Found problems: 410
Let $H, K, L$ be the feet from the altitudes from vertices $A, B, C$ of the triangle $ABC$, respectively. Prove that
$| AK | \cdot | BL | \cdot| CH | = | HK | \cdot | KL | \cdot | LH | = | AL | \cdot | BH | \cdot | CK | $.
Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$. The circumcircle of $ADE$ cuts the circumcircle of $ABC$ at $F \neq A$. Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ concur at a point on $BC.$
The diagonals of a cyclic quadrilateral $ABCD$ meet at point $P$. The bisector of angle $ABD$ meets $AC$ at point $E$, and the bisector of angle $ACD$ meets $BD$ at point $F$. Prove that the lines $AF$ and $DE$ meet on the median of triangle $APD$.
$M$ is any point on the side $AB$ of the triangle $ABC$. $r,r_1,r_2$ are the radii of the circles inscribed in $ABC,AMC,BMC$. $q$ is the radius of the circle on the opposite side of $AB$ to $C$, touching the three sides of $AB$ and the extensions of $CA$ and $CB$. Similarly, $q_1$ and $q_2$. Prove that $r_1r_2q=rq_1q_2$.
On the sides $AB$,$BC$, and $CA$ of $\triangle ABC$ are chosen points $C_1$, $A_1$, and $B_1$ respectively,
in such way that $AA_1$, $BB_1$, and $CC_1$ intersect in one point $X$. If $\angle A_1C_1B = \angle B_1C_1A$, prove that $CC_1$ is perpendicular to $AB$.
Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle.
[i]Proposed by Dimitris Kontogiannis, Greece[/i]
Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that
\[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC.
\]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1.
\]
The inner bisections of the $ \angle ABC $ and $ \angle ACB $ angles of the $ ABC $ triangle are at $ I $. The $ BI $ parallel line that runs through the point $ A $ finds the $ CI $ line at the point $ D $. The $ CI $ parallel line for $ A $ finds the $ BI $ line at the point $ E $. The lines $ BD $ and $ CE $ are at the point $ F $. Show that $ F, A $, and $ I $ are collinear if and only if $ AB = AC. $
Let $ABC$ be a triangle with incenter $I$. The incircle of the triangle is tangent to the sides $BC$ and $AC$ in points $D$ and $E$, respectively. Let $P$ denote the common point of lines $AI$ and $DE$, and let $M$ and $N$ denote the midpoints of sides $BC$ and $AB$, respectively. Prove that points $M, N$ and $P$ are collinear.
[i](Proposed by Karl Czakler)[/i]
Consider 2 concentric circle radii $ R$ and $ r$ ($ R > r$) with centre $ O.$ Fix $ P$ on the small circle and consider the variable chord $ PA$ of the small circle. Points $ B$ and $ C$ lie on the large circle; $ B,P,C$ are collinear and $ BC$ is perpendicular to $ AP.$
[b]i.)[/b] For which values of $ \angle OPA$ is the sum $ BC^2 \plus{} CA^2 \plus{} AB^2$ extremal?
[b]ii.)[/b] What are the possible positions of the midpoints $ U$ of $ BA$ and $ V$ of $ AC$ as $ \angle OPA$ varies?
Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC] \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$
touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent.
In isosceles triangle $ABC(AC=BC)$ the point $M$ is in the segment $AB$ such that $AM=2MB,$ $F$ is the midpoint of $BC$ and $H$ is the orthogonal projection of $M$ in $AF.$ Prove that $\angle BHF=\angle ABC.$
Consider a triangle $ABC$, with the points $A_1$, $A_2$ on side $BC$, $B_1,B_2\in\overline{AC}$, $C_1,C_2\in\overline{AB}$ such that $AC_1<AC_2$, $BA_1<BA_2$, $CB_1<CB_2$. Let the circles $AB_1C_1$ and $AB_2C_2$ meet at $A$ and $A^*$. Similarly, let the circles $BC_1A_1$ and $BC_2A_2$ intersect at $B^*\neq B$, let $CA_1B_1$ and $CA_2B_2$ intersect at $C^*\neq C$. Prove that the lines $AA^*$, $BB^*$, $CC^*$ are concurrent.
