How is the source from 2025?!

WheatNeat wrote: How is the source from 2025?! Since CMO 2024 is used to address the competition in the last year.

How can a point on segment $BC$ satisfy $BC = BD$ if $C \neq D$?

Seungjun_Lee wrote: How can a point on segment $BC$ satisfy $BC = BD$ if $C \neq D$? Sorry for the dumb typo…I have fixed it.

I am using phone rn so i will only post the sketch. (Now editted) Let $DI$ intersect $LN$ at $T$ and $R$ be the midpoint of $AD$. Let $O_1$ be the center of $\omega$, so the midpoint of $DJ$. Here, $T$ is the midpoint of $DI$. First, we will prove that $T$ lies on $EF$. Consider the homothety with center $D$ and ratio $2$. $T \mapsto I$ and $EF \mapsto E_1F_1$ when $E$ and $F$ are the midpoints of $DE_1$, $DF_1$, respectively. Since $DE_1 = 2DE = DA + DC - AC$, we have that $BE_1 = BD + DE_1 = BA + BC - AC$. This implies that the tangency point of $(I)$ and $AB$ is the midpoint of $BE_1$. This is equivalent to $IB = IE_1$, which implies that $\angle IBE_1 = \angle BE_1I$. Here, as $\angle IBE_1 = \angle DEF$, we have that $EF \parallel E_1I$. Therefore, $F_1$ lies on $E_1I$, or, $T$ lies on $EF$. Since $O_1T = \frac{1}{2}JI = \frac{1}{2}JA = O_1R$ and $TR \parallel AI$, it follows that the perpendicular bisector $\ell$ of $PR$, is a line passing through $O_1$ which is perpendicular to $AI$. This is the perpendicular bisector of $MP$, since $MP \parallel AI$. Also, $DQ \perp JL$ and $JL \perp AI$ gives that $DQ \parallel MP$. Hence, if we reflect $PQ$ wrt $\ell$, it becomes $MD$, and $R$ lies on it. Since $P$ is the reflection of $R$ wrt $\ell$, it follows that $T$ lies on $PQ$.

Seungjun_Lee wrote: I am using phone rn so i will only post the sketch. (Now editted) Let $DI$ intersect $LN$ at $T$ and $R$ be the midpoint of $AD$. Let $O_1$ be the center of $\omega$, so the midpoint of $DJ$. Here, $T$ is the midpoint of $DI$. First, we will prove that $T$ lies on $EF$. Consider the homothety with center $D$ and ratio $2$. $T \mapsto I$ and $EF \mapsto E_1F_1$ when $E$ and $F$ are the midpoints of $DE_1$, $DF_1$, respectively. Since $DE_1 = 2DE = DA + DC - AC$, we have that $BE_1 = BD + DE_1 = BA + BC - AC$. This implies that the tangency point of $(I)$ and $AB$ is the midpoint of $BE_1$. This is equivalent to $IB = IE_1$, which implies that $\angle IBE_1 = \angle BE_1I$. Here, as $\angle IBE_1 = \angle DEF$, we have that $EF \parallel E_1I$. Therefore, $F_1$ lies on $E_1I$, or, $T$ lies on $EF$. Since $O_1T = \frac{1}{2}JI = \frac{1}{2}JA = O_1R$ and $TR \parallel AI$, it follows that the perpendicular bisector $\ell$ of $PR$, is a line passing through $O_1$ which is perpendicular to $AI$. This is the perpendicular bisector of $MP$, since $MP \parallel AI$. Also, $DQ \perp JL$ and $JL \perp AI$ gives that $DQ \parallel MP$. Hence, if we reflect $PQ$ wrt $\ell$, it becomes $MD$, and $R$ lies on it. Since $P$ is the reflection of $R$ wrt $\ell$, it follows that $T$ lies on $PQ$. Just out of curiousity, how long did it take you to type that, because I find it especially hard to type in Latex on mobile.

WheatNeat wrote: Just out of curiousity, how long did it take you to type that, because I find it especially hard to type in Latex on mobile. Ah, it took about 30 minutes. I think that I am quite fast at typing in mobile.

