Easy problem. It's pure angle chasing and similar triangles. (I'll let someone else post the solution)

Way too many points for a good IMO problem.

I quite liked it... much more innovative than problem 5, even if a little clunky.

$RS\cap CQ=X$. $PQ \cap SD=Y$. $\triangle ETD \equiv \triangle CTB$. By angle chasing , $\angle DSE=\angle CQB$. $\therefore XYSQ$ is cyclic. By angle chasing, $\triangle XTE \sim \triangle YTB$. $\therefore {\frac{TX}{TE}}={\frac{TY}{TB}} \Rightarrow {\frac{TX}{TC}}={\frac{TY}{TD}}\therefore XY//CD//PR$.$\therefore \angle RPQ=\angle XYQ=\angle XSQ=\angle RSQ \Rightarrow PSQR $ is cyclic.

Haha this year's IMO Geometry is so messed up.

what does imo mean souradipclash

Let $CT,DT$ meet $SR,QP$ at $M,N$ respectively It is clear that $\triangle{TDE}\cong \triangle{TBC}$, so $\angle{ETD}=\angle{CTB}$, and we know $\angle{MTD}=\angle{NTC}, \angle{MTE}=\angle{NTB}$, given $\angle{AET}=\angle{ABT}, \angle{MET}=\angle{NBT}$, so we get $\angle{QMS}=\angle{SNQ}$it sufficates to prove $M,N,S,Q$ are concyclic As sol1 showed $Q,S,C,Q$ are concyclic, $MN||RP$ by Reim, the rest is easy, $\angle{SQP}=\angle{SMN}=\angle{SRP}$ implies $Q,S,P,R$ are concyclic

This problem isn't bad per se but I was extremely disappointed to see it as the only geometry problem this IMO. Notice that $\triangle TBC \cong \triangle TDE$ since their sidelengths are all equal. Now $\angle BTC = \angle DTE$ and therefore $\angle BTQ = \angle ETS$. This plus $\angle ABT = \angle TEA$ gives $\triangle BTQ \sim \triangle ETS$ and therefore $$\frac{TQ}{TS} = \frac{TB}{TE} = \frac{TD}{TC} \implies TC \cdot TQ = TD \cdot TS$$ Therefore $Q, S, C, D$ are concylcic. Finally write $$\angle QPR = \angle TCD - \angle BQT = \angle TSQ - \angle TSE = \angle RSQ$$ And we are done.

Let $U = CT \cap AE$ and $V = DT \cap AB$ By side lengths, $\triangle TBC \cong \triangle TDE$. So, $\angle BTC = \angle DTE$. \[ \angle VQU = \angle BTC - \angle TBQ = \angle DTE - \angle TES = \angle VSU \rightarrow SVUQ \text{ cyclic}\] Notice also that $\triangle TVB \sim \triangle TUE$. Hence, \[ \frac{TV}{TD} = \frac{TV}{TB} = \frac{TU}{TE} = \frac{TU}{TC} \rightarrow VU || CD || PR \]and we are done by Reim's theorem.

Basically similar triangles and Reim's, same as the above ones, but I will post it since this is a really nice geo. Here it is: Let $DT$ meet $AB$ at $X$ and let $CT$ meet $AE$ at $Y$. To begin with, note that triangles $TBC$ and $TDE$ are congruent (3 equal sides), so there is a spiral similarity (actually rotation) that takes $B$ to $D$ and $C$ to $E$. Hence triangles $TBD$ and $TCE$ are similar isosceles triangles, so $\angle DBT=\angle DBT= \angle TCE=\angle TEC=\varphi$. In addition, $\angle BTC=\angle DTE$ (due to the congruent triangles), so $\angle BQT =\angle STE$ and $XYQS$ is cyclic. So we are left to prove $XY||PR$, then we will be done by Reim's theorem. Now note that $\angle TBX=\angle TEY = \theta$, $\angle BXD=\angle CYE=180-2\varphi-\theta$, so $BTX$ and $TYE$ are similar. Hence $\frac {TY} {TX} = \frac {TE} {TB}=\frac {TC} {TD}$, hence $XY||PR$ by Thales, done.

