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BdMO Online Forum • Search

## Search found 137 matches

### Re: Beginner's Marathon

As this is the Beginner's Marathon, I request everyone to not give shortlist problems. Problem $25$ Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\Omega$ is a circle passing through $A,B,C,D$. Let $\omega$ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ ...
Mon Jun 26, 2017 4:04 am

Forum: Junior Level
Topic: Beginner's Marathon
Replies: 66
Views: 2178

Solution to Problem $24$ ( NOT OVERKILL ) As the previous solution we solve by contradiction. So, assume that $a_1a_k \equiv a_k (mod n)$. $a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k-2}\cdot a_k\equiv\dots\equiv a_1\cdo... Mon Jun 26, 2017 3:58 am Forum: Junior Level Topic: Beginner's Marathon Replies: 66 Views: 2178 ### Re: Beginner's Marathon Solution to problem$24$( OVERKILL ) We prove by contradiction. Assume that$n$divides$a_k(a_1-1)$. Assume that$(a_i, n)=1$. Then$n| a_i(a_{i+1}-1)$implies$n|a_{i+1}-1$which implies$a_{i+1}=1$. Then$n|1(a_{i+2}-1)$which is only possible if$a_{i+2}=1$but this contradicts the distinction.... Mon Jun 26, 2017 3:44 am Forum: Junior Level Topic: Beginner's Marathon Replies: 66 Views: 2178 ### Re: Good inequality.. By AM-GM inequality,$ac+bc+ab \geq 3(abc)^{2/3}$. Also by AM-GM inequality,$a+b+c \geq 3(abc)^{1/3}$. So,$3 \geq 2+ \frac{3 (abc)^{1/3}}{(a+b+c)}$. Or,$ac+bc+ab \geq 3(abc)^{2/3} \geq (abc)^{2/3} (2+ \frac{3 (abc)^{1/3}}{(a+b+c)})$. Or,$ac+bc+ab \geq 2(abc)^{2/3} + \frac {3abc}{a+b+c}$. Or,$ac...
Mon Jun 26, 2017 1:51 am

Forum: Algebra
Topic: Good inequality..
Replies: 1
Views: 59

### Re: $x^2 \equiv x (mod n)$

If we denote $f(n)$ the desired number, then CRT implies $f(mn)=f(m)f(n)$ whenever $(m;n)=1$. So it suffices to find $f(n)$ when $n$ is a prime power. It's quite trivial that there are $2$ solutions for $x$ if $n$ is a prime power. So, the answer is $2^z$ where $z$ is the number of distinct prim...
Thu Jun 22, 2017 6:51 am

Forum: Number Theory
Topic: $x^2 \equiv x (mod n)$
Replies: 1
Views: 92

We rearrange the equation as, $n^5+n^4+1= 7^m$. Observe that $n^5+n^4+1=(n^2+n+1)(n^3-n+1)$. So, each of $n^2+n+1$ and $n^3-n+1$ is $7^x$ for some integer $x \geq 0$. But using the euclidean algorithm, we find that $gcd(n^2+n+1, n^3-n+1)$ divides $7$. So we consider the $4$ equations $n^2+n+1=1$ $... Thu Jun 22, 2017 6:20 am Forum: Number Theory Topic: n^5+n^4 Replies: 1 Views: 55 ### Re: Divisibility... with x,y The left hand side is even, so$ x^4+y^2$is even, and$ x, y$has the same parity. However, if$ x, y$are odd, we have that$ 4$divides$ 7^x-3^y$, but$ 4$doesn't divide$ x^4+y^2$. So$ x, y$are even, and let's say$ x=2x_1,y=2y_1$. You'll get$ (7^{x_1}-3^{y_1})(7^{x_1}+3^{y_1})$divides$ 4...
Thu Jun 22, 2017 5:42 am

Forum: Number Theory
Topic: Divisibility... with x,y
Replies: 1
Views: 60

### Re: IMO 2009 SL(G-1)

Ugly problem tbh. Let us assign $\angle BAC = 4a$. Then quick angle chases give us, $\angle CAK = a$, $\angle DCI = 45-a$, $\angle CEK = 3a$. We can apply sine law on $\triangle IEK$ and $\triangle CEK$ to get $\frac{IK}{KC} = \frac{sin(45)*sin (45-a)}{sin(3a)*sin(90-2a)}$ ........(relation 1)...
Thu Jun 22, 2017 4:53 am

Forum: Geometry
Topic: IMO 2009 SL(G-1)
Replies: 1
Views: 66

### Re: Beginner's Marathon

Problem $23$

Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
চিপা চিপা
Wed Jun 21, 2017 5:26 pm

Forum: Junior Level
Topic: Beginner's Marathon
Replies: 66
Views: 2178

### Re: Beginner's Marathon

They start, from Mirpur, one with the bike($A$) and the other($B$) on foot, and from Lalmatia($C$) on foot of course. After one hour $A$ will give the bike to $C$, and $B$ will stop, and wait for the bike. $A$ will reach Lamatia after 2 hours. $C$ rides the bike until he meets $B$ , and giv...
Tue Jun 20, 2017 10:39 pm

Forum: Junior Level
Topic: Beginner's Marathon
Replies: 66
Views: 2178
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