Search found 65 matches
- Tue Jan 31, 2017 5:35 pm
- Forum: Secondary Level
- Topic: Find (m , n)
- Replies: 1
- Views: 2312
Find (m , n)
Find all natural number $m,n$ such that $m^2 = 5.2^n + 1$
- Mon Jan 30, 2017 11:17 pm
- Forum: Divisional Math Olympiad
- Topic: regional mo 2015
- Replies: 3
- Views: 3798
Re: regional mo 2015
Use the Power of Point theorem to prove that $PA\times PD=PE^2=PB\times PC$ $PA \times PD = 12 \times (PA + AB + BC + CD) = 12 \times (12 + AB + BC + 2AB)$ $ = 36AB + 12BC + 144...(i)$ $PB \times PC = (PA+AB) \times (PA + AB + BC) = (12+AB) \times (12 + AB + BC)$ $ = AB^2 + 24AB + 12AB + AB \times ...
- Sun Jan 29, 2017 10:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/5
- Replies: 4
- Views: 3877
Re: BDMO NATIONAL JUNIOR 2015/5
Let there are $m$ boys and $n$ girls.Thus the number of Boys' handshake = $mn$.And the number of girls' handshake = $\frac{(n-1)^2+(n-1)}{2} = \frac{n^2 - n}{2}$
Thus,$mn + \frac{n^2-n}{2} = 40$
or,$2mn + n^2 - n = 80$
The only solutions of $(m,n)$ is $(40 , 1)$ and $(6,5)$
Thus,$mn + \frac{n^2-n}{2} = 40$
or,$2mn + n^2 - n = 80$
The only solutions of $(m,n)$ is $(40 , 1)$ and $(6,5)$
- Sun Jan 29, 2017 10:25 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/02
- Replies: 3
- Views: 3286
Re: BDMO NATIONAL JUNIOR 2015/02
The answer will be $\frac{25}{2}$
- Sun Jan 29, 2017 1:02 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/5
- Replies: 4
- Views: 3877
Re: BDMO NATIONAL JUNIOR 2015/5
The girls have already handshaked with boys.Will they handshake again?Moreover,$1$ boy makes $m$ handshakes.Isn't $n$ boys make in total $mn$ handshakes?
- Sat Jan 28, 2017 11:10 pm
- Forum: Junior Level
- Topic: Dhaka '15 \10
- Replies: 3
- Views: 3655
Re: Dhaka '15 \10
Tasnood's solution(latexed): Let $DF$ and $BC$ (Extended) meet at point $Q$. $FB:AB=1:7$, $FB:CD=1:8$. $AE:ED=1:3$, $AD:ED=4:3$. Between $\triangle DQC$ and $\triangle FQB$, $\angle Q$ is common and $\angle BFQ = \angle CDQ$ as $AB||CD$ and $QD$ is bisector. So, $\triangle DQC$~$\triangle FQB$. Then...
- Fri Jan 27, 2017 5:46 pm
- Forum: Secondary Level
- Topic: Dhaka secondary '16 \6
- Replies: 1
- Views: 2363
Re: Dhaka secondary '16 \6
My solution: Let the point of tangency of $PQ$ is $T$ and the point of tangency of $PQ$ is $T'$. Note that $\triangle PQO \cong \triangle PRO$.Thus $\angle POQ = \angle POR = 90$ degree. Thus $QR || XY$ , $\triangle PXY$and $\triangle PQR$ similar , $PX = PY$ and $QX = YR$.Moreover $XT = XT'$. Thus ...
- Fri Jan 27, 2017 5:26 pm
- Forum: Secondary Level
- Topic: Dhaka secondary '16 \6
- Replies: 1
- Views: 2363
Dhaka secondary '16 \6
Two tangent $PQ$ and $PR$ are drawn from external point $P$ to a circle with center $O$; where $Q, R$ are not the point of tangency. $Q, R$ are two points such that $PQ=PR$ and $O$ is the midpoint of the line $QR$. $X, Y$ are two points situated on $PQ$ and $PR$ respectively in such a way so that $X...
- Thu Jan 19, 2017 12:16 pm
- Forum: Junior Level
- Topic: Dhaka '15 \10
- Replies: 3
- Views: 3655
Dhaka '15 \10
$ABCD$ is a parallelogram. $E$ intersects $AD$ as $AE:ED =1:3$ and $F$ intersects $AB$ as $AF:FB=7:1$. $CE$ and $DF$ meets at point $P$. $CP:PE =$?
- Wed Jan 18, 2017 11:01 pm
- Forum: Junior Level
- Topic: Kushtia '15 \9
- Replies: 2
- Views: 2838
Re: Kushtia '15 \9
My solution: Just the use of pythagorus' theorem and $(a+b)^2 - (a-b)^2 = 4ab$. $E$ is a point on $CD$ such that $BE \perp CD$.Thus $CE = CD - AB,BE = AD$. Using the pythagorus' theorem,$BC^2 = BE^2 + CE^2 \rightarrow (CD+AB)^2 = (CD-AB)^2 + AD^2 $ $\rightarrow (CD+AB)^2 - (CD-AB)^2 = 8^2 \rightarro...