## Search found 17 matches

- Sat Jul 27, 2019 12:02 pm
- Forum: Geometry
- Topic: EGMO 2018 P5
- Replies:
**3** - Views:
**9874**

### 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
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P1
- Replies:
**1** - Views:
**6686**

### 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:
**6036**

### 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:
**6336**

### 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...

### 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:
**6336**

### Re: Placing Bishop in a chess board

The answer is 14

- Mon Dec 03, 2018 5:37 pm
- Forum: Geometry
- Topic: Secondary Special Camp 2011: Geometry P 4
- Replies:
**6** - Views:
**11279**

### 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:
**7781**

### 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.

Again AM=AD/2=BE/2.

So, BE is the diameter of the circumcircle of triangle BAE. So,<BAE=90.So,<BAC=90.

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.

Again AM=AD/2=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:
**27334**

### Re: A hard Geometry Problem

<CAH= 90 - <C

<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C

So, <CAH=<BAO

<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:
**2020**

### 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...