### St. Petersburg 2008-19 IX-XI (Russia) 62p

geometry problems from Saint Petersburg Mathematical Olympiads
with aops links in the names

started in 1934 as Leningrad MO,
renamed in 1992 as St. Petersburg

2008 - 2019

Pentagon $ABCDE$ has circle $S$ inscribed into it. Side $BC$ is tangent to $S$ at point $K$. If $AB=BC=CD$, prove that angle $EKB$ is a right angle.

Point $O$ is the center of the circle into which quadrilateral $ABCD$ is inscribed. If angles $AOC$ and $BAD$ are both equal to $110$ degrees and angle $ABC$ is greater than angle $ADC$, prove that $AB+AD>CD$.

2008 St. Petersburg  MO grade X P5
In cyclic quadrilateral $ABCD$ rays $AB$ and $DC$ intersect at point $E$, while segments $AC$ and $BD$ intersect at $F$. Point $P$ is on ray $EF$ such that angles $BPE$ and $CPE$ are congruent. Prove that angles $APB$ and $DPC$ are also equal.

2008 St. Petersburg  MO grade XI P5
All faces of the tetrahedron $ABCD$ are acute-angled triangles.$AK$ and $AL$ -are altitudes in faces $ABC$ and $ABD$. Points $C,D,K,L$ lies on circle. Prove, that $AB \perp CD$

Points $A_1$ and $C_1$ are on $BC$ and $AB$ of acute-angled triangle $ABC$ . $AA_1$ and $CC_1$ intersect in $K$. Circumcircles of $AA_1B,CC_1B$ intersect in $P$ - incenter of $AKC$.
Prove that $P$ - orthocenter of $ABC$ .

2009 St. Petersburg  MO grade IX P5
$ABC$ is acute-angled triangle. $AA_1,BB_1,CC_1$ are altitudes. $X,Y$ - midpoints of $AC_1,A_1C$. $XY=BB_1$. Prove that one side of $ABC$ in $\sqrt{2}$ greater than other side.

2009 St. Petersburg  MO grade X P2
$ABCD$ is convex quadrilateral with $AB=CD$. $AC$ and $BD$ intersect in $O$. $X,Y,Z,T$ are midpoints of $BC,AD,AC,BD$. Prove, that circumcenter of $OZT$ lies on $XY$.

Streets of Moscow are some circles (rings) with common center $O$ and some straight lines from center $O$ to external ring. Point $A,B$ - two crossroads on external ring. Three friends want to move from $A$ to $B$. Dima goes by external ring, Kostya goes from $A$ to $O$ then to $B$. Sergey says, that there is another way, that is shortest. Prove, that he is wrong.

2009 St. Petersburg  MO grade X P7
Points $Y,X$ lies on $AB,BC$ of $\triangle ABC$ and $X,Y,A,C$ are concyclic. $AX$ and $CY$ intersect in $O$. Points $M,N$ are midpoints of $AC$ and $XY$. Prove, that $BO$ is tangent to circumcircle of $\triangle MON$

2009 St. Petersburg  MO grade XI P5
$O$ -circumcenter of $ABCD$. $AC$ and $BD$ intersect in $E$, $AD$ and $BC$ in $F$. $X,Y$ - midpoints of $AD$ and $BC$. $O_1$ -circumcenter of $EXY$. Prove that $OF \parallel O_1E$

$M,N$ are midpoints of $AB$ and $CD$ for convex quadrilateral $ABCD$. Points $X$ and $Y$ are on $AD$ and $BC$ and $XD=3AX,YC=3BY$. $\angle MXA=\angle MYB = 90^o$.  Prove that $\angle XMN=\angle ABC$

2010 St. Petersburg  MO grade IX P7
Incircle  of $ABC$ tangent $AB,AC,BC$ in $C_1,B_1,A_1$. $AA_1$ intersect incircle in $E$. $N$ is midpoint $B_1A_1$. $M$ is symmetric to $N$ relatively $AA_1$. Prove that $\angle EMC= 90^o$

$ABC$ is triangle with $AB=BC$. $X,Y$ are midpoints of $AC$ and $AB$. $Z$  is base of perpendicular from $B$ to $CY$. Prove, that circumcenter of $XYZ$ lies on $AC$

