Search found 98 matches
- Sat Feb 11, 2017 11:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2017 : Junior 8
- Replies: 8
- Views: 5188
Re: National BDMO 2017 : Junior 8
Are there anyone who solve this in the exam time? Did you @dshasan?
- Thu Feb 02, 2017 7:11 pm
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2016 : Junior 8
- Replies: 4
- Views: 11033
Re: National BDMO 2016 : Junior 8
I had made the reflection but didn't get that it was a right triangle. Have to draw diagram as scale from now on.
- Tue Jan 31, 2017 9:05 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2016/05
- Replies: 2
- Views: 2663
Re: BDMO NATIONAL Junior 2016/05
Let $N = (d_3,d_2,d_1,d_0)$ mean $N = \overline{d_3d_2d_1d_0}$. For $P = (a,b,c,d)$ with $1 \leq a \leq b \leq c \leq d$ we consider the following three cases: Case 1: $d=a$. Then $a=b=c=d$, yielding $Q=P$, which means $X = P - Q = 0$ and $Y=0$. Hence $X + Y = 0$. Case 2: $d>a$ and $c=b$. Then $X = ...
- Tue Jan 31, 2017 8:11 pm
- Forum: Junior Level
- Topic: BDMO National Junior 2016/6
- Replies: 9
- Views: 11089
Re: BDMO National Junior 2016/6
It is no my solution. The credit goes to @thanicsamin $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$ Now, write this down in trinary form. It would be $ 1\overbrace{000 \cdots 000}^{2^{2w}} + 1\overbrace{000 \cdots 000}^{2^{3x}} + 1\overbrace{000 \cdots 000}^{2^{5y}} = 1\overbrace{000 \cdots 000}^{2^{7x}} $ No...
- Wed Jan 18, 2017 12:28 am
- Forum: Social Lounge
- Topic: Best of Luck :)
- Replies: 2
- Views: 15625
Re: Best of Luck :)
Thanks. (3 years later). I am taking this as 2017 :p
- Tue Jan 17, 2017 8:40 pm
- Forum: Geometry
- Topic: Sharygin Geometry Olympiad 2016 grade 9/2
- Replies: 0
- Views: 2242
Sharygin Geometry Olympiad 2016 grade 9/2
Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $X_A$ lying on the tangent at $H$ to the circumcircle of triangle $BHC$ is such that $AH=AX_A$ and $X_A \not= H$. Points $X_B,X_C$ are defined similarly. Prove that the triangle $X_AX_BX_C$ and the orthotriangle of $ABC$ are similar.
- Sun Jan 15, 2017 11:23 pm
- Forum: Geometry
- Topic: Sharygin(Russia) Geometry Olympiad 2016 grade 9/1
- Replies: 1
- Views: 2673
Re: Sharygin(Russia) Geometry Olympiad 2016 grade 9/1
We know that, $\angle DBC =\angle ADB =\angle FOC =\angle a $[last one is for tangency] $\angle CDB = \angle ABD =\angle b$ $\angle AGB = \angle DAG =\angle i$ $\angle GAC = \angle u$ $\angle AGD =\angle o$ and,$\angle DAC+\angle GAC =\angle DAC = \angle ACB =\angle i+\angle u$ Now from triangle $\t...
- Sun Jan 15, 2017 10:58 pm
- Forum: Geometry
- Topic: Sharygin(Russia) Geometry Olympiad 2016 grade 9/1
- Replies: 1
- Views: 2673
Sharygin(Russia) Geometry Olympiad 2016 grade 9/1
The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.
- Sun Jan 15, 2017 1:50 am
- Forum: Algebra
- Topic: Seems ugly, But cute
- Replies: 5
- Views: 4065
Re: Seems ugly, But cute
It took me more than 9 hours to do. Now the solution: If we multiple the first equation and the the third equation, we get: $(x_1+2x_2+3x_3+4x_4+5x_5)(x_1+2x_2+3x_3+4x_4+5x_5)$ $=$ $x^6_1+2x_2x^5_1+3x_3 x^5_1+4x_4 x^5_1+5x_5 x^5_1+$ $2x_1 x^5_2+4x^6_2+6x_3 x^5_2+8x_4 x^5_2+10x_5 x^5_2$ $+$ $3x_1 x^5...
- Sun Jan 15, 2017 12:18 am
- Forum: Algebra
- Topic: Seems ugly, But cute
- Replies: 5
- Views: 4065
Seems ugly, But cute
Find all positive real solutions $(x_1, x_2, x_3, x_4, x_5, a)$ to the following set of equations:
$$ \displaystyle\sum_{k=1}^{5} kx_k=a $$
$$ \displaystyle\sum_{k=1}^{5} kx^3_k=a^2 $$
$$ \displaystyle\sum_{k=1}^{5} kx^5_k=a^3 $$
$$ \displaystyle\sum_{k=1}^{5} kx_k=a $$
$$ \displaystyle\sum_{k=1}^{5} kx^3_k=a^2 $$
$$ \displaystyle\sum_{k=1}^{5} kx^5_k=a^3 $$