## Search found 66 matches

Wed Jun 28, 2017 4:23 pm
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 7853

### Re: Beginner's Marathon

Let $AD \cap \Omega = P$. Now, note that $AP = BB_1, AA_1 = CA_2$ and $BB_1 = CB_2$. Now, power of point implies that $AP.AD = AC.AA_1 \Rightarrow BB_1.BC = CA_2.AC \Rightarrow CA_2.AC=CB_2.BC$. Therefore, $A,B,A_2,B_2$ are cyclic.

Somebody post the next problem.
Fri Jun 23, 2017 6:17 pm
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 7853

### Re: Beginner's Marathon

$\text{Problem 24}$
Let $n$ be a positive integer and let $a_1, a_2,.....a_k$(here $k$ > 1) be distinct integers in the set {${1,2.....n}$} such that $n$ divides $a_i(a_{i+1}-1)$ for $i = 1,2,.....k-1$. Prove that $n$ does not divide $a_k(a_1 - 1)$
Fri Jun 23, 2017 6:07 pm
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 68
Views: 7853

$\text{Problem 23}$ Let $a^2 + b + c = (a + x)^2$$b^2 + c + a = (b + y)^2$$c^2 + a + b = (c + z)^2$where $x,y,z$ are positive integers.Now, from this 3 equations we get $b + c = x(2a + x)$$c + a = y(2b + y)$$a + b = z(2c + z)$Adding them yields $2(a + b + c) = 2(ax + by + cz) + x^2 + y^... Sat Jun 10, 2017 11:52 pm Forum: Number Theory Topic: x^2 \equiv x (mod n) Replies: 1 Views: 497 ### x^2 \equiv x (mod n) Let n be a positive integer. Determine, in terms of n, the number of x such that x \in {1,2,...n} and \[x^2 \equiv x(mod n)$
Mon Jun 05, 2017 10:32 am
Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 11
Views: 1846

$\text{Problem 3}$ Let $4n^2 - 6n + 45 = (2k+1)^2$ $\Rightarrow (2n-3)^2 +36 + 6n = (2k + 1)^2$ $\Rightarrow 36 + 6n = (2k + 1)^2 - (2n-3)^2$ $\Rightarrow 6(6 + n) = (2k + 2n -2)(2k -2n + 4)$ $\Rightarrow 3(6 + n) = (k + n -1)(k - n + 2)$ Now, for odd $n$, $3(6 + n)$ is odd. And $(k + n -1)$ and $(... Sun Jun 04, 2017 10:45 pm Forum: News / Announcements Topic: MPMS Problem Solving Marathon Replies: 11 Views: 1846 ### Re: MPMS Problem Solving Marathon$\text{Problem 1}$Stronger claim: There exists an integer$a$and prime$p$such that$p|a^2 + 1$if and only if$ p\equiv 1(mod 4)$Proof: For the first part, we have$a^2 + 1 \equiv -1(mod p)\Rightarrow (a^2)^{\dfrac{p-1}{2}} \equiv -1^{\dfrac{p-1}{2}}(mod p)\Rightarrow 1 \equiv -1^{\dfrac{...
Sat May 27, 2017 1:45 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2017 P5
Replies: 0
Views: 420

### APMO 2017 P5

Let $n$ be a positive integer. A pair of $n$-tuples $(a_1,....a_n)$ and $(b_1,...b_n)$ with integer entries is called an exquisite pair if $|a_1b_1 +...+ a_nb_n| \leq 1$. Determine the maximum number of distinct $n$-tuples with integer entries such that any two of them form an exquisite pair.
Sat May 27, 2017 1:40 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2017 P4
Replies: 0
Views: 484

### APMO 2017 P4

Call a rational number $r$ powerful if $r$ can be expressed in the form $\dfrac{p^k}{q}$ for some relatively prime positive integers $p,q$ and some integer $k > 1$. Let $a,b,c$ be positive rational numbers such that $abc = 1$. Suppose there exist positive integers $x,y,z$ such that $a^x + b^y + c^z$...
Sat May 27, 2017 1:36 pm
Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2017 P3
Replies: 0
Views: 366