All Russian Olympiad Day 1 P4

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Source: Grade 11 P4

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252 posts

#1 h Apr 28, 2018, 10:36 AM • 2 Y

Y by Adventure10, Mango247

On the sides $AB$ and $AC$ of the triangle $ABC$ , the points $P$ and $Q$ are chosen, respectively, so that $PQ\parallel BC$ . Segments $BQ$ and $CP$ intersect at point $O$ . Point $A'$ is symmetric to point $A$ relative to line $BC$ . The segment $A'O$ intersects the circumcircle $w$ of the triangle $APQ$ at the point $S$ . Prove that circumcircle of $BSC$ is tangent to the circle $w$ .

This post has been edited 3 times. Last edited by djmathman, Dec 18, 2018, 7:56 PM
Reason: formatting + wording

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#2 h Apr 28, 2018, 11:00 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

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Dr.Sop

#3 h Apr 28, 2018, 11:26 AM • 4 Y

Y by AlastorMoody, Yusuf3563_, Adventure10, Mango247

Lemma. $ABC$ given and $P, Q$ on $AB, AC$ , $T= CP\cap BQ$ , $R = AT\cap PQ$ . Let $RH$ be the perpendicular from $R$ on $BC$ , then $RH$ is angle $AHT$ bissector.

Now, in the main problem let $W$ is the circumcenter of $(APQ)$ . Let the circle $\omega$ through $B, C$ is tangent to $(APQ)$ at $S'\not= A$ . Consider the tangent line to $(ABC)$ at $A$ and intersect it with $BC$ at $L$ . From radical center theorem we get that $S'L$ is tangent to $\omega$ , $(APQ)$ at $S'$ . Let the circle $(AS'LW)$ meet $BC$ at $L, J$ . So $OJ\perp BC$ . $WA = WS'$ and $AWS'J$ is cyclic, so $JW$ is angle $\angle S'JA$ bisector. Note that $OJ$ is passing through the middle of $PQ$ which coincides with $PQ\cap AO$ , so from lemma we get that $J, S', O$ are collinear. From symmetry $A'\in JO$ and $S' = S$ . $\Box$

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#4 h Apr 28, 2018, 1:23 PM • 5 Y

Y by DanDumitrescu, Yusuf3563_, Adventure10, Mango247, ehuseyinyigit

Replace label $O$ with $K$ and let $O$ be the center of $\odot(ABC)$ instead. Let a line through $A$ parallel to $BC$ cut $\omega$ and $\odot(ABC)$ at $X,Y$ resp. Let $T$ be the circumcenter of $\Delta ASA'$ , which clearly lies on $BC$ .

First, note that $\Delta XPQ$ and $\Delta A'CB$ are hmothetic so $A', K, X$ are colinear. Moreover, if we let $A_1$ be the point diametrically opposite to $A$ w.r.t. $\odot(ABC)$ . Then,
$\measuredangle ASA' = \measuredangle ASX = \measuredangle AA_1Y = \measuredangle(AO, AA').$ But $\measuredangle TAA' = \measuredangle ASA' - 90^{\circ}$ so $AT\perp AO$ . Hence $TS^2=TA^2 = TB\cdot TC$ and we are done.

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#5 h Apr 28, 2018, 9:56 PM • 2 Y

Y by Adventure10, Mango247

Davrbek wrote:

On the sides $AB$ and $AC$ of the triangle $ABC$ , the points $P$ and $Q$ are chosen, respectively, so that $PQ||BC$ . Segments of $BQ$
and $CP$ intersect at point $O$ . Point $A'$ is symmetric to point $A$ relative to line $BC$ . The segment $A'O$ intersects
circle $w$ circumcircle of the triangle $APQ$ , at the point $S$ .
Prove that circumcircle of $BSC$ is tangent to the circle $w$ .