I think it is pure calculation problem, call $EF,PQ$ intersect $LN$ at $T_1,T_2$ and prove $|T_1N|=|T_2N|$ by calculating power.

We claim that these lines pass through the midpoint of $ID$, call it $R$. Lemma: $ABC$ is a triangle with $\measuredangle A>90$. The incircle with circumcenter $I$ of $ABC$ touches $AC,AB$ at $E,F$ respectively. $S$ is taken on $BI$ such that $\measuredangle CSB=180-\frac{\measuredangle A}{2}$. Then, $EF$ bisects $AS$. Proof: $EF\cap AS=N$.Let $H$ be the altitude from $C$ to $BS$. By Iran Lemma, $E,F,H$ are collinear. \[\measuredangle HSC=180-\measuredangle CSB=\frac{\measuredangle A}{2}=\measuredangle EAP\]Also $\measuredangle APE=90=\measuredangle CHS$ hence $HSC\sim PAE$. \[\measuredangle HIC=180-\measuredangle CIB=90-\frac{\measuredangle A}{2}=\measuredangle PIE\]Similarily we see that $\measuredangle EPI=90=\measuredangle CHI$ thus, $HIC\sim PIE$. Combining these similarilties, we get the following equations \[\frac{HS}{PA}=\frac{HC}{PE}=\frac{HI}{PI}\]By Menelaus at $\triangle AIS$, \[1=\frac{HS}{HI}.\frac{PI}{PA}.\frac{NA}{NS}=\frac{NA}{NS}\implies AN=NS\]As we have claimed.$\square$ Lemma: $ABC$ is a triangle with incenter $I$. An arbitrary point $D$ is taken on $AC$. $E$ is the altitude from $D$ to the perpendicular bisector of $AI$. $I'$ is the point on $AI$ where $II'ED$ is an isosceles trapezoid. Then, $I'A=DE$. Proof: Let $L$ be the midpoint of $AI$. Let $H$ be the altitude from $D$ to $AI$. Since $II'ED$ is an isosceles trapezoid, we see that $IH=LI'$. Combining this with $IL=LA$ implies $H'L=I'A$. Since $DELH$ is a rectangle, $DE=HL=I'A$.$\square$ Applying the first lemma on triangle $BDA$, we get that $E,F,R$ are collinear. Also by the latter lemma, if $I'$ is the reflection of $I$ with respect to the perpendicular bisector of $AI$, then $AC$ passes through the midpoint of $I'Q$. Reflect over the perpendicular bisector of $AI$ which swaps $Q,D$ and $M,P$. $AC$ swaps with $PQ$ and $ID$ swaps $I'Q$. Thus, $PQ$ bisects $ID$ as desired.$\blacksquare$

Funny geo, the trick is simply keep composture to make sense of everything. Reflect $D$ over $E,F$ and let those points be $E',F'$ respectively, from I-E Lemma $BI \cap (ABC)=J$ and let $X$ midpoint of $DI$, we aim to connect all the points in different ways. Claim 1: $X$ lies on $EF$. Proof: take homothety with scale factor $2$ from $D$ to get that we need $E',F',I$ colinear, now notice from tangency lenghts that $DE'=DF'=AD+DB-AB$ which means that $AE'=AB-DB$ so $CE'=AC+DB-AB=AC+BC-AB$ which is twice the lenght of tangent points of inicircle close to $C$ from $C$, therefore from the altitude of $I$ we conclude that $IE'=IC$ and thus $\angle CE'I=0.5 \angle C=\angle DE'F'$ therefore the colinearity is true. The finish: Again take homothety scale factor $2$ from $D$, while noticing that $DJ$ is diameter of $\omega$ and from I-E Lemma we have that $JL$ is perpendicular bisector of $AI$ which means that $LJ \perp AI \parallel MN$, so from angles and arcs we get $DQ \parallel PM$ therefore $DM,QP$ are symetric w.r.t. the prependicular bisector of $DQ$, so in the homothety all that's left to prove is that $I$ lies the reflection of $DM$ w.r.t. $JL$ but this is clear as $A,I$ are symetric in this reflection and $A$ lies on $DM$, therefore taking the homothety back gives $P,Q,X$ colinear as well so $PQ,LN,EF$ are concurrent at $X$ thus we are done .

