## Search found 48 matches

Wed Feb 14, 2018 11:01 am
Forum: Geometry
Topic: Angles of tetrahedron
Replies: 3
Views: 8267

### 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: 61489

### 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: 61489

### 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: 61489

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: 61489 ### 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: 61489 ### 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: 61489 ### 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: 61489

### 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: 61489

### 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: 61489

### 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 $M$....