## Search found 57 matches

- Thu Jan 31, 2019 7:52 am
- Forum: Junior Level
- Topic: কোণের মান
- Replies:
**8** - Views:
**6499**

### Re: কোণের মান

There is something called Google.

- Wed Jan 30, 2019 9:05 pm
- Forum: Junior Level
- Topic: কোণের মান
- Replies:
**8** - Views:
**6499**

### Re: কোণের মান

The above solution is wrong. Isolate $\bigtriangleup ABC$ from the parallelogram $ABCD$.Your task is finding $\angle CPQ$.Now,notice that this problem has completely been reduced to the famous langley's adventetious angles.So,it follows that $\angle CPQ=70^{\circ}$.And finally your desired answers a...

- Tue Jan 08, 2019 4:19 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies:
**11** - Views:
**9873**

### Re: MPMS Problem Solving Marathon

Produce $DA$ upto $E$ such that $AE=BC$.It follows that $AEBC$ is a parallelogram.So,$BE=AC=BD=d$.By Stewart's theorem on $\bigtriangleup BED,d^2b+d^2c=m(a^2+bc)$ where $m=b+c$.Cancelling $b+c=m$ from both sides we finally have $d^2=a^2+bc$.

- Sat Mar 03, 2018 5:11 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2006/10
- Replies:
**5** - Views:
**2124**

### Re: BdMO National Secondary 2006/10

It suffices to prove that $\bigtriangleup PFQ$ is a right triangle.Now,let $OF$ and $CF$ extended meet the $2$ circles for the second time at $A$ and $B$ respectively.$\bigtriangleup APF$ and $\bigtriangleup BQF$ both are right triangles. So,by Alternate Segment Theorem,angle $FPQ$=angle $FAP$=$90-A...

- Sat Mar 03, 2018 4:37 pm
- Forum: Secondary Level
- Topic: Right Triangle Geometry
- Replies:
**7** - Views:
**8604**

### Re: Right Triangle Geometry

Let $F$ be such a point so that $ACDF$ is a rectangle.So,angle $MDF$=$60$°.Since $M$ is the circumcenter of $\bigtriangleup ABC$,$MC=MA$.Now,using perpendicularity lemma $MC^2+MF^2$=$MA^2+MD^2$ implying $MD=MF$.Since,angle $MDF$=$60$°,$\bigtriangleup MDF$ is equilateral.

- Tue Feb 27, 2018 6:12 pm
- Forum: Junior Level
- Topic: Diagonals of 8-gon
- Replies:
**2** - Views:
**7529**

### Re: Diagonals of 8-gon

The answer is $\frac{8(8-3)}{2}$=$20$.

- Tue Feb 27, 2018 5:58 pm
- Forum: Secondary Level
- Topic: Unsolvable equation
- Replies:
**1** - Views:
**1289**

### Re: Unsolvable equation

Multiply the whole equation by $a^2b^2$.The equation turns to $a^2+ab+b^2$=$a^2b^2$.Add $ab$ to both sides and so $(a+b)^2-(ab)^2=ab$.Observe that $a+b$>$ab$.Notice that $(ab+1)^2-(ab)^2$=$2ab+1$>$ab$.The rest is trivial.

- Fri Feb 09, 2018 9:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2008/1
- Replies:
**2** - Views:
**1749**

### Re: BdMO National Higher Secondary 2008/1

Tasnood,$0$ is not a positive integer. Now,the sum of the first $n$ even numbers is $n(n+1)$ and for the case of odd numbers,that is $n^2$.So,by the question, the answer is $2008(2008+1)-2008^2$=$2008$.

- Thu Jan 25, 2018 8:44 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 4
- Replies:
**5** - Views:
**2465**

### Re: BDMO 2017 National round Secondary 4

ahmedittihad meant $AD$ is perpendicular to $CD$.Let $C'$ be the reflection of $C$ in $M$.So,$CC'$=$13$.$C',A,D$ are collinear.Again,[$ABCD$]=[$AMCD$]+[$BMC$]=[$AMCD$]+[$AC'M$]=[$CC'D$].Suppose $BC$=$a$,$CD$=$b$,$AD$=$c$.Since $\bigtriangleup CC'D$ is a right triangle,it suffices to find $b(a+c)$.Fr...

- Sat Dec 16, 2017 7:37 pm
- Forum: Number Theory
- Topic: Prime numbers
- Replies:
**4** - Views:
**4948**

### Re: Prime numbers

$n$=$2,5,7$.