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Source: 2017 Olympic Revenge, Problem 2

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49 posts

#1 h Feb 20, 2017, 1:03 PM • 3 Y

Y by baopbc, Tawan, Adventure10

Let $\triangle$ $ABC$ a triangle with circumcircle $\Gamma$ . Suppose there exist points $R$ and $S$ on sides $AB$ and $AC$ , respectively, such that $BR=RS=SC$ . A tangent line through $A$ to $\Gamma$ meet the line $RS$ at $P$ . Let $I$ the incenter of triangle $\triangle$ $ARS$ . Prove that $PA=PI$

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#3 h Feb 21, 2017, 2:55 AM • 6 Y

Y by SAM8, LittleGlequius, Tawan, nguyendangkhoa17112003, Adventure10, Mango247

Assume WLOG $AB < AC$ . $\measuredangle RBS = \measuredangle RSB= \measuredangle IRS$ . On the other hand $\measuredangle PRA= \measuredangle BRS$ . And $\measuredangle AIP+ \measuredangle RIP = \measuredangle AIR= \measuredangle PAI+ \measuredangle SBC$ , $( \star )$ .

Law of sines in $\triangle PAI \Longrightarrow \frac{PA}{PI} = \frac{ \sin \angle AIP}{ \sin \angle PAI}$

Law of sines in $\triangle PIR \Longrightarrow \frac{PI}{PR} = \frac{ \sin \angle PRI}{ \sin \angle RIP}$

Law of sines in $\triangle PRA \Longrightarrow \frac{PR}{PA} = \frac{ \sin \angle PAR}{ \sin \angle PRA}$

Law of sines in $\triangle RBS \Longrightarrow \frac{BS}{BR} = \frac{ \sin \angle BRS}{ \sin \angle RSB}$

Law of sines in $\triangle BSC \Longrightarrow \frac{SC}{BS} = \frac{ \sin \angle SBC}{ \sin \angle ACB}$

Multiplying all the equations we get: $\frac{ \sin \angle PAI}{ \sin \angle SBC} = \frac{ \sin \angle AIP}{ \sin \angle RIP}$ , $( \star \star )$ .

By $( \star )$ , $( \star \star )$ and Two Equal Angles Lemma it follows $\measuredangle PAI= \measuredangle AIP$ . We conclude $PI=PA$ .

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49 posts

#4 h Mar 3, 2017, 1:11 AM • 3 Y

Y by Tawan, Adventure10, Mango247

Here's another solution:

Let $E = RI \cap AS$ , $F = SI \cap AR$ , $O$ and $J$ the circumcenter and $A$ -excenter of triangle $\triangle ARS$ , respectively.
Let $\phi$ a inversion centered at $A$ and radius $\sqrt{AR \times AS}$ followed by a reflection on angle bissector of $\angle RAS$
Since $AB=AR+RS$ and $AE=\frac{AR \times AS}{AR+RS} \Longrightarrow \phi(B)=E$ . Analogously, $\phi(C)=F$ .
Also we have $BC \parallel EF, \phi(I)=J$ and $\phi(AP)$ is the line $l$ through $A$ parallel to $BC$ .
Let $K \neq A$ the intersection of $l$ and the circumcircle of triangle $\triangle ARS$ . Then, $\phi(P)=K$

Then, $PA=PI \Longleftrightarrow \angle PAI=\angle PIA \Longleftrightarrow \angle KAJ=\angle AKJ \Longleftrightarrow JA=JK \Longleftrightarrow JO \perp AK \Longleftrightarrow OJ \perp EF$ . This is well-known, see here: https://www.artofproblemsolving.com/community/c5h1230491p6213572
We are Done!

