## Search found 138 matches

- Tue Feb 21, 2017 2:52 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 30: }$ Easy angle chase gives that $K$ is the midpoint of $\text{arc BC}$ not containing $A$.Now, Let $M$ be the midpoint of $BC$ and $L$ be the midpoint of $\text{arc BC}$ containing $A$. We get, $S,E,L$ are collinear and $ALME$ is cyclic. So radical axis theorem gives us...

- Mon Feb 20, 2017 4:16 pm
- Forum: Geometry
- Topic: A Problem for Dadu
- Replies:
**2** - Views:
**1721**

### A Problem for Dadu

Let $ABCD$ be a cyclic quadrilateral. Let $H_A, H_B, H_C, H_D$ denote the orthocenters of triangles $BCD, CDA, DAB$ and $ABC$ respectively. Prove that $AH_A, BH_B, CH_C$ and $DH_D$ concur.

- Fri Feb 17, 2017 1:35 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{Solution of Problem 28 :}$ Let, $T$ be the nine-point circle of $\bigtriangleup ABC$. Let $l_y$ be the radical axis of $T,(Y)$ and $l_z$ be the radical axis of $T,(Z)$.And let $A_1$=$l_y \cap l_z$. And the nagel point and centroid of $\bigtriangleup DEF$ be $M$ and $P$ respectively. $\text{L...

- Tue Feb 14, 2017 1:28 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{Problem 27:}$ Let $ABC$ be a scalene triangle, let $I$ be its incentre , and let $A_1,B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC,CA$ and $AB$ respectively. Prove that the circumcircles of the triangles $AIA_1,BIB_1$ and $CIC_1$ have a common point different fr...

- Mon Feb 13, 2017 11:24 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{**The Problem 24 has been posted before here as Problem no 4.}$ $\text{So I'll just give the solution of Problem 25**}$ $\text{Solution of Problem 25 :}$ Let, $A_1$ be a point such that $\bigtriangleup A_1BC$ is equilateral where $A_1$ lies on the side of $BC$ oppostie to $A$. We define $B_1,...

- Mon Feb 13, 2017 1:59 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{Problem 23:}$ Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ agai...

- Sat Feb 04, 2017 1:02 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

$\text{Problem 21:}$ Consider a circle $(O)$ and two fixed points $B,C$ on $(O)$ such that $BC$ is not the diameter of $(O)$. $A$ is an arbitrary point on $(O)$, distinct from $B,C$. Let $D,K,J$ be the midpoints of $BC,CA,AB$, respectively, $E,M,N$ be the feet of perpendiculars from $A$ to $BC$, $B...

- Sat Feb 04, 2017 12:58 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

That's the solution of problem 20 actually.rah4927 wrote:$\text{Solution of Problem } 21$

- Fri Feb 03, 2017 12:32 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

**$\text{Problem 20:}$**

Let $ABC$ be an arbitrary triangle,$P$ is the intersection point of the altitude from $C$ and the tangent line from $A$ to the circumcircle. The bisector of angle $A$ intersects $BC$ at $D$ . $PD$ intersects $AB$ at $K$, if $H$ is the orthocenter then prove : $HK\perp AD$

- Wed Feb 01, 2017 12:27 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**63421**

### Re: Geometry Marathon : Season 3

Aaaaa, sry but the coordinates of $A_2=(0:s-c:s-b)$ .... ur solu is correct though...rah4927 wrote:$A_2=(0:s-b:s-c)$