Cyclic quadrilateral

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Cyclic quadrilateral

G H J

Reveal topic tags

cyclic quadrilateral

L

G H BBookmark kLocked kLocked NReply

Source: Iran MO 3rd round 2016 finals - Geometry P2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

178 posts

bgn

#1 h Sep 5, 2016, 7:25 AM • 3 Y

Y by Davi-8191, anantmudgal09, Adventure10

Given $\triangle ABC$ inscribed in $(O)$ an let $I$ and $I_a$ be it's incenter and $A$ -excenter ,respectively.
Tangent lines to $(O)$ at $C,B$ intersect the angle bisector of $A$ at $M,N$ ,respectively.
Second tangent lines through $M,N$ intersect $(O)$ at $X,Y$ .

Prove that $XYII_a$ is cyclic.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

824 posts

andria

#2 h Sep 5, 2016, 9:11 AM • 5 Y

Y by anhtaitran, Math-Ninja, mrjava, Adventure10, Mango247

Let $BX\cap CY=T,XY\cap BC=D$ .

$\left.\begin{array}{lll} \text{paskal's theorem for}\ YYXBBC\Longrightarrow T,N,D\ \text{are collinear}\\ \\ \text{paskal's theorem for}\ XXYCCB\Longrightarrow T,M,D\ \text{are collinear} \end{array}\right\}\Longrightarrow T,N,M,D\ \text{are collinear}\Longrightarrow D\in II_a$

Hence:

$\left.\begin{array}{ccc} B,Y,C,X\in (O)\Longrightarrow BD.DC=DY.DX\\ \\ C,B,I,I_a\in (\text{circle with diameter} II_a)\Longrightarrow BD.DC=DI.DI_a \end{array}\right\}\Longrightarrow DI.DI_a=DY.DX\Longrightarrow X,Y,I,I_a\ \text{lie on a circle}$
Q.E.D

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1246 posts

#3 h Dec 8, 2016, 6:43 PM • 2 Y

Y by Stupke, Adventure10

Let $AD\cap \odot ABC=\{F\}$ .

$XX,CC,AF$ are concurrent $\implies$ $AXFC$ is harmonic,analogously so is $ABFY$ .Consider an inversion $\Psi$ with center $A$ radius
$\sqrt{AB\cdot AC}$ composed with reflection over $A$ -angle bisctor.Let $\Psi(Z)=Z^*$ .

Obviously $\Psi(B)=C$ and $\Psi(I)=I_A$ .Inversion preserves harmonic ratios, as $(A,F;X,C)=-1$ $\implies$ $(\infty,F^*;X^*,B)=-1$ $\implies$
$F^*$ is a midpoint of $BX^*$ but $F^*\equiv D$ $\implies$ $BD=DX^*$ ,analogously $CD=DY^*$ .

$IBI_AC$ is cyclic $\implies$ $ID\cdot DI_A=BD\cdot CD=DX^*\cdot DY^*$ $\implies$ $IX^*I_AY^*$ is cyclic hence we're done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1979 posts

#4 h Apr 8, 2018, 6:58 PM • 2 Y

Y by Adventure10, Mango247

bgn wrote:

Given $\triangle ABC$ inscribed in $(O)$ an let $I$ and $I_a$ be it's incenter and $A$ -excenter ,respectively.
Tangent lines to $(O)$ at $C,B$ intersect the angle bisector of $A$ at $M,N$ ,respectively.
Second tangent lines through $M,N$ intersect $(O)$ at $X,Y$ .

Prove that $XYII_a$ is cyclic.

Let $D$ be the foot of $A$ -angle bisector and $L$ be the mid-point of arc $BC$ . Then we claim that $\overline{XY}$ passes through $D$ . If true, this tells us that $DX \cdot DY=DA \cdot DL=DI \cdot DI_a$ since $(AD, II_a)=-1$ and $L$ is the midpoint of $II_a$ ; thus proving that $XYII_a$ is cyclic.

Apply inversion at $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in line $\overline{AI}$ . Let $\omega_B, \omega_C$ be the circles through $B', C'$ respectively, tangent to line $B'C'$ and passing through $A$ . Then $M'$ lies at the intersection of $\overline{AI}$ and $\omega_B$ while $N'$ lies at the intersection of $\omega_C$ and $\overline{AI}$ . The other tangents map to the other circles passing through $A$ and touching line $\overline{B'C'}$ . Thus, $L'X'^2=L'M' \cdot L'A=L'B'^2 \implies L'X'=L'B'$ and similarly $L'C'=L'Y'$ . Thus, $L'A' \cdot L'D'=L'B' \cdot L'C'=L'X' \cdot L'Y'$ proving $AX'D'Y'$ cyclic; thus $D$ lies on $\overline{XY}$ .

This post has been edited 1 time. Last edited by anantmudgal09, Apr 8, 2018, 6:59 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

475 posts

Anar24

#5 h Apr 9, 2018, 10:46 AM • 2 Y

Y by tenplusten, Adventure10

Pole-Polar kills this problem by considering the common point of BY and CX along within proving the famous claim(previous Aopsers did that) by Brocard.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

475 posts

#6 h May 20, 2018, 6:37 AM • 2 Y

Y by Adventure10, Mango247

Let $XY\cap BC=D$ , but consider complete quadrilateral $XYBCD\cup \{Z=BX\cap CY\}$ , $M, N$ lies on the line $DZ$ , so $D$ lies on $AI_a$ . Then with $(BI_aCI), (BCXY)$ we conclude that $XYII_a$ is cyclic by converse of Radical Axis Theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

214 posts

#7 h Nov 13, 2019, 9:50 AM • 2 Y

Y by Myriam2003, Adventure10

Let $S=XY \cap BC$ and $K=AI \cap (ABC)$ and $R=AA \cap KK$
we should prove that $S \in AD$ if it does then
$SX.SY=SB.SC=SI.SI_a$ (the last equality comes from $BCII_a$ is cyclic)
claim:
polar $M=XC$
polar $N=YB$
then the pole of $MN$ is $XC \cap YB$ and also
pole $AK=R \implies R=XC \cap YB$
the new problem :
let $ABKC$ a quadrilateral and Let $R=AA \cap KK$
$X=RB \cap (ABKC)$
$Y=RC \cap (ABCD)$
prove that $BC,AK,XY$ are concurrent

take a projective transformation that fixes $ABKC$ and send $AK \cap BC$ to the center of $(ABKC)$ $O$ then $ABKC$ is a rectangle and
$AA \cap KK=P_\infty$
easily get $X,Y,O$ are collinear

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a