Parallel Line Makes Tanget Circumcircle

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War-Hammer

670 posts

War-Hammer

#1 h Sep 5, 2013, 7:19 PM • 1 Y

Y by Adventure10

Let $ABCDE$ be a pentagon inscribe in a circle $(O)$ . Let $BE \cap AD = T$ . Suppose the parallel line with $CD$ which passes through $T$ which cut $AB,CE$ at $X,Y$ . If $\omega$ be the circumcircle of triangle $AXY$ then prove that $\omega$ is tangent to $(O)$ .

This post has been edited 2 times. Last edited by War-Hammer, Sep 6, 2013, 10:04 PM

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djmathman

7938 posts

djmathman

#2 h Sep 5, 2013, 8:19 PM • 2 Y

Y by Adventure10, Mango247

EDIT: There was originally a typo in this question. Now that it has been edited, I can go ahead and write a correct proof.

First, note that $\angle AEC=\angle ADC=\angle YTD$ by a combination of cyclic quads and parallel lines, so quadrilateral $AEYT$ must be cyclic. Therefore $\angle AYX=\angle AEB=\angle APB$ and $BP\parallel XY$ . This means that triangles $\triangle AXY$ and $\triangle ABP$ are homothetic with center of homothety $A$ , implying that their circumcircles are homothetic. Therefore, since the circumcircle of $\triangle ABP$ is simply $(O)$ , circles $(O)$ and $\omega$ are internally tangent, as desired. $\blacksquare$

EDIT2: Oops, I forgot the case where $X$ and $Y$ are outside $(O)$ . I'm sure the solution is very similar, although several changes need to be made as a result of configuration issues.

This post has been edited 4 times. Last edited by djmathman, Sep 6, 2013, 2:58 PM

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War-Hammer

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War-Hammer

#3 h Sep 5, 2013, 8:35 PM • 1 Y

Y by Adventure10

Thanks , typo edited.

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sunken rock

4401 posts

sunken rock

#4 h Sep 7, 2013, 3:04 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

As above, $A,T,E,Y$ are concyclic, hence $\angle AYT=\angle AET$ , let $\alpha$ being the common angle value. Seeing that $\angle XAO=\angle BAO=90^\circ-\alpha$ we infer that the circumcenter of $\Delta AXY$ belongs to $(AO$ , so being done.

Best regards,
sunken rock

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acupofmath

325 posts

acupofmath

#5 h Sep 17, 2013, 3:05 PM • 2 Y

Y by Adventure10, Mango247

@sunken : rock very very beautiful!!!!!!

can any body solve with menelaus i tried but i didn't success

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vanu1996

607 posts

vanu1996

#6 h Nov 22, 2013, 2:49 AM • 2 Y

Y by Adventure10, Mango247

Let $AY$ meet the circumcircle of $ABCDE$ at $F$ .clearly $ATYE$ is cyclic but $\angle AEB=\angle AFB$ , hence we get $\angle AYX=\angle AFB$ ,now if we draw a tangent $MA$ to $\omega$ at $A$ then we get $\angle MAX=\angle AYX=\angle AFB$ ,so $MA$ is also tangent to the circumcircle of $ABCDE$ ,hence the result.

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fmasroor

779 posts

fmasroor

#7 h Nov 23, 2013, 3:20 AM • 1 Y

Y by Adventure10

djmathman wrote:

EDIT: There was originally a typo in this question. Now that it has been edited, I can go ahead and write a correct proof.

First, note that $\angle AEC=\angle ADC=\angle YTD$ by a combination of cyclic quads and parallel lines, so quadrilateral $AEYT$ must be cyclic. Therefore $\angle AYX=\angle AEB=\angle APB$ and $BP\parallel XY$ . This means that triangles $\triangle AXY$ and $\triangle ABP$ are homothetic with center of homothety $A$ , implying that their circumcircles are homothetic. Therefore, since the circumcircle of $\triangle ABP$ is simply $(O)$ , circles $(O)$ and $\omega$ are internally tangent, as desired. $\blacksquare$

EDIT2: Oops, I forgot the case where $X$ and $Y$ are outside $(O)$ . I'm sure the solution is very similar, although several changes need to be made as a result of configuration issues.

What's P?

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djmathman

7938 posts

djmathman

#8 h Nov 23, 2013, 3:01 PM • 2 Y

Y by Adventure10, Mango247

Oops. I forgot to mention that $P$ is the second intersection point of $\omega$ and $AY$ .

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sayantanchakraborty

505 posts

sayantanchakraborty

#9 h Apr 7, 2014, 6:33 AM • 1 Y

Y by Adventure10

From $A$ ,draw a tangent $AL$ to circle $(O)$ .

$\angle{TAE}=\angle{DAE}=\angle{DCE}=\angle{CYX}=\angle{CYT} \Rightarrow T,A,Y,E$ concyclic.So $\angle{LAB}=\angle{AEB}=\angle{AET}=\angle{AYT}=\angle{AYX}$ or $AL$ is also tangent to $\omega$ .The result follows..

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AlastorMoody

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AlastorMoody

#10 h Jan 7, 2020, 4:11 PM • 3 Y

Y by amar_04, BinomialMoriarty, Adventure10

Solution