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Source: Iranian TST 2018, second exam day 2, problem 5

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24 posts

#1 h Apr 17, 2018, 3:34 PM • 7 Y

Y by Mathuzb, GeoMetrix, o_i-SNAKE-i_o, itslumi, Adventure10, Mango247, sami1618

Let $\omega$ be the circumcircle of isosceles triangle $ABC$ ( $AB=AC$ ). Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP=AQ$ . $AP$ and $BC$ intersect at $R$ . Prove that the tangents from $B$ and $C$ to the incircle of $\triangle AQR$ (different from $BC$ ) are concurrent on $\omega$ .

Proposed by Ali Zamani, Hooman Fattahi

This post has been edited 6 times. Last edited by Etemadi, Apr 21, 2018, 3:43 PM

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#2 h Apr 17, 2018, 3:57 PM • 6 Y

Y by microsoft_office_word, sholly, o_i-SNAKE-i_o, Adventure10, Mango247, sami1618

Suppose that $P$ is closer to $B$ . Let the bisector of $\angle{PAQ}$ intersect $\overline{BC}$ at $H$ . Let $\overline{PH}$ intersect $\overline{AQ}$ at $G$ . Clearly $\overline{PG}$ is tangent to $\omega$ . Note that the segment $QG$ is the reflection of $PR$ over $\overline{AH}$ .
$\angle{PGA}=\angle{PGQ}=\angle{ARQ}=180-\angle{RPB}-\angle{RBP}=180-\angle{ACB}-(180-\angle{PBA}-\angle{ABC})=\angle{PBA}$ Thus $G$ lies on $\omega$ , and the result follows from Poncelet's Porism.

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155 posts

#3 h Apr 17, 2018, 4:25 PM • 4 Y

Y by o_i-SNAKE-i_o, Adventure10, Mango247, sami1618

Take AQ $\cap \omega$ =K
Angles : AKP = 180-ABP = APB+PAB=ABC+BAP= $\frac{arcAC+arcBP} {2}$ = ARC
So, the incircle of ARQ is also of AKP
Since BC is tangent to $\omega$ , tangents from $B$ and $C$ must lie on the circumcirlce (by Poncelet's porism)

This post has been edited 1 time. Last edited by govind7701, Apr 27, 2018, 3:29 PM

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412 posts

Yaghi

#4 h Apr 17, 2018, 4:29 PM • 4 Y

Y by o_i-SNAKE-i_o, Adventure10, Mango247, sami1618

My solution:
Problem is equivalent to below:(We just reverse the problem statement)
In $\triangle ABC$ , $M$ is the midpoint of arc $BAC$ .tangents from $M$ to incircle of $\triangle ABC$ intersect $BC$ and $(ABC)$ at $K,L,T,S$ ( $K,L,M$ and $M,T,S$ are collinear, $K,T$ are on $BC$ , $L,S$ are on $(ABC)$ ) then $ML=MT$ and $MK=MS$ .
Proof:
We use following two lemmas:
Lemma 1: $KLTS$ is cyclic.
proof is trivial by inversion with center $M$ radius $MB^2$ .
Lemma 2: $LS$ is tangent to the incircle of $\triangle ABC$
This is well-known and true for any point on $(ABC)$ replaced with $M$ .

Back to OP,Let $LS \cap KT =X$ ,then by lemma 2, $MLXT$ is inscribed.Also,by lemma 1, $\angle MLX =\angle MTX$ which means that $MLXT$ is kite,finishing our proof.

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475 posts

Anar24

#5 h Apr 27, 2018, 3:15 PM • 4 Y

Y by o_i-SNAKE-i_o, Adventure10, Mango247, sami1618

govind7701 wrote:

Take AQ $\cap \omega$ =K
Angles : AKP = 180-ABP = APB+PAB=ABC+BAP= $\frac{arcAC+arcBP} {2}$ = ARC
So, the incircle of ARQ is also of AKP
By Poncelet's porism, we are done

Can you elaborate where do you use Poncelet Porism?

