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Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :)

Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end!


Problems discussed in the blog post

Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic.

Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

Problem 3: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$.

Problem 4: Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Problem 5: In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.



STEMS Mathematics 2023 Category B P3

Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic.

Proof: Define $G$ as midpoint of $LD$, $J$ as midpoint of $FD$, $K$ as midpoint of $ED$. Note that $G-J-K$ collinear.

Claim: $BJKC$ cyclic

Proof: Note that$$\angle JKD=\angle FED$$Also $BJ \perp FD, CK\perp ED$. So$$\angle JKC=90+\angle FED=90+\angle FDB=90+90-\angle JBD\implies BJKC \text{ cyclic }.$$

Note that $(L,D;B,C)=-1$. So $GB\cdot GD=GD^2$.


Claim: $PXJK$ cyclic

Proof: Invert $PBC$ about the incirlce, we get $B\rightarrow J, C\rightarrow K$.


Claim: $G-X-P$ collinear

Proof: Radical axis on $(DFE),(BJKC),(XPJK)$


So $GD^2=GB\cdot GC=GJ\cdot GK=GX\cdot GP$.

So $GD$ is tangent $(XPD)$. So $\angle XPD=\angle XDL$.


To show $XHDL$ cyclic, it is enough to show $LH$ is tangent to $(XPH)$.


Note that $LP=LH$ as $EF$ is perpendicular bisector of $PH$.


Now, we have the reduced problem.

Problem: Given $ABC$ a triangle, define $E=AA\cap BC$. Define $H$ as the orthocenter and $D'=AH\cap (ABC)$. Define $X=(AHE)\cap (ABC)$. Prove that $EH$ is tangent to $(HXD')$.

Proof:$$\angle EHX=\angle EAX=\angle ACX=\angle AD'X=\angle HD'X.$$So $EH$ is tangent to $(HXD')$.

We are done~~

Brazilian Math Olympiad 2007, Problem 5

Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

Proof: Define $QO\cap AB=X,PO\cap BC=Y$. Note that by ceva-menalaus, we have $(P,X;A,B)=-1=(B,C;F,E)$. Since $\angle POX=90\implies \angle AOX=\angle XOB$. By angle bisector harmonic lemma. Similarly, we have $\angle BOF=\angle COF$. Done!


USA TSTST 2016 Problem 2, by Evan Chen

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$.

Proof: Note that $Q$ is the humpty point. So $AG,HQ,BC$ concur at say $X$. Note that by brocards $N-H-G$ collinear and $HG\perp AG$. So $O-M-P||N-H-G$. Also note that $Q\in (GNX)$.

Claim: $PGMA$ cyclic

Proof: Since $M$ is centre of $(GMH)$, we get$$\angle GMA=2\angle GHA=2\angle PGA=2(90-\angle GPM)=180-\angle GPA.$$

Claim: $MP$ diameter of $AGMP$

Proof:$$\angle MAG=\angle MGA=90-\angle GMP= 90-\angle PAG.$$

Define $Y=(APGM)\cap (BMC)$. By radicals on $(MBC),(APGM),(ABCG)$ we get $BC,AG,MY$ concurring at $X$.

Define $E,F$ as feet of the altitude on $AC,AB$. Radicals on $(AH),(FEBC),(ABC)\implies E-F-X$ Note that $MYEF$ cyclic as$$XY\cdot XM=XB\cdot XC=XE\cdot XF.$$

Claim: $P-Y-N$ collinear

Proof: As $MN$ is the diameter of $N_9$ we get$$90=\angle MEN=\angle MYN.$$Since $\angle MYP=90\implies P-Y-N.$

So $Y\in (XGQN)$. Hence $Y$ is the desired concurrency.

Iranian TST 2020, second exam day 2, problem 4

Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proof: Define $X=AD\cap BC$. 

Claim: $N$ is the centre of $\omega$. 

Proof: Define $N'$ such that $N'I\perp AI$ and $N'\in AC$. Note that $$\angle N'CI=\angle BCI=\angle N'IC\implies N'\text{ is the center of }\omega\implies N'=N.$$

Define $P=BC\cap \omega$. Define $P'$ as the reflection of $P$ wrt $N$. Then note that $$\angle QP'C=90\implies P'\in \omega\implies QP'||BC.$$

Claim: $P'\in AC$

Proof: $$\angle AP'Q=\angle QPD=\angle 90-\angle DPX=\angle AXB\implies P'\in AD.$$

So by homothety, we get $X-N-M$ collinear. 


2022 IZHO P3

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

Proof: We begin with the claim. 