In a convex quadrilateral $ABCD$, $E$ is the intersection of $AB$ and $CD$, $F$ is the intersection of $AD$ and $BC$ and $G$ is the intersection of $AC$ and $EF$. Prove that the following two claims are equivalent:
$(i)$ $BD$ and $EF$ are parallel.
$(ii)$ $G$ is the midpoint of $EF$.
Point $D$ is chosen on the Euler line of triangle $ABC$ and it is inside of the triangle. Points $E,F$ are were the line $BD,CD$ intersect with $AC,AB$ respectively. Point $X$ is on the line $AD$ such that $\angle EXF =180 - \angle A$, also $A,X$ are on the same side of $EF$. If $P$ is the second intersection of circumcircles of $CXF,BXE$ then prove the lines $XP,EF$ meet on the altitude of $A$
Proposed by [i]Alireza Danaie[/i]
In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$.
Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.
Let $AA_1$, $BB_1$, $CC_1$ be the altitudes of acute angled triangle $ABC$. $A_2$ be the touching point of the incircle of triangle $AB_1C_1$ with $B_1C_1$, points $B_2$, $C_2$ be defined similarly. Prove that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ concur.
Let $ ABC$ be a triangle and let $ P$ be a point on the angle bisector $ AD$, with $ D$ on $ BC$. Let $ E$, $ F$ and $ G$ be the intersections of $ AP$, $ BP$ and $ CP$ with the circumcircle of the triangle, respectively. Let $ H$ be the intersection of $ EF$ and $ AC$, and let $ I$ be the intersection of $ EG$ and $ AB$. Determine the geometric place of the intersection of $ BH$ and $ CI$ when $ P$ varies.
Let triangle $ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of its angle $\angle B$ meets the side $AC$ at a point $D$ and that $BC = BD+AD$.
Determine $\angle A$.
The circle $\omega_1$ intersects the circle $\omega_2$ in the points $A$ and $B$, a tangent line to this circles intersects $\omega_1$ and $\omega_2$ in the points $E$ and $F$ respectively. Suppose that $A$ is inside of the triangle $BEF$, let $H$ be the orthocenter of $BEF$ and $M$ is the midpoint of $BH$. Prove that the centers of the circles $\omega_1$ and $\omega_2$ and the point $M$ are collinears.
Let $ABC$, $AB<AC$ be an acute triangle inscribed in circle $\Gamma$ with center $O$. The altitude from $A$ to $BC$ intersects $\Gamma$ again at $A_1$. $OA_1$ intersects $BC$ at $A_2$ Similarly define $B_1$, $B_2$, $C_1$, and $C_2$. Then $B_2C_2=2\sqrt{2}$. If $B_2C_2$ intersects $AA_2$ at $X$ and $BC$ at $Y$, then $XB_2=2$ and $YB_2=k$. Find $k^2$.
[i]2017 CCA Math Bonanza Individual Round #15[/i]
Let $ABCD$ be a convex quadrilateral with $AB$ parallel to $CD$. Let $P$ and $Q$ be the midpoints of $AC$ and $BD$, respectively. Prove that if $\angle ABP=\angle CBD$, then $\angle BCQ=\angle ACD$.
Given a nonisosceles, nonright triangle ABC, let O denote the center of its circumscribed circle, and let $A_1$, $B_1$, and $C_1$ be the midpoints of sides BC, CA, and AB, respectively. Point $A_2$ is located on the ray $OA_1$ so that $OAA_1$ is similar to $OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1$, respectively, are defined similarly. Prove that lines $AA_2$, $BB_2$, and $CC_2$ are concurrent, i.e. these three lines intersect at a point.
Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$. Let $M$ be a point on the side $BC$ such that $\angle BAM = \angle DAC$. Further, let $L$ be the second intersection point of the circumcircle of the triangle $CAM$ with the side $AB$, and let $K$ be the second intersection point of the circumcircle of the triangle $BAM$ with the side $AC$. Prove that $KL \parallel BC$.