Oh well. I think everyone should try to finish this problem in no more than 5 mins. Solution. Let $AB\cap SD = X, AE\cap QC = Y$. By really obvious $\triangle QBT\sim\triangle SET$ we have $\angle S=\angle Q \implies (S,X,Y,Q)$. Henceforth we want $XY//CD$. This is also easy once you notice $\triangle XBT\sim\triangle YTE$ and $\triangle TBC\cong\triangle TDE$: $$\frac{XT}{YT} = \frac{TB}{TE} = \frac{TD}{TC}\qquad\blacksquare$$ UPD: I have a bunch of 7 graders in our school solving this...

Clearly $\triangle BTC \cong \triangle DTE$ since their sides are equal by the problem condition, so $\angle BTC = \angle DTE$ which together with $\angle ABT = \angle AET$ yields $\angle TSE = \angle TQB$. In particular, if $X = DS \cap AP$, $Y = AR \cap CQ$, then $XYQS$ is cyclic and hence the assertion is equivalent to $XY \parallel PR$ - that is, $XY \parallel CD$. By Thales' theorem it suffices to have $\frac{XT}{TD} = \frac{YT}{TC}$, i.e. $\frac{XT}{TB} = \frac{YT}{TE}$ which would follow from $\triangle BXT \sim \triangle EYT$. The latter holds since $\angle ABT = \angle AET$ (given) and $\angle BXT = \angle SXA = \angle QYA = \angle TYE$ (as $XYQS$ is cyclic).

Let $$ M = TD \cap AB $$$$ N = TC \cap AE $$One can deduce that quadrilateral $QSMN$ is cyclic by simple angle chasing. Thus, it suffices to show that $$ MN \parallel DC $$Note that $\triangle TMB \sim \triangle TNE$ which implies $$ \frac{TM}{TN} = \frac{TB}{TE} = \frac{TD}{TC} \implies MN \parallel DC $$as required. $\blacksquare$

This is a rather cute problem. I'd say it's M5

mathisreaI wrote: Let $ABCDE$ be a convex pentagon such that $BC=DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB=TD,TC=TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P,B,A,Q$ occur on their line in that order. Let line $AE$ intersect $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R,E,A,S$ occur on their line in that order. Prove that the points $P,S,Q,R$ lie on a circle. Wow! Nice geo! First, \(TBC\) and \(TDE\) are congruent. Now, we prove \(C,D,Q,S\) concyclic. To show that we prove that \(BTQ\) and \(STE\) are similar. This is because \(\angle TES=\angle TBQ\) and \[\angle STE=180-\angle DTE=180-\angle BTC=\angle BTQ\]Now, that means \[\frac{CT}{DT}=\frac{BT}{TE}=\frac{QT}{TS}\]and so \(C,D,Q,S\) are concyclic as desired. Now, we have \begin{align*} \angle PQS &= \angle BQS \\ &= \angle CQS-\angle CQB \\ &= \angle CDS-\angle TSE \end{align*}And also: \begin{align*} \angle PRS &= \angle DRS \\ &= 180-\angle TSE-\angle SDR \\ &= \angle CDS-\angle TSE \end{align*}and so we are done. $\blacksquare$

Love how there's a monster-ish P1 and a very cute P4 on the test.

Let $DT$ and $CT$ intersect $AB$ and $AE$ at $X$ and $Y$ .$\triangle TBC \simeq \triangle TDE$. So, $\angle BTD=\angle CTE$. So, $\triangle BTD \sim \triangle CTE$. As, $\angle BTX=\angle ETY$ and $\angle ABT=\angle AET$,so, $\triangle BTX \sim \triangle ETY$. So, $\frac{TX}{TY}=\frac{TB}{TE}=\frac{TD}{TC}$ $\implies XY//CD$. Also, $\angle ESD=\angle BQC$ bcuz of equal angles . So, $SQYX$ is cyclic. By Reim's Theorem $PQRS$ is cyclic. I didn't cheat from above solutions.

W just ak-ed elem geo It is easy to see that $\triangle{TBC}\cong \triangle{TDE}$. Because $\angle{QBT} = \angle{SET}$ and $\angle{QTB} = \angle{STE}$, we get that $\triangle{QTB}\sim \triangle{STE}$, which implies that $QT\cdot TC = ST\cdot TD$, which means that $SQDC$ is cyclic. We can use this fact and set a ton of angle variables to get that $\angle{SQP} = \angle{SRP}$, implying that $PSQR$ is cyclic.