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924 posts

#6 h Oct 23, 2019, 12:37 PM • 5 Y

Y by amar_04, o_i-SNAKE-i_o, DPS, Adventure10, sami1618

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.452355633764675, xmax = 7.206350371537186, ymin = -11.60879940294072, ymax = 9.842227310515382; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-14.074411844006573,6.1856950212950075)--(-13.624244305053052,6.248953392507186)--(-13.68750267626523,6.6991209314607065)--(-14.137670215218751,6.635862560248529)--cycle, linewidth(1.6)); 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dot((-11.666002891786134,4.517805346546533),linewidth(4pt) + dotstyle); label("$S$", (-11.587406299492825,4.699124002593838), NE * labelscalefactor); dot((-15.463483961649452,4.978940434021899),linewidth(4pt) + dotstyle); label("$K$", (-15.38044498908497,5.149145542036973), NE * labelscalefactor); dot((-12.406552331736776,5.408506093653417),linewidth(4pt) + dotstyle); label("$L$", (-12.316012601448378,5.577737484363769), NE * labelscalefactor); dot((-14.137670215218751,6.635862560248529),linewidth(4pt) + dotstyle); label("$J$", (-14.051809967871902,6.7992245199951356), NE * labelscalefactor); dot((-13.935018146693112,5.193723263837656),linewidth(4pt) + dotstyle); label("$N$", (-13.858943593824844,5.363441513200371), NE * labelscalefactor); label("$90^\circ$", (-13.901802788057523,6.306343786319321), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

We begin with a crucial lemma
Lemma $CJ \perp JG$ at $J$ .
Proof Let $LK \cap HS=N$ . By la-hires since $N$ lies on the polars of $J,G$ we have that $JG$ is precicely the polar of $N$ . Now let $J'=GN \cap JG$ we have that $GJ'$ is tangent to $\odot(JKLG) \implies J\equiv J'$ and we are done with the claim. $\blacksquare$

Now back to the main problem. We use phantom points . Let $F=GJ \cap \omega$ . We'll prove that $F \equiv J$ . It is trivial to see that $GF$ bisects $\angle KFL$ . Now $\angle AFG=\angle CBG=90- \frac{\angle BGA}{2}=90-\frac{\angle BFA}{2}$ and so $\angle CFG=90^\circ \implies F \equiv J$ and we are done. $\blacksquare$ .

This post has been edited 3 times. Last edited by GeoMetrix, Oct 23, 2019, 2:41 PM

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#8 h Oct 23, 2019, 7:33 PM • 8 Y

Y by GeoMetrix, o_i-SNAKE-i_o, Pakistan, mueller.25, Don6Bradman9, Adventure10, Mango247, sami1618

Iran TST 2018 Day 2 P5 wrote:

Let $\omega$ be the circumcircle of isosceles triangle $ABC$ ( $AB=AC$ ). Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP=AQ$ . $AP$ and $BC$ intersect at $R$ . Prove that the tangents from $B$ and $C$ to the incircle of $\triangle AQR$ (different from $BC$ ) are concurrent on $\omega$ .

Proposed by Ali Zamani, Hooman Fattahi

Solution:- Notations:- Let $AQ\cap\odot(ABC)=T$ and let the incircle of $\triangle AQR$ be $\gamma$ .

Claim 1:- $PRTQ$ is a cyclic quadrilateral, moreover $PRTQ$ is an isosceles trapezoid.

Let $\angle ABC=\theta$ and $\angle BAT=\alpha\implies\angle AQR=\theta+\alpha$ . Also $\angle QAC=180^\circ-2\theta-\alpha=\angle CBT\implies\angle APT=\theta+\alpha$ as $AB=AC$ .So $TQPR$ is a cyclic quadrilateral. So, $AQ.AT=AP.AR\implies AT=AR\implies QT=PR$ Hence, $PRTQ$ is also an isosceles trapezoid.

