Found problems: 3585
On the plane, a Cartesian coordinate system is chosen. Given points $ A_1,A_2,A_3,A_4$ on the parabola $ y \equal{} x^2$, and points $ B_1,B_2,B_3,B_4$ on the parabola $ y \equal{} 2009x^2$. Points $ A_1,A_2,A_3,A_4$ are concyclic, and points $ A_i$ and $ B_i$ have equal abscissas for each $ i \equal{} 1,2,3,4$.
Prove that points $ B_1,B_2,B_3,B_4$ are also concyclic.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$.
Remark: the original question has missed the condition $ AB \neq AC$
Let $ ABC$ be triangle, $ I$ its in-center; $ A_1,B_1,C_1$ be the reflections of $ I$ in $ BC, CA, AB$ respectively. Suppose the circum-circle of triangle $ A_1B_1C_1$ passes through $ A$. Prove that $ B_1,C_1,I,I_1$ are concylic, where $ I_1$ is the in-center of triangle $ A_1,B_1,C_1$.
$I$ is the incenter of a given triangle $\triangle ABC$. The angle bisectors of $ABC$ meet the sides at $D,E,F$, and $EF$ meets $(ABC)$ at $L$ and $T$ ($F$ is on segment $LE$.). Suppose $M$ is the midpoint of $BC$. Prove that if $DT$ is tangent to the incircle of $ABC$, then $IL$ bisects $\angle MLT$.
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.
[i]Author: Andrey Gavrilyuk, Russia[/i]
Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$.
[i]Proposed by Victor Domínguez and Ariel García[/i]
Let $ABC$ be a triangle such that $\angle BAC = 45^{\circ}$. Let $H,O$ be the orthocenter and circumcenter of $ABC$, respectively. Let $\omega$ be the circumcircle of $ABC$ and $P$ the point on $\omega$ such that the circumcircle of $PBH$ is tangent to $BC$. Let $X$ and $Y$ be the circumcenters of $PHB$ and $PHC$ respectively. Let $O_1,O_2$ be the circumcenters of $PXO$ and $PYO$ respectively. Prove that $O_1$ and $O_2$ lie on $AB$ and $AC$, respectively.
In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI = 2$ and $LD = 3$, then $IC = \tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
Cyclic quadrilateral $ABCD$ is such that $\angle BAD = 2\angle ADC$ and $CD = 2BC$. Let $H$ be the projection of $C$ onto $AD$. Prove that $BH \parallel CD$.
[i]Proposed by Fedir Yudin, Anton Trygub[/i]
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.
[i]Proposed by El Salvador[/i]
Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP=BC$ and $CQ=AB$. Let $M$ be the midpoint of $PQ$. Show that $MA\perp MC$.
Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$.
Given a quadrilateral $ABCD$, around which you can circumscribe a circle. The perpendicular bisectors of sides $AD$ and $CD$ intersect at point $Q$ and intersect sides $BC$ and $AB$ at points $P$ and $K$ resepctively. It turned out that the points $K, B, P, Q$ lie on the same circle. Prove that the points $A, Q, C$ lie on one line.
(Olena Artemchuk)
Let $ABC$ be a triangle and let $AD,BE,CF$ be its altitudes . $FA_{1},DB_{1},EC_{1}$ are perpendicular segments to $BC,AC,AB$ respectively.
Prove that : $ABC$~$A_{1}B_{1}C_{1}$
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. 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 $\angle ACB = 90^{\circ}$.
Point $M$ is a midpoint of side $BC$ of a triangle $ABC$ and $H$ is the orthocenter of $ABC$. $MH$ intersects the $A$-angle bisector at $Q$. Points $X$ and $Y$ are the projections of $Q$ on sides $AB$ and $AC$. Prove that $XY$ passes through $H$.
Quadrilateral $ALEX,$ pictured below (but not necessarily to scale!)
can be inscribed in a circle; with $\angle LAX = 20^{\circ}$ and $\angle AXE = 100^{\circ}:$
Let the circumcenter of triangle $ABC$ be $O$. $H_A$ is the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $BOC$ at $A'$. The projections of $A'$ onto $AB, AC$ are $D,E$, and $O_A$ is the circumcentre of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly.
Prove: $H_AO_A, H_BO_B, H_CO_C$ are concurrent
In an acute triangle $ABC$, let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $C$ to $AM$. Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$. Let $N$ be the midpoint of $\overline{AQ}$. Prove that $NB=NC$.
[i]Proposed by Holden Mui[/i]
The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are colored red and blue in an alternating fashion.
Suppose the extensions of the red sides determine a regular $100$-gon, as do the extensions of the blue sides.
Prove that the $50$ diagonals $\overline{A_1A_{101}},\ \overline{A_3A_{103}},\ \dots,
\ \overline{A_{99}A_{199}}$ are concurrent.
[i]Proposed by: [b]Ankan Bhattacharya[/b][/i]
In triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. Let $AI$ intersects $(ABC)$ second time at $P$ .
The line passes through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The feet of the perpendicular from $X$ to $IO$ is $Y$. Prove that $A,P,X,Y$ cyclic.
Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$.
Prove that lines $AD, PM$, and $BC$ are concurrent.
Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$.
[asy]
unitsize (1.2 cm);
pair A, B, C, D, M, N, O;
A = (0,0);
B = (2,0);
D = (1,3);
C = B + D - A;
M = extension(A, incenter(A,B,D), B, C);
N = extension(A, incenter(A,B,D), D, C);
O = circumcenter(C,M,N);
draw(D--A--B--C);
draw(interp(D,N,-0.1)--interp(D,N,1.1));
draw(A--interp(A,N,1.1));
draw(circumcircle(M,C,N));
label("$\circ$", A + (0.45,0.15));
label("$\circ$", A + (0.25,0.35));
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$M$", M, SE);
dot("$N$", N, dir(90));
dot("$O$", O, SE);
[/asy]
Let $ABC$ be a acute, non-isosceles triangle. $D,\ E,\ F$ are the midpoints of sides $AB,\ BC,\ AC$, resp. Denote by $(O),\ (O')$ the circumcircle and Euler circle of $ABC$. An arbitrary point $P$ lies inside triangle $DEF$ and $DP,\ EP,\ FP$ intersect $(O')$ at $D',\ E',\ F'$, resp. Point $A'$ is the point such that $D'$ is the midpoint of $AA'$. Points $B',\ C'$ are defined similarly.
a. Prove that if $PO=PO'$ then $O\in(A'B'C')$;
b. Point $A'$ is mirrored by $OD$, its image is $X$. $Y,\ Z$ are created in the same manner. $H$ is the orthocenter of $ABC$ and $XH,\ YH,\ ZH$ intersect $BC, AC, AB$ at $M,\ N,\ L$ resp. Prove that $M,\ N,\ L$ are collinear.