Clearly, we have $\triangle TBC\cong\triangle TDE$ and $\triangle TBQ\sim\triangle TES$. Thus, \[\frac{TQ}{TS} = \frac{TB}{TE} = \frac{TD}{TC}\]so $SQDC$ is cyclic. Now, \[\angle PQS = \angle CQS-\angle CQP=\angle CDS-\angle DSR=\angle PRS\]as desired.

$\triangle EDT$ is congruent to $\triangle CBT$ by SSS, so $\angle DET=\angle BCT=b, \angle ETD=\angle BTC=d, \angle EDT = \angle CBT = o$. Let $\angle ABT=\angle TEA=a$, where $a,b,d,o$ are variables $d+\angle QTE+\angle QTS=d+\angle STB+\angle QTS=180$, so $\angle QTE=\angle STB=c$, where $c$ is a variable $\angle ESD = 180-a-b-o = \angle CQP$ And $\triangle STE$ and $\triangle QTB$ are similar by AA So $ST/TE=QT/TB$ $ST/TC=QT/TD$ $\triangle QTD$ and $\triangle STC$ are similar by SAS $QSCD$ is cyclic Let $\angle QPR=x$ $\angle QCR = \angle QPR + \angle CQP = 180+x-a-b-o$ By same arc, $\angle QSD = \angle QCR = 180+x-a-b-o$ $\angle QSR = 180+x-a-b-o-\angle RSD = 180+x-a-b-o-(180-a-b-o)=x$ So $\angle QSR = \angle QPR$ and $QSPR$ is cyclic

This feels... really alien for some reason First notice that $\triangle BCT \cong \triangle DET$. Then notice that since $\angle BTC = \angle DTE$ we have $\angle BQC = \angle DSE$. Furthermore since $\angle BTD = \angle CTE$ we have $\angle STB = \angle QTE$ so if $X, Y$ are $TD \cap AB$ and $CT \cap AE$ respectively, then $\triangle XTB \sim \triangle YTE$ so $\frac{XT}{YT} = \frac{BT}{TE} = \frac{TD}{TC}$ so it follows that $XY \parallel CD$. Furthermore since $\triangle SAX \sim \triangle QAY$ we have $XYQS$ cyclic so by Reim's theorem we have $PRQS$ cyclic, finishing.

We have using the initial condition that $TB \cdot TC = TD \cdot TE \implies TB/TD = TE/TC \implies \triangle TBD \sim \triangle TEC$. Now observe that $\triangle QTB \sim \triangle STE$. Indeed, $\angle STE = \angle QTB$ and $\angle TES = \angle TBQ$. Thus \begin{align*} \frac{TB}{TQ} &= \frac{TE}{TS} \\ \implies \frac{TB}{TE} &= \frac{TQ}{TS} = \frac{TD}{TC} \implies \text{$D, C, S, Q$ are concylic.} \end{align*} We finish off with a simple angle chase: \begin{align*} \angle CPQ &= 180 - (\angle PCQ + \angle CQP) \\ &= 180 - (\angle PCS + \angle SCQ + \angle CQP) \\ &= 180 - (\angle DQS + \angle SDQ + \angle RSD) \\ &= 180 - (\angle SDQ + \angle DQS) - \angle RSD \\ &= \angle QSD - \angle RSD \\ &= QSR \implies \text{$R, P, S, Q$ are concyclic. $\blacksquare$} \end{align*}

Let $X = \overline{AB} \cap \overline{TD}$ and $Y = \overline{AE} \cap \overline{TC}$. $\newline$ Clearly we have $\triangle TBC \cong \triangle TDE$ because of $SSS$. We also have $\triangle BXT \sim \triangle EYT$, due to $\angle ABT = \angle TEA$ and $\angle CTB = \angle DTE$(which implies $\angle BTX = \angle ETY$. $\newline$ From this, we can deduce that $\triangle TBQ \sim TES$ due to $\angle STE = \angle QTB$, and that $\triangle TXY \sim TCD$. $\newline$ This implies that $\angle Q = \angle S$, which then implies $XYQS$ cyclic. Because we have $\triangle TXY \sim TCD$, $XY \parallel PR$, which then implies that $\triangle AXY \sim \triangle APR$. $\newline$ From here we can show that $\frac{AS}{AQ} = \frac{AX}{AY} = \frac{AP}{AR}$, so $AS \cdot AR = AP \cdot AQ$, which proves $PSQR$ cyclic, as desired. Geo isn't the only thing that is dead. Also sorry for the bad write up