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4147 posts

#5 h Mar 10, 2017, 4:57 AM • 3 Y

Y by Tawan, Adventure10, Mango247

Lemma: In a $\triangle ABC$ with orthocenter $H$ and 9-point center $N,$ let $D,E,F$ be the feet of the altitudes on $BC,CA,AB.$ The perpendicular bisector of $\overline{DH}$ cuts $EF$ at $L.$ Then $LH \perp AN.$

Proof: Let the perpendicular from $H$ to $AN$ cuts $EF$ at $L'$ and $\odot(HBC)$ again at $X.$ As the inversion with center $H$ and power $HA \cdot HD=HB \cdot HE=HC \cdot HF$ takes $\odot(HBC)$ into $EF,$ it follows that $AXDL'$ is cyclic $\Longrightarrow$ $\triangle HAX \sim \triangle HL'D.$ But since $AN$ passes through the center of $\odot(HBC)$ (reflection of $A$ on $N$ ), then $\triangle HAX$ is A-isoscles $\Longrightarrow$ $\triangle HL'D$ is L'-isosceles $\Longrightarrow$ $L \equiv L'$ $\Longrightarrow$ $LH \perp AN. \ \blacksquare$

Back to the proposed problem. Redefine $P$ as the intersection of $RS$ with the perpendicular bisector of $\overline{AI},$ thus we shall show that $PA$ is tangent to $\odot(ABC).$ Let $U$ and $J$ be the circumcenter and A-excenter of $\triangle ARS.$ Using the previous lemma on the excentral triangle of $\triangle ARS,$ it follows that $PI \perp JU.$ But by extraversion of a well-known property discussed at https://artofproblemsolving.com/community/c6h321293 (see post #5), we have $JU \perp BC,$ $\Longrightarrow$ $IP \parallel BC$ $\Longrightarrow$ $\angle PAI=\angle PAB+\tfrac{1}{2}\angle BAC=\tfrac{1}{2}\angle BAC+\angle ACB$ $\Longrightarrow$ $\angle ACB=\angle PAB$ $\Longrightarrow$ $PA$ is tangent to $\odot(ABC).$

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154 posts

#6 h Mar 10, 2017, 4:59 AM • 3 Y

Y by Tawan, Adventure10, Mango247

uh..... what problem is this?

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225 posts

baopbc

#7 h Mar 10, 2017, 8:54 AM • 3 Y

Y by buratinogigle, Tawan, Adventure10

Here is a simple solution.

Let $\triangle DEF$ be the cevian triangle of $I$ w.r.t $\triangle ARS$ , $EF$ meets $RS$ at $T$ . Since $SF$ is the bisector of $\angle ASR$ so $SF$ $\parallel$ $RC$ , thus $\tfrac{AF}{AR}$ $=$ $\tfrac{AS}{AC}$ imply $AF\cdot AC$ $=$ $AR\cdot AS$ . Similarly, $AE\cdot AB$ $=$ $AR\cdot AS$ , hence $\tfrac{AF}{AB}=\tfrac{AE}{AC}$ i.e $EF$ $\parallel$ $BC$ . Note that circle $(AEF)$ tangent to $\Gamma$ so $PA$ is the tangent line of $(AEF)$ . $K$ is the intersection point of $AI$ with $EF$ , we get $AP$ passes through the midpoint of $TK$ . So by Menelaus theorem for $\triangle SKD$ we imply $\tfrac{PD}{PT}$ $=$ $\tfrac{AD}{AK}$ $=$ $\tfrac{ID}{IK}$ . Thus, $PI$ $\parallel EF$ so $\angle AIP$ $=$ $\angle AKT$ $=$ $\angle IAP$ which imply $\triangle IPA$ is isosceles i.e $PA$ $=$ $PI$ . $\square$

This post has been edited 1 time. Last edited by baopbc, Mar 10, 2017, 9:28 AM
Reason: fixed

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2327 posts

#8 h Apr 22, 2018, 2:23 PM • 1 Y

Y by Adventure10

Actually, I proposed a similar problem in August 2015, in problem weekly, see

https://www.dropbox.com/s/crx5o1dyt2mmhca/derakynay1187.pdf

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297 posts

#9 h Jul 22, 2018, 1:45 AM • 2 Y

Y by Adventure10, Mango247

buratinogigle wrote:

Actually, I proposed a similar problem in August 2015, in problem weekly, see

https://www.dropbox.com/s/crx5o1dyt2mmhca/derakynay1187.pdf

Please make files available. Thanks.

