BdMO Online Forum

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...

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. $Proof:$ ...

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{\...

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...

but before impling f(n+2)=f(n)+k you have to proof that f(n) makes an arithmetic progression.

The answer is 14

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...

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.

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

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...