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$$\angle XZW=\angle HZA=90\implies H\in X-Z-Y.$$
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