Let \(ABCDE\) be a regular pentagon. Let \(P\) be a variable point on the interior of segment \(AB\) such that \(PA\ne PB\). The circumcircles of \(\triangle PAE\) and \(\triangle PBC\) meet again at \(Q\). Let \(R\) be the circumcenter of \(\triangle DPQ\). Show that as \(P\) varies, \(R\) lies on a fixed line. Proposed by Karthik Vedula
Problem
Source: ELMO Shortlist 2023 G1
Tags: Elmo, geometry
29.06.2023 05:47
Actually a very nice problem.
30.06.2023 11:32
Let $R$ be the intersection of lines $AE$ and $BC$. Point $Q$ lies on $(CRE)$ due to $\angle CQE = \angle CBA + \angle BAE = \angle CRE$. Clearly $R$ is the radical centre of circles $(PAE), (CBP),$ and $(ABCDE)$, hence $R$ lies on $PQ$. Now $R$ is also the radical centre of circles $(DPQ)$, $(PAE)$ and $(ABCDE)$, so the second intersection $K$ of circles $(DPQ)$ and $(ABCDE)$ lies on $DR$. Since $DR$ and $(ABCDE)$ are fixed, so is $K$. The circumcentre then lies on the bisector of $DK$, which is fixed.
30.06.2023 13:44
The internal angles of this pentagon are all $72^{\circ}$. Let $T$ be the intersection of lines $AE$ and $BC$. Let $\omega$ be the circumcircle of triangle $TEC$. We claim that $Q\in\omega$. Note that $\angle CTE=36^{\circ}$ and $$\angle EQP=\angle EAP=72^{\circ}=\angle PBC=\angle PQC\Longrightarrow\angle EQC=144^{\circ}=\angle ETC\Longrightarrow Q\in\omega.$$Note that since $TA\cdot TE=TB\cdot TC$, point $T$ has same power w.r.t circles $(PAE)$ and $(PBC)$. So, $T,P,Q$ are collinear. Moreover, $TP\cdot TQ=TA\cdot TB$. Let $F$ be the center of $\omega$. Then,$$\angle CFE=2\angle CTE=72^{\circ}=\angle CDE.$$So, $F\in(CDE)$. Since $ABCDE$ is regular, it is cyclic as well. So, $F\in(ABCDE)$. Moreover, points $T,F,D$ all lie on the perpendicular bisector of $AB$. So, $$TF\cdot TD=TA\cdot TE=TP\cdot TQ.$$We conclude that $F\in(DPQ)$. So, $R$ lies on the perpendicular bisector of $FD$, which is the required (fixed) line.
30.06.2023 22:37
Let $M$ be the midpoint of small arc $AB$. Then note that $DPQM$ is cyclic so the circumcenter of $\triangle DPQ$ is just the circumcenter of $\triangle DPM$ which clearly lies on the perpendicular bisector of $DM$.
01.07.2023 23:51
I claim the fixed line is the one passing through the center of $(ABCDE)$ parallel to $\overline{AB}$. Let $X \neq D:=(DPQ) \cap (ABCDE)$. First, by radical center on $(APE)$, $(BCP)$, and $(ABCDE)$, we find that $\overline{AE}, \overline{BC}, \overline{PQ}$ concur. Then, by radical center on $(APE), (DPQ), (ABCDE)$, we find that $\overline{AE}, \overline{PQ}, \overline{DX}$ concur, so these two concurrency points must be identical. Then by symmetry it follows that $X$ is the point on $(ABCDE)$ diametrically opposite $D$, and the conclusion follows. $\blacksquare$ Remark: $P$ lying on $\overline{AB}$ is completely irrelevant lol
06.07.2023 09:26
Let $X=\overline{AE}\cap\overline{BC}$, $\omega_B=(BCQP)$, $\omega_E=(AEQP)$, $\omega=(ABCDE)$, and $\gamma=(DPQ)$. Note by radical axis on $\omega_E$, $\omega_B$, and $\omega$ we have $X\in\overline{PQ}$. Also, radical axis on $\gamma$, $\omega$, $\omega_B$ yields $X\in\overline{DY}$ where $Y=\omega\cap\gamma$. Hence, $R$ lies on the perpendicular bisector of $\overline{DY}$, which is fixed. $\square$
23.09.2023 07:44
Solution from Twitch Solves ISL: In fact, the condition that $ABCDE$ is regular is needlessly strong. The conclusion holds for any cyclic pentagon. [asy][asy] size(12cm); pair M = dir(260); pair D = dir(90); pair A = dir(240); pair E = dir(170); pair B = dir(300); pair T = extension(E, A, M, D); pair O = origin; pair C = -B+2*foot(O, T, B); draw(unitcircle); filldraw(A--B--C--D--E--cycle, invisible, blue); filldraw(unitcircle, invisible, blue); draw(A--T--B, blue); draw(D--T, heavycyan); pair P = 0.65*B+0.35*A; pair Q = -P+2*foot(circumcenter(E, A, P), P, T); draw(circumcircle(E, A, P), deepgreen); draw(circumcircle(B, C, P), deepgreen); filldraw(circumcircle(D, P, Q), invisible, red); pair R = circumcenter(D, P, Q); draw(T--Q, dashed); draw(R--O, lightred+dashed); dot("$M$", M, dir(315)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$E$", E, dir(E)); dot("$B$", B, dir(B)); dot("$T$", T, dir(T)); dot("$O$", O, dir(0)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); M 315 = dir 260 D = dir 90 A = dir 240 E = dir 170 B = dir 300 T = extension E A M D O 0 = origin C = -B+2*foot O T B unitcircle A--B--C--D--E--cycle / 0.1 cyan / blue unitcircle / 0.1 lightcyan / blue A--T--B / blue D--T / heavycyan P = 0.65*B+0.35*A Q = -P+2*foot (circumcenter E A P) P T circumcircle E A P / deepgreen circumcircle B C P / deepgreen circumcircle D P Q / 0.1 palered / red R = circumcenter D P Q T--Q / dashed R--O / lightred dashed */ [/asy][/asy] Claim: $D$, $P$, $Q$, $M$ are cyclic. Proof. Let $T \coloneqq \overline{AE} \cap \overline{BC}$ and let $\overline{DT}$ meet the circumcircle of the pentagon again at $M$. Since $TA \cdot TE = TB \cdot TC$, it follows $T$ lies on the radical axis $\overline{PQ}$. Then $TP \cdot TQ = TM \cdot TD$. $\blacksquare$ Hence $R$ lies on the perpendicular bisector of $\overline{DM}$, which does not depend on the choice of $P$ and is thus the desired fixed line.
09.05.2024 18:55
Let $X$ be the intersection of lines $EA$ and $BC$, and let $M$ be the midpoint of minor arc $AB$ on $(ABCDE)$. Since $XA \cdot XE = XB \cdot XC$, we have that $X$ lies on the radical axis of $(AEPQ)$ and $(BCPQ)$, which is line $PQ$. So, \[XP \dot XQ = XA \cdot XE = XM \cdot XD,\]which implies $PQDM$ is cyclic. Therefore, the center of $(PQDM)$ always lies on the perpendicular bisector of line $DM$.
09.12.2024 18:31
Let $X$ be the intersection of lines $AE$ and $BC$, and let $M$ be the intersection of circles of $ABCDE$ and $PQD $ Observe $ XA.XE = XP.XQ = XM.XD$. Since$ D $ is fixed it means that $M$ is fixed.( It is the midpoint of $arc$ $ AB$). Which implies that the radical axis is fixed which implies that $R$ varies as a straight line since center of$ ABCDE $ is fixed