Same as @above and #6, easy for CMO ig. Homothety of factor 2 and guessing the point were the important part, rest of it followed nicely. My approach just used some phantom point arguments as well, which were not very useful, but still worth mentioning.

Let $X=ID\cap LN$ (note that it's the midpoint of $ID$). We show that $X$ lies on both $PQ$ and $EF$. Part 1: $X$ lies on $PQ$ Let $X'=PQ\cap LN$. We have \[ \measuredangle X'QD = \measuredangle PQD = \measuredangle PMD = \measuredangle NMC = \measuredangle IAD \]Note that $JL\perp AI$, and we redefine $Q$ as feet of $D$ onto the perpendicular bisector of $AI$. Let $L_1 = LQ\cap AD$ and $L_2$ be midpoint of $L_1D$, then $IL_1\parallel XL_2$, hence \[ \measuredangle QL_2D + \measuredangle DL_2X = -2\measuredangle L_2DQ + \measuredangle DL_1I = -2\measuredangle L_1AI+\measuredangle AL_1I = 0 \]or $X, L_2, Q$ collinear, hence $\measuredangle XQD = \measuredangle IAD$ and $X=X'$. Part 2: $X$ lies on $EF$ Let $I_A$ be the incenter of $\triangle ABD$, $W$ be the feet of perpendicular from $B$ to $AC$ and $Y$ be the feet of $I$ on $BW$ (note that $W$ is the angle bisector of $\angle DBC$). Then, \[ \measuredangle IBW = \measuredangle I_ABW - \measuredangle I_ABI = \frac{\measuredangle ABC}{2} - \left(\frac{\measuredangle ABC}{2} - \measuredangle ABI_A\right) = \measuredangle ABI_A \]Also note that $I_AD$ is perpendicular to the internal angle bisector of $\angle BDC$ and $EF$. Hence \[ \frac{I_AE}{YI} = \frac{I_AF}{YI} = \frac{BI_A}{BI} = \frac{\sin\angle BII_A}{\sin\angle BI_AI} = \frac{\sin(\angle ABI + \angle BAI)}{\sin(\angle BAI + \angle ABI_A)} = \frac{\cos(\angle ABI_A + \angle BAI)}{\sin(\angle BAI + \angle ABI_A)} \]\[ = \frac{1}{\tan\left(\frac{\angle BAC}{2} + \frac{\angle ABC}{2} - (90 - \angle BCA)\right)} = \frac{1}{\tan\left(\frac{\angle BCA}{2}\right)} = \frac{1}{\tan\left(\frac{\angle BDC}{2}\right)} = \frac{I_AE}{ED} \]or $ED=YI$, and since $I$ and $I_A$ are isogonal conjugates in $\triangle ABW$, \[ \tan\angle YEW = \frac{YW}{EW} = \frac{\frac{IY}{\tan\angle YWI}}{\frac{I_AE}{\tan\angle EWI_A}} = \frac{IY}{I_AE} = \tan\left(\frac{\angle BDC}{2}\right) \]thus $YE$ is parallel to the perpendicular bisector of $\angle BDC$ and hence perpendicular to $I_AD$, i.e. $E, F, Y$ collinear. This is enough to conclude that $EF$ and $DI$ must intersect at a point on $LN$, which is $X$.