Note that by SSS congruency $ \triangle TED \cong \triangle TCB $ Note that $\angle ESD = \angle BQC$ since $$\angle BQC =180^{\circ} - \angle QBT- \angle TBC -\angle TCB= 180^{\circ} - \angle SET - \angle TED - \angle TDE= \angle ESD $$Note by LoS $$\frac{ST}{\sin \angle SET} = \frac{ET}{\sin \angle EST}$$$$\frac{QT}{\sin \angle TBA} = \frac{BT}{ \sin \angle BQC}$$Dividing gives $\frac{ST}{QT}= \frac{CT}{TD}$, which implies $\triangle STQ \sim \triangle CTD$ Now \begin{align*} \angle RSQ &= \angle DSQ -\angle DSE \\ &= \angle TCD - \angle DSE \\ &=\angle TCD - ( 180^{\circ} - \angle SET - \angle TED - \angle TDE) \\ &= \angle TCD + \angle SET - \angle ETD \\ \end{align*}$$ $$\begin{align*} \angle QPR &= 180^{\circ} - (\angle PBC - \angle PCB) \\ &= 180^{\circ} - ((180^{\circ} - \angle CBQ) + (180^{\circ} - \angle BCD )) \\ &= \angle CBQ + \angle BCD - 180^{\circ} \\ &= \angle TBQ + \angle TCD - \angle ETD \\ &= \angle TCD + \angle SET - \angle ETD \end{align*}Which implies the cyclicity $\square$

g2 should've been chosen instead of this one The side conditions equate to $\triangle TBC \cong \triangle TDE$. Let $P_1$ be the intersection of $\overline{SD}$ with $\overline{PQ}$, and let $R_1$ be the intersection of $\overline{QC}$ with $\overline{RS}$. Then, we have \[ \angle BQC = 180^{\circ} - \angle QBC - \angle TCB = 180^{\circ} - \angle TBC - \angle QBT - \angle TCB = \angle BTC - \angle QBT.\]Since $\angle BTC = \angle DTE$ and $\angle QBT = \angle SET$, it follows that $\angle BQC = \angle ESD$, so $SQR_1P_1$ is cyclic. By Reim's, it suffices to prove that $\overline{P_1R_1} \parallel \overline{PR}$. Since $\angle P_1BT = \angle R_1ET$ and \[\angle P_1TB = \angle P_1TC - \angle BTC = \angle R_1TD - \angle ETD = \angle R_1 TE,\]it follows that $\triangle P_1BT \sim \triangle R_1ET$. So, \[ \frac{P_1T}{R_1T} = \frac{BT}{TE} = \frac{TD}{TC}, \]as desired.

Pretty straightforward. The given equates to $\triangle EDT \cong \triangle CBT$. Let $X = \overline{CQ} \cap \overline{AR}$ and $Y = \overline{RS}\cap \overline{AP}$. Then $\triangle EXT\sim \triangle BYT$, hence $\angle QXS = \angle QYS$ and $QXYS$ is cyclic. However, as $\frac{TX}{TC } =\frac{TY}{TD}$, we have $\overline{XY} \parallel \overline{PR}$, so the result follows by Reim's theorem.

Nice Problem! Let $TS$ and $TQ$ intersect $AB$ and $AE$ at $S'$ and $Q'$. Claim: $SS'QQ'$ is cyclic Triangles $BCT$ and $TDE$ are similar leading to $\angle BTQ=\angle STE$ since $\angle ABT=\angle TEA$ the result follows. Claim: $S'Q'||PR$ Since triangles $BS'T$ and $EQ'T$ are also similar we get that $\frac{TS'}{TQ'}=\frac{TD}{TC}$ impying the result. The conclusion follows by Reim Theorem.

