Inscribed pentagon

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Dadgarnia

164 posts

Dadgarnia

#1 h Aug 14, 2019, 7:29 PM • 1 Y

Y by Adventure10

Given an inscribed pentagon $ABCDE$ with circumcircle $\Gamma$ . Line $\ell$ passes through vertex $A$ and is tangent to $\Gamma$ . Points $X,Y$ lie on $\ell$ so that $A$ lies between $X$ and $Y$ . Circumcircle of triangle $XED$ intersects segment $AD$ at $Q$ and circumcircle of triangle $YBC$ intersects segment $AC$ at $P$ . Lines $XE,YB$ intersects each other at $S$ and lines $XQ, Y P$ at $Z$ . Prove that circumcircle of triangles $XY Z$ and $BES$ are tangent.

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bjh0411

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bjh0411

#2 h Aug 15, 2019, 10:31 AM • 5 Y

Y by guptaamitu1, Infinityfun, Adventure10, Mango247, Fatemeh06

Assume the circumcircles of $ABY$ and $AEX$ meet at $A$ and $K$ .
$\angle KEX=\angle KAX=180^\circ-\angle KAY=180^\circ-\angle KBY=\angle KBS$ .
So, $(KBES)$ is concyclic.
$\angle YKB=\angle YAB=\angle BCA=\angle PYB$
$\angle EKX=\angle EAX=\angle ADE=\angle QXE$
In solution, $\angle YZX=\angle YSX-\angle SYZ-\angle SXZ=180^\circ-(\angle BKE+\angle YKB+\angle EKX)=180^\circ-\angle YKX$
So, $(YKXZ)$ is concyclic. Let this circle $\Omega$ .
Let's say the tangent of $\Omega$ at $K$ meet $XY$ at $T$ .
$\angle TKB=\angle TKY+\angle YKB=\angle KXT+\angle YAB=\angle KEA+\angle AEB=\angle KEB$
As a result, $TK$ is a tangent of the circumcircle of $KEB$ .
So, the to circles are tangent at $T$ .

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huricane

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huricane

#3 h Aug 15, 2019, 10:32 AM • 2 Y

Y by Adventure10, Mango247

Kind of easy for #3.

Invert in $A$ . Then $\overline{A-X-Y}\parallel\overline{B-C-D-E-F}.$ $Q$ becomes the second intersection of $AD$ with $(XED)$ , $P$ becomes the second intersection of $AC$ and $(YBC)$ , while $Z$ becomes the second intersection of $(XAQ)$ and $(AYP)$ and $S$ becomes the second intersection of $(AEX)$ and $(ABY).$ Take $T=XE\cap YB.$ All angles are oriented.

Note that $\angle{XZA}=\angle{XQA}=\angle{XQD}=\angle{XEB}$ and in the same way $\angle{YZA}=\angle{YBE}$ , so we can easily see that $T\in (XYZ)$ .
Also note that $\angle{ESA}=\angle{EXA}=\angle{XEB}$ and in the same way $\angle{BSA}=\angle{YBE}$ , so we can easily see that $T\in (BES).$
The porblem thus reduces to showing that $(TXY)$ and $(TBE)$ are tangent, which is clear since $XY\parallel BE.$

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GeoMetrix

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GeoMetrix

#4 h Nov 14, 2019, 4:06 PM • 1 Y

Y by Adventure10

Nice problem.We'll prove a more general problem that is we'll prove this problem for any position of $X,Y$ .
solution

