parallel and perpendicular related, starting with a cyclic ABCD

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0 replies

jlacosta

Apr 2, 2025

0 replies

Arbitrary point on BC and its relation with orthocenter

falantrng 21

N 11 minutes ago by Mapism

Source: Balkan MO 2025 P2

In an acute-angled triangle $ABC$ , $H$ be the orthocenter of it and $D$ be any point on the side $BC$ . The points $E, F$ are on the segments $AB, AC$ , respectively, such that the points $A, B, D, F$ and $A, C, D, E$ are cyclic. The segments $BF$ and $CE$ intersect at $P.$ $L$ is a point on $HA$ such that $LC$ is tangent to the circumcircle of triangle $PBC$ at $C.$ $BH$ and $CP$ intersect at $X$ . Prove that the points $D, X,$ and $L$ lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus

21 replies

falantrng

Yesterday at 11:47 AM

Mapism

11 minutes ago

Integral-Summation Duality

Mathandski 0

31 minutes ago

Source: Friend at school gave it to me

Given a continuous function $f$ such that $f(2x) = 3 f(x)$ and $\int_0^1 f(x) \, dx = 1$ , evaluate $\int_1^2 f(x) \, dx$ .

0 replies

Mathandski

31 minutes ago

0 replies

Fixed point in a small configuration

Assassino9931 3

N 39 minutes ago by dno1467

Source: Balkan MO Shortlist 2024 G3

Let $A, B, C, D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$ . Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$ . Prove that $ZT$ passes through a fixed point independent of $\omega$ .

3 replies

Assassino9931

Yesterday at 10:23 PM

dno1467

39 minutes ago

hard problem

Cobedangiu 12

N an hour ago by Edward_Tur

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

12 replies

Cobedangiu

Apr 21, 2025

Edward_Tur

an hour ago

No more topics!

parallel and perpendicular related, starting with a cyclic ABCD

parmenides51 1

N Sep 21, 2020 by DNCT1

Source: 2017 Saudi Arabia IMO TST III p2

Let $ABCD$ be a quadrilateral inscribed a circle $(O)$ . Assume that $AB$ and $CD$ intersect at $E, AC$ and $BD$ intersect at $K$ , and $O$ does not belong to the line $KE$ . Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively. Let $(I)$ be the circumcircle of the triangle $GKH$ . Let $(I)$ and $(O)$ intersect at $M, N$ such that $MGHN$ is convex quadrilateral. Let $P$ be the intersection of $MG$ and $HN,Q$ be the intersection of $MN$ and $GH$ .
a) Prove that $IK$ and $OE$ are parallel.
b) Prove that $PK$ is perpendicular to $IQ$ .

1 reply

parmenides51

Jul 27, 2020

DNCT1

Sep 21, 2020

parallel and perpendicular related, starting with a cyclic ABCD

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Source: 2017 Saudi Arabia IMO TST III p2

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parmenides51

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#1 h Jul 27, 2020, 10:33 AM

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DNCT1

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DNCT1

#2 h Sep 21, 2020, 5:39 PM • 1 Y

Y by amar_04

solution part a

a) First, we have that $\bigtriangleup KAB\sim\bigtriangleup KDC$ . Hence $G,H$ is the midpoint of segment $AB,CD$ respectively so we have
$\bigtriangleup KAG\sim\bigtriangleup KDH$ so $\angle KGA=\angle KHD$ and since $\angle OGA=\angle OHD=90^o$ so we have $\angle KGO=\angle KHO$ too.Now, since $OGEH$ is cyclic quadrilateral so by angle chasing we have
$\angle (\overline{IK},\overline{GO})=180^o-\angle GKI-\angle KGO=180^o-(90^o-\angle GHK)-\angle KHO=90^o-(\angle KHO-\angle GHK)=90^o-\angle GHO=\angle GHE=\angle GOE=\angle (\overline{OE},\overline{GO})\Rightarrow \overline{IK}\parallel\overline{OE}$ (complete the proof).

solution part b

Let $\overline{DA}\cap\overline{CB}=Z$ . $\overline{ZC}\cap(O),\overline{AB},\overline{DC}=U,V,X,Y$ respectively
Then we easily get that $(E,X,D,C)\overset{AB}{=}(E,V,A,B)=C(E,V,A,B)\overset{ZK}{=}(Z,K,V,X)=-1(1)$ So by Maclaurin we have $\overline{EV}.\overline{EG}=\overline{EA}.\overline{EB}=\overline{ED}.\overline{EC}=\overline{EX}.\overline{EH}$ Implies that $V,G,H,X$ are concyclic.Let $\overline{HG}\cap\overline{ZX}=W$ ,by Gauss's line in quadrilateral $ABCD$ we conclude that $W$ is the midpoint of $ZX$
From $(1)$ and Newton's relation we have $WK^2=\overline{WV}.\overline{WX}=\overline{WG}.\overline{WH}$ So $\overline{ZK}$ is tagent to $(GKH)$ at $K$ .Hence $Z$ is the union point with $K$ WRT $(O)$ so we have that $(Z,K,U,Y)=-1$
$\Rightarrow WK^2=\overline{WU}.\overline{WY}=\overline{WG}.\overline{WH}$ so we get that $W$ lies on the radical axis of $(GHK)$ and $(O)$ or $\overline{N,M,W}$ this implies that in the circumscribed $(GHK)$ we have $\overline{KK},\overline{NM},\overline{HG}$ are concurrent at $W$ or $W\equiv Q$ .
Let $QS$ be the secend tagents to $(GHK)$ with $S$ lies on $(GHK)$
Let $\overline{KS}\cap \overline{GH},\overline{MN},\overline{GN}=X_1,Y_1,T$ respectively hence we have $GKHS,MKNS$ are harmonic so we get that $K(K,S,G,H)\overset{QH}{=}(Q,X_1,G,H)=(Q,Y_1,M,N)=-1\Rightarrow \overline{SK}\cap\overline{MG}\cap\overline{NH}=P$ and hence $N(H,G,K,S)\overset{PK}{=}(P,K,T,S)=-1$ so we can easily get that $\overline{PK},\overline{GM},\overline{NH}$ are concurrent at $T$
By Brocard's theorem on cyclic quadrilateral $MGHN.PQ$ we get that $\overline{PK}\bot\overline{IQ}$ (complete the proof).

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