Fix $k \in \mathbb{R}$ with $\overrightarrow{BC}=(1+k)\overrightarrow{PQ}$ . Redefine $S \ne A$ as the point on $\omega$ with $\odot(BSC)$ tangent to $\omega$ . Let tangent at $A$ to $\odot(ABC)$ meet $BC$ at $T$ . By radical axis theorem applied to $\odot(BSC), \odot(ABC), \odot(APQ)$ we conclude that $ST$ is tangent to $\omega$ .

Let $AA_1$ be an altitude in $\triangle ABC$ ; $S_1$ be the midpoint of $AS$ ; $O_1$ be the midpoint of $AO$ . Now we need to show $A_1,S_1,O_1$ are collinear. Let $Y$ be the circumcenter of $\omega$ . Suppose $AA_1MM_1$ is a rectangle. Let $X=AM_1 \cap A_1O_1$ .

Claim. $\triangle AA_1X \sim \triangle ATY$

(Proof) Let $N$ be the circumcenter of $\triangle ABC$ . Then $\triangle AM_1N \sim \triangle AA_1T$ so by spiral similarity, $\triangle ATN \sim \triangle AA_1M_1$ Now $\overrightarrow{AN}=(1+k)\overrightarrow{AY}$ . Thus, we need $\overrightarrow{AM_1}=(1+k)\overrightarrow{AX}$ to conclude. Now observe by Van-Aubel's that $\tfrac{AO}{OM}=\tfrac{2}{k}$ hence $\tfrac{AX}{A_1M}=\tfrac{AO_1}{O_1M}=\tfrac{1}{(k+1)}$ which together with $A_1M=AM_1$ and directions proves the claim. $\blacksquare$

Finally, we observe $\angle AA_1X=\angle ATY=\angle ATS_1=\angle AA_1S_1$ as $ATA_1S_1$ is cyclic so $A_1,S_1,O_1$ are collinear.

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IstekOlympiadTeam

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IstekOlympiadTeam

#6 h May 16, 2018, 4:21 PM • 4 Y

Y by tenplusten, Anar24, Adventure10, Mango247

Let the line parallel to $BC$ through $A$ and $A'O$ intersect at $S'$ and $\omega$ and $T=S'Q\cap BC$ .

Claim: $APQS'$ is isosceles trapezoid and $A'BS'T$ is parallelogram.

Proof: Let $M=A'S\cap BC$ . Let's show that $BM=MT$ . By Menelaus's and Thales's Theorem $\frac{TM}{MB}=\frac{OQ}{BO}\times\frac{S'T}{S'Q}=\frac{AP}{AB}\times\frac{AB}{AP}=1 \ (\bigstar)$ $\rightarrow$ $BM=MT$ . On the other hand, parallelity of $AS'$ and $BC$ implies that $BC$ bisects the segment $A'S'$ as well. Therefore $MA=\boxed{MA'=MS'} \ (\bigstar\bigstar)$ in the right $A'AS'$ triangle $\rightarrow$ $\boxed{A'BS'T}$ is parallelogram.

Combining $(\bigstar)$ , $(\bigstar\bigstar)$ and $AS'\parallel BT$ $\rightarrow$ $ABTS'$ and $\boxed{APQS'}$ isosceles trapezoid.
Back to main Problem: Let the tangent line to $\omega$ at $A$ and $BC$ intersect at $X$ .It is enough to show that $XA=XS$ $\iff$ $X$ is center of circumcircle of triangle $AA'S$ $\iff$ $\frac{\angle AXA'}{2}+\angle ASA'=180^\circ$ .But we have already got $\angle SAQ=\angle SS'Q=\angle BA'S'$ using the fact that $TQ\parallel BA'$ (See above). Finally, the last property of angles makes it easy to prove that $\frac{\angle AXA'}{2}+\angle ASA'=180^\circ$ with angle chasing. $\blacksquare$

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bobthesmartypants

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bobthesmartypants

#7 h Jul 25, 2018, 6:13 PM • 2 Y

Y by Adventure10, Mango247

solution

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#8 h Jul 12, 2019, 5:27 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

