Search found 312 matches
- Mon Feb 20, 2017 12:05 am
- Forum: Combinatorics
- Topic: Combi Solution Writing Threadie
- Replies: 10
- Views: 13345
Re: Combi Solution Writing Threadie
Problem 1 Let $n > 3$ be a fixed positive integer. Given a set $S$ of $n$ points $P_1, P_2,\cdots, P_n$ in the plane such that no three are collinear and no four concyclic, let $a_t$ be the number of circles $P_i P_j P_k$ that contain $P_t$ in their interior, and let $m(S) = a_1 + a_2 +\cdots + a_n...
- Sun Feb 19, 2017 10:19 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 44669
Re: Combi Marathon
Problem 1 Let $n > 3$ be a fixed positive integer. Given a set $S$ of $n$ points $P_1, P_2,\cdots, P_n$ in the plane such that no three are collinear and no four concyclic, let $a_t$ be the number of circles $P_i P_j P_k$ that contain $P_t$ in their interior, and let $m(S) = a_1 + a_2 +\cdots + a_n...
- Sat Jan 07, 2017 10:56 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
Problem 10:
Let $I$ be the incenter of $\triangle ABC$. The incircle touches $BC$ at $D$ and $K$ is the antipode of $D$ in $(I)$.
Let $M$ be the midpoint of $AI$. Prove that $KM$ passes through the Feuerbach Point.
Let $I$ be the incenter of $\triangle ABC$. The incircle touches $BC$ at $D$ and $K$ is the antipode of $D$ in $(I)$.
Let $M$ be the midpoint of $AI$. Prove that $KM$ passes through the Feuerbach Point.
- Sat Jan 07, 2017 11:13 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
Problem 8: Given a cyclic quadrilateral $ABCD$ with circumcircle $(O)$. Let $AB \cap CD=E, \ AD \cap BC=F, \ AC \cap BD=G, \ AC \cap EF=P, \ BD \cap EF=Q$. Let $M, \ N$ be midpoints of $AC, \ BD$, respectively and let $MN \cap EF=H$. (i) Prove that $M, \ N, \ P, \ Q$ are concyclic. (ii) Let $K$ be ...
- Fri Jan 06, 2017 10:09 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
Problem 5:
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ (ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ (ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.
- Fri Jan 06, 2017 9:46 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
$\text{Problem 3:}$ In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear. $\text{Another Solution:...
- Fri Jan 06, 2017 11:58 am
- Forum: Secondary Level
- Topic: TJMO 1996/2
- Replies: 4
- Views: 3815
Re: TJMO 1996/2
Our counting system is Decimal. It has $10$ digits: $0,1,2,3,4,5,6,7,8,9$.siwomcre wrote:What is trinary?
Trinary is another system of number counting. But it has only three digits: $0,1$ and $2$.
- Fri Jan 06, 2017 11:52 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
Problem 2 In $\triangle ABC$, $\angle ABC=90^{\circ}$. Let $D$ be any point on side $AC$, $D \neq A,C$. The circumcircle of $\triangle BDC$ and the circle with center $C$ and radius $CD$ intersect at $D,E$. Let $F$ be a point on side $BC$ so that $AF \parallel DE$. $X$ is another point on $BC$(Diff...
- Fri Jan 06, 2017 11:31 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 192598
Re: Geometry Marathon : Season 3
Problem 1 $ \triangle ABC,$ ,a right triangle with $ \angle A = 90^0 $, is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (ot...
- Wed Jan 04, 2017 11:14 pm
- Forum: Geometry
- Topic: ISL 2012 G4: angle biscector with circumcenter
- Replies: 3
- Views: 3565
Re: ISL 2012 G4: angle biscector with circumcenter
A synthetic solution: Our goal is to show that $\angle XCY=90^{\circ}$ (Samely, we can show that $\angle XBY=90^{\circ}$). To do this, we have to show that $\angle XCD=\angle EYC$ or $\dfrac {EC} {XD}=\dfrac {EY} {CD}$ or $XD \times EY=EC \times CD$. Let $AD \cap (O)=Q$. As $\triangle AXD \sim \tria...