Geometry ( Finally!!!)

 This is just such an unfair blog. 

Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!! 

These problems are randomly from A Beautiful Journey through Olympiad Geometry. Also perhaps I will post geo after March, because I am studying combi. 

Problem: Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral.

Solution: Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral.

Problem: A convex quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$, $BD$ of $ABCD$ meet at $P$. Circumcircles of $\triangle ABP$ and $\triangle CDP$ meet at $P$ and $Q$ ($O$, $P$ and $Q$ are pairwise distinct). Show that $\angle OQP = 90^{\circ}.$

Solution: ( Thenku JIB!  ) We will complete the quad and introduce Miquel point $M$.

Let $AD\cap BC=F $ and $CD\cap BA=G$ , let $M$ be the miquel point of quad $ABCD.$

Note that $QP,CD,AB$ concur at $G$ because of radical axis lemma. Also it is well known that $O,P,M$ are collinear and $OM\perp FG.$ By POP on $G$ we get $GP\cdot GQ=GA\cdot GB=GM\cdot GF\implies FMPQ $ is cyclic. So $\angle PQF=90.$

Now by Brokard's theorem , we already know that $O$ is the orthocentre of $FPG.$ So we know that $PG\perp FO $ at a point say $Q'.$ But then $Q'$ has to be $Q$, since there can't be two points on the same liken $PG$ such that $\angle FQG=\angle FQ'G=90.$

Hence $Q\in OF.$ So $\angle OQP=90^{\circ}.$

Problem: Let the incircle and the $A$-mixtilinear incircle of a triangle $ABC$ touch $AC$, $AB$ at $E$, $F$ and $K$, $J$ respectively. $EF$ and $JK$ meet $BC$ at $X$, $Y$ respectively. The $A$-mixtilinear incircle touches the circumcircle of $ABC$ at $T$ and the reflection of $A$ in $O$, the circumcenter, is $A'$. The midpoint of arc $BAC$ is $M$. Prove that the lines $TA'$, $OY$, $MX$ are concurrent.

Solution: (With JIB and RG) Let $G$ be the intouch miquel point i.e $(ABC)\cap (AI).$ It is well known that $A,G,Y$ are collinear. Also if $M_A$ is the midpoint of minor arc $BC$, then we also have $M_A,T,Y$ collinear.

Then if $M$ is the midpoint of $BC$ containing $A$, then we have $X,G,M$ collinear.

Define $L:=TA' \cap GM.$ Then enough to show that $ L,Y,O$ is collinear, which is just Pascal on $M_AMGAA'T$

Problem: Let $ABC$ be a triangle and $O$ be its circumcenter. Let $T$ be the intersection of the circle through $A$ and $C$ tangent to $AB$ and the circumcircle of $\Delta BOC.$ Let $K$ be the intersection of the lines $TO$ and $BC.$ Prove that $KA$ is tangent to the circumcircle of $\Delta ABC.$

Solution: We will use $ = \sqrt{bc/2}$ inversion followed by reflection wrt Angle bisector centred at $A.$

Then it is well known that this inversion takes $B$ and $C$ to $M_{CA}$ and $M_{BA}$, $O$ to $H_A.$ The circle through $A$ and $C$ tangent to $AB$ becomes parallel line to $BA$ from $M_{CA}$ and $(BOC)$ goes to $(N_9).$

Hence $T$ goes to $M_{BC}.$ Now as $K=BC\cap OT$, we get $K'=(AM_{CA}M_{BA})\cap (AM_{BC}M_{CA}).$

In the question we were asked to show that $\angle OAK=90.$

So after inverting, it's enough to show that $\angle H_AAK'=90.$ 

Problem: Given $ABC$ a triangle, define $K':=(AM_{BA}M_{CA})\cap (AH_AM_{BC}).$ Then prove that $\angle H_AMK'=90.$

Proof: Note that $O\in (AM_{BA}M_{CA}).$ So $\angle AK'O=90.$ Also $M_{BC}K'A=90.$ Hence $ M_{BC},O,K'$ are collinear. But we have $OM_{BC}\perp BC.$ Hence $AK'II BC\implies \angle H_AAK'=90.$

Problem: Let $MN$ be a line parallel to the side $BC$ of a triangle $ABC$, with $M$ on the side $AB$ and $N$ on the side $AC$. The lines $BN$ and $CM$ meet at point $P$. The circumcircles of $\triangle BMP$ and $\triangle CNP$ meet at two distinct points $P$ and $Q$. Prove that $\angle BAQ = \angle CAP$.

Solution: Claim: $AP\cap BC=M_{BC},$ where $M_{BC}$ is the midpoint of $BC.$

Proof: Let $AP\cap BC=X.$ Note that by menelaus, we get $\frac {AM}{MB}\cdot \frac{BX}{XC}\cdot \frac {CN}{AN}=1. $ Using the parallel condition, the claim follows.

Claim: $AQ$ is the symmedian.

Proof: Well.. it's well known. By spiral similarity (or simple angle chasing), we get $MBQ\sim CNQ.$ Dropping perps from $Q$ to $AB$ ( say $R$) and $AC$( say $S$) and using $\frac{MB}{NC}=\frac{AB}{AC}.$ We get $\frac{QR}{QS}=\frac{AB}{AC}.$

Since symmedian and median is reflection about angle bisector, the result follows.

Problem: Let $ABC$ be an acute-angled scalene triangle, with centroid $G$ and orthocenter $H.$ The circle with diameter $AH$ cuts the circumcircle of $BHC$ at $A',$ distinct from $H.$ Analogously define $B'$ and $C'.$ Prove that $A', B', C'$ and $G$ are concyclic.

