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Let $\bigtriangleup$ $ABC$ be an isosceles triangle with $AB$=$AC$.The bisectors of $\angle BAC$ and $\angle ABC$ intersect $BC$ and $AC$ at $D$ and $E$ respectively.Suppose $\angle BEK$ to be $45$ degree,where $K$ is the incenter of $\bigtriangleup$ $ACD$.Determine all possible values of $\angle BA... Mon Jun 19, 2017 3:02 pm Forum: Geometry Topic: IMO 2009 SL(G-1) Replies: 1 Views: 66 ### Re: Beginner's Marathon What has been meant by 'after 2 hours 30 min'? Thu Jun 15, 2017 2:44 pm Forum: Junior Level Topic: Beginner's Marathon Replies: 66 Views: 2149 ### Re: Beginner's Marathon An easy problem:$\text{Problem 20}$The fractions$\frac{7x+1}{2}$,$\frac{7x+2}{3}$,.......,$\frac{7x+2016}{2017}$are irreducible.Find all possible values of$x$such that$x$is less than or equal to$300$.This Problem was recommended by Zawad bhai in a mock test at CMC. Fri Jun 09, 2017 4:27 pm Forum: Junior Level Topic: Beginner's Marathon Replies: 66 Views: 2149 ### Re: Beginner's Marathon$\text{Problem 19}$Let,the number of games played by a person$i$be$g_i$.Therefore,the sum$g_1$+$g_2$+......+$g_{127}$is even as each game is counted twice.We,assume that the number of people who played an odd numbered games is odd.In that case,if we eliminate their played games from the sum,we... Thu Jun 08, 2017 11:42 pm Forum: Junior Level Topic: Beginner's Marathon Replies: 66 Views: 2149 ### Re: 2017 Regional no.9 Dhaka Check the parity of the equation and the rest should be clear. Mon Jun 05, 2017 2:31 pm Forum: Divisional Math Olympiad Topic: 2017 Regional no.9 Dhaka Replies: 8 Views: 201 ### Re: MPMS Problem Solving Marathon Simplier solution to problem$1$,I used this after getting to know what Wilson's Theorem is from dshasan's post.Let,$a$=$1$.$2$......$2k$.Telescope$a^2$as$1$.$2$....$2k$.($-2k$)....($-2$)($-1$).From the telescoping,it follows$a^2\equiv1$.$2$....$2k$($p-2k$)....($p-2$)($p-1$) (mod$p...
Mon Jun 05, 2017 2:09 pm

Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 8
Views: 348

### Re: MPMS Problem Solving Marathon

$Problem 4$: If $7^m$+$11^n$ is a perfect square,then $m$ must be even and $n$ must be odd. (By using mod $3$ and mod $4$);let $m$=$2q$ and $n$=$2k$+$1$ and $49^q$+$121^k.11$=$a^2$ $\Rightarrow$ ($x$+$7^q$)($x$-$7^q$)=$121^k$.$11$.Now make a total of $2$ cases along with $6$ subcases each.$1$ case w...
Mon Jun 05, 2017 1:47 pm

Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 8
Views: 348

### Re: MPMS Problem Solving Marathon

$Alternative Solution to 3$:It is obvious that if $4n^2$-$6n$+$45$ is a perfect square,then $4n^2$-$6n$+$45$ $\equiv$ $1$ (mod $2$) $\Rightarrow$ $4n^2$-$6n$+$45$ $\equiv$ $1$ (mod $4$) $\Rightarrow$ $1$-$6n$ $\equiv$ $1$ (mod $4$) $\Rightarrow$ $n$ $\equiv$ $0$ (mod $2$);so $n$ cannot be odd.
Mon Jun 05, 2017 1:17 pm

Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 8
Views: 348

### Re: MPMS Problem Solving Marathon

For problem 1,if we substitute $a^2$=$4k$ where $k$ is an integer,we can easily find that $p$ |$a^2$+$1$.
Mon May 29, 2017 3:53 pm

Forum: News / Announcements
Topic: MPMS Problem Solving Marathon
Replies: 8
Views: 348

### Re: APMO 2017 P2

Let the reflection of $DM$ over $M$ be $D'M$.Thus we concurr that $AD'BD$ is a parallelogram as $M$ is designed to be the midpoint of $AB$.Let the midpoint of $AC$ be $X$.Thus we have $AX$=$CX$; $ZX$ as a common side of triangles $ZXA$ and $ZXC$;$\angle ZXA$=$\angle ZXC$.So,by $SAS$,we have \$\bigtri...
Mon May 29, 2017 12:34 am

Forum: Asian Pacific Math Olympiad (APMO)
Topic: APMO 2017 P2
Replies: 1
Views: 98
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