Search found 17 matches

by thczarif
Sat Jul 27, 2019 12:02 pm
Forum: Geometry
Topic: EGMO 2018 P5
Replies: 3
Views: 9615

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...
by thczarif
Fri Jul 19, 2019 4:19 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2019/P1
Replies: 1
Views: 6499

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. $Proof:$ ...
by thczarif
Fri Jul 19, 2019 12:11 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2018 Problem 1
Replies: 1
Views: 5858

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{\...
by thczarif
Thu Feb 21, 2019 5:46 pm
Forum: Combinatorics
Topic: Placing Bishop in a chess board
Replies: 6
Views: 6008

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...
by thczarif
Tue Dec 04, 2018 8:00 pm
Forum: Algebra
Topic: Easy FE
Replies: 2
Views: 4212

Re: Easy FE

but before impling f(n+2)=f(n)+k you have to proof that f(n) makes an arithmetic progression.
by thczarif
Tue Dec 04, 2018 3:20 pm
Forum: Combinatorics
Topic: Placing Bishop in a chess board
Replies: 6
Views: 6008

Re: Placing Bishop in a chess board

The answer is 14 :D :D
by thczarif
Mon Dec 03, 2018 5:37 pm
Forum: Geometry
Topic: Secondary Special Camp 2011: Geometry P 4
Replies: 6
Views: 10966

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...
by thczarif
Mon Dec 03, 2018 3:26 pm
Forum: Geometry
Topic: EGMO 2013/1
Replies: 2
Views: 7576

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.
by thczarif
Mon Dec 03, 2018 3:10 pm
Forum: Geometry
Topic: A hard Geometry Problem
Replies: 4
Views: 26961

Re: A hard Geometry Problem

<CAH= 90 - <C
<BAO = (180 - <BOA)/2=(180 - 2<C)/2=90 - <C
So, <CAH=<BAO
by thczarif
Wed Dec 20, 2017 3:16 pm
Forum: Junior Level
Topic: so big series
Replies: 2
Views: 1849

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...