Solution: Note that $A',B',C'$ are just humpty points ( denote it by $X_A,X_B,X_C$ ) and it is well known that that $HX_AG=90, $ and similarly for others. All these points lie on circle with diametre $HG.$

Problem: Let $\omega$ be a semicircle with diameter $PQ$. A circle $k$ is tangent internally to $\omega$ and to the segment $PQ$ at $C$. Let $AB$ be the tangent to $k$ perpendicular to $PQ$, with $A$ on $\omega$ and $B$ on the segment $CQ$. Show that $AC$ bisects $\angle PAB$.

Solution: Define $T :=$ touch point of $k$ and $\omega $ , $D:=$ touch point of $AB$ and $k$ .

Note that $T-D-Q$ is collinear by homothety . Now invert wrt $Q$ of radius $QB\cdot QP.$ Then note that $P\leftrightarrow B$, $T\leftrightarrow D. $

Here $A=(QTP)\cap DB\implies A'= DB\cap (QTP)=A.$ Also $C=$ Touch point of $k$ and $PQ,$ but are fixed so $C\leftrightarrow C.$ Hence $QB\cdot QP=QA^2=QC^2\implies QA=QC.$

Now by angle chase we get $\angle BAC=90-\angle QAB=\angle PAC.$

Problem: Triangle $ABC$ is inscribed into the circle $\omega_{}$. The circle $\omega_{1}$ touches the circle $\omega_{}$ internally and touches sides $AB$ and $AC$ in the points $M$ and $N,$ respectively. The circle $\omega_{2}$ also touches the circle $\omega_{}$ internally and touches sides $AB$ and $BC$ in the points $P$ and $K,$ respectively.

Prove that $NKMP$ is a parallelogram.

Solution:Let $I$ be the incentre, then we have $AI\perp MN$ and $IM=IN.$ Similary, we have $IP=IK.$ Also we have $I,K,P$ and $I,M,N$ collinear. Hence $NKMP$ is a parallelogram.

Problem: The diagonals of the quadrilateral $ABCD$ intersect at $P.$ Let $O_{1}$ and $O_{2}$ be the circumcenters of $\Delta APD$ and $\Delta BPC,$ respectively. Let $M, N$ and $O$ be the midpoints of $AC, BD$ and $O_{1}O_{2}$, respectively. Prove that $O$ is the circumcenter of $\Delta MPN.$

Solution: Let $E:=(APD)\cap (BPC).$ Note that $E$ is the spiral centre of the spiral similarity $\chi : AC\rightarrow DB.$ As $M,N$ are midpoints , we get that $\chi :M\rightarrow N. $ Hence $ \chi :CM\rightarrow BN\implies \chi :  MN\rightarrow CB\implies (EMNP)$ is cyclic.

Now note that $O_1O_2\perp PE.$ Hence $O_1O_2$ is perpendicular bisector of $PE.$ Note that $\angle OO_1E=\angle PDE$ and $\angle EO_2O=\angle ECM.$ So $EO_1O_2\sim EAC.$ But $M,O$ are midpoints, so $\angle EOO_2=\angle EMC\implies \angle O_1OE=\angle PME \implies 2\cdot \angle PME=\angle POE. $( As $O$ lies on perp bisector.)

Hence $O$ is the circumcenter of $\Delta MPN.$

Problem(APMO 2013): Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

Solution: We will show $[OEB]=[AOF],$ rest follows by similarly. So using sin formula for areas and fact that $OB=OA,$ enough to show that $\frac12\cdot \sin \angle {OBC}\cdot BE= \frac12\cdot \sin \angle {OAC}\cdot AF$ or using the fact that $O $ and $H$ are isogonal conjugates, enough to show $ \frac{AF}{AB}\cdot BE=\frac{AF}{AB}\cdot BE.$ done!

Problem(Macedonia MO 2018): Given is an acute $\triangle ABC$ with orthocenter $H$. The point $H'$ is symmetric to $H$ over the side $AB$. Let $N$ be the intersection point of $HH'$ and $AB$. The circle passing through $A$, $N$ and $H'$ intersects $AC$ for the second time in $M$, and the circle passing through $B$, $N$ and $H'$ intersects $BC$ for the second time in $P$. Prove that $M$, $N$ and $P$ are collinear.

Solution: We just have to use the fact that when $H$ is reflected wrt a side, the point lies on the circumcircle. So $H'\in(ABC).$ Again note that $H'N \perp AB,$ since it's just the extension of the altitude. So by cyclicity, $\angle H'MA=\angle H'PB=90.$ Hence $M,N,P$ are the foot of the altitude and we are done by simson lines.

Problem(JBMO 2016): A trapezoid $ABCD$ ($AB || CF$,$AB > CD$) is circumscribed.The incircle of the triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$,respectively.Prove that the incenter of the trapezoid $ABCD$ lies on the line $MN$

Solution: Define $I$ as incentre of $ABCD,$ $J$ as incentre $ABC.$

For now define $I:= JB\cap NM.$ But note that ( by iran lemma) , we get $\angle AI'B=90.$ Now, we know that incentre bisect angles of trapezoid , so by angle chase we get $\angle AIB=90.$ But incentre also lies on angle bisector of angle $B.$

Hence $I'=I.$ And we are done!

So yeah! That's it for this time!! Maybe a combo post next time. 

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I also compiled them in one pdf as a problem set. Here's the google drive link https://drive.google.com/file/d/1OpcNYdvyjeBZvJ-3IZKcx8A09nSBQPSK/view?usp=sharing. Hope you like it!

Sunaina💜