Search found 136 matches
- Mon Nov 10, 2014 7:50 pm
- Forum: Number Theory
- Topic: Perfect Square ratio
- Replies: 5
- Views: 5831
Re: Perfect Square ratio
And that can be shown in two different ways.
- Mon Nov 10, 2014 7:46 pm
- Forum: Number Theory
- Topic: Perfect Square ratio
- Replies: 5
- Views: 5831
Re: Perfect Square ratio
which is impossible.mutasimmim wrote:The reasoning falls apart if $ x $ is greater than $ y $
- Mon Nov 10, 2014 3:32 am
- Forum: Number Theory
- Topic: Perfect Square ratio
- Replies: 5
- Views: 5831
Re: Perfect Square ratio
If $x\neq y$ then \[\mathbb{N}\ni\dfrac{x^2+2y^2}{2x^2+y^2}=1+\dfrac{y^2-x^2}{2x^2+y^2}~\Rightarrow ~ 2x^2+y^2\mid y^2-x^2\] \[\Rightarrow ~ 2x^2+y^2\le y^2-x^2 ~\Rightarrow ~ 3x^2\le 0 ~\Rightarrow ~ x=0\not\in\mathbb{N} \] hence we must have $x=y$, thus all pairs $\mathbb{N}^2\ni (x,y)=(n,n)$ works.
- Sat Nov 08, 2014 10:36 pm
- Forum: Geometry
- Topic: A Self Posed Geo Prob
- Replies: 2
- Views: 2594
Re: A Self Posed Geo Prob
Yes, I noticed later that the problem is trivial. Actually I started with a triangle with three altitudes, and inverted everything wrt a vertex as center. So my intended solution was: invert everything wrt $\omega$, then the problem becomes to prove that altitudes of a triangle concur. Missed that a...
- Sat Nov 08, 2014 8:06 pm
- Forum: Junior Level
- Topic: Sequence problem
- Replies: 2
- Views: 3559
Re: Sequence problem
Convert the terms to base $4$. You'll get the sequence of consecutive binary naturals.
So to find the $n$-th term, convert $n$ to binary, consider the result to be in base $4$ and convert to decimal.
So to find the $n$-th term, convert $n$ to binary, consider the result to be in base $4$ and convert to decimal.
- Sat Nov 08, 2014 2:54 pm
- Forum: Geometry
- Topic: A Self Posed Geo Prob
- Replies: 2
- Views: 2594
A Self Posed Geo Prob
Consider a circle $\omega$ with center $O$. A chord $AB$ subtends an acute angle at the center. Points $P$ and $Q$ are taken on segments $OA$ and $OB$ respectively, and two circles $\omega_1$ and $\omega_2$ are drawn with diameters $OP$ and $OQ$ respectively. Suppose $\omega_1\cap OB=M$ and $\omega_...
- Sat Nov 08, 2014 2:48 pm
- Forum: Secondary Level
- Topic: Line in 2D figure
- Replies: 1
- Views: 2654
- Wed Oct 22, 2014 5:44 pm
- Forum: Secondary Level
- Topic: Functional divisibility
- Replies: 3
- Views: 3874
Re: Functional divisibility
Really? :? Clearly $f^{100}(n)=3^{100}n+a_{100}$ where $a_n=3a_{n-1}+2~\forall ~n\in\mathbb{N}$ and $a_1=2$. So we have to prove that $1998m-3^{100}n=a_{100}$ has natural solutions. But taking mod $3$ on both side shows that this is absurd as $3\mid 1998$ and $a_{100}=3a_{99}+2\equiv 2~(\bmod~3)$. A...
- Tue Oct 21, 2014 11:19 pm
- Forum: Algebra
- Topic: (Angle)^(Side)[Inequality]
- Replies: 7
- Views: 6029
Re: (Angle)^(Side)[Inequality]
Inverting and taking $\ln$ gives us to prove \[\dfrac{1}{a} \ln\dfrac{\pi}{2A}+\dfrac{1}{b} \ln\dfrac{\pi}{2B}+\dfrac{1}{c} \ln\dfrac{\pi}{2C}\ge \dfrac{\sqrt 3}{R}\ln \dfrac{3}{2}.\] Jensen's on convex $f(x)=\ln\dfrac{\pi}{2x}$ gives \[\ln\dfrac{\pi}{2A}+\ln\dfrac{\pi}{2B}+\ln\dfrac{\pi}{2C}\ge 3\...
- Sun Oct 19, 2014 2:05 pm
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 6738
Re: Regional Mathematical Olympiad(India) 1994,P6
Easier approach: let $\overleftrightarrow{LM}\cap AB=X$ and $\overleftrightarrow{KM}\cap CD=Y$. Then $\angle LDM=\angle LMD=\angle XMB$ and $\angle XBM=\angle DCM$. So $\angle XMB+\angle XBM=\angle LDM+\angle DCM=90^\circ$, therefore $MX\perp AB$. Similarly $MY\perp CD$. Hence $OK\parallel ML$ and $...