . $QH$ passes through the midpoint of $BC$ , and the $A$ -antipode wrt $(ABC)$ , so if $H'$ is the reflection of $H$ over $BC$ then $QBH'C$ is harmonic, and $(Z, J; B, C) \overset D'= (D'Z \cap (ABC), H'; B, C) \implies D', Q, Z$ are collinear as desired. $\square$

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#3 h Dec 18, 2023, 8:57 PM • 1 Y

Y by Infinityfun

Definitions: $AD\cap ABC=E$ , $A'$ is the antipode of $A$ . $D'$ is the reflection of $D$ over $BC$ .

$Q,H,M,A'$ are collinear. $\angle XBY= \angle XCY= 90^{\circ} \implies XBCY$ is cyclic. If we prove that $XY, BC, QD'$ is concurrent at $Z$ , then $ZX \cdot ZY = ZB \cdot ZC= ZQ \cdot ZD'$ gives the desired result.

To prove it, note that by butterfly theorem, midpoint of segment $(QE\cap BC)(AA'\cap BC)$ is $M$ . In addition, $HDA'A$ is a parallelogram so $M$ is the midpoint of segment $(HD \cap BC)(AA'\cap BC)$ . Thus $HD, QE,BC$ are concurrent.

We have $(XY \cap BC,DH \cap BC; B,C)=-1$ (well-known Ceva + Menelaus config) and $-1=(\infty_{\parallel BC}, M;B,C)=(D',E;B,C)=(QE \cap BC, QD'\cap BC; B,C)$ . Because $HD \cap BC = QE\cap BC$ , we get $XY \cap BC=QD'\cap BC \implies XY, BC, QD'$ is concurrent.

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#5 h Dec 20, 2023, 6:33 PM • 2 Y

Y by Noob_at_math_69_level, MathLuis

A non-projective solution (ft. spiral)
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Let $L = SD \cap BC$ where $D$ is intersection of $(ABC)$ with line through $A$ parallel to $BC$ . Notice that it suffices to prove that $Y-Z-L$ To prove this Notice the spiral similarity at $S :$ $\triangle SBL \sim \triangle SAC$ and $\triangle SKL \sim \triangle SEB \sim \triangle SFC \sim \triangle SAD$ then do some ratio chasing to get the desired conclusion.

This post has been edited 1 time. Last edited by math_comb01, Dec 20, 2023, 6:34 PM

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#6 h Dec 29, 2024, 2:04 PM • 1 Y

Y by ehuseyinyigit

Let $Q'$ be the reflection of $A-$ queue point over $BC$ which lies on $(DBCH)$ . Since $H$ is the orthocenter of $\triangle DXY$ , $Q'$ is $D-$ queue point on $\triangle DXY$ . $QD',Q'D,BC$ are concurrent and $XY,BC,Q'D$ concur thus, power of point at the concurrency point finishes. $\blacksquare$

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awesomeming327.

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awesomeming327.

#7 h Yesterday at 9:29 PM

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Note that $\angle HBD =\angle HBC+\angle CBD=\angle HBC+\angle BCA=90^\circ$ and similarly $\angle HCD=90^\circ$ . Thus, $H$ is the orthocenter of $\triangle DXY$ .

Let $Q'$ be the Queue Point of $\triangle DXY$ .

Claim 1: $Q$ and $Q'$ are reflections across $BC$ .

Let $H'$ be the reflection of $H$ across $BC$ . Then $\triangle H'BC\stackrel{+}{\sim}\triangle HXY$ , so there is a spiral similarity from $Q$ taking $\triangle H'BC$ to $HXY$ . Thus, since $Q'H$ bisects $XY$ , $Q'H'$ must bisect $BC$ . Therefore, if $Q''$ is the reflection of $Q'$ across $BC$ then $Q''$ lies on $(ABC)$ and also $HM$ where $M$ is the midpoint of $BC$ , so $Q=Q''$ , meaning that $Q'$ is the reflection of $Q$ across $BC$ .

Note that by Miquel Point of a cyclic quadrilateral, $XY$ , $BC$ , and $Q'D$ concur. $QD'$ is the reflection of $Q'D$ across $BC$ , so it is also concurrent with the other lines. By Radical Center we are done.

This post has been edited 1 time. Last edited by awesomeming327., Yesterday at 9:35 PM

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