Claim: $AMNK$ cyclic

Proof: Let $\angle MAB=\theta\implies \angle ABM=180-2\theta.$ Let $$\angle MBC=\alpha \implies \angle ABC=180-2\theta+\alpha$$ $$\implies \angle MCD=\angle BCD-\angle MCB=2\theta-\alpha-\alpha=2\theta-2\alpha.$$ $$\angle MDC=90-\theta+\alpha\implies \angle ADM=\angle ADC-\angle MDC$$ $$=180-2\theta+\alpha-(90-\theta+\alpha)=90-\theta\implies \angle DNK=\theta\text{ as } DM\perp NK.$$

So $$\angle MNK=\angle DNK-\angle DNC=\angle DNK-(180-\angle NCD-\angle NDC)$$ $$=\angle DNK-(180-2\theta+2\alpha-180+2\theta-\alpha)=\theta-\alpha.$$

Moreover, $$\angle MAN=\angle BCD-\angle MAB=2\theta-\alpha-\theta=\theta-\alpha.$$

So $$\angle MAN=\angle MNK\implies AMNK\text{ is cyclic }.$$

Claim: $MK||AB||BC$

Proof: $$\angle MKD=\angle MND=\angle DNC=\alpha.$$ And $$\angle KDC=\angle ADC-\angle NDK=180-2\theta+\alpha-(180-2\theta)=\alpha.$$

So $$\angle MKD=\angle KDC=\alpha\implies MK||AB||BC.$$

Define $O$ as the circumcenter of $(AMD)$. 

Claim: $O-L-M-K$ collinear

Proof: $$\angle NMO=\angle AMO-\angle AMN=\frac{180-\angle AOM}{2}-\angle AMN$$ $$=90-\angle ADM-\angle AMN$$ $$=\angle DNK-\angle ANK=\angle DNK-(\angle DNM-\angle MNK)=2\angle MNK.$$

Note that $AB=BM=MC=MD$. Note that $OC$ is the perpendicular bisector of $MOD$.

Claim: $CKNO$ cyclic

Proof: $$\angle KOC=\angle MAD=\angle MAN=\angle MKN=\angle MNK.$$

Infact, $MO=MC=CD$ and $NOCK$ is isosceles trapezoid. 

Define $(AMD)\cap (CNKO)=P,Q$. We use radical axis on $(CKNOP),(AKDO),(AMDP)$. We get $PQ,OK,AD$ concurring at $L$.

We use radicals on $(ANMK),(AMD),(NKC)$ we get $NK,PQ,AM$ concurring at say $X$. 

Claim: $M$ is incenter of $AKD$

Proof: $$\angle AKM=\angle DNM=\angle MKD$$ and $MN=MK$. 

But note that $AMDP$ is appollinius circle i.e define $M'$ as antipode of $M$ in $(AMD)$, we have $(M',M;L,K)=-1$. So $\angle MPK=\angle MPQ$.

To finish, note that $$\angle CPM=\angle CPK+\angle MPK$$ $$=\angle CNK+\angle MPK$$ $$=\angle MKN+\angle MPK$$ $$=\angle MAD+\angle MPK$$ $$=\angle MPD+\angle MPK$$ $$=\angle MPD+\angle MPQ=\angle DPL.$$

Some random problems I did

Problem 1: Let $ABCD$ a convex quadrilateral. Suppose that the circumference with center $B$ and radius $BC$ is tangent to $AD$ in $F$ and the circumference with center $A$ and radius $AD$ is tangent to $BC$ in $E$. Prove that $DE$ and $CF$ are perpendicular.

Proof: Note that $BF\perp AF, AE\perp BE\implies AFEB \text{ is cyclic}$.

Note that$$\angle FCE=\angle CFB=90-\frac{\angle FBE}{2}=90-\frac{\angle DAE}{2}=\angle FDE\implies DCEF\text{ is cyclic}.$$To finish,$$\angle DEC=180-(\angle DEA+\angle AFD)=180-(\angle ADE+90)=90-\angle ADE=90-\angle FCE\implies DE\perp FC.$$


Problem 2: Consider an acutangle triangle $ABC$ with circumcenter $O$. A circumference that passes through $B$ and $O$ intersect sides $BC$ and $AB$ in points $P$ and $Q$. Prove that the orthocenter of triangle $OPQ$ is on $AC$.

Proof: Define $R$ as the $(AQO)\cap AC.$ We claim that $R$ is the orthocenter of $OPQ$.

For $RO\perp PQ$, note that $\angle OQP=90-A$ and $A=\angle QAR=180-\angle QOR$. So $RO\perp PQ$. For $RP\perp QO$, note that $\angle ORP=\angle OCP=90-A$ and$$\angle QPR=\angle QPO+\angle RPQ=\angle OBA+\angle OCA=90-C+90-B\implies RP\perp QO.$$

Random configurations pics




Diagram creds to mogmog8

We use a lot of stuffs in USATST 2013 P2. Firtly, we use prism lemma, which states like, if two lines $l_1$ and $l_2$ meet at say $X$ and $A_1,B_1,C_1$ are points on $l_1$ and $A_2,B_2,C_2$ are points on $l_2$. We have $A_1A_2,B_1B_2,C_1C_2$ concur iff $(A_1,B_1;C_1,X)=(A_2,B_2;C_2,X)$. Concurrency and prism lemma goes well with each other.


We also use pole and polars/brokards.  And we use radicals too, Pretty good problem.

I even tried desargues, which led to thinking of cross ratios. 



Nine point circle configuration is pretty cute but I tend to forget it. I think most important is knowing the circle exists and the radius is $R/2$. 



 


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