### Silk Road 2002-19 (SRMC) (Kazakhstan) 17p

geometry problems from Silk Roal Mathematics Competitions (SRMC)
with aops links in the names

2002 - 2019

Let $\triangle ABC$ be a triangle with incircle $\omega(I,r)$and circumcircle $\zeta(O,R)$.Let $l_{a}$ be the angle bisector of $\angle BAC$.Denote $P=l_{a}\cap\zeta$.Let $D$ be the point of tangency $\omega$ with $[BC]$.Denote $Q=PD\cap\zeta$.Show that $PI=QI$ if $PD=r$.

Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and      $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$.

In-circle of $ABC$ with center $I$ touch $AB$ and $AC$ at $P$ and $Q$ respectively. $BI$ and $CI$ intersect $PQ$ at $K$ and $L$ respectively. Prove, that circumcircle of $ILK$ touch incircle of $ABC$ iff $|AB|+|AC|=3|BC|$

Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$ are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$

Let $\omega$ be the incircle  of triangle $ABC$ touches $BC$  at point $K$ . Draw a circle passing through points $B$ and $C$ , and touching  $\omega$ at the point $S$ . Prove that $S K$ passes through the center of the exscribed circle of triangle $A B C$ , tangent to side $B C$ .

Let $ABC$ be a triangle and $A_0,B_0,C_0$ be the midpoints of $BC,CA,AB$,respectively.And $A_1,B_1,C_1$ be the midpoints of broken lines $BAC$,$ABC$,$ACB$,respectively.Prove that $A_0A_1,B_0B_1,C_0C_1$ are concurrent.

Bisectors of triangle $ABC$ of an angles $A$ and $C$ intersect with $BC$ and $AB$ at points $A_1$ and $C_1$ respectively. Lines $AA_1$ and $CC_1$ intersect circumcircle of triangle $ABC$ at points $A_2$ and $C_2$ respectively.$K$ is intersection point of $C_1A_2$ and $A_1C_2. I$ is incenter of $ABC$. Prove that the line $KI$ divides $AC$ into two equal parts.

In a convex quadrilateral  it is known $ABCD$ that  $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^{\circ}$  and $AD = BC$. Prove that from the lengths $DB$, $CA$ and  $DC$, you can make a right triangle.

Given an isosceles triangle $ABC$ with base $AB$. Point $K$ is taken on the extenstion of the side $AC$ (beyond the point $C$ ) so that  $\angle KBC = \angle ABC$. Denote  $S$ the  intersection point of angle - bisectors of $\angle BKC$ and $\angle ACB$. Lines  $AB$ and  $KS$ intersect at  point $L$,  lines $BS$ and  $CL$  intersect at point $M$ . Prove that line $KM$ passes through the middle of the segment $BC$.

Trapezium $ABCD$, where $BC||AD$, is inscribed in a circle,  $E$ is middle of the arc  $AD$  of this circle not containing point $C$ . Let $F$ be the base of the perpendicular dropped from   $E$ on the line tangent to the circle at the point $C$ . Prove that $BC=2CF$.

Circle with center  $I$, inscribed in a triangle $ABC$ , touches the sides $BC$ and  $AC$ at  points  $A_1$ and $B_1$ respectively. On rays $A_1I$  and $B_1I$, respectively,  let be the points  $A_2$ and  $B_2$ such that $IA_2=IB_2=R$, where $R$is the radius of the circumscribed circle of the triangle $ABC$. Prove that:
a) $AA_2 = BB_2 = OI$ where $O$ is the center of the circumscribed circle of the triangle $ABC$,
b) lines  $AA_2$ and $BB_2$ intersect on the circumcircle of the triangle $ABC$.

Let $w$ be the circumcircle of non-isosceles acute triangle $ABC$. Tangent lines to $w$ in $A$ and $B$ intersect at point $S$. Let M be the midpoint of $AB$, and $H$ be the orthocenter of triangle $ABC$. The line $HA$ intersects lines $CM$ and $CS$ at points $M_a$ and $S_a$, respectively. The points $M_b$ and $S_b$ are defined analogously. Prove that $M_aS_b$ and $M_bS_a$ are the altitudes of triangle $M_aM_bH$.

Let $O$ be a circumcenter of an acute-angled triangle$ABC$. Consider two circles $\omega$ and $\Omega$ inscribed in the angle B$AC$ in such way that ω is tangent from the outside to the arc $BOC$ of a circle circumscribed about the triangle $BOC$, and the circle $\Omega$ is tangent internally to a circumcircle of triangle $ABC$. Prove that the radius of $\Omega$ is twice the radius $\omega$.

Around the acute-angled triangle  $ABC$ ($AC>CB$)  a circle is circumscribed, and the point $N$ is midpoint of the arc $ACB$ of this circle. Let the points $A_1$  and  $B_1$ be the bases of perpendiculars on the straight line $NC$, drawn from points  $A$ and $B$  respectively (segment $NC$  lies inside the segment $A_1B_1$). Altitude $A_1A_2$ of triangle $A_1AC$  and altitude $B_1B_2$ of triangle $B_1BC$  intersect at a point $K$ . Prove that $\angle A_1KN=\angle B_1KM$, where $M$ is midpoint of the segment $A_2B_2$ .

Quadrilateral $ABCD$ inscribed in a circle $\omega$. Diagonals $AC$ and $BD$ intersect at a point $O$. On segments $AO$ and $DO$ points $E$ and $F$  are selected respectively. Straight $EF$ crosses $\omega$ in points $E_{1}$ and $F_{1}$. The circumscribed circles of triangles $ADE$ and $BCF$ cross the segment $EF$ in points $E_{2}$ and $F_{2}$ respectively (assume that all points $E$, $E_{1}$, $E_{2}$, $F$, $F_{1}$, $F_{2}$ are different). Prove that $E_{1}E_{2}=F_{1}F_{2}$

In an acute-angled triangle  $ABC$ on the sides $AB$, $BC$, $AC$   the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices  $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and   $H$ lie on the line $PQ$.
The altitudes of the acute-angled non-isosceles triangle $ABC$ intersect at the point $H$. On the segment $C_1H$, where $CC_1$ is the altitude of the triangle, the point $K$ is marked. Points $L$ and $M$ are the bases of perpendiculars from point $K$ to straight lines $AC$ and $BC$, respectively. The lines $AM$ and $BL$ intersect at $N$. Prove that $\angle ANK = \angle HNL$.