Found problems: 433
Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same half-plane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too.
[i]Proposed by Géza Kós, Hungary[/i]
[asy]
pathpen=black;
size(400);
pair A=(0,0), B=(4,0), C=(10,0);
draw(L(A,C,0.3));
MP("A",A); MP("B",B); MP("C",C);
pair X=(5,-7);
path G1=D(arc(X,C,A));
pair Y=(5,7), Z=(9,6);
draw(Z--B--Y);
struct T {pair C;real r;};
T f(pair X, pair B, pair Y, pair Z)
{
pair S=unit(Y-B)+unit(Z-B);
real s=abs(sin(angle((Y-B)/(Z-B))/2));
real t=10, r=abs(X-A);
pair Q;
for(int k=0;k<30;++k)
{
Q=B+t*S;
t-=(abs(X-Q)-r)/abs(S)-s*t;
}
T T=new T;
T.C=Q; T.r=s*t*abs(S);
return T;
}
void g(pair Q, real r)
{
real t=0;
for(int k=0;k<30;++k)
{
X=(5,t);
t+=(abs(X-Q)+r-abs(X-A));
}
}
pair Z1=(1.07,6);
draw(B--Z1);
T T=f(X,B,Y,Z1);
draw(CR(T.C,T.r));
T T=f(X,B,Y,Z);
draw(CR(T.C,T.r));
g(T.C,T.r);
path G2=D(arc(X,C,A));
T T=f(X,B,Y,Z1);
draw(CR(T.C,T.r));
T=f(X,B,Y,Z);
draw(CR(T.C,T.r));
g(T.C,T.r);
path G3=D(arc(X,C,A));
pen p=black+fontsize(8);
MC("\gamma_1",G1,0.85,p);
MC("\gamma_2",G2,0.85,NNW,p);
MC("\gamma_3",G3,0.85,WNW,p);
MC("h_1",B--Z1,0.95,E,p);
MC("h_2",B--Y,0.95,E,p);
MC("h_3",B--Z,0.95,E,p);
path[] G={G1,G2,G3};
path[] H={B--Z1,B--Y,B--Z};
pair[][] al={{S+SSW,S+SSW,3*S},{SE,NE,NW},{2*SSE,2*SSE,2*E}};
for(int i=0;i<3;++i)
for(int j=0;j<3;++j)
MP("V_{"+string(i+1)+string(j+1)+"}",IP(H[i],G[j]),al[i][j],fontsize(8));[/asy]
Let $ ABC$ be an isosceles triangle with $ AC \equal{} BC.$ Its incircle touches $ AB$ in $ D$ and $ BC$ in $ E.$ A line distinct of $ AE$ goes through $ A$ and intersects the incircle in $ F$ and $ G.$ Line $ AB$ intersects line $ EF$ and $ EG$ in $ K$ and $ L,$ respectively. Prove that $ DK \equal{} DL.$
As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$.
[asy]
size(200);
pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C);
draw(F--C--A--B--C^^A--D^^B--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("84", centroid(H, C, E), fontsize(9.5));
label("35", centroid(H, B, D), fontsize(9.5));
label("30", centroid(H, F, B), fontsize(9.5));
label("40", centroid(H, A, F), fontsize(9.5));[/asy]
Let $ABCD$ be a convex quadrilateral. Construct $S$ and $T$ on the side $AD$ and $AB$ respectively such that $AS=AT$. Construct $U$ and $V$ on the side $BC$ and $CD$ respectively such that $CU=CV$. Assume that $BT=BU$ and $ST, UV, BD$ are concurrent, prove that $AB+CD=BC+AD$.
Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.
Let $ABC$ be triangle such that $|AB| = 5$, $|BC| = 9$ and $|AC| = 8$. The angle bisector of $\widehat{BCA}$ meets $BA$ at $X$ and the angle bisector of $\widehat{CAB}$ meets $BC$ at $Y$. Let $Z$ be the intersection of lines $XY$ and $AC$. What is $|AZ|$?
$
\textbf{a)}\ \sqrt{104}
\qquad\textbf{b)}\ \sqrt{145}
\qquad\textbf{c)}\ \sqrt{89}
\qquad\textbf{d)}\ 9
\qquad\textbf{e)}\ 10
$
Let the tangent line passing through a point $A$ outside the circle with center $O$ touches the circle at $B$ and $C$. Let $[BD]$ be the diameter of the circle. Let the lines $CD$ and $AB$ meet at $E$. If the lines $AD$ and $OE$ meet at $F$, find $|AF|/|FD|$.
Given is a triangle $ABC$ with the property that $|AB| + |AC| = 3|BC|$. Let $T$ be the point on segment $AC$ such that $|AC| = 4|AT|$. Let $K$ and $L$ be points on the interior of line segments $AB$ and $AC$ respectively such that $KL \parallel BC$ and $KL$ is tangent to the inscribed circle of $\vartriangle ABC$. Let $S$ be the intersection of $BT$ and $KL$. Determine the ratio $\frac{|SL|}{|KL|}$
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
The exinscribed circle of a triangle $ABC$ corresponding to its vertex $A$ touches the sidelines $AB$ and $AC$ in the points $M$ and $P$, respectively, and touches its side $BC$ in the point $N$. Show that if the midpoint of the segment $MP$ lies on the circumcircle of triangle $ABC$, then the points $O$, $N$, $I$ are collinear, where $I$ is the incenter and $O$ is the circumcenter of triangle $ABC$.