2010 St. Petersburg  MO grade XI P5
$SABCD$ is quadrangular pyramid. Lateral faces are acute triangles with orthocenters lying in one plane. $ABCD$ is base of pyramid and $AC$ and $BD$ intersects at $P$, where $SP$ is height of pyramid. Prove that $AC \perp BD$

2011 St. Petersburg  MO grade IX  P3
Point $D$ is inside $\triangle ABC$ and $AD=DC$. $BD$ intersect $AC$ in $E$. $\frac{BD}{BE}=\frac{AE}{EC}$. Prove, that $BE=BC$

$ABCD$ - convex quadrilateral. $\angle A+ \angle D=150^o, \angle B<150^o, \angle C<150^o$ Prove, that area $ABCD$ is greater than $\frac{1}{4}(AB\cdot CD+AB\cdot BC+BC\cdot CD)$

$ABC$-triangle with circumcenter $O$ and $\angle B=30$. $BO$ intersect $AC$ at $K$. $L$ - midpoint of arc $OC$ of circumcircle $KOC$, that does not contains $K$. Prove, that $A,B,L,K$ are concyclic.

$ABCD$ - convex quadrilateral.  $P$ is such point on $AC$ and inside $\triangle ABD$, that $\angle ACD+\angle BDP = \angle ACB+ \angle DBP = 90^o-\angle BAD$. Prove that $\angle BAD+ \angle BCD =90^o$ or $\angle BDA + \angle CAB = 90^o$ .

$ABCD$ - convex quadrilateral. $M$ -midpoint $AC$ and $\angle MCB=\angle CMD =\angle MBA=\angle MBC -\angle MDC$.  Prove that $AD=DC+AB$.

$ABCD$ is inscribed. Bisector of angle between diagonals intersect $AB$ anc $CD$ at $X$ and $Y$. $M,N$ are midpoints of $AD,BC$. $XM=YM$ Prove, that $XN=YN$.

$ABC$ is triangle. Point $L$ is inside $ABC$ and lies on bisector of $\angle B$. $K$ is on $BL$. $\angle KAB=\angle LCB= \alpha$. Point $P$ inside triangle is such, that $AP=PC$ and $\angle APC=2\angle AKL$.  Prove that $\angle KPL=2\alpha$

Points $C,D$ are on side $BE$ of triangle $ABE$, such that $BC=CD=DE$. Points $X,Y,Z,T$ are circumcenters of $ABE,ABC,ADE,ACD$. Prove, that $T$ - centroid of $XYZ$ .

$ABCD$ is parallelogram. Line $l$ is perpendicular to $BC$ at $B$. Two circles passes through $D,C$, such that $l$ is tangent in points $P$ and $Q$. $M$ - midpoint $AB$. Prove that $\angle DMP=\angle DMQ$.

At the base of the pyramid $SABCD$ lies a convex quadrilateral $ABCD$, such that $BC \cdot AD = BD \cdot AC$. Also $\angle ADS =\angle BDS ,\angle ACS =\angle BCS$. Prove that the plane $SAB$ is perpendicular to the plane of the base.

$ABC$ is triangle. $l_1$- line passes through $A$ and parallel to $BC$, $l_2$ -  line passes through $C$ and parallel to $AB$. Bisector of $\angle B$ intersect $l_1$ and $l_2$ at $X,Y$. $XY=AC$. What  value can take $\angle A- \angle C$ ?

Given quadrilateral $ABCD$ with $AB=BC=CD$. Let $AC\cap BD=O$, $X,Y$ are symmetry points of $O$ respect to midpoints of $BC$, $AD$, and $Z$ is intersection point of lines, which perpendicular bisects of  $AC$, $BD$. Prove that $X,Y,Z$ are collinear.

