Search found 17 matches

by prottoy das
Tue Feb 19, 2019 10:29 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2017 National Round Secondary 7
Replies: 13
Views: 4193

Re: BdMO 2017 National Round Secondary 7

The problem is from Turkey Junior National math olympiad 2015,Q2!!! At least 3 of the problems of secondary category of nationa bdmo 2017 was stolen!!!
by prottoy das
Mon Feb 18, 2019 1:35 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National Secondary: Problem Collection(2016)
Replies: 17
Views: 5276

Re: BdMO National Secondary: Problem Collection(2016)

Let $CH\perp AB$. $BI$ & $DO$ meet at $E$. Now $BIOD$ cyclic so, $\angle IBD=\angle EOI$. Again, as $\angle ABC$ is bisected by $BI$ so $\angle IBD= \angle IBH$. So,$\angle IBH=\angle EBH=\angle EOH$, so the quadrangle $EDBH$ is cyclic. So $\angle BEO=\angle BHO=90^\circ$
by prottoy das
Fri Feb 15, 2019 11:52 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2017 National Round Secondary 10
Replies: 5
Views: 2267

Re: BdMO 2017 National Round Secondary 10

now the question is what's the exact answer? Some say it is $p-1$ & some say it is $p-2$!!!!!!! I think the answer is $p-1$
by prottoy das
Thu Feb 14, 2019 11:32 am
Forum: National Math Olympiad (BdMO)
Topic: BdMO 2017 National Round Secondary 10
Replies: 5
Views: 2267

Re: BdMO 2017 National Round Secondary 10

So sorry that it was an stolen problem. The 9th problem of National BDMO 2017, secondary, was also an stolen problem. This means at least two problems of the national olympiad was stolen!!!!
by prottoy das
Thu Jan 17, 2019 7:33 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2018 junior 10
Replies: 12
Views: 3500

Re: BDMO 2018 junior 10

And the first line of my solution has a typing mistake. Correct: As, $\triangle ABC$ is equilateral, So,$\angle CBD:\angle ABD=2:1$. So, similarly you can find $\angle BCE$. As, $\angle COD$ is the exterior angle of $\triangle BCO$, so, $\angle COD=\angle CBD+\angle BCE$. Then you will get $\angle C...
by prottoy das
Thu Jan 17, 2019 7:22 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2018 junior 10
Replies: 12
Views: 3500

Re: BDMO 2018 junior 10

I forgot to write in the problem $D$ and $E$ lies on $AC$ and $AB$.
by prottoy das
Wed Jan 16, 2019 11:22 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO National Junior 2018/7
Replies: 3
Views: 1181

Re: BDMO National Junior 2018/7

Probably it will be $4^4$ permutations. Because here, it is not said that a digit will be written one time in a number
by prottoy das
Wed Jan 16, 2019 11:17 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2018 junior 10
Replies: 12
Views: 3500

Re: BDMO 2018 junior 10

NABILA wrote:
Tue Jan 15, 2019 7:51 pm
@prottoy das,
Post the complete answer.
I posted the complete answer!!!!!!!!!! Did you mean to explain it?
by prottoy das
Tue Mar 20, 2018 2:59 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2018 junior 10
Replies: 12
Views: 3500

Re: BDMO 2018 junior 10

$\angle COD=20$ and $\angle BOE=40$ So $\angle COD=60$. Now $\angle EAD=\angle COD$. so, $ADOE$ is a cyclic quadrilateral. Join $D,E$. Here $AE={2/3}AB$ and $AD$ is half of $AE$ and their intersected angle is $60$. SO, $AED$ is a $30-60-90$ triangle. So,$ \angle AED=\angle AOD=30$ So,$\angle AOC= 60...
by prottoy das
Tue Mar 20, 2018 2:51 pm
Forum: National Math Olympiad (BdMO)
Topic: BDMO 2018 junior 10
Replies: 12
Views: 3500

BDMO 2018 junior 10

$ABC$ is an equilateral triangle. $D$ and $E$ is such a point that, $AD:CD=1:2$ and $BE:AE=1:2$. If $O$ is the intersection point of $BD$ and $CE$ find $ \angle AOC$ .