Let \( X \) be the midpoint of \( ID \), which we claim is the desired intersection point. Clearly, \( X \) lies on \( LN \), the \( I \)-midline of triangle \( AIC \). First, we show that \( X \), \( E \), and \( F \) are collinear. Let \( K \) be the midpoint of \( CD \). Notice that \( CNXK \) is a parallelogram. Observe that the length of the projection of \( XK \) onto \( AC \) is equal to the length of the projection of \( NC \) onto \( AC \), which is half the length of the tangent from \( C \) to the incircle of triangle \( ABC \). Furthermore, \[ KE = \frac{2DE + CD}{2} = \frac{(AB + BD - AB) + CD}{2} = \frac{AC + BC - AB}{2}, \]which is the length of the tangent from \( C \) to the incircle. Thus, \( X \) lies on the perpendicular bisector of \( KE \), so \( XK = XE \), and therefore \( CNXE \) is an isosceles trapezoid. We are now done, as \( \angle DEX = \angle DEF = \frac{1}{2} \angle C \). Next, we show that \( X \), \( P \), and \( Q \) are collinear. Notice that \( DQ \perp QL \) and \( IL \perp QL \). Thus, \( X \) lies on the perpendicular bisector of \( QL \), so \( XQ = XL \). To finish: \[ \angle PQJ = \angle PMJ = 180^\circ - \angle JMN = 90^\circ - \angle CMN = 90^\circ - \angle NLI = \angle XLQ = \angle XQL. \]

Let $D'$ be the reflection of $D$ over $EF$. Let $I'$ be the incenter of $\triangle ABD$. Claim: $ID'\perp I'D$ $D'$ is on $I'D$ is immediate. We proceed with length bash. Let $B'$ be the foot of the altitude from $I$ to $AC$. It suffices to show that \[DD'=\frac{I'D^2+ID^2-II'^2}{2I'D}\]We have $II'=EB'\sec\left(\frac{\angle A}{2}\right)$, so \begin{align*} &\phantom{\quad~}I'D^2+ID^2-II'^2 \\ &= IE^2+ED^2+IB'^2+DB'^2-EB'^2\sec^2\left(\frac{\angle A}{2}\right)\\ &= AE^2\tan^2\left(\frac{\angle A}{2}\right)+AB'^2\tan^2\left(\frac{\angle A}{2}\right)+ED^2+DB'^2-EB'^2 \left(\tan^2\left(\frac{\angle A}{2}\right)+1\right)\\ &=\tan^2\left(\frac{\angle A}{2}\right)\left(AE^2+AB'^2-(AE-AB')^2\right)+(ED^2+DB'^2-(ED+DB')^2)\\ &=\tan^2\left(\frac{\angle A}{2}\right)(2AE\cdot AB')-(2ED\cdot DB')\\ &=2(I'E\cdot IB'-ED\cdot DB')\\ &=2\left(ED\cdot \tan\left(90^\circ-\frac{\angle C}{2}\right)\cdot CB'\cdot \tan\left(\frac{\angle C}{2}\right)-ED\cdot DB'\right)\\ &=2ED(CB'-DB')\\ &=2ED(2CB'-CD)\\ &=2ED(AC+BC-AB-AC+AD)\\ &=2ED(BD-AB+AD)\\ &=4ED^2\\ &=2DD'\cdot I'D \end{align*}where the last step is by power of a point from $D$ to the incircle of $\triangle ABD$. Note that since $ID'\perp I'D$, $EF$ is the perpendicular bisector of a chord of the circle $(ID'D)$ which means it passes through the center of this circle, which happens to be the midpoint of $ID$. As a result, it also passes through $LN$. It suffices to show that $PQ$ also bisects $ID$. To do this, we take the homothety at $D$ with factor $2$. Since the circle originally had diameter $DJ$, it now has center $J$. $J$ maps to $G$, the $D$-antipode. $M$ maps to $H$, the reflection of $D$ across $M$. $L$ and $N$ maps to $Y$ and $X$, respectively, such that $AYID$ and $CXID$ are parallelograms. $P$ and $Q$ map to $P'$ and $Q'$, the intersection of $XH$ and $YG$ with $(DGH)$, respectively. It suffices to show that $I$, $P$, $Q$ are collinear. Claim: $ADQI$ is an isosceles trapezoid. Since $JL\parallel GY$, $GY\perp DQ\implies GY\perp JL$ so $JL$ is the perpendicular bisector of both $AI$ and $DQ$. Claim: $IHPA$ is an isosceles trapezoid. Since $AH=DC=IX$ and $AH\parallel IX$, $AHIX$ is a parallelogram. Therefore, $PH\parallel AI$. Again this implies that $JL$ is the common perpendicular bisector. We therefore have $\angle AIP=\angle IAH=\angle IAD=\angle AIQ$ so we're done.