Some Geometry Problems for everyone to try!

 These problems are INMO~ish level. So trying this would be a good practice for INMO! 

Let $ABCD$ be a quadrilateral. Let $M,N,P,Q$ be the midpoints of sides $AB,BC,CD,DA$. Prove that $MNPQ$ is a parallelogram.

Consider $\Delta ABD$ and $\Delta BDC$ .Note that $NP||BD||MQ$. Similarly, $NM||AC||PQ$. Hence the parallelogram.

In $\Delta ABC$, $\angle A$ be right. Let $D$ be the foot of the altitude from $A$ onto $BC$. Prove that $AD^2=BD\cdot CD$.

Note that $\Delta ADB\sim \Delta CDA$. So by similarity, we have $$\frac{AD}{BD}=\frac{CD}{AD}.$$

In $\Delta ABC$, $\angle A$ be right. Let $D$ be the foot of the altitude from $A$ onto $BC$. Prove that $AD^2=BD\cdot CD$.

Let $D\in CA$, such that $AD = AB$.Note that $BD||AS$. So by the Thales’ Proportionality Theorem, we are done!

Given $\Delta ABC$, construct equilateral triangles $\Delta BCD,\Delta CAE,\Delta ABF$ outside of $\Delta ABC$. Prove that $AD=BE=CF$.

This is just congruence. Note that in $\Delta ABD, \Delta FBC$ we have $$\angle FBC=\angle ABC+60^{\circ}=\angle ABD.$$ And $FB=AB,BC=BD$. So we get $\Delta ABD \cong \Delta FBC$. So $AD=FC$. Similarly, we can show $BE=FC$.

If line $PQ$ intersecting $AB$ on $\triangle ABC$, where $P$ is on $BC$, $Q$ is on the extension of $AC$, and $R$ on the intersection of $PQ$ and $AB$, then $$\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.$$

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$. $$\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}$$ $$\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}$$ Multiplying the two equalities together to eliminate the $PK$ factor, we get: $\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1$

In $\Delta ABC$, choose points $D,E,F$ on sides $BC,CA,AB$ respectively. Prove that circles $(AEF),(BFD),(CDE)$ share a point known as the miquel point.

Define $M=(AEF)\cap (BFD)$. So note that $$\measuredangle FME=\measuredangle A,\measuredangle EMD=\measuredangle C\implies \measuredangle FMD=B\implies M\in (DFB).$$

Let $\omega_1, \omega_2$ be two circles intersecting at $M,N$. Let line $\ell_M$ through $M$ intersect $\omega_1, \omega_2$ at $A_1, A_2$. Let $B_1, B_2$ be points on $\omega_1, \omega_2$ respectively, Then $A_1B_1\parallel A_2B_2$ if , and only if, $B_1,N,B_2$ are collinear on a line $\ell_N$.

Suppose that B_1NB_2 is a straight line. Then $$\measuredangle MA_1B_1 = \measuredangle MNB_1 = \measuredangle MA_2B_2 \implies A_1B_1 \parallel A_2B_2.$$

Let a triangle $\triangle ABC$ and a point $P$ be given. Let $D, E,$ and $F$ be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively. Then points $D, E,$ and $F$ are collinear iff the point $P$ lies on circumcircle of $\triangle ABC.$

Let the point $P$ be on the circumcircle of $\triangle ABC.$ So $$\angle BFP = \angle BDP = 90^\circ \implies BPDF \text{ is cyclic }\implies \angle PDF = 180^\circ – \angle CBP.$$ So $$\angle ADP = \angle AEP = 90^\circ \implies AEPD \text{ is cyclic }$\implies \angle PDE = \angle PAE.$$ And $$ACBP \text{ is cyclic } \implies \angle PBC = \angle PAE \implies \angle PDF + \angle PDE = 180^\circ$$ $\implies D, E,$ and $F$ are collinear as desired.

Let $AB$ be a chord in $\omega(O, r)$ and let $TA$ be a tangent to $\omega$ at $A$. Let $\angle BAT = \alpha$. Let $\angle APB$ be any inscribed angle over the arc $AB$. Then $$\angle BAT=\angle APB.$$

Since $TA$ is a tangent, then it must be perpendicular to $OA$. So $\angle OAT = 90^{\circ}$. So $\angle OAB = \angle OAT − \angle BAT = 90− \alpha$. Note that $\Delta OAB$ is isosceles. So $$\angle OAB = \angle OBA = 90^{\circ} − \alpha,\angle AOB = 180^{\circ} −2(90^{\circ} −\alpha) = 2\alpha.$$ And $\angle APB=\angle AOB/2=\alpha$. So done.