In a convex quadrilateral $ABCD$ , $M,N$ are midpoints of $BC,AD$ respectively. If $AM=BN$ and  $DM=CN$ then prove that $AC=BD$.

by S. Berlov
Let $M$ and $N$ are midpoint of edges $AB$ and $CD$ of the tetrahedron $ABCD$, $AN=DM$ and $CM=BN$. Prove that $AC=BD$.

by S. Berlov
Let $(I_b)$, $(I_c)$ are excircles of a triangle $ABC$. Given a circle $\omega$ passes through $A$ and externally tangents to the circles $(I_b)$ and $(I_c)$ such that it intersects with $BC$ at points $M$, $N$. Prove that $\angle BAM=\angle CAN$.

by A. Smirnov
All angles of $ABC$ are in $(30,90)$. Circumcenter of $ABC$ is $O$ and circumradius is $R$. Point $K$ is projection of $O$ to angle bisector of $\angle B$, point $M$ is midpoint $AC$. It is known, that $2KM=R$. Find $\angle B$ .

Points $A,B$ are on circle $\omega$. Points $C$ and $D$ are moved on the arc $AB$, such that $CD$ has constant length. $I_1,I_2$ - incenters of $ABC$ and $ABD$. Prove that line $I_1I_2$ is tangent to some fixed circle.

$D$ is inner point of triangle $ABC$. $E$ is on $BD$ and $CE=BD$. $\angle ABD=\angle ECD=10^o,\angle BAD=40^o,\angle CED=60^o$ . Prove that $AB>AC$.

Incircle $\omega$ of $ABC$ touch $AC$ at $B_1$. Point $E,F$ on the $\omega$ such that $\angle AEB_1=\angle B_1FC=90$. Tangents to $\omega$ at $E,F$ intersects in $D$, and $B$ and $D$ are on different sides for line $AC$.  $M$- midpoint of $AC$. Prove, that $AE,CF,DM$ intersects at one point.

Points $B_1,C_1$  are on $AC$ and $AB$ and $B_1C_1 \parallel BC$. Circumcircle of $ABB_1$ intersect $CC_1$ at $L$. Circumcircle $CLB_1$ is tangent to $AL$. Prove $AL \leq \frac{AC+AC_1}{2}$.

$I$ - incenter , $M$- midpoint of arc $BAC$ of circumcircle, $AL$ - angle bisector of triangle $ABC$. $MI$ intersect circumcircle in $K$. Circumcircle of $AKL$ intersect $BC$ at $L$ and $P$.
Prove that $\angle AIP=90^o$.

$AB=CD,AD \parallel BC$ and $AD>BC$. $\Omega$ is circumcircle of $ABCD$. Point $E$ is on $\Omega$ such that $BE \perp AD$. Prove that $AE+BC>DE$.

$ABCD$ is convex quadrilateral. Circumcircle of $ABC$ intersect $AD$ and $DC$ at points $P$ and $Q$. Circumcircle of $ADC$ intersect $AB$ and $BC$ at points $S$ and $R$. Prove that if $PQRS$ is parallelogram then $ABCD$ is parallelogram.

$ABCD$ - convex quadrilateral. Bisectors of angles $A$ and $D$ intersect in $K$,  Bisectors of angles $B$ and $C$ intersect in $L$. Prove $2KL \geq |AB-BC+CD-DA|$.

$ABCDE$ is convex pentagon. $\angle BCA=\angle BEA = \frac{\angle BDA}{2}, \angle BDC =\angle EDA$.  Prove, that $\angle DEB=\angle DAC$.

Let $BL$ be angle bisector of acute triangle $ABC$.Point $K$ choosen on $BL$ such that $\measuredangle AKC-\measuredangle ABC=90º$.point $S$ lies on the extention of $BL$ from $L$ such that $\measuredangle ASC=90º$.Point $T$ is diametrically opposite the point $K$  on the circumcircle of $\triangle AKC$.Prove that $ST$ passes through midpoint of arc $ABC$.

by S. Berlov
2016 St. Petersburg  MO grade IX P3
On the side $AB$ of the non-isosceles triangle $ABC$, let the points $P$ and $Q$ be so that $AC = AP$ and $BC = BQ$. The perpendicular bisector of the segment $PQ$ intersects the angle bisector of the $\angle C$ at the point $R$ (inside the triangle). Prove that $\angle ACB + \angle PRQ = 180^o$.

2016 St. Petersburg  MO grade IX P6
Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that angle bisector of $\angle GDF$ passes though the midpoint of $I_1I_2$.

