Search found 181 matches
- Mon Jan 30, 2017 10:49 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2014/10
- Replies: 2
- Views: 2339
Re: BDMO NATIONAL JUNIOR 2014/10
Let $a_i$ be the number of chocolates eaten in total to day $i$. So, $0<a_1<a_2<...<a_{58}<101$ implies $15<a_1+15<a_2+15<...<a_{58}+15<116$. Now, among the $116$ numbers $a_1$,$...$, $a_{58}$, $a_1+15$,$...$, $a_{58}+15$. So there are indices $x$, $y$ such that $a_x+15=a_y$. Implying that Oindri at...
- Mon Jan 30, 2017 10:12 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2016/04
- Replies: 8
- Views: 5909
Re: BDMO NATIONAL Junior 2016/04
This is known as the perpendicular lemma. It is quite handy in proving perpendicularity. PROOF Let $AC\cap BD=E$. Now, apply the Law of Cosines on triangles $ABE$,$CBE$,$CDE$,$DAE$. Let $\angle AEB=\theta$. We have $$\begin{aligned} BA^2&=AE^2+BE^2-2(AE)(BE)(\cos \theta)\\ BC^2&=CE^2+BE^2+2(BE)(CE)(...
- Mon Jan 30, 2017 10:02 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2013/04
- Replies: 4
- Views: 3029
Re: BDMO NATIONAL Junior 2013/04
$a^{2012} + a^{2013}+.....+a^{3012}=a^{2012}(1+a+a^2+.....+a^{1000})$.
The number in the bracket is odd. So, the highest power of two dividing the whole number is the highest power of two dividing $a^{2012}$. Which is clearly $2012$. So, $n=2012$.
The number in the bracket is odd. So, the highest power of two dividing the whole number is the highest power of two dividing $a^{2012}$. Which is clearly $2012$. So, $n=2012$.
- Mon Jan 30, 2017 9:55 pm
- Forum: Divisional Math Olympiad
- Topic: Geometry
- Replies: 10
- Views: 7176
Re: Geometry
Oh, right. I don't know much English terms. Sorry about that. We are actually used to consider sides as diagonals too.
- Mon Jan 30, 2017 9:48 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL Junior 2016/05
- Replies: 2
- Views: 2555
Re: BDMO NATIONAL Junior 2016/05
$P = 1000a+100b+10c+d$, $Q = 1000d+100c+10b+a$.
So, $Q-P = 999(d-a)+90(c-b)$. Now, $0<=(c-b)<=(d-a)<=9$.
Plugging in values will solve the problem now.
So, $Q-P = 999(d-a)+90(c-b)$. Now, $0<=(c-b)<=(d-a)<=9$.
Plugging in values will solve the problem now.
- Mon Jan 30, 2017 9:37 pm
- Forum: Divisional Math Olympiad
- Topic: Geometry
- Replies: 10
- Views: 7176
Re: Geometry
It depends on your consideration of a diagonal. I computed the total number of intersection points.
- Mon Jan 30, 2017 6:37 pm
- Forum: Divisional Math Olympiad
- Topic: Geometry
- Replies: 10
- Views: 7176
Re: Geometry
Let $ABCD$ be the four points. $AB$ and $CD$ intersect at point $1$.
$AC$ and $BD$ intersect at point $2$.
$AD$ and $BC$ intersect at point $3$.
A total of $3$ points.
$AC$ and $BD$ intersect at point $2$.
$AD$ and $BC$ intersect at point $3$.
A total of $3$ points.
- Mon Jan 30, 2017 12:26 pm
- Forum: Geometry
- Topic: USAMO (GEOMETRY) 1995
- Replies: 1
- Views: 2226
Re: USAMO (GEOMETRY) 1995
From the definition of inversions, we get $A_2$ is the inverted $A_1$ wrt circle $ABC$.
So, $AA_2$ is just the $A$-symmedian of $\triangle ABC$. We know that the symmedians of a triangle meet at the symmedian point. The problem is solved.
So, $AA_2$ is just the $A$-symmedian of $\triangle ABC$. We know that the symmedians of a triangle meet at the symmedian point. The problem is solved.
- Sun Jan 29, 2017 2:18 am
- Forum: Geometry
- Topic: USA TST 2017
- Replies: 2
- Views: 2926
USA TST 2017
Let $ABC$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $\overline{AT}$ intersects the circumcircle of $\triangle BOC$ at two points $A_1$ and $A_2$, where $OA_1 < OA_2$. Points $B_1$, $B_2$, $C_1$, $C_2$...
- Fri Jan 27, 2017 1:08 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2015/02
- Replies: 3
- Views: 3262
Re: BDMO NATIONAL JUNIOR 2015/02
You should look up the method of slicing and dicing. These type of geometry problems require the figures to be cut and thenn have their area found. As the smaller circle is $1/2$ in diameter wrt the bigger circle, it's area is $1/4$ of the bigger one. Also, each region cut by the two lines are $1/4$...