Olympiad Problems

Let $ABC$ be a triangle. Let $I$ be the Incenter of $ABC$ and $S$ be the midpoint of arc $BAC$. Define $IA$ as the $A$-excenter wrt $ABC$. Define $\omega$ to be the circle centred at $S$ with radius $SB$. Let $AI_A \cap \omega = X$, $Y$. Show that $\angle BCX = \angle ACY$.

Note that $$\angle AYC=\angle XYC=\angle XBC.$$ And we have $$\angle BXC=180-\frac{BSC}{2} = 180 - \frac{A}{2}.$$ And note that $X-A-Y$ is the angle bisector of $\angle BAC$. So $$\angle YAC=180-\frac{A}{2}.$$ So we get that $$\Delta BXC\sim\Delta YAC\implies \angle BCX=\angle ACY.$$

Let $ABC$ be a triangle. The incircle of $ABC$ has center $I$ and is tangent to $AB$ and $AC$ at $D$ and $E$ respectively. Let $O$ denote the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.

Note that $A,I,O$ are collinear. Now $$\triangle ADO \cong \triangle AEO \implies \angle ODB = \angle OEC$$

Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying $DA < AB = BC < CD$. Points $E$ and $F$ are chosen on sides $CD$ and $AB$ such that $BE \perp AC$ and $EF \parallel BC$. Prove that $FB = FD$.

Let $P = AC \cap BE$. Since $\triangle{ABC}$ is isosceles we know that $P$ is the mid point of $AC$. Let $\angle{CBE} = \angle{ABE} = \alpha$. We also know that $\angle{BCA} = \angle{BAC} = 90 - \alpha$. By parallel lines we know that $\angle{FEB} = \alpha$. Hence $\triangle{FEB}$ is isosceles or $FB = FE$. Let $\angle{DCA} = \angle{DBA} = \theta$. By parallel lines we see that $\angle{DEF} = 90 - \alpha + \theta$. Since $EP \perp AC$ and $P$ is the midpoint of $AC$ as previously stated, we know that $\triangle{AEC}$ is isosceles or $\angle{EAC} = \angle{ECA} = \theta$. Note that $AFED$ is cyclic due to $$\angle{AFE} + \angle{ADE} = 180 - \angle{BFE} + 180 - \angle{ABC} = 180 - (180 - 2\alpha) + 180 - 2\alpha = 180.$$ Hence we know that $\angle{FAE} = \angle{FAC} + \angle{CAE} = 90 - \alpha + \theta = \angle{FDE}$ meaning that $\angle{FDE} = \angle{FED}$ or $\triangle{FED}$ is isosceles. Therefore we know that $FE = FD$. Previously we found that $FB = FE$, thus we get $FB = FE$ as desired.

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$. Let $AB$ be the line tangent to these circles at $A$ and $B$, respectively, so that $M$ lies closer to $AB$ than $N$. Let $CD$ be the line parallel to $AB$ and passing through the point $M$, with $C$ on $G_1$ and $D$ on $G_2$. Lines $AC$ and $BD$ meet at $E$; lines $AN$ and $CD$ meet at $P$; lines $BN$ and $CD$ meet at $Q$. Show that $EP = EQ$.

Let $MN\cap AB=P \implies M$ is midpoint of $PQ.$ It is enough to show that $EM\cap CD$ or show that $EM\cap AB.$ Note that $$\angle ACM=\angle,~~\angle EAB=\angle ECD.$$ So $EAMB$ is kite. And we are done.

Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

Redefine $T$ such that $T\in (ABC)$ and $AT||BC$.

$TF$ tangent to $(BDF)$.

Note that is enough to show that $180-\angle BDF=\angle BFT$. But note that $$\angle BDF=\angle TFG=\angle FTA.$$

Similarly, we get that $TG$ is tangent to $(EGC)$. And we are done.

Two circles $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.

Note that the composition of homotheties gives us $$\omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2.$$ Moreover, since the product of scales are $-1$ ( not $1$, else it will be a transformation), so the composition is a homothety. But the homothety taking $\omega_1\rightarrow \omega_2$ is simply $1$ or $-1$ ratio as the radius is the same and the centre lies on $O_1O_2$. But ratio $1$ is absurd. Hence,the ratio is $-1$ with the centre of the homothety being the midpoint of $X_1X_2,O_1O_2$. Let $M$ be the midpoint of $O_1O_2,X_1X_2$. So note that $$\omega_1\xrightarrow{O} \omega_2=\omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2.$$ But the homothety $\omega_1\xrightarrow{O} \omega_2$ takes $X_1\rightarrow X_2$. Now, we consider the homothety $\omega_1 \xrightarrow{T_1} \omega \xrightarrow{T_2} \omega_2$, this takes $$X_1\rightarrow X_1T_1\cap \omega \text { say } A \rightarrow AT_2\cap \omega_2.$$ But we should have $AT_2\cap \omega_2=X_2$. So $T_2-A-X_2$. And we are done!

Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.) Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Note that $\angle FKE=2\angle A$ and $$\angle OBC=\angle OCB=90-\angle A\implies \angle FDB=\angle EDC=A\implies \angle FDE=180-2\angle A\implies KFDE \text{ cyclic}.$$ As $$KF=KE\implies \angle KEF=90-A\implies \angle KDF=90-A\implies$$ $$\angle KDB=\angle KDF+\angle FDB=90$$ $$\implies KD\perp BC.$$ And we are done.

Let $ABC$ be an acute-angled triangle with $AB\le AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.

$L-B-E$ collinear

Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \angle SPA'=90-(90-A/2-C)=A/2+C\implies \angle LBD=A/2+C\implies \angle LBD=A/2+C.$$ By angle chase, $$\angle DBC=90-A/2,\angle EBC=90-C.$$ So $E-B-L$ collinear.

$LD||BC$

Define $F$ as midpoint of arc $BC$ not containing $A$. Now define $X=PP\cap AF$. Define $O$ as the circumcenter of $(ABC)$. We know that $$\angle EBC=90-C,\angle ELD=\angle BLD=\angle BPD=\angle BPA'=\angle BAO=90-C.$$ So $LD||BC$.

$PXOF$ is cyclic

Let $\angle PAB=\theta$. So $\angle PSB=\theta$ and $\angle PSF=\theta+A/2$. So $\angle POF=2\theta+A$. And $\angle PXF=\angle PAX+\angle APX$. But $\angle PAX=\theta+A/2$. And $$\angle APX=\angle APS+\angle SPX=\angle ACS+\angle SPX=90-A/2-C+\angle SPX.$$ So to show cyclicity, we need to show that $$2\theta+A=\angle PXF=90+\theta-C+\angle SPX.$$ Or enough to show $$\angle SPX=\theta+A-90+C.$$ But $$\angle SPX=\angle SPB-\angle XPB$$ $$=\angle SAB-\angle XPD-\angle DPB= 90+A/2-\angle DBP-\angle DLB$$ $$=90+A/2-\angle DLP-(90-C)$$ $$=C+A/2-\angle DLP$$ as $SF||AD,BC||LD, AD\perp BC$ we get $$C+A/2-\angle DLP=C+A/2-(90-A/2-\theta)=C+A-90+\theta.$$ And so we get $PXOF$ is cyclic.

$XA=XP$

Note that $$\angle XPA=\angle PXF-\angle PAX=2\theta+A-\theta-A/2=\theta+A/2$$ . So $\angle XPA=\angle XAP$. And hence the claim. But $$XA^2=XP^2=XD\cdot XB.$$ So $XA$ is tangent to $(DBA)$. Hence $$\angle ABD=\angle DAS=\angle DAF=\angle AFS=\angle ABS.$$

And hence we get $B-X-S$ collinear. And we are done!

Let $H$ be the orthocenter of an acute-angled triangle $ABC$. The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$. Similarly, define the points $B_{1}$, $B_{2}$, $C_{1}$ and $C_{2}$. Prove that the six points $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$ and $C_{2}$ are concyclic.

$A_1A_2C_1C-2$ is cyclic

Note that $H$ lies on the radical axis of $(C_1C_2)$ and $(A_1A_2)$. Moreover, we know that $BH$ is perpendicular to line joining the centre. So $BC_2\cdot BC_1=BA_1\cdot BA_2\implies A_1A_2C_1C_2$ cyclic.

Similarly, we get $A_1A_2B_1B_2$ and $B_1B_2C_1C_2$ cyclic. If these three circles are not the same then, consider the pairwise radical axis which should concur( but they don't).

Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that $$\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.$$ Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.

Circumcircles of $B_1A_2AB, AA_1C_2C,$ and $BCC_1B_2$ mutually intersect at some point $P.$

Let $P$ be $(BCC_1B_2)\cap (CAA_2C_2)$. Note that $$\angle AB_1B=180^\circ-(\angle BC_1C+\angle CA_1A)$$ $$=\angle BPC+\angle APC-180^\circ=180^\circ - \angle APB$$

Lines $B_1C_2, C_1A_2,$ and $A_1B_2$ concur at $P.$

But $$\angle A_2PB=\angle BPC_1=90^\circ.$$

So done!