2016 St. Petersburg  MO grade X P3
The circle inscribed in the triangle $ABC$ is tangent to side $AC$ at point $B_1$, and to side $BC$ at point $A_1$. On the side $AB$ there is a point $K$ such that $AK = KB_1, BK = KA_1$. Prove that $\angle ACB\ge 60$

2016 St. Petersburg  MO grade X P5
Points $A$ and $P$ are marked in the plane not lying on the line $\ell$. For all right triangles $ABC$ with hypotenuse on $\ell$, show that the circumcircle of triangle $BPC$ passes through a fixed point other than $P$.

2016 St. Petersburg  MO grade XI P3
In a tetrahedron, the midpoints of all the edges lie on the same sphere. Prove that it's altitudes intersect at one point.

2016 St. Petersburg  MO grade XI P5
Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that $I_1I_2 \perp$ bisector of $\angle ABC$

Given a triangle $ABC$, there’s a point $X$ on the side $AB$ such that $2BX = BA + BC$. Let $Y$ be the point symmetric to the incenter $I$ of triangle $ABC$, with respect to point $X$. Prove that $YI_B\perp AB$ where $I_B$ is the $B$-excenter of triangle $ABC$.

2017 St. Petersburg  MO grade IX P5
Given a scalene triangle $ABC$ with $\angle B=130^{\circ}$. Let $H$ be the foot of altitude from $B$. $D$ and $E$ are points on the sides $AB$ and $BC$, respectively, such that $DH=EH$ and $ADEC$ is a cyclic quadrilateral. Find $\angle{DHE}$.

2017 St. Petersburg  MO grade  X P3
Let $ABC$ be an acute triangle, with median $AM$, height $AH$ and internal angle bisector $AL$. Suppose that $B, H, L, M, C$ are collinear in that order, and $LH<LM$. Prove that $BC>2AL$.

2017 St. Petersburg  MO grade X P6
In acute-angled triangle $ABC$, the height $AH$ and median $BM$ were drawn. Point $D$ lies on the circumcircle of triangle $BHM$ such that $AD \parallel BM$ and $B, D$ are on opposite sides of line $AC$. Prove that $BC=BD$.

2017 St. Petersburg  MO grade XI P2
A circle passing through vertices $A$ and $B$ of triangle $ABC$ intersects the sides $AC$ and $BC$ again at points $P$ and $Q$, respectively. Given that the median from vertex $C$ bisect the arc $PQ$ of the circle. Prove that $ABC$ is an isosceles triangle.

Given a tetrahedron $PABC$, draw the height $PH$ from vertex $P$ to $ABC$. From point $H$, draw perpendiculars $HA’,HB’,HC’$ to the lines $PA,PB,PC$. Suppose the planes $ABC$ and $A’B’C’$ intersects at line $\ell$. Let $O$ be the circumcenter of triangle $ABC$. Prove that $OH\perp \ell$.

2018 St. Petersburg  MO grade IX  P3
$ABC$ is acuteangled triangle. Variable point $X$ lies on segment $AC$, and variable point $Y$ lies on the ray $BC$ but not segment $BC$, such that $\angle ABX+\angle CXY =90^o$. $T$ is projection of $B$ on the $XY$. Prove that all points $T$ lies on the line.

2018 St. Petersburg  MO grade IX  P5
Can we draw $\triangle ABC$ and points $X,Y$, such that $AX=BY=AB$, $BX = CY = BC$,
$CX = AY = CA$?

$ABCD$ is inscribed quadrilateral. Line, that perpendicular to $BD$ intersects segments $AB$ and $BC$ and rays $DA,DC$ at $P,Q,R,S$ . $PR=QS$. $M$ is midpoint of $PQ$. Prove that $AM=CM$

Point $T$ lies on the bisector of $\angle B$ of acuteangled $\triangle ABC$. Circle $S$ with diameter $BT$ intersects $AB$ and $BC$ at points $P$ and $Q$. Circle, that goes through point $A$ and tangent to $S$ at $P$ intersects line $AC$ at $X$.  Circle, that goes through point $C$ and tangent to $S$ at $Q$ intersects line $AC$ at $Y$. Prove, that $TX=TY$

Points $A,B$ lies on the circle $S$. Tangent lines to $S$ at $A$ and $B$ intersects at $C$. $M$ -midpoint of $AB$. Circle $S_1$ goes through $M,C$ and intersects $AB$ at $D$ and $S$ at $K$ and $L$. Prove, that tangent lines to $S$ at $K$ and $L$ intersects at point on the segment $CD$.