Claim 2:- $PT$ is tangent to $\gamma$ .

So, if $PT\cap QR=K$ . Then $PK=QK$ , hence by the converse of Pilot's Theorem $AQKP$ must have an inscribed circle which is $\gamma$ . Hence, $PT$ is tangent to $\gamma$ .

So $\triangle APT$ has both an inconic as well as circumconic.So, by Poncelet's Porism we get that tangents at $B$ and $C$ to $\gamma$ must intersect on $\omega$ . $\blacksquare$ .

This post has been edited 13 times. Last edited by amar_04, Oct 24, 2019, 9:27 PM

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308 posts

#9 h Jul 21, 2020, 7:02 AM • 1 Y

Y by sami1618

Let $I$ and $I_A$ be the incenter and $A-excenter$ of $\triangle ARQ$ . Now notice that $\angle ARI=\angle IRQ=\angle II_AQ=\angle PI_AI$ . Hence $I, I_A,P,R$ are concyclic. Now by shooting lemma,
$AB^2=AC^2=AR\times AP=AI\times AI_A$ Therefore $\angle ACI=\angle IBC$ .
$\angle BIC=180^{\circ}-\angle ACB=\angle ACQ=90^{\circ}+\frac{A}{2}$ Suppose the tangents from $B$ and $C$ to the incircle meet at $X$ . Then $I$ is the incenter of $\triangle XBC$ . Hence
$\angle BXC=2(\angle BIC-90^{\circ})=\angle BAC$ Hence $B,A,X,C$ are concyclic.

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993 posts

khina

#10 h Jan 3, 2021, 4:04 AM • 2 Y

Y by Mango247, sami1618

Let $AQ$ intersect $\omega$ at $S$ . Note that by the shooting lemma, $AQ \cdot AS = AB^2 = AP \cdot AR$ , so $AS = AR$ . Thus $PS$ and $QR$ are reflections across the angle bisector of $\angle{QAR}$ , and thus $PS$ is also tangent to the incircle of $AQR$ . This finishes the problem by poncelet's porism.

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170 posts

#11 h Sep 20, 2022, 5:30 PM • 1 Y

Y by sami1618

Consider the inversion $\varphi: \mathbb{R}^2\to \mathbb{R}^2$ centered at $A$ with power $AB^2$ . Then $P$ and $R$ are interchanged and so are $Q$ and $\varphi(Q)=AQ\cap (ABC)\neq A$ .

Since $AP=AQ$ , then $AR=AQ'$ , and by symmetry $\triangle AQR$ and $\triangle AP\varphi(Q)$ share an incircle $\omega$ .

By Poncelet's Porism, there is a triangle with side $BC$ inscribed in $(ABC)$ and circumscribed about $\omega$ . Hence done. $\blacksquare$

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#12 h Apr 3, 2025, 5:05 PM

Y by

I suppose it's easier than usual Iran TST P2; here is my solution:
Let $I$ be incenter of $ARQ$ and $AI$ intersects with $BC$ and $w$ in $N$ and $S$ respectively. Let $T$ be midpoint of arc $BC$ in $w$ (other than $A$ ). We know $AR \times AP = AB^2 = AR \times AQ = AN \times AS$ so we have $S$ is on $ARQ$ and so $S$ is midpoint arc of $RQ$ in the circle $ARQ$ . Let $I_a$ be A-excenter of $AQR$ . We know $S$ is midpoint of $II_a$ and since $A$ and $T$ are opposite in $ABC$ so $TS$ is perpendicular bisector of $II_a$ . Let $u$ be circle centered at $T$ with radius $TI$ ; power of $A$ to this circle is: $AI \times AI_a = AT^2 - TI^2 = AB^2 = AT^2 - TB^2$ so we have $TB=TI$ so if reflection of BC over $BI$ and $CI$ intersects at $X$ we have $I$ is incenter of $XBC$ and $T$ is circumcenter of $BIC$ so $X$ lies on $TBC$ as desired..

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