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1915 posts

#10 h Dec 4, 2019, 9:59 AM • 3 Y

Y by GeoMetrix, strawberry_circle, Adventure10

2017 Olympic Revenge P2 wrote:

Let $\triangle$ $ABC$ a triangle with circumcircle $\Gamma$ . Suppose there exist points $R$ and $S$ on sides $AB$ and $AC$ , respectively, such that $BR=RS=SC$ . A tangent line through $A$ to $\Gamma$ meet the line $RS$ at $P$ . Let $I$ the incenter of triangle $\triangle$ $ARS$ . Prove that $PA=PI$

Solution:- Let $RI\cap AC=X$ and $SI\cap AB=Y$ and $AI\cap RS=Z$ . Just notice that $SY\|RC$ and $XR\|BS$ . So, $\begin{cases} \frac{AR}{RB}=\frac{AX}{XS} \\ \frac{AY}{AR}=\frac{AS}{AC}\end{cases}\implies \frac{AY}{AB}=\frac{AX}{AC}\implies YX\|BC$ .

Let $RS\cap YX=T$ and $AI\cap XY=K$ and $AP\cap XY=J$ . So, $-1=(R,S;Z,T)\overset{A}{=}(Y,X;K,T)$ Also as $\odot(AYX)$ and $\odot(ABC)$ are homothetic we get that $AP$ is tangent to $\odot(AYX)$ . So, $JY.JX=JA^2\implies JK=JA$ as $(Y,X;K,T)=-1$ .

Now by Menelaus we get $\frac{ZA}{AK}=\frac{PZ}{PT}$ . Also $(I,A;K,Z)=-1\implies \frac{ZA}{KA}=\frac{ZI}{KI}=\frac{ZP}{PT}\implies IP\|KJ$ . So as $AJ=JK$ we get that $PA=PI$ . $\blacksquare$ .

This post has been edited 2 times. Last edited by amar_04, Dec 4, 2019, 4:39 PM

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Arefe

#11 h May 29, 2020, 3:28 PM

Y by

Let $Q$ on the $RS$ such that $IQ||BC$ , It's enough to say that $P=Q$ . Note that if $K$ is the intersection of $BS,RC$ then it's easy to see that $IRKS$ is parallelogram .
$\frac{QR}{QS}=\frac{IR}{IS}\times{\frac{KC}{KB}}=\frac{sin\angle{ACR}}{sin\angle{BCR}}\times{\frac{sin\angle{CBS}}{sin\angle{ABS}}}=\frac{sin\angle{CAK}}{sin\angle{BAK}}=\frac{CS.AR}{SA.RB}\times{\frac{AB}{AC}}=\frac{PR}{PS}$ $\implies$ $P=Q$ and we are done $\blacksquare$

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1979 posts

#12 h Jun 6, 2020, 6:38 AM

Y by

Cute!

LittleGlequius wrote:

Let $\triangle$ $ABC$ a triangle with circumcircle $\Gamma$ . Suppose there exist points $R$ and $S$ on sides $AB$ and $AC$ , respectively, such that $BR=RS=SC$ . A tangent line through $A$ to $\Gamma$ meet the line $RS$ at $P$ . Let $I$ the incenter of triangle $\triangle$ $ARS$ . Prove that $PA=PI$

Let $A'$ be the reflection of $A$ in $\triangle ARS$ and $O$ be its circumcentre. Let bisectors of $\angle R$ and $\angle S$ meet $AS$ and $AR$ at $D$ and $E$ respectively. Note that the problem is equivalent to $\odot(AIA')$ and $\odot(ABC)$ being orthogonal. Invert at $A$ with radius $\sqrt{AR \cdot AS}$ followed by reflection in $AI$ , we see that $\{D, B\}, \{E, C\}, \{A', O\}$ swap and $I$ maps to the $A$ -excenter $J$ of $\triangle ARS$ . Thus we require $JO \perp DE$ , which is well-known.

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520 posts

zuss77

#13 h Jun 6, 2020, 6:48 PM

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With $RE \parallel BS$ and $SF \parallel CR$ by Pappus we have $EF \parallel BC$ .
With $(T,K;E,F)=-1$ and $MA^2=MF\cdot ME$ we have $MK=MT=MA$ .
Then $PT/PD=AK/AD=IK/ID$ (first by Menelaus on $\triangle DKT$ , second by harmonic division).
Hence $PI\parallel EF \implies PI/PA=MK/MA=1$ .
This problem also here and here.

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