## Search found 17 matches

Sat Jul 27, 2019 12:02 pm
Forum: Geometry
Topic: EGMO 2018 P5
Replies: 3
Views: 9968

### Re: EGMO 2018 P5

Let $D$ be the second intersection of $\tau$ and the bisector $\angle BCA$.Let $E,F$ be the tangency point of $\Omega$ with $AB$ and $\tau$ respectively. $Lemma$ 1: $DE \cdot DF=DB^2$ $Proof$ : $$\angle DFB= \frac{\angle C}{2}= \angle DBE$$. So, $\triangle DFB$ is similar to $\triangle DBE$ \right...
Fri Jul 19, 2019 4:19 pm
Topic: IMO 2019/P1
Replies: 1
Views: 6752

### Re: IMO 2019/P1

Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $f(2a)+2f(b)=f(f(a+b)).$ Proposed by Liam Baker, South Africa $f(x)=0$ or $2x+c$ are the only solution to this function where $c$ is an integer. $Proof:$ ...
Fri Jul 19, 2019 12:11 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2018 Problem 1
Replies: 1
Views: 6090

### Re: APMO 2018 Problem 1

Let $P$ be a point on the common tangent to $\left(BMH\right)$ and $\left(CMH\right)$ such that $P$ is further from $BC$ than $H$. Then $\angle{MHP}=\angle{MKH}=\angle{MBH}=\frac{\pi}{2}-A$and similarly $\angle{NHP}=\frac{\pi}{2}-A$, so $\angle{MHN}=\pi-2A$. Then \$\angle{MJN}=\frac{\pi}{2}+\frac{\...
Thu Feb 21, 2019 5:46 pm
Forum: Combinatorics
Topic: Placing Bishop in a chess board
Replies: 6
Views: 6466

### Re: Placing Bishop in a chess board

As bishops in the black squares don't attack the bishops on the white squares so, we can count at most how many bishops we can place on the white squares and then double the number to get the total. Now, lets divide the chess board into 8 white diagonals. We can place 8 bishops in those but we can't...
Tue Dec 04, 2018 8:00 pm
Forum: Algebra
Topic: Easy FE
Replies: 2
Views: 4468

### Re: Easy FE

but before impling f(n+2)=f(n)+k you have to proof that f(n) makes an arithmetic progression.
Tue Dec 04, 2018 3:20 pm
Forum: Combinatorics
Topic: Placing Bishop in a chess board
Replies: 6
Views: 6466

### Re: Placing Bishop in a chess board

Mon Dec 03, 2018 5:37 pm
Forum: Geometry
Topic: Secondary Special Camp 2011: Geometry P 4
Replies: 6
Views: 11416

### Re: Secondary Special Camp 2011: Geometry P 4

Let M be the midpoint of BC. Now,M,D,E,F are cyclic (nine point circle). So, PD.PM=PF.PE=PB.PC=PM^2 -BM^2 (as BFEC is cyclic). So,PM.PD=PM^2 - BM^2 PM^2- PD.DM - DM^2=PM^2-BM^2 => PD.DM=BM^2 - DM^2=BD.CD Now,<RCB= <RCD= <ACD= <EHA= <EFA= <RQA= <RQB. So, RCQB is cyclic. So, QD.RD=BD.CD=PD.DM So, PQMR...
Mon Dec 03, 2018 3:26 pm
Forum: Geometry
Topic: EGMO 2013/1
Replies: 2
Views: 7841

### Re: EGMO 2013/1

Join D,E.Let AD meets BE at M.
Now, C is the midpoint of BD and
EA:AC=2:1. So, A if the centroid of triangle EDB . So, M is the midpoint of BE. BM=EM=BE/2.
So, BE is the diameter of the circumcircle of triangle BAE. So,<BAE=90.So,<BAC=90.
Mon Dec 03, 2018 3:10 pm
Forum: Geometry
Topic: A hard Geometry Problem
Replies: 4
Views: 27441

### Re: A hard Geometry Problem

<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO
Wed Dec 20, 2017 3:16 pm
Forum: Junior Level
Topic: so big series
Replies: 2
Views: 2072

### Re: so big series

We can divide all the numbers in 7 parts as (mod 7),7n,7n+1,7n+2,7n+3,7n+4,7n+5,7n+6. now see, 7n+1+7n+6=14n+7,7n+2+7n+5=14n+7 and 7n+3+7n+4=14n+7.so if we take those number which are 1(mod 7) we cannot take those which are 6(mod 7) and so on. there are floor of (400-1)/7 or 57 numbers from 1 to 400...