Search found 66 matches
- Tue Jan 10, 2017 11:39 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2016 : Junior 8
- Replies: 4
- Views: 6628
National BDMO 2016 : Junior 8
In $\bigtriangleup ABC$ , $\angle A = 20$, $\angle B = 80$, $\angle C = 80$, $BC = 12$ units. Perpendicular $BP$ is drawn on $AC$ from from $B$ which intersects $AC$ at the point $P$. $Q$ is a point on $AB$ in such a way that $QB = 6$ units. Find the value of $\angle CPQ$.
- Tue Jan 10, 2017 2:53 pm
- Forum: Geometry
- Topic: Similarity
- Replies: 1
- Views: 2600
Similarity
In isosceles $\bigtriangleup ABC (AB = AC), CB $ is extended through $B$ to $P$. A line form $P$ , parallel to altitude $BF$, meets $AC$ at $D$ (where $D$ is between $A$ and $F$). From $P$, a perpendicular is drawn to meet the extension of $AB$ at $E$ so that $B$ is between $E$ and $A$. Express $BF$...
- Thu Jan 05, 2017 8:07 pm
- Forum: Junior Level
- Topic: 2014 chittagong BDMO question 10
- Replies: 2
- Views: 3522
Re: 2014 chittagong BDMO question 10
Note that $ADCP$ is a square and $CD = \frac{1}{2}CB = CP = DA$
$BP^2 = CB^2 + CP^2 = CD^2 + (2CD)^2 = 5CD^2 = 340 \Longrightarrow CD^2 = 68 \Longrightarrow CD = 2\sqrt{17}$
So, $\bigtriangleup ABC = \frac{1}{2} * CB * AD = CD * CD = 68$
$BP^2 = CB^2 + CP^2 = CD^2 + (2CD)^2 = 5CD^2 = 340 \Longrightarrow CD^2 = 68 \Longrightarrow CD = 2\sqrt{17}$
So, $\bigtriangleup ABC = \frac{1}{2} * CB * AD = CD * CD = 68$
- Tue Jan 26, 2016 11:28 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Junior 2010/5
- Replies: 3
- Views: 3746
Re: Junior 2010/5
Solving without using Fermat's last theorem. $n^3-m^3 = 1331$ $=(n-m)(n^2+nm+m^2)$ And,$1331 = 11^3 $ Now,we have two cases, $(1).(n-m)=11, (n^2+nm+ m^2)=11^2 $ But if, $(n-m)=11, then 11^2=(n^2-2mn+m^2)$ and thus, case $1$ is obviously not true. $(2). (n-m)=1, (n^2+nm+m^2)= 11^3 $ Then,$(n-m)^2+3mn...
- Sun Jan 24, 2016 11:09 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2011/9
- Replies: 3
- Views: 3629
Re: BdMO National Junior 2011/9
The only prime $p$ possible is $3$. I proved that there cannot be any other $p$ more than $3$ using $Bertrand's $ $postulate$.
- Thu Jan 21, 2016 3:00 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Primary 6
- Replies: 3
- Views: 4615
Re: BdMO National 2013: Primary 6
A regular hexagon also meets the condition.