Found problems: 114
Points $A, B$ move with equal speeds along two equal circles.
Prove that the perpendicular bisector of $AB$ passes through a fixed point.
Let $\Gamma_1$ be a circle with centre $A$ and $\Gamma_2$ be a circle with centre $B$, with $A$ lying on $\Gamma_2$. On $\Gamma_2$ there is a (variable) point $P$ not lying on $AB$. A line through $P$ is a tangent of $\Gamma_1$ at $S$, and it intersects $\Gamma_2$ again in $Q$, with $P$ and $Q$ lying on the same side of $AB$. A different line through $Q$ is tangent to $\Gamma_1$ at $T$. Moreover, let $M$ be the foot of the perpendicular to $AB$ through $P$. Let $N$ be the intersection of $AQ$ and $MT$.
Show that $N$ lies on a line independent of the position of $P$ on $\Gamma_2$.
In the circle $\omega$, we draw a chord $BC$, which is not a diameter. Point $A$ moves in a circle $\omega$. $H$ is the orthocenter triangle $ABC$. Prove that for any location of point $A$, a circle constructed on $AH$ as on diameter, touches two fixed circles $\omega_1$ and $\omega_2$.
(Dmitry Prokopenko)
Different points $A, B, C, D$ lie on a circle with a center at the point $O$ at such way that $\angle AOB$ $= \angle BOC =$ $\angle COD =$ $60^o$. Point $P$ lies on the shorter arc $BC$ of this circle. Points $K, L, M$ are projections of $P$ on lines $AO, BO, CO$ respectively . Show that
(a) the triangle $KLM$ is equilateral,
(b) the area of triangle $KLM$ does not depend on the choice of the position of point $P$ on the shorter arc $BC$
Let $A$ be a point and let k be a circle through $A$. Let $B$ and $C$ be two more points on $k$. Let $X$ be the intersection of the bisector of $\angle ABC$ with $k$. Let $Y$ be the reflection of $A$ wrt point $X$, and $D$ the intersection of the straight line $YC$ with $k$. Prove that point $D$ is independent of the choice of $B$ and $C$ on the circle $k$.
Inside the convex quadrilateral $ABCD$ lies the point $'M$. Reflect it symmetrically with respect to the midpoints of the sides of the quadrilateral and connect the obtained points so that they form a convex quadrilateral. Prove that the area of this quadrilateral does not depend on the choice of the point $M$.
Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.
Take an arbitrary point $D$ on side $BC$ of triangle $ABC$ and draw a circle through point $D$ and the centers of the circles inscribed in triangles $ABD$ and $ACD$. Prove that all circles obtained for different points $D$ of side $BC$ have a common point.
Two circles $\omega_1$ and $\omega_2$, which have centers $O_1$ and $O_2$, respectively, intersect at $A$ and $B$. A line $\ell$ that passes through $B$ cuts to $\omega_1$ again at $C$ and cuts to $\omega_2$ again in $D$, so that points $C, B, D$ appear in that order. The tangents of $\omega_1$ and $\omega_2$ in $C$ and $D$, respectively, intersect in $E$. Line $AE$ intersects again to the circumscribed circumference of the triangle $AO_1O_2$ in $F$. Try that the length of the $EF$ segment is constant, that is, it does not depend on the choice of $\ell$.
Given a minor sector $AOB$ (Here minor means that $ \angle AOB <90$). $O$ is the centre , chose a point $C$ on arc $AB$ ,Let $P$ be a point on segment $OC$ , join $AP$ , $BP$ , draw a line through $B$ parallel to $AP$ , the line meet $OC$ at point $Q$ ,join $AQ$ . Prove that the area of polygon $AQPBO$ does not change when points $P,C$ move .
[img]https://cdn.artofproblemsolving.com/attachments/3/e/4bdd3a20fe1df3fce0719463b55ef93e8b5d7b.png[/img]
Let $ABC$ be a triangle with no right angle, $E$ on the line $BC$ such that $\angle AEB = \angle BAC$ and $\Delta_A$ the perpendicular to $BC$ at $E$. Let the circle $\gamma$ with diameter $BC$ intersect $BA$ again at $D$. For each point $M$ on $\gamma$ ($M$ is distinct from $B$), the line $BM$ meets $\Delta_A$ at $M'$ and the line $AM$ meets $\gamma$ again at $M''$.
(a) Show that $p(A) = AM' \times DM''$ is independent of the chosen $M$.
(b) Keeping $B,C$ fixed, and let $A$ vary. Show that $\frac{p(A)}{d(A,\Delta_A)}$ is independent of $A$.
Michel Bataille
Let $ABC$ be a triangle inscribed in a circle $(O)$, and $M$ be some point on the perpendicular bisector of $BC$. Let $I_1, I_2$ be the incenters of triangles $MAB,MAC$. Prove that the incenters of triangles $A_II_1I_2$ are on a fixed line when $M$ varies on the perpendicular bisector.
