Search found 4 matches
- Wed Nov 02, 2022 5:04 pm
- Forum: Combinatorics
- Topic: Dhaka H. Secondary 2014/7
- Replies: 1
- Views: 8822
Re: Dhaka H. Secondary 2014/7
Subsets containing 2 elements: Let the elements be ${a_1,a_2}$. It is obvious that, $1 \leq a_1 < a_2 -1 \leq 11$ . Set $a_1=b_1 , a_2-1=b_2$ . So we get, ,$1 \leq b_1 < b_2 \leq 11$ . We can choose $b_1,b_2$ in $11 \choose 2$ Subsets containing 3 elements: Let the elements be ${a_1,a_2,a_3}$. It is...
- Thu Oct 20, 2022 9:14 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2007#11
- Replies: 1
- Views: 6047
Re: BdMO National Secondary 2007#11
There are 3 possibles squares for 3 different bases. Let the base BC be $m$ and side of the square be $x$ . Now by heron's formula, $[ABC]=6 \sqrt{6}$ Let the vertex of the square B'C'LF in $AB$ be $B'$ and in $AC$ be $C'$ (and F,L in base BC) Now $\angle ABC=90-\angle BB'F =\angle AB'C'$ . Simillar...
- Sat Apr 23, 2022 11:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2009/3
- Replies: 4
- Views: 7386
Re: BdMO National Higher Secondary 2009/3
$CF$ isn't Perpendicular bisector Alternative solution: Let the intersection point of them be $H$, Perpendicular bisector of $AB$ be $HF$ and altitude from $B$ to $AC$ be $BD$. Now in $BF=AF$ , $HF$ common side , $\angle{BFH}=\angle{AFH}=90$. So, $\triangle{AFH}=\triangle{BFH}$ [RHS]. So $\angle{FBH...
- Fri Apr 22, 2022 1:51 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2008/8
- Replies: 4
- Views: 7287
Re: BdMO National Higher Secondary 2008/8
Let $\angle{CAB}=x$ $\angle{ABD}=f$ $\angle{CAD}=y$ $\angle{ADB}=z$ As $ABCD$ a cyclic quadrilateral . So $ \frac{56}{sin(x)}=\frac{60}{sin(f)}=\frac{39}{sin(z)}=\frac{CD}{sin(y)}=2R$ Now in $\triangle ABE$, $\frac{sin(f)}{45}=\frac{sin(x)}{BE}$ So, We get $BE=42$ Now in $\triangle ADE$ $\frac{sin(z...