Let $\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C.$ The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle ADE$ again at $F.$ If $\angle ADF = 45^{\circ}$, show that $CF$ is tangent to $\omega .$

Note that $\angle FDE=45$. As $F$ lies on the angle bisector and $FD$ is the external bisector of $\angle CDB$, we get that $F$ is the $B-$ excenter of $\Delta CDB$. Note that $$FCD=\frac{180-(\angle DCB)}{2}=\frac{180-90+\angle B}{2}=45+\frac{\angle B}{2}$$ and $$\angle AED=180-\angle FEA-\angle DEB=180-45-90+\frac{\angle B}{2}=45+ \frac{\angle B}{2}$$ $$\implies CF||AF$$ So $$\angle FEA=\angle FAE=\angle AFX, X\in \overrightarrow{ CF}.$$

Let $ABC$ be a triangle inscribed in circle $\omega$ centered at $O$. Let $H$ be the orthocenter of $\triangle ABC$. Let $Q$ be a point on $\omega$ such that $\angle AQH = 90^\circ$. Let $N$ be the nine point center of $\triangle QBC$. Show that $HO= 2 HN$.

Let the orthocentre of $QBC$ be $J$. Let $L$ be the $N_9$ center of $ABC$. Let $M$ be the midpoint of $BC.$ Note that $Q-H-M$

$AQJH$ is a parallelogram $\implies QM\perp NL$

Note that $NL||HJ$. Note that $$AH\perp BC,QJ\perp BC\implies AH||QJ, AH=2R\cos (A)=QJ\implies AQJH\text{ is a parallelogram }.$$

$MN=ML$

As the radius of nine point circle is $1/2$ radius of the circumcircle, but $(QBC)=(ABC).$

So $Q-H-M$ is the perpendicular bisector of $NL$. Hence $$HN=HL\implies HO=2HN.$$

Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.

$GBHC$ cyclic

Note that $HC\perp HG$ as $HC\perp AB$. And $GB\perp BC$ as $AH\perp BC$. So $GBHC$ cyclic.

Note that $$\angle HBG=\angle HBC+\angle CBG=90-C+90=180-C\implies \angle HCG=C\implies 90-C=\angle IGC=\angle IJC.$$ But $$\angle IA'A=\angle A'AH=90-C\implies IJ=IA'=AH.$$

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD\le DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$

Let $N$ be the midpoint of $AB$, $M$ be the midpoint of $PQ$. Note that $$\angle CND=90\implies N\in (CPDQF).$$ Also, $CN$ is the angle bisector $\angle BCA \implies CN$ is the angle bisector of $\angle PCQ\implies NP=NQ\implies C-M-E$. But $$\angle PNC=\angle CQP=\angle FPQ=\angle FNQ.$$ So $\triangle PNC \cong \triangle FNQ \implies M\in$ perpendicular bisector of $CF$. So $M$ is the circumcentre of $(ACB)$ as perpendicular bisector of $AB$ and $CF$ concur at $M$.

Diagonals ${AC}$ and ${BD}$ of convex quadrilateral $ABCD$ meet at $P$. Prove that the incenters of the triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$ are concyclic if and only if their $P$-excenters are also concyclic.

Note that $AI_A\cdot AJ_A=AB\cdot AC.$ We then use POP $PA\cdot PB=PI_1\cdot PJ_1$ and we do it cyclically. Doing manipulations, we get that $$\frac{(PI_2\cdot PE_2)(PI_4\cdot PE_4)}{(PI_1\cdot PE_1)(PI_3\cdot PE_3)}=1$$ And then we take POP on $P$ in concyclicity etc.

Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Define $Q=GE\cap (ABC),P=DF\cap (ABC)$

$Q\in (ADE)$

Note that $$\angle AQE=\angle AQG=\angle GCA=\angle GCE=\angle GEC=\angle AEQ.$$

Hence, if $D$ is the midpoint of major arc $BC$, note that $A-I-D$ and $D$ is the centre of $(BIC).$ Note that by triangle inequality we have $$AD\le AP+PD\implies AI+ID\le AP+PD\implies AI\le AP.$$

Define $Z=(BWX)\cap (CWY)$

$X-Z-Y$ collinear

As $XW$ is the diameter, we have $\angle XZW=90$, similarly, we have $\angle WZY=90$

$Z\in (ANHM)$

Note that $Z$ is the miquel point. So $Z\in (ANM)$, But $H\in (ANM)\implies (ANHMZ)$ is cyclic.

Note that $NMCB$ is cyclic

$Z\in AW$

Note that $$\angle NZW=180-\angle NBW=\angle NMC=180-\angle NMA=180-\angle NZA.$$

Now, note that $$\angle XZW=\angle HZA=90\implies H\in X-Z-Y.$$