2019 IGO Intermediate P4

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Source: 6th Iranian Geometry Olympiad (Intermediate) P4

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164 posts

#1 h Sep 20, 2019, 9:43 PM • 2 Y

Y by Adventure10, Mango247

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

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578 posts

#2 h Sep 21, 2019, 1:54 AM • 4 Y

Y by Purple_Planet, Greenleaf5002, Neo-Pythagorean, Adventure10

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.

$\square$

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464 posts

#3 h Sep 21, 2019, 1:56 AM • 1 Y

Y by Adventure10

When can we expect the results of igo??

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InternetPerson10

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InternetPerson10

#4 h Sep 21, 2019, 3:12 AM • 1 Y

Y by Adventure10

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.

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142 posts

#5 h Dec 5, 2019, 3:44 PM • 2 Y

Y by Purple_Planet, Adventure10

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 $a,x,y,d$ .

This post has been edited 1 time. Last edited by Euler365, Dec 5, 2019, 3:47 PM

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#6 h Dec 5, 2019, 6:26 PM • 5 Y

Y by a_simple_guy, Math5000, Maxito12345, Purple_Planet, Adventure10

My solution on the test

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#7 h Dec 29, 2021, 12:57 PM

Y by

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.

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lian_the_noob12

173 posts

lian_the_noob12

#8 h Sep 6, 2023, 2:17 PM • 1 Y

Y by David2663

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?

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