Search found 5 matches
- Tue May 18, 2021 9:09 am
- Forum: National Math Camp
- Topic: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
- Replies: 6
- Views: 7507
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Let $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$. My solution is a bi...
- Tue May 11, 2021 10:15 pm
- Forum: National Math Camp
- Topic: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
- Replies: 6
- Views: 7507
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Let $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$. My solution is a bi...
- Mon May 03, 2021 8:31 pm
- Forum: National Math Camp
- Topic: National Math Camp 2020 Exam 1 Problem 1
- Replies: 4
- Views: 7234
Re: National Math Camp 2020 Exam 1 Problem 1
Let $m, n, k$ be integers such that $(m-n)^2 + (n-k)^2 + (k-m)^2 = mnk$. Prove that, $m^3 + n^3 + k^3$ is divisible by $m + n + k + 6$. mnk=$(m-n)^2+(n-k)^2+(k-m)^2$ => mnk=$2(m^2+n^2+k^2-mn-nk-mk)$ => 3mnk=$6(m^2+n^2+k^2-mn-nk-mk)$ Now, $m^3+n^3+k^3-3mnk$=(m+n+k)($m^2+n^2+k^2-mn-nk-mk$) =>$m^3+n^3...
- Wed Apr 14, 2021 7:09 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2021 Higher Secondary Problem 9
- Replies: 3
- Views: 4087
Re: Solution :
Let me start with an exercise. Find three distinct positive integers $a,b,c$ where $a<b<c$ such that \[\frac1a+\frac1b+\frac1c=1\] You may post the solution below. Well, the only solution to this equation is $(a,b,c)=(2,3,6)$. Let's assume, $n$ is a nice integer. Let $\frac{n}{x},\frac{n}{y},\frac{...
- Fri Apr 02, 2021 10:20 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2015/2
- Replies: 5
- Views: 3738
Re: BdMO National Secondary 2015/2
Solution $m+n=mn$ $\Rightarrow m+n=m+m+m+m+m+m+...+m $ (n times) $\Rightarrow n=m+m+m+m+m....+m+m$ {(n-1)times} $\Rightarrow n=m(n-1)\Rightarrow \dfrac {n}{n-1}=m$ But $n$ and $n-1$ are co-prime . So, $m$ cannot be an integer. A Contrudiction So,there are no such $(m,n) $that satisfies the equation...