At long last after over 2 years of procrastination I finally solved this problem which many people had previously claimed to be hilariously easy. I personally did not believe them until I actually tried it. I'm actually not sure which kind of idiot put this on the actual IMO. Anyways, for the solution, we let $M = \overline{AB} \cap \overline{DT}$ and $N = \overline{AE} \cap \overline{CT}$. Before we get started note that since $TB=TD$ and $TC=TE$, combined with the fact that $DE=BC$ we have $\triangle TED \cong \triangle TBC$. Thus, $\measuredangle ETD = \measuredangle CTB$. The entirety of the solution is the following two claims. Claim : Quadrilateral $SQNM$ is cyclic. Proof : Simply note that, \[\measuredangle NSM = \measuredangle EST = \measuredangle ETD + \measuredangle SET = \measuredangle CTB + \measuredangle TBQ = \measuredangle TQB = \measuredangle NQM \]from which the claim clearly follows. Claim : Lines $\overline{MN}$ and $\overline{CD}$ are parallel. Proof : Simply note that, \[\measuredangle NTE = \measuredangle NTD + \measuredangle DTE = \measuredangle CTM + \measuredangle BTC = \measuredangle BTM\]Thus, combined with the condition that $\measuredangle ABT = \measuredangle TEA$ we conclude that $\triangle MTB \sim \triangle NTE$. Now, it is easy to see that, \[\frac{MT}{NT} = \frac{TB}{TE} = \frac{TD}{TC}\]so, since $\measuredangle MTN = \measuredangle DTC$ quite clearly, $\triangle TMN \sim \triangle DTC$. Thus, it follows that \[\measuredangle CNM = \measuredangle TNM = \measuredangle TCD= \measuredangle NCD\]which proves the claim. Now we are done since the result follows from Reim's Theorem.

Let $DT \cap AB = X, CT \cap AE = Y$. Angle chasing gives $\angle XSY = \angle DSE = 180 - \angle ADE - \angle AED = 180 - \angle TDE - \angle TED - \angle AET = 180 - \angle TBC - \angle TCB - \angle ABT = 180 - \angle QBC - \angle QCB = \angle BQC = \angle XQY$, so $(XSQY)$ is cyclic. Then Note $STE$ is similar to $QTB$, so $\frac{ST}{QT} = \frac{TE}{TB} = \frac{TC}{TD}$, so $(SQCD)$ is cyclic, giving $CD$ is parallel to $XY$ by Reim's. Thus $PQ$ is parallel to $XY$, and we are done by Reim's.

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Therefore we get angle condition that $$\boxed{\angle BCT=\angle DET.}$$Now Angle Chasing we get, $$\angle TBD=\angle TDB=\angle TCE=\angle TEC.$$Also we get, $\angle CTD+\angle BTC=180^o-2\angle TDB.$ Now From Angle condition $$\angle ABT=\angle TEA \implies \boxed{\triangle BTQ \sim \triangle ETS.}$$Therefore we get, $\frac{TD}{TC}=\frac{BT}{ET}=\frac{TQ}{TS}=\frac{BQ}{ES}\implies C,S,Q,R$ are concyclic points by PoP. $\color{red}\textbf{Claim:-}$ Since, $C,S,Q,D \implies P,S,Q,R$ are also cyclic points. $\color{blue}\textbf{Proof:-}$ From Angle Condition we get, $$DSQ=DCQ\implies \angle BCP=180^o-(\angle BCT+\angle DCQ)$$We Know that $\boxed{\angle PCB=180^o-(\angle CBT+\angle ABT).}$ Now in $\triangle BPC$ we get by angle chasing that $$\angle BPC+180^o-\angle BCT+180^o-\angle CBT-\angle ABT=180^o$$Now in $\triangle CBQ$ we get, $\boxed{\angle BCT+\angle TBC+\angle BTC=180^o}$ Equating these we get, $$\angle BPC+\angle BTC=\angle ABT+\angle DCQ$$Now by a little bit Angle Chasing we get that $\boxed{\angle RSQ=\angle RPQ}\implies P,S,Q,R$ are concyclic points.

This is not a new solution, but I have to post for the record. Let $X,Y$ be the intersections of $TD,TC$ with $AB,AC$ respectively. Claim 1: $S,Q,X,Y$ are concyclic. Note that $\angle XQY = \angle BQT = \angle BTC - \angle ABT = \angle DTC - \angle AET = \angle TSE = \angle XSY$. Therefore, $S,Q,X,Y$ are concyclic. Claim 2: $XY \parallel CD$. By angle chasing, $\triangle XBT \sim \triangle YCT$. Thus, $$\frac{XT}{TD} = \frac{XT}{TB} = \frac{YT}{TC} = \frac{YT}{TC},$$implying the claim. By both claims and Reim's theorem, it can be concluded that $P,Q,R,S$ are concyclic.