In the non-isosceles triangle $ABC$,$D$ is the midpoint of side $BC$,$E$ is the midpoint of side $CA$,$F$ is the midpoint of side $AB$.The line(different from line $BC$) that is tangent to the inscribed circle of triangle $ABC$ and passing through point $D$ intersect line $EF$ at $X$.Define $Y,Z$ similarly.Prove that $X,Y,Z$ are collinear.
In convex quadrilateral $ABCD$, let diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $E$. Let the circumcircles of $ADE$ and $BCE$ intersect $\overline{AB}$ again at $P \neq A$ and $Q \neq B$, respectively. Let the circumcircle of $ACP$ intersect $\overline{AD}$ again at $R \neq A$, and let the circumcircle of $BDQ$ intersect $\overline{BC}$ again at $S \neq B$. Prove that $A$, $B$, $R$, and $S$ are concyclic.
[i]Tiger Zhang[/i]
In an acute-angled triangle $ABC, CD$ is the altitude. A line through the midpoint $M$ of side $AB$ meets the rays $CA$ and $CB$ at $K$ and $L$ respectively such that $CK = CL.$ Point $S$ is the circumcenter of the triangle $CKL.$ Prove that $SD = SM.$
Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$
Let $P$ be a point inside the acute triangle $ABC,$ and let $Q$ be the isogonal conjugate of $P.$ Let $L,M$ and $N$ be the midpoints of the shorter arcs $BC,CA$ and $AB$ of the circumcircle of $ABC,$ respectively. Let $X_A$ be the intersection of ray $LQ$ and circle $(PBC),$ let $X_B$ be the intersection of ray $MQ$ and circle $PCA,$ and let $X_C$ be the intersection of ray $NQ$ and circle $(PAB).$ Prove that $P,X_A,X_B$ and $X_C$ are concyclic or coincide.
[i]Proposed by Gustavo Cruz (São Paulo)[/i]
The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.
In the acute angle triangle $ABC$ the point $O$ is the center of the circumscribed circle and the lines $OA,OB,OC$ intersect sides $BC,CA,AB$ respectively in points $M,N,P$ such that $\angle NMP=90^o$.
[b](a)[/b] Find the ratios $\frac{\angle AMN}{\angle NMC}$,$\frac{\angle AMP}{\angle PMB}$.
[b](b)[/b] If any of the angles of the triangle $ABC$ is $60^o$, find the two other angles.
Point $ P$ lies on side $ AB$ of a convex quadrilateral $ ABCD$. Let $ \omega$ be the incircle of triangle $ CPD$, and let $ I$ be its incenter. Suppose that $ \omega$ is tangent to the incircles of triangles $ APD$ and $ BPC$ at points $ K$ and $ L$, respectively. Let lines $ AC$ and $ BD$ meet at $ E$, and let lines $ AK$ and $ BL$ meet at $ F$. Prove that points $ E$, $ I$, and $ F$ are collinear.
[i]Author: Waldemar Pompe, Poland[/i]
In the non-isosceles triangle $ABC$,$D$ is the midpoint of side $BC$,$E$ is the midpoint of side $CA$,$F$ is the midpoint of side $AB$.The line(different from line $BC$) that is tangent to the inscribed circle of triangle $ABC$ and passing through point $D$ intersect line $EF$ at $X$.Define $Y,Z$ similarly.Prove that $X,Y,Z$ are collinear.
(A.Myakishev, 9--10) Given a triangle $ ABC$. Point $ A_1$ is chosen on the ray $ BA$ so that segments $ BA_1$ and $ BC$ are equal. Point $ A_2$ is chosen on the ray $ CA$ so that segments $ CA_2$ and $ BC$ are equal. Points $ B_1$, $ B_2$ and $ C_1$, $ C_2$ are chosen similarly. Prove that lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ are parallel.
Let $ABCD$ be a convex quadrilateral such that $\angle{BAD} = 90^{\circ}$ and its diagonals $AC$ and $BD$ are perpendicular. Let $M$ be the midpoint of side $CD$, and $E$ be the intersection of $BM$ and $AC$. Let $F$ be a point on side $AD$ such that $BM$ and $EF$ are perpendicular. If $CE = AF\sqrt{2}$ and $FD = CE\sqrt{2}$, show that $ABCD$ is a square.
Betal marks $2021$ points on the plane such that no three are collinear, and draws all possible segments joining these. He then chooses any $1011$ of these segments, and marks their midpoints. Finally, he chooses a segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint using [b]only[/b] a straightedge. Can Vikram always complete this challenge?
[i]Note.[/i] A straightedge is an infinitely long ruler without markings, which can only be used to draw the line joining any two given distinct points.
[i]Proposed by Prithwijit De and Sutanay Bhattacharya[/i]
In a non isoceles triangle $ABC$, let the perpendicular bisector of $[BC]$ intersect $(ABC)$ at $M$ and $N$ respectively. Let the midpoints of $[AM]$ and $[AN]$ be $K$ and $L$ respectively. Let $(ABK)$ and $(ABL)$ intersect $AC$ again at $D$ and $E$ respectively, let $(ACK)$ and $(ACL)$ intersect $AB$ again at $F$ and $G$ respectively.
Prove that the lines $DF$, $EG$ and $MN$ are concurrent.
The incircle of $\vartriangle ABC$ is tangent to sides $\overline{BC}, \overline{AC}$, and $\overline{AB}$ at $D, E$, and $F$, respectively. Point $G$ is the intersection of lines $AC$ and $DF$ as shown. The sides of $\vartriangle ABC$ have lengths $AB = 73, BC = 123$, and $AC = 120$. Find the length $EG$.
[img]https://cdn.artofproblemsolving.com/attachments/d/a/aede28071a1a6b94bbe3ad8e1e104822b89439.png[/img]
Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.