Search found 79 matches

by Fm Jakaria
Tue Jan 08, 2019 3:59 pm
Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 11
Views: 8872

Re: MPMS Problem Solving Marathon

Problem 5 A Hydra has 2019 heads and is immune to damage from conventional weapons. However, with one blow of a magical sword, Hercules can cut off its 9, 10, 11 or 12 heads. In each of these cases, 5, 18, 7 and 0 heads grow on its shoulder. The Hydra will die only if all the heads are cut off. Can...
by Fm Jakaria
Fri Jan 04, 2019 10:22 pm
Forum: News / Announcements
Topic: Spam
Replies: 1
Views: 3411

Spam

Where are our admins? This is clearly a spam and no one seems to care about it! And this and this are also spam posts, and these are all in our front page! Who knows how many of those have been posted here in our forum, and overlooked. Please do something about it! I just scrolled down our front pag...
by Fm Jakaria
Sat Jul 08, 2017 10:45 pm
Forum: Number Theory
Topic: Pairing up consecutive numbers may give a prime..(Self-made)
Replies: 1
Views: 3850

Pairing up consecutive numbers may give a prime..(Self-made)

Is it true that for each even positive integer $n$, the integers $1$ through $n$ can be paired with each other into $\frac{n}{2}$ pairs - so that the product of each pairs, when summed up - gives a prime number? For example, for $n = 8$, we can pair up $1,7$; $2,8$; $3,6$; $4,5$. Then $1*7+ 2*8+ 3*6...
by Fm Jakaria
Sat Aug 01, 2015 9:26 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2015 - Problem 1
Replies: 1
Views: 1511

Re: IMO 2015 - Problem 1

a) Consider any circle with it’s center as one point A. If n is odd, take any $\frac {n-1}{2}$ different equilateral triangles $A_i$ with one vertice A, other two on the circle; such that any two $A_i$ do not share vertices except A. If n is even, consider similar configuration of $\frac {n}{2}$ tri...
by Fm Jakaria
Sat Aug 01, 2015 8:31 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO 2015 - Problem 3
Replies: 1
Views: 1389

Re: IMO 2015 - Problem 3

Let $O$ be the center of $\gamma$, the circumcircle of ABC. $ AH = 2OM$, because$ O$ is the orthocenter of the medial triangle of $\triangle ABC$. So $P := AO \cap HM$ sends $OM$ to $AH$ by a homothety of ratio $2:1$. So $P$ lies on $\gamma$, and it is diametrically opposite to $ A$. As the projecti...
by Fm Jakaria
Sun Apr 12, 2015 4:41 am
Forum: Number Theory
Topic: USA TSTST 2012/3
Replies: 1
Views: 1537

Re: USA TSTST 2012/3

I solve the problem for a fixed general nonnegative integer $ c$. My solution: For $c = 0; f(n) = n$ for all $n$; so done. Assume $c > 0$. One thing is obvious, $f(n) > 1$ for all $n > 1$. Definitions: Call a finite nonempty sequence $A$ of positive integers ‘discrete’ if it contains no repeatetion ...
by Fm Jakaria
Thu Apr 02, 2015 8:50 pm
Forum: Number Theory
Topic: Diophantine-ness Preserving Functional equation(Self-made)
Replies: 0
Views: 1035

Diophantine-ness Preserving Functional equation(Self-made)

Suppose that a positive integer $n$ is given. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any polynomial $P$ with positive integer coefficients, with at least $n$ nonzero coefficients; and for any pair of positive integers $a,b$; if $P(a)$ divides $P(b)$; then $Q(a)$ also...
by Fm Jakaria
Wed Mar 04, 2015 2:40 pm
Forum: Number Theory
Topic: Form of the divisors
Replies: 2
Views: 1436

Re: Form of the divisors

It suffices to show that all prime divisors $p$ of $4n^2+1$ is of the form $4k+1$. This is obvious because $p$ must be odd and has $-1$ as a quadratic residue.
by Fm Jakaria
Fri Feb 27, 2015 8:14 pm
Forum: Geometry
Topic: Bulgaria 1996
Replies: 4
Views: 1838

Re: Bulgaria 1996

As mentioned in the solution, $IC$ is the line $l$.
by Fm Jakaria
Fri Feb 27, 2015 9:13 am
Forum: Geometry
Topic: Bulgaria 1996
Replies: 4
Views: 1838

Re: Bulgaria 1996

Let $C,Y,Z$ be pairwise touchpoints of $k_1,k_2,k$. Note that the circle $CYZ$ with center $I$(say) is the incircle of triangle $O_1O_2O$.Here $C,Y,Z$ are the points where incircle touches the sides. So $CI$ is the common tangent of $k_1,k_2$; so is both perpendicular to $O_1O_2$ and $AB$. Then $O_1...