## Search found 73 matches

Sun Feb 24, 2019 11:17 pm
Topic: BDMO National Secondary/Higher Secondary 2018/5
Replies: 5
Views: 625

### Re: BDMO National Secondary/Higher Secondary 2018/5

Bangla version of my solution! Mon Jan 21, 2019 8:58 pm
Forum: Secondary Level
Topic: 4-letter Code Words
Replies: 3
Views: 349

### Re: 4-letter Code Words

Doesn't need to have a meaning of the new words
Mon Jan 21, 2019 1:10 am
Forum: Secondary Level
Topic: Devide by $6$
Replies: 8
Views: 600

### Re: Devide by $6$

This answer is wrong . Nabila was right. We can write any number in this form: $(6n+x) : x \in {(1,2,3,4,5)}$ So, by squaring them, we get: $(6n+1)^2 \equiv 1$ $(6n+1)^2 \equiv 4$ $(6n+3)^2 \equiv 3$ $(6n+4)^2 \equiv 4$ $(6n+5)^2 \equiv 1$ $(6n)^2 \equiv 0$ mod $(6)$ So the pattern of the remainder ...
Mon Jan 21, 2019 12:14 am
Forum: Secondary Level
Topic: 4-letter Code Words
Replies: 3
Views: 349

### 4-letter Code Words

How many $4$-letter code words can be formed from the letters in $CALCULUS$?
Thu Jan 17, 2019 5:32 pm
Topic: BdMO regional 2018 set 4 Secondary P 06
Replies: 2
Views: 255

### Re: BdMO regional 2018 set 4 Secondary P 06

$\angle DCA=\angle DBA$
So, $ABCD$ is cyclic quadrilateral.
$\angle BAC=\angle BDC=60^{\circ}$
So, $\triangle BDC$ is isosceles and $BD=2$ Wed Jan 16, 2019 11:18 pm
Topic: BdMO regional 2018 set 4 Secondary P 10
Replies: 1
Views: 250

I set $D$ as $(0,0)$ and worked with analytic geometry. $AF\rightarrow y=-4x+2$ $FB\rightarrow y=4x-2$ $DH\rightarrow y=2x$ $HC\rightarrow y=-2x-2$ Then i called $I$ the intersection of $AF$ and $DH$ and $J$ the intersection of $FB$ and $HC$. Hence, $I=(\frac{1}{3},\frac{2}{3})$ and $J=(\frac{2}{3},... Thu Jan 10, 2019 10:06 am Forum: National Math Olympiad (BdMO) Topic: BDMO National Secondary/Higher Secondary 2018/5 Replies: 5 Views: 625 ### Re: BDMO National Secondary/Higher Secondary 2018/5 Capture.PNG$AB$and$BC$are diameter. So,$\angle AB'B=\angle BB'C=90$So,$A,B',C$are collinear. Easily it can be proved:$D',B' \in AC, C'.A' \in BD$From construction,$BC'B'C$is cyclic quadrilateral. So,$\angle AB'C'=\angle C'BC$On$\widehat {AA'}, \angle AB'A'=\angle ABA'\angle A'B'C'...
Thu Jan 10, 2019 9:46 am
Topic: IMO 2018 P1
Replies: 1
Views: 360

Let's use our contradiction! ;) Capture.PNG Assume, $FG$ and $DE$ are not parallel. So let $G'$ be on the circle such that $FG' \parallel DE$ and $E'$ on $AC$ such that $G'E'=G'C$. Let $FG'$ intersects $AB$ and $AC$ at $X$ and $Y$ respectively. We have these property: $FD=FB, GE=GC$ $\blacktriangler... Wed Jan 09, 2019 2:26 pm Forum: National Math Olympiad (BdMO) Topic: BdMO National Secondary/Higher Secondary 2018/6 Replies: 3 Views: 736 ### Re: BdMO National Secondary/Higher Secondary 2018/6$3(m^2+n^2)-7(m+n)=-4\Rightarrow 6m^2+6n^2-14m-14n=-8\Rightarrow 36m^2+36n^2-84m-84n=-48\Rightarrow (6m)^2-84m+7^2+(6n)^2-84n+7^2=-48+49+49\Rightarrow (6m-7)^2+(6n-7)^2=50\blacktriangleright(6m-7,6n-7)\in\{(\pm 1,\pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)\}\blacktriangleright\boxed{(...
Thu Apr 19, 2018 9:35 am
Forum: Secondary Level
Topic: Easy Projective Geo
Replies: 3
Views: 740

### Re: Easy Projective Geo

4.PNG I just made it harder ! :oops: We will use this lemma: Lemma: A point $P$ is outside or on a circle $\omega$. Let $PC$ and $PD$ be tangents to $\omega$, and $\iota$ be a line through $P$ intersecting $\omega$ at $A,B$. Let $AB$ intersect $CD$ at $Q$. Then $ABCD$ is a harmonic quadrilateral an...