P',Q' sym P,Q wrt BC,PQ'-P'Q=E by Desargues for BQP'-CPQ': O lies on A'E. (E,EP) cuts AC,AB at H,I then I,H on (EP'B),(EQ'C) but AP.AI=AH.AC so A lies on radical center of (EP'B),(QE'C). Reflect wrt BC so A'E is radical center of (EPB),(EQC) so D lies on 2 circle but by Miquel D lies on (APQ) so D=O.
Thus PQO+OCB=PQE=PEB=POB so (APQ) tangent to (BSC)

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L567

#9 h Jan 2, 2023, 5:33 PM • 8 Y

Y by mathscrazy, jelena_ivanchic, TheProblemIsSolved, mxlcv, PROA200, Mango247, Mango247, Resolut1on07

This solution seems to be new. Solved with Sunaina Pati, Krutarth Shah, and Malay Mahajan.

Define $X$ to be the intersection of lines $OA'$ and $BC$ and $M$ be the midpoint of $PQ$ . Supose $BC= s \cdot PQ$ . By Ceva, we have that $A,M,O$ are collinear. But note that $\frac{OX}{XA'} = \frac{d(O,BC)}{d(A',BC)} = \frac{d(O,BC)}{d(A,BC)} = \frac{s \cdot d(O,PQ)}{s \cdot d(A,PQ)} = \frac{OM}{MA}$ which means $XM$ and $AA'$ are parallel, giving $XP = XQ$ .

Now, define $Y$ to be the reflection of $A'$ over $X$ . Since $MX$ is parallel to $AA'$ , we have that $d(Y,MX) = d(A',MX) = d(A,MX)$ . But also, if $D$ is the foot of the $A$ -altitude, $AY$ is parallel to $DX$ because of midpoints, and so $AY$ is parallel to $PQ$ . These two together imply that $A$ is the reflection of $A$ over the perpendicular bisector of $PQ$ and hence lies on the circumcircle of $\triangle APQ$ .

This gives that $\angle YSP = \angle YQP = \angle APQ = \angle ABC = \angle PQX$ , implying that points $P,S,X,B$ are concyclic. Analogously, $Q,S,X,C$ are concyclic, so $S$ is the miquel point of $P,Q$ and $X$ . To finish, we have that $\angle PSB = \angle PXB = \angle PQX = \angle PQS + \angle SQX = \angle PQS + \angle SCB$ so the circumcircles of $BSC$ and $APQ$ are tangent to each other, as desired. $\blacksquare$

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1162 posts

#10 h Apr 9, 2023, 7:30 PM

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Let $K$ be the point on $w$ such that $AK \parallel PQ$ and $T$ lie on $BC$ such that $TA$ is tangent to $(ABC)$ . It's easy to see that $(ABC)$ and $w$ are tangent at $A$ , so $TA$ is clearly the common tangent of $(ABC)$ and $w$ .

Consider the homothety taking $QP$ to $BC$ . Because $QP \parallel BC$ , this homothety must be centered at $O$ . Now, observe $\measuredangle (KP, BC) = \measuredangle KPQ = \measuredangle KAQ = \measuredangle KAC = \measuredangle BCA = \measuredangle A'CB$ which means $KP \parallel A'C$ . An analogous process yields $KQ \parallel A'B$ . Combining these two results with $QP \parallel BC$ , we conclude that the aforementioned homothety maps $KPQ$ to $A'CB$ , so $K \in \overline{A'OS}$ follows immediately.

Since $AK \parallel \overline{TBC}$ follows from transitivity, symmetry implies $\measuredangle ATA' = \measuredangle ATB + \measuredangle BTA' = 2 \measuredangle ATB$ $= 2 \measuredangle TAK = 2 \measuredangle ASK = 2 \measuredangle ASA'.$ Hence, because $TA = TA'$ by symmetry, we know $T$ is the center of $(ASA')$ .

