Search found 98 matches
- Fri Jan 13, 2017 11:50 pm
- Forum: Junior Level
- Topic: An old jr Balkan problem
- Replies: 1
- Views: 2623
Re: An old jr Balkan problem
It is trivial that the condition is true when RHS is even. And if the RHS is odd, then LHS have to contain even number of even number. So, only hard case is to proof that it is impossible to have all integer odd in LHS. We know that every odd number can be written as $2k+1$ form. So every square of ...
- Fri Jan 13, 2017 11:24 pm
- Forum: Junior Level
- Topic: An old jr Balkan problem
- Replies: 1
- Views: 2623
An old jr Balkan problem
Let $n_1,n_2,\cdots ,n_{1998}$ be positive integers such that
$$\ n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2 $$
Show that at least two of the numbers are even.
$$\ n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2 $$
Show that at least two of the numbers are even.
- Thu Jan 12, 2017 9:27 pm
- Forum: Junior Level
- Topic: Coloring chessboard
- Replies: 1
- Views: 2367
Coloring chessboard
Assume on an $8 \times 8$ chessboard with the usual coloring. You may repaint all squares:
(a) of a row or colum.
(b) of a $2\times 2$ square
The goal is to attain just one black square. Can you reach the goal?
(a) of a row or colum.
(b) of a $2\times 2$ square
The goal is to attain just one black square. Can you reach the goal?
- Thu Jan 12, 2017 9:23 pm
- Forum: Junior Level
- Topic: Mod 8
- Replies: 1
- Views: 2374
Mod 8
If the $x^2+2y^2$ is an odd prime, then it has the form $8n+1$ or $8n+3$.
- Thu Jan 12, 2017 7:14 pm
- Forum: Secondary Level
- Topic: 2014-national
- Replies: 15
- Views: 12648
Re: 2014-national
I had mistakenly put this here. But i can't delete it. so i had edited that to something unnecessary.Thamim Zahin wrote:qqqqqqqqqqq
![Embarrassed :oops:](./images/smilies/icon_redface.gif)
Sorry for wasting your (and mine) time.
![Very Happy :D](./images/smilies/icon_e_biggrin.gif)
- Thu Jan 12, 2017 7:04 pm
- Forum: Secondary Level
- Topic: 2014-national
- Replies: 15
- Views: 12648
Re: 2014-national
qqqqqqqqqqq
- Thu Jan 12, 2017 3:45 pm
- Forum: Geometry
- Topic: IGO 2016 Medium/2
- Replies: 1
- Views: 2563
Re: IGO 2016 Medium/2
Extend the segment $QD$. Name it $C'$. $\angle CQB=\angle BAC=\angle a$ $\angle BAD=\angle DPB=\angle b$ [cyclic quad] $\angle DAP=\angle C'DP=\angle a$ [$DC'$ is tangent to $(DAP)$] Now, notice that $\angle DAC=\angle b+\angle c$ and $\angle DAP=\angle a$ So, we have to proof that $\angle c+\angle ...
- Wed Jan 11, 2017 7:57 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/3
- Replies: 1
- Views: 2339
Re: IGO 2016 Elementary/3
It is obvious that, if the angle of one side (the adjacent two angle of side) less than $180^o$, than the two parallelogram made by that side and the adjacent side are outside. Cause their angle sum is $180^o$. But the angle sum is less than $180^o$. Now take one more side adjacent to the side we ta...
- Wed Jan 11, 2017 4:06 pm
- Forum: Geometry
- Topic: IGO 2016 Medium/3
- Replies: 1
- Views: 2719
Re: IGO 2016 Mediam/3
It is possible for all integer $N \ge 3$ It is obvious that a triangle can't be partitioned in $1$ quadrilateral. Also if we divide a triangle into $2$ quadrilaterals, one is convex but other one is not. For $N=3$, take a equilateral triangle. And divide the triangle into $3$ congruent quadrilateral...
- Wed Jan 11, 2017 12:06 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies: 5
- Views: 4800
Re: IGO 2016 Elementary/2
$\angle CAP = \angle CBP= \angle CYP= \angle CXY= \angle x$ $\angle BAC =\angle BPC =\angle a = \angle YCA$ [Cause, $YCAB$ is a cyclic quad, and $BA=YC$] $\angle APB=\angle ACB=\angle b$ $\angle PCA=\angle PBA=\angle n$ [All of them by angle chasing] Now, $180- (\angle a + \angle x)= \angle n+ \angl...