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Idio-logy

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Idio-logy

#5 h Jan 29, 2020, 8:14 PM • 1 Y

Y by Adventure10

Sketch

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alinazarboland

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alinazarboland

#6 h May 24, 2022, 11:41 AM

Y by

At first I was like " A pentagon with lot of points on it and a tangency from nowhere?!" but I've got better when it turns $C,D$ are actually unnecessary; In fact , by changing $C,D$ over $(ABCDE)$ and fixing $X,Y,A,B,E$ , not only the tangency point but the whole $(XYZ) , (BES)$ remain unchanged...
We have $\angle EXZ = \angle EDA = \angle EAX$ . Similarly , $\angle BYZ = \angle BAY$ .So $XZ,YZ$ are tangent to $(AXE),(AYB)$ respectively. Let $T = (AYB) \cap (AXE)$ . Now $\angle TXZ + \angle TYZ = \angle TAX + \angle TAY = \pi$ and $TXYZ$ is cyclic. Also , $\angle BTE = \angle BTA + \angle ATE =\angle AYS+\angle AXS = \pi - \angle BSE$ which implies that $TBSE$ is cyclic. So $T = (BSE) \cap (XYZ)$ . Now the angle between $TX$ and the tangent through $T$ to $(TXZY)$ is $\angle TYX = \angle TBA$ and the angle between $TB$ and the tangent through $T$ to $(TBSE)$ is $\angle TEB = \angle TEA + \angle AEB$ . Now since $\angle TXY + \angle TYX = \angle XZY$ we have the difference between the recent values is in fact $\angle XTB$ which implies that the tangents through $T$ are equal lines and we're done.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#7 h Jul 17, 2022, 12:29 PM

Y by

Claim $: ZY,ZX$ are tangent to $ABY$ and $AEX$ .
Proof $:$ Note that $\angle ZYS = \angle PYS = \angle PCB = \angle ACB = \angle YAB \implies ZY$ is tangent to $ABY$ . we prove the other one with same approach.
Note that $\angle BSE = \angle 180 - (\angle 180 - \angle AYB + \angle 180 - \angle AEX)$ and $\angle YZX = \angle 180 - (\angle ZYX + \angle ZXY) = \angle 180 - (\angle YBA + \angle XEA)$ so if $AEX$ and $ABX$ meet at $R$ we have $BSER$ and $XZYR$ are cyclic so Now we need to show they are tangent at $R$ . Let $N$ be arbitrary point such that $RN$ is tangent to $XZYR$ , we have $\angle ERN = \angle XRN - \angle XRE = \angle XYR - \angle XAE = \angle AYR - \angle EBA = \angle ABR - \angle ABE = \angle EBR \implies ESBR$ is tangent to $RN$ at $R$ .
One can also prove that $S,Z,R$ are collinear.

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kiemsibongtoi

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kiemsibongtoi

#9 h Aug 16, 2024, 6:08 AM

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Dadgarnia wrote:

Given an inscribed pentagon $ABCDE$ with circumcircle $\Gamma$ . Line $\ell$ passes through vertex $A$ and is tangent to $\Gamma$ . Points $X,Y$ lie on $\ell$ so that $A$ lies between $X$ and $Y$ . Circumcircle of triangle $XED$ intersects segment $AD$ at $Q$ and circumcircle of triangle $YBC$ intersects segment $AC$ at $P$ . Lines $XE,YB$ intersects each other at $S$ and lines $XQ, Y P$ at $Z$ . Prove that circumcircle of triangles $XY Z$ and $BES$ are tangent.

Let $X'$ , $Y'$ in order are intersection of lines $AB$ , $AE$ with circle $(SBE)$ ( $X' \neq B$ , $Y' \neq C$ )
First, we'll prove that there exist a homothety centered taking $\triangle X'SY'$ to $\triangle XZY$ :
$\, \,$ Cuz $XY$ is tangent to $\Gamma$ at $A$ so $(XY, AB) = (EA, EB) = (X'Y', X'B)$ (mod $\pi$ ). Lead to $X'Y'$ $\|$ $XY$
$\, \,$ Cuz $Y$ , $P$ , $B$ , $C$ are concyclic, so $(YZ, YB) = (CP, CB) = (EA, EB) = (SY', SB)$ (mod $\pi$ ). Lead to $ZY$ $\|$ $SY'$ . Similar, $ZX$ $\|$ $SX'$
$\, \,$ Therefore, $\triangle SX'Y'$ and $\triangle ZXY$ are similar and oriented in the same way.
$\, \,$ Which means there exist a homothety centered taking $\triangle X'SY'$ to $\triangle XZY$
Thus, If intersection $T$ of lines $XX'$ and $YY'$ lies on circle $(SBE)$ ,
we ez to see that cirles $(SBE)$ , $(ZXY)$ are tangent at $T$
Hence, we just need to prove that intersection of lines $XX'$ and $YY'$ lies on circle $(SBE)$ :
$\, \,$ Let $T$ is the intersection of line $XX'$ and circle $(SBE)$
$\, \,$ Use Pascal theorem for $\binom{S\ X'\ Y'}{T\ E\ B}$ , we see that $Y'$ , $Y$ , $T$ are colinear (Cuz $Y$ is the intersection of lines $AX$ , $SB$ )
Done.[/quote]

Attachments:

iran-mo-r3-2019.pdf (71kb)

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