Trần Quang Hùng
Points $A ,B ,C ,D$ lie on a line in this order. Draw parallel lines $a$ and $b$ through $A$ and $B$, respectively, and parallel lines $c$ and $d$ through $C$ and $D$, respectively, such that their points of intersection are vertices of a square. Prove that the side length of this square does not depend on the length of segment $BC$.
The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ meet at $O$. Let $m$ be the measure of the acute angle formed by these diagonals. A variable angle $xOy$ of measure $m$ intersects the quadrilateral by a convex quadrilateral of constant area. Prove that $ABCD$ is a square.
The equilateral triangle $ABC$ is inscribed in the circle $w$. Points $F$ and $E$ on the sides $AB$ and $AC$, respectively, are chosen such that $\angle ABE+ \angle ACF = 60^o$. The circumscribed circle of $\vartriangle AFE$ intersects the circle $w$ at the point $D$ for the second time. The rays $DE$ and $DF$ intersect the line $BC$ at the points $X$ and $Y$, respectively. Prove that the center of the inscribed circle of $\vartriangle DXY$ does not depend on the choice of points $F$ and $E$.
(Hilko Danilo)
On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $K$ are taken, respectively, so that $S_{KMC} + S_{KAC}=S_{ABC}$. Prove that all such lines $MK$ pass through one point.
Let $P$ be an arbitrary interior point of the equilateral triangle $ABC$. From $P$ draw parallel to the sides: $A'_1A_1 \parallel AB$, $B' _1B_1 \parallel BC$ and $C'_1C_1 \parallel CA$. Prove that the sum of legths $| AC_1 | + | BA_1 | + | CB_1 |$ is independent of the choice of point $P$.
[img]https://cdn.artofproblemsolving.com/attachments/5/a/15b06706c09e2458fb5938807b9f3833ffb62e.png[/img]
The trapezoid $ABCD$, of bases $AB$ and $CD$, is inscribed in a circumference $\Gamma$. Let $X$ a variable point of the arc $AB$ of $\Gamma$ that does not contain $C$ or $D$. We denote $Y$ to the point of intersection of $AB$ and $DX$, and let Z be the point of the segment $CX$ such that $\frac{XZ}{XC}=\frac{AY}{AB}$ . Prove that the measure of $\angle AZX$ does not depend on the choice of $X.$
Consider all the tetrahedrons $AXBY$, circumscribed around the sphere. Let $A$ and $B$ points be fixed. Prove that the sum of angles in the non-plane quadrangle $AXBY$ doesn't depend on points $X$ and $Y$ .
Let $BC$ be a chord not passing through the center of a circle $\omega$. Point $A$ varies on the major arc $BC$. Let $E$ and $F$ be the projection of $B$ onto $AC$, and of $C$ onto $AB$ respectively. The tangents to the circumcircle of $\vartriangle AEF$ at $E, F$ intersect at $P$.
(a) Prove that $P$ is independent of the choice of $A$.
(b) Let $H$ be the orthocenter of $\vartriangle ABC$, and let $T$ be the intersection of $EF$ and $BC$. Prove that $TH \perp AP$.
The chord $MN$ on the circle is fixed. For every diameter $AB$ of the circle consider the intersection point $C$ of the lines $AM$ and $BN$ and construct the line $\ell$ passing through $C$ perpendicularly to $AB$. Prove that all the lines $\ell$ pass through a fixed point.
(E. Kulanin, Moscow)
A circle is fixed, point $A$ is on it and point $K$ outside the circle. The secant passing through $K$ intersects circle at points $P$ and $Q$. Prove that the orthocenters of the triangle $APQ$ lie on a fixed circle.
In an equilateral triangle $ ABC $, point $ O $ is chosen and perpendiculars $ OM $, $ ON $, $ OP $ are dropped to the sides $ BC $, $ CA $, $ AB $, respectively. Prove that the sum of the segments $ AP $, $ BM $, $ CN $ does not depend on the position of point $ O $.
Let $ABCD$ be a convex quadrilateral with shortest side $AB$ and longest side $CD$, and suppose that $AB < CD$. Show that there is a point $E \ne C, D$ on segment $CD$ with the following property:
For all points $P \ne E$ on side $CD$, if we define $O_1$ and $O_2$ to be the circumcenters of $\vartriangle APD$ and $\vartriangle BPE$ respectively, then the length of $O_1O_2$ does not depend on $P$.
The quadrilateral $ABCD$ is inscribed in a fixed circle. It has $AB$ parallel to $CD$ and the length $AC$ is fixed, but it is otherwise allowed to vary. If $h$ is the distance between the midpoints of $AC$ and $BD$ and $k$ is the distance between the midpoints of $AB$ and $CD$, show that the ratio $h/k$ remains constant.