Now, $TA = TS$ means $TS$ is tangent to $w$ by equal tangents and $TS^2 = TA^2 = Pow_{(ABC)}(T) = TB \cdot TC$ implying $TS$ is tangent to $(BSC)$ , which finishes. $\blacksquare$

This post has been edited 1 time. Last edited by ike.chen, Apr 14, 2023, 7:53 AM

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#11 h Aug 16, 2023, 11:56 AM

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always a fun day to do moving points!
solution

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#12 h Mar 25, 2024, 10:43 AM

Y by

Let $T=OS \cap \omega$ .
Claim: $AT \parallel BC$ .
If so, let tangent to $\omega$ at $A$ intersect $BC$ at $X$ . By easy angle chasing we can get that $X$ is center of $(ASA')$ and so $XS^2=XA^2=XB\cdot XC$ (the last is since $(APQ)$ and $(ABC)$ are tangent) and we are done.
Sketch for proof of claim: fix $A, T, P, Q$ and move linearity $B, C$ . Then points $O, A'$ also move linearity. We want to prove that $T, O, A'$ are collinear, so it's enough to check 3 cases.
1) $B=P$ , $C=Q$ , $O$ is a midpoint of $PQ$ .
2) $B=A=C=A'=O$ .
3) $BC$ is midline of $\Delta APQ$ .
All cases are obvious (but in last I bashed in complex numbers...).

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#13 h Jul 23, 2024, 12:35 PM

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Let $A''$ be point on $APQ$ such that $AA'' \parallel BC$ . Since $A''PQ$ and $A'CB$ are homothetic we have that $A''$ lies on $A'O$ . Let $A'O$ meet $BC$ at $T$ .
Claim $1 :PBTS$ and $QCTS$ are cyclic.
Proof $:$ Note that $\angle BTA' = \angle AA''S = \angle SPB$ so $PBTS$ is cyclic. we prove the other part with same approach.
Claim $2 :TP = TQ$ .
Proof $:$ Note that $BC \parallel AA''$ and midpoint of $AA'$ lies on $BC$ so $T$ is midpoint of $A'A''$ and since $\angle A'AA'' = 90$ we have that $T$ lies on perpendicular bisector of $AA'$ and since $AA'QP$ is isosceles trapezoid we have that $T$ lies on perpendicular bisector of $PQ$ as well.
Now note that $\angle PSB = \angle PTB = \angle PQT = \angle PQS + \angle SQT = \angle PAS + \angle SCT$ so $BSC$ and $APQ$ are tangent.

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#15 h Apr 7, 2025, 7:45 AM

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Let $AO\cap BC=M,T\in (ABC)$ and $AT\parallel BC,\ AT\cap (APQ)=W$ . Since $M$ is the midpoint of $BC$ , we have $MA'=MA=MT$ and $\measuredangle TAA'=90$ hence $A',M,T$ are collinear. $\frac{A'M}{A'T}.\frac{WT}{WA}.\frac{OA}{OM}=\frac{1}{2}.\frac{QC}{QA}.\frac{2QA}{QC}=1$ so by Menelaus we get that $A',O,W$ are collinear. Thus, $\measuredangle A'SA=180-\measuredangle ASW=180-\measuredangle B+\measuredangle C$ which implies $S$ lies on $A-$ apollonius circle, under $\sqrt{bc}$ inversion $S^*$ lies on the perpendicular bisector of $BC$ hence $(BSC)$ and $P^*Q^*$ are tangent as desired. $\blacksquare$

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#16 h Apr 7, 2025, 2:32 PM

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let $A'O \cap (APQ), BC$ be $X, Y$

the structures $XPQ$ and $A'CB$ are spun 180 degrees wrt $O$ , so angle chasing gives $\angle PBY = \angle PSX$ , so $BPSY$ cyclic.

Similarly, $CQSY$ cyclic. Also, $Y$ lies on the perpendicular bisector of $AX$ , $PQ$ and $AA'$ , so $\angle QPY = \angle PQY$ , $\triangle PQY$ is isosceles.

Finally, $\angle QSC = \angle QYC = \angle PQY = \angle QPY = \angle QPS + \angle SPY = \angle QPS + \angle SBC$ , so the two circles are tangent.

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