## Search found 14 matches

- Thu Apr 20, 2017 10:17 pm
- Forum: Geometry
- Topic: USA TST 2011/1
- Replies:
**3** - Views:
**1098**

### Re: USA TST 2011/1

Well known- $A$ and $H$ are the excenter and incenter of $\bigtriangleup DEF$ respectively. So, $Q$ and $P$ are respectively incircle and excircle touchpoint on EF of $\bigtriangleup DEF$. By a well known lemma $R$ lies on the incircle of $\bigtriangleup DEF$. Since H is the incenter of $\bigtriangl...

- Sat Mar 25, 2017 1:41 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**7823**

### Re: Combi Marathon

Solution of problem17 If n is odd then the graph has an eularian path. Since the total number of edges is $1+2+...+n-1$, we can divide the eularian path into path of size $1,2,...,n-1$. Now if n is even then we use induction. In base case, when n=2, it is obvious. Now, suppose that the statement is...

- Tue Feb 21, 2017 2:22 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**15483**

### Re: Geometry Marathon : Season 3

Problem 30: Let $ABC$ be a triangle. Let $D$ be a point on $AB$ such that the intersection point of perpendicular bisector of AC and the line through $D$ parallel to the angle bisector of $\angle BAC$ is $P$ and it is inside $\bigtriangleup ABC$. Circumcircle of $ABC$ and $ADP$ intersects at $K$ (o...

- Tue Feb 21, 2017 1:52 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**15483**

### Re: Geometry Marathon : Season 3

solution to problem 29: Lemma: O,H,R are collinear and quadrilateral $HRCP$ is cyclic. proof: Assume that O,H,R are not collinear. Let $OH \cap BC= R_1$. Let $\angle APO= \theta$. Now, $\angle OR_1C= 180-\angle R_1OP-\angle OPA-\angle ACB= 180-\angle HOP-\theta-\angle ACB$ $= 180-2\angle DAC-\theta...

- Mon Feb 20, 2017 10:40 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**7823**

### Re: Combi Marathon

**Problem 4:**

Each edge of a polyhedron is oriented with an arrow such that at each vertex, there is at least on arrow leaving the vertex and at least one arrow entering the vertex. Does there always exists two faces on the polyhedron such that the edges on each of it's boundary form a directed cycle?

- Mon Feb 20, 2017 10:12 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**7823**

### Re: Combi Marathon

Solution to problem 3 We will show that all initial position with an odd number of black disks works. First we show that the number of black disks must be odd. Assign each of the black disks $-1$ and the white $1$. let M be the product of all of these numbers. M must be invariant since in each step...

- Sat Jan 07, 2017 1:22 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**15483**

### Re: Geometry Marathon : Season 3

**problem 7:**

Let $I$ be the incenter of triangle $ABC$. Prove that the nine point circles of triangle $AIB$, $BIC$ and $CIA$ are concurrent at the feuerbach point of triangle $ABC$.

- Fri Jan 06, 2017 11:41 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**15483**

### Re: Geometry Marathon : Season 3

Problem 6 In an acute triangle $ABC$ the points $D$,$E$ and $F$ are the feet of the altitudes through A,B,C respectively. The incenters of the triangle $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove th...

- Fri Jan 06, 2017 11:25 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**15483**

### Re: Geometry Marathon : Season 3

Solution of problem 5: we know,$SB=SI=SC$ since $K$ is on the angle bisector of $\angle$BSI $KB=KI$ And since $P$ is the reflection of $I$ across $KL$ $KP=KI$ SO $K$ is the circumcentre of $\triangle$BPI Similarly $L$ is the circumcentre of $\triangle$IPC Now, $\angle$BPC= $\angle$BPI+$\angle$IPC= ...

- Tue Aug 16, 2016 9:03 pm
- Forum: Geometry
- Topic: A self-made geo
- Replies:
**3** - Views:
**1047**

### Re: A self-made geo

Solution: apply an inversion centred A that takes the incircle to the A-excircle (let it be $\alpha$).we call the inversion $\sigma$.We define $P'$ as the inversion of any point, $P$. $\sigma$ takes BC to $\bigodot AB'C'$ which is tangent to $\alpha$.Obviously The point of tangency is $D$' .So, $\al...