Prove that the distance between the midpoint of side $BC$ of triangle $ABC$ and the midpoint of the arc $ABC$ of its circumscribed circle is not less than $AB / 2$

The bisectors $BB_1$ and $CC_1$ of the acute triangle $ABC$ intersect in point $I$. On the extensions of the segments $BB_1$ and $CC_1$, the points $B'$ and $C'$ are marked, respectively So, the quadrilateral $AB'IC'$ is a parallelogram. Prove that if $\angle BAC = 60^o$, then the straight line $B'C'$ passes through the intersection point of the circumscribed circles of the triangles $BC_1B'$ and $CB_1C'$.

Given a convex quadrilateral $ABCD$. The medians of the triangle $ABC$ intersect at point $M$, and the medians of the triangle $ACD$ at point$N$. The circle, circumscibed around the triangle $ACM$, intersects the segment $BD$ at the point $K$ lying inside the triangle $AMB$ . It is known that $\angle MAN = \angle ANC = 90^o$. Prove that $\angle AKD = \angle MKC$.

A non-equilateral triangle $\triangle ABC$ of perimeter $12$ is inscribed in circle $\omega$ .Points $P$ and $Q$ are arc midpoints of arcs $ABC$ and $ACB$ , respectively. Tangent to $\omega$ at $A$ intersects line $PQ$ at $R$. It turns out that the midpoint of segment $AR$ lies on line $BC$ . Find the length of the segment $BC$.

oldies

1997
Let $B'$ be the antipode of $B$ on the circumcircle of triangle $ABC,$ let $I$ be the incenter of triangle $ABC,$ and let $M$ be the point where the incircle touches $AC.$ The points $K$ and $L$ are chosen on sides $AB$ and $BC,$ respectively, so that $KB = MC,$ $LB=AM.$ Prove that the lines $B'I$ and $KL$ are perpendicular.

The line S is tangent to the circumcircle of acute triangle ABC at B. Let K be the projection of the
orthocenter of triangle ABC onto line S (i.e. K is the foot of perpendicular from the orthocenter of triangle ABC to S). Let L be the midpoint of side AC. Show that triangle BKL is isosceles.

2002
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$ respectively. The perpendicular to the line $AA_{1}$ through the point $A_{1}$ intersects the line $B_{1}C_{1}$ at a point $X$. Prove that the line $BC$ bisects the segment $AX$.

Year Unknown
The point $I$ is the incenter of triangle $ABC$. A circle centered at $I$ meets $BC$ at $A_{1}$ and $A_{2}$, $CA$ at $B_{1}$ and $B_{2}$, and $AB$ at $C_{1}$ and $C_{2}$, where the points occur around the circle in the order $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$. Let $A_{3}$, $B_{3}$, $C_{3}$ be the midpounts of the arcs $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$, respectively. The lines $A_{2}A_{3}$ and $B_{1}B_{3}$ meet at $C_{4}$, $B_{2}B_{3}$ and $C_{1}C_{3}$ meet at $A_{4}$, and $C_{2}C_{3}$ and $A_{1}A_{3}$ meet at $B_{4}$. Prove that the lines $A_{3}A_{4}$, $B_{3}B_{4}$, $C_{3}C_{4}$ are concurrent.

Year Unknown2
Let ABCD be an isosceles trapezoid with bases AD and BC. Let a circle which is tangent to both AB and AC intersect segment BC at points M and N. Now, consider $\omega$, the incircle of triangle BCD. Let X and Y be the intersections (closer to D) of DM and DN with $\omega$. Prove that XY is parallel to BC.

Year Unknown3
Let S be a square, Q be the perimeter of the square, and P be the perimeter of a quadrilateral T inscribed within S such that each of its vertices lies on a different edge of S. What is the smallest possible ratio of P to Q?

official page: www.pdmi.ras.ru/~olymp/