AoPS - Problem

Source: 6th Iranian Geometry Olympiad (Intermediate) P4

Tags: IGO, Iran, geometry

Dadgarnia

21.09.2019 00:43

Let $ABCD$ be a parallelogram and let $K$ be a point on line $AD$ such that $BK=AB$. Suppose that $P$ is an arbitrary point on $AB$, and the perpendicular bisector of $PC$ intersects the circumcircle of triangle $APD$ at points $X$, $Y$. Prove that the circumcircle of triangle $ABK$ passes through the orthocenter of triangle $AXY$. Proposed by Iman Maghsoudi

Pathological

21.09.2019 04:54

Let $T = PC \cap (\triangle APD)$ and $T'$ be the orthocenter of $\triangle AXY.$ Let $AT' \cap (\triangle APD) = T_1.$ We claim that $ATCT'$ is a parallelogram. Indeed, $AT' \perp XY$ and $PC \perp XY$ yield that $TC || AT'.$ We'll now show that $AT || CT'.$ First of all, note that since $ATPT_1$ is cyclic with $AT_1 || PT$, we've that $ATPT_1$ is an isosceles trapezoid. Since $T'$ is the orthocenter of $\triangle AXY$, it's also well-known that $T', T_1$ are reflections over $XY.$ Since $P, C$ are also reflections over $XY$, we've that $T'CPT_1$ is an isosceles trapezoid. These are enough to imply that $AT || CT',$ and so we have that $ATCT'$ is a parallelogram. Since $\angle DTC = 180 - \angle PTD = 180 - \angle PAD = \angle BAD$, we have by symmetry that $\angle BT'A = \angle DAB.$ This implies that $\angle BT'A = \angle BAD = \angle BAK = \angle BKA,$ and so hence $BT'KA$ is cyclic. $\square$

GammaBetaAlpha

21.09.2019 04:56

When can we expect the results of igo??

InternetPerson10

21.09.2019 06:12

GammaBetaAlpha wrote: When can we expect the results of igo?? Last year, the results were released a bit over one week after the problems were released, so expect the results near Sep 27-28.

Euler365

05.12.2019 18:44

We first see that the problem can be restated as follows: Let $X,Y$ be 2 arbitrary pts. on $\odot(APD)$ and let $C$ be reflection of $P$ in $XY$. Let $B$ be s.t. $ABCD$ is a parallelogram. Let $K\in AD$ s.t. $AB=BK$. Then $\odot (ABK)$ passes through the orthocentre of $\triangle AXY$. We'll use complex no.s. Lets set up the complex system. Let $\odot (APD)$ be the unit circle with centre at the origin. Also let wlog $p=1$. $C$ is reflection of $P$ in $XY$. $\therefore c=x+y-xy\overline{p}=x+y-xy$ Now $ABCD$ is a parallelogram. $\therefore a+c=b+d$. So we obtain $b=a+x+y-xy-d$ Now let $Z$ be foot of $\perp$ from $B$ onto $AD$. Then $z=\frac{b+a+d-ad\overline{b}}{2}$. Note that $Z$ is midpoint of $KA$. $\therefore k=b+d-ad\overline{b}$ Now let orthocentre of $\triangle AXY$ be $H$. Then $h=a+x+y$. We want to prove that $A,B,K,H$ are concyclic. This is true iff $\frac{(h-b)(k-a)}{(h-a)(k-b)}={\frac{\overline{(h-b)}\overline{(k-a)}}{\overline{(h-a)}\overline{(k-b)}}}$ This can be verified by brute force by plugging in expressions for $a,b,k,h$ in terms of $a,x,y,d$.

AlastorMoody

05.12.2019 21:26

IGO Intermediate 2019 P4 wrote: Let $ABCD$ be a parallelogram and let $K$ be a point on line $AD$ such that $BK=AB$. Suppose that $P$ is an arbitrary point on $AB$, and the perpendicular bisector of $PC$ intersects the circumcircle of triangle $APD$ at points $X$, $Y$. Prove that the circumcircle of triangle $ABK$ passes through the orthocenter of triangle $AXY$. Proposed by Iman Maghsoudi Solution: Let the perpendicular from $A$ to $XY$ Intersect $\odot (ABK)$ $=$ $H$ and $\odot (AXY)$ $=$ $H'$. Let $\odot (ABK)$ $\cap$ $\odot (PCB)$ $=$ $E$, $\odot (PCB)$ $\cap$ $BK$ $=$ $F$ and $\odot (PCB)$ $\cap$ $CD$ $=$ $R$. Clearly, $R$ $\in$ $\odot (APD)$. $$\angle CFB=\angle CPB=\angle HAB=\angle HKB \implies HK||CF$$By Converse of Reim's Theorem, $E \in HC$ and $E \in AR$. 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Mahdi_Mashayekhi

29.12.2021 15:57

Nice problem Let AF be altitude of AXY. Let H be intersection of AKB and AF. Let CP meet APD at G and AH meet APD at S and meet CD at J. XF is perpendicular to both PC and AJ so AJ || PC. we will prove S is reflection of H about XY. step1 : ABH and CGD are congruent. Notice that APCJ is parallelogram. ∠CGD = 180 - ∠GDA = 180 - ∠BAK = 180 - ∠BKA = ∠BHA ∠GCD = ∠HAB AB = CD step2 : AGCH is parallelogram. AH = CG and AH || CG. step3 : PHSC is isosceles trapezoid. CH = AG = PS ---> CH = PS and PC || HS. so S is reflection of H about XY. we're Done.

lian_the_noob12

06.09.2023 17:17

Pathological wrote: Let $T = PC \cap (\triangle APD)$ and $T'$ be the orthocenter of $\triangle AXY.$ Let $AT' \cap (\triangle APD) = T_1.$ We claim that $ATCT'$ is a parallelogram. Indeed, $AT' \perp XY$ and $PC \perp XY$ yield that $TC || AT'.$ We'll now show that $AT || CT'.$ First of all, note that since $ATPT_1$ is cyclic with $AT_1 || PT$, we've that $ATPT_1$ is an isosceles trapezoid. Since $T'$ is the orthocenter of $\triangle AXY$, it's also well-known that $T', T_1$ are reflections over $XY.$ Since $P, C$ are also reflections over $XY$, we've that $T'CPT_1$ is an isosceles trapezoid. These are enough to imply that $AT || CT',$ and so we have that $ATCT'$ is a parallelogram. Since $\angle DTC = 180 - \angle PTD = 180 - \angle PAD = \angle BAD$, we have by symmetry that $\angle BT'A = \angle DAB.$ This implies that $\angle BT'A = \angle BAD = \angle BAK = \angle BKA,$ and so hence $BT'KA$ is cyclic. $\square$ How did you claim them? I coud find anything seeing the figure Why you thought these claims? what are the ideas or motivations?