## Search found 48 matches

- Wed Feb 14, 2018 11:01 am
- Forum: Geometry
- Topic: Angles of tetrahedron
- Replies:
**3** - Views:
**8263**

### Re: Angles of tetrahedron

yep, my counter example was wrong. Here's my solution : The 4 points are not coplaner. There are 4 triangles, each can have atmost one non-acute angle. So there can be atmost 4 non-acute angles in total. So if we assume there is no such vertex as stated above, then each triangle and each vertex shou...

- Fri Dec 29, 2017 12:45 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Lemma 5: Let $F_e$ be the feuerbach point of $\triangle ABC$. Then $F_e , A_o$ and $L_a$ are colinear . Better proof : $F_e , A_o$ and $L_a$ are the pairwise center of similitudes of the incircle , ninepoint circle and the $A$-excircle . So by d'Alembert's Theorem $F_e , A_o$ and $L_a$ are colinear .

- Thu Dec 14, 2017 5:57 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

As i promised before , here i am going to give my solution for Problem 38 . Solution of Problem 38 : Lemma 1: In a triangle $\triangle ABC$ ,let $M$ be the midpoint of $BC$ .Let $D$ be the projection of $A$ on $BC$ .Let $E$ be a point such that ,$E$ and $A$ are on the same side of $BC$ , $ED \| AM$ ...

- Thu Nov 30, 2017 6:15 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Solution of Problem 50 : Let $H$ be the orthocenter and $(N)$ be the nine point circle of $ \triangle ABC$. $NA'\cap (N)=\{A',L\}$. Let $J$ be a point on $LA'$ such that $JH_c$ is tangent to $(N)$ at $H_c$.$K$ is a point on $JH_c$ such that $KN \|A'H_c$ .Let $M$ be the midpoint of $AH_b$.Then $LM \...

- Sun Nov 26, 2017 1:40 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

We can also prove Problem $48$ by using the fact that (we are not giving a proof for this here) , the isogonal conjugate of a line wrt a triangle is a circumconic of that triangle .So the isogonal conjugate of that line must be a circumconic .But $ABCPP'QQ'$ cannot be a circumconic, as a line can cu...

- Sun Nov 26, 2017 1:37 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Solution of problem 48 : We will prove that for any line $L$ there can be at most one pair of isogonal conjugate point on that line . Let $I$ be the incenter and $I_b$ and $I_c$ be the B and C excenters respectively . Let $$L\cap BI_c=M$$ , $$L\cap CI_b=N$$ , $$L\cap BI=X$$ ,$$L\cap CI=Y$$ . We wan...

- Sat Nov 18, 2017 3:25 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Solution of Problem 46 : lets prove a generalization : "Given $ \triangle ABC $ and a point $ P $. Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE$ . Let $M$ be the midpoint of $BC$ . Let $ EF \c...

- Mon Oct 30, 2017 11:54 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Problem 46: Given a $ \triangle ABC $ with a point $ P $ lying on the A-bisector of $ \triangle ABC. $ Let $ \triangle DEF $ be the cevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be the point such that $ TB \parallel DF, $ $ TC \parallel DE. $ Prove that the perpendicular from $ T $ t...

- Mon Oct 30, 2017 11:15 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

Solution of problem 45: Let $OH \cap \omega = \{X,Y\}$ , $OH \cap P_aQ_a=J$ .Let $G$ be the centroid . $A'$ be the antipode of $A$ wrt $\omega$ , then $Q_a ,M_a,A'$ are collinear. Then ,$$(X,Y;J,G)\stackrel{P_a}{=}(X,Y;Q_a,A)\stackrel{A'}{=}(X,Y;H,O)$$ , which is symmetric for $A, B,C$ . So they co...

- Sat Apr 29, 2017 10:09 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**61191**

### Re: Geometry Marathon : Season 3

As problem 38 remained unsolved for many days , i will give another problem. (I will try to find a solution for this later when I get enough time . ) Problem 39 : Let $ABC$ be a triangle with circumcenter $O$.Let $BC$ be the smallest side. $(BOC)$ cuts $CA, AB$ at $E$ and $F$. $BE$ cuts $CF$ at $M$....