Search found 66 matches
- Sat May 27, 2017 1:07 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2017 P1
- Replies: 0
- Views: 6122
APMO 2017 P1
We call a $5$-tuple of integers arrangeable if its elements can be labeled $a, b, c, d, e$ in some order so that $a-b+c-d+e = 29$. Determine all $2017$-tuples of integers $n_1,n_2,...,n_{2017}$ such that if we place them in a circle in clockwise order, then any $5$-tuple of numbers in consecutive po...
- Mon May 15, 2017 6:02 pm
- Forum: Geometry
- Topic: ISL 2006 G3
- Replies: 2
- Views: 9061
Re: ISL 2006 G3
My solution Let $BD$ and $CD$ meet $AC$ and $AD$ at $X$ and $Y$ resp. lets define $\angle ACE = \beta$, $\angle DCE = \alpha$, $\angle BDA = \theta$, $\angle CDB = \gamma$ Now, we can easily get that $\bigtriangleup ABD \cong \bigtriangleup ACE$ and $\bigtriangleup BCD \cong \bigtriangleup CDE$ this...
- Mon May 15, 2017 5:30 pm
- Forum: Geometry
- Topic: ISL 2006 G3
- Replies: 2
- Views: 9061
ISL 2006 G3
let $ABCDE$ be a convex pentagon such that
$\angle BAC =\angle CAD =\angle DAE$ and $\angle ABC =\angle ACD =\angle ADE$.
Diagonals $BD$ and $CE$ meet at $P$. Prove that ray $AP$ bisects $CD$.
$\angle BAC =\angle CAD =\angle DAE$ and $\angle ABC =\angle ACD =\angle ADE$.
Diagonals $BD$ and $CE$ meet at $P$. Prove that ray $AP$ bisects $CD$.
- Thu May 11, 2017 5:34 pm
- Forum: Secondary Level
- Topic: BDMO 2017/09
- Replies: 1
- Views: 2625
Re: BDMO 2017/09
double post. the topic is discussed here viewtopic.php?f=13&t=3908
- Sat Apr 22, 2017 6:23 pm
- Forum: Algebra
- Topic: USA(J)MO 2017 #2
- Replies: 1
- Views: 4939
USA(J)MO 2017 #2
Consider the equation $(3x^3 + xy^2)(x^2y + 3y^3) = (x - y)^7$
a. Prove that there are infinitely many pairs$(x,y)$ of positive integers satisfying the equation.
b. Describe all pairs $(x,y)$ of positive integers satisfying the equation.
a. Prove that there are infinitely many pairs$(x,y)$ of positive integers satisfying the equation.
b. Describe all pairs $(x,y)$ of positive integers satisfying the equation.
- Sat Apr 22, 2017 6:06 pm
- Forum: Geometry
- Topic: USA(J)MO 2017 #3
- Replies: 6
- Views: 13305
USA(J)MO 2017 #3
Let $ABC$ be an equilateral triangle, and point $P$ on it's circumcircle. Let $PA$ and $BC$ intersect at $D$, $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$. Prove that the area of $\bigtriangleup DEF$ is twice the area of $\bigtriangleup ABC$
- Thu Apr 20, 2017 7:50 pm
- Forum: Geometry
- Topic: USA TST 2011/1
- Replies: 3
- Views: 4344
USA TST 2011/1
In acute scalene triangle $ABC$, $D,E,F$ be the feet of the perpendicular from $A,B,C$ to $BC,AC$ and $AB$ respectively.Let $H$ be the orthocenter. Points $P$ a.d $Q$ lie on segment $EF$ such that $AP$ and $HQ$ are perpendicular on $EF$. Lines $DP$ and $QH$ intersects at point $R$. Compute $\dfrac{H...
- Tue Apr 18, 2017 12:35 am
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies: 30
- Views: 47755
Re: BDMO Forum Mafia #2
I didn't understand the role darij grinberg. Please explain.
- Sat Apr 15, 2017 12:38 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 44492
Re: Beginner's Marathon
$\text{PROBLEM 13 :}$
Let $AD, BE, CF$ be concurrent cevians in a triangle, meeting at $P$. Prove that $\dfrac{PD}{AD} + \dfrac{PE}{BE} + \dfrac{PF}{CF} =1$.
Let $AD, BE, CF$ be concurrent cevians in a triangle, meeting at $P$. Prove that $\dfrac{PD}{AD} + \dfrac{PE}{BE} + \dfrac{PF}{CF} =1$.
- Sat Apr 15, 2017 12:33 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 44492
Re: Beginner's Marathon
$\text{SOLUTION TO PROBLEM 12}$ $f(1) + f(2) +..... f(2^n) = 2^0 + 2^1+...... 2^{n-1}$ PROOF: lemma 1:$ f(2k +1) = 0$ proof: trivial Lemma : $f(2^m + 1) + f(2^m + 2) +........f(2^{m+1}) = 2^m$ Proof : We use induction. Base case m = 1 is true. Now, let $g(m) = f(2^m + 1) + f(2^m + 2) +........f(2^{m...