Search found 665 matches
- Sun Apr 14, 2013 12:03 am
- Forum: Geometry
- Topic: Line obtained by reflection passes through orthocentre
- Replies: 0
- Views: 1772
Line obtained by reflection passes through orthocentre
Some time ago, I was requested by Fahim vai to find a synthetic solution to the following problem: Let $P$ be a point on the circumcircle of $\triangle ABC$. Then the reflections of $P$ wrt the sides of $\triangle ABC$ lie on a line that passes through the orthocentre of $\triangle ABC$. I thought I...
- Sat Apr 13, 2013 11:40 pm
- Forum: Number Theory
- Topic: 2002 Czech-Polish-Slovak
- Replies: 4
- Views: 4148
Re: 2002 Czech-Polish-Slovak
$n|p-1 \Rightarrow n \le p-1 \Rightarrow n-1 \le p-2<p \Rightarrow p \nmid n-1$. Hence $p|n^3-1 \Rightarrow p|n^2+n+1$. Let $n^2+n+1=pk[k \in \mathbb{N}] \Rightarrow n|pk-1=k(p-1)+k-1 \Rightarrow n|k-1$. Let $k-1=nt[t \in \mathbb{N}_0]$. So $n^2+n+1=p(nt+1) \Rightarrow n(n+1-pt)=p-1 \Rightarrow n+1-...
- Thu Apr 11, 2013 11:59 pm
- Forum: Number Theory
- Topic: I Love Mr.Green
- Replies: 5
- Views: 4391
Re: I Love Mr.Green
$a,b \in N$ such that $\forall n \in N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$. :mrgreen: I guess you meant $a,b \in \mathbb{N}_0$. If b=0, then $a,2a$ are both perfect square but $\frac {2a}{a}=2$, which is not a perfect square, so a contradiction. So we assume $a,b \neq 0$. We define ...
- Thu Apr 04, 2013 12:15 am
- Forum: Higher Secondary Level
- Topic: Where is my incenter [self-made]
- Replies: 5
- Views: 5516
- Wed Apr 03, 2013 11:50 pm
- Forum: Geometry
- Topic: CMO 2013#5
- Replies: 4
- Views: 3960
Re: CMO 2013#5
Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\angle BOP = \angle ABC$, and let point $Q$ on side $AC$ be such that $\angle COQ = \angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle ...
- Wed Apr 03, 2013 11:45 pm
- Forum: Junior Level
- Topic: phi (n) divides n [self-made???]
- Replies: 1
- Views: 2419
- Thu Mar 28, 2013 11:58 pm
- Forum: Geometry
- Topic: IMO Shortlist 2005 G5
- Replies: 8
- Views: 5821
Re: IMO Shortlist 2005 G5
Actually a few days ago I proved a more generalized version of your lemma- Let $P,Q$ be on sides $AB,CD$ respectively of a convex quadrilateral $ABCD$(not necessarily cyclic) such that $\frac{AP}{PB}=\frac{DQ}{CQ}$. Then $A,P,I,Q$(where $I$ is the Miquel point) is cyclic. Proof: Use Zhao-2 and then...
- Thu Mar 28, 2013 11:11 pm
- Forum: Geometry
- Topic: IMO Shortlist 2005 G5
- Replies: 8
- Views: 5821
Re: IMO Shortlist 2005 G5
Lemma 2: (The main crux move)Let $ABCD$ be a cyclic quad with $AB\nparallel CD$. Suppose $AB\cap CD=E$. The circumcircles of triangles $EAD$ and $EBC$ meet again at $I$. (i.e. $I$ is the Miquel point of $ABCD$.) The angle bisector $\angle AFB;\; F=AC\cap BD$ meets $AB$ at $P$, $CD$ at $Q$. Then $P,...
- Thu Mar 28, 2013 10:53 pm
- Forum: Geometry
- Topic: IMO Shortlist 2005 G5
- Replies: 8
- Views: 5821
Re: IMO Shortlist 2005 G5
Let $\odot ABC \cap \odot ADE=I \neq A$ If we can prove $\angle AIH=90^{\circ}$, notice what happens. $IH$ goes through $A'$.[অর্ধবৃত্তস্থ কোণ] A half turn wrt $M$ sends $H$ to $A'$, so $M$ lies on $IH$ which means $MH \bot AI$, hence the proof is complete. Now to prove the claim. Note that $A$ is t...
- Wed Mar 27, 2013 8:18 pm
- Forum: Geometry
- Topic: One of the Coolest Problems I've Ever Seen (though many may
- Replies: 7
- Views: 5914
Re: One of the Coolest Problems I've Ever Seen (though many
Let me just finish Tahmid's proof from the part $MP = \frac 1 2 |b-c| $. Let $M'$ be the midpoint of $XW$. Then $2KL \leq 2(M'K + M'L) = (|YX - YW| + |XZ- XW|) = XY + XZ - YZ = XY$. So $XY \geq 2KL$ (also, I just noticed that the question was posted wrong :P edited now, sorry tahmid) Now I see what...