## Search found 186 matches

Fri Aug 30, 2013 10:39 pm
Forum: National Math Camp
Topic: [OGC1] Online Geometry Camp: Day 6 (EXAM!)
Replies: 36
Views: 13459

### Re: [OGC1] Online Geometry Camp: Day 6 (EXAM!)

I am unable to understand - why it is showing 'outbox' still [after near 1.5 hour] instead of sent messages ? I sent a test message to someone and it was also in 'outbox' .
Wed Aug 28, 2013 9:57 pm
Forum: National Math Camp
Topic: [OGC1] Online Geometry Camp: Day 4
Replies: 32
Views: 12751

### Re: [OGC1] Online Geometry Camp: Day 4

@ Samiun Fateeha Ira , apply power of a point any other way.

By the way , did it occur to you that $AB=48$ ? Wed Aug 28, 2013 5:22 pm
Forum: National Math Camp
Topic: [OGC1] Online Geometry Camp: Day 4
Replies: 32
Views: 12751

### Re: [OGC1] Online Geometry Camp: Day 4

Problem 6 Two circles intersect at $AB$ . A line through $B$ intersect the first circle at $C$ and the second circle at $D$. The tangents to the first circle at $C$ and the second at $D$ intersect at $M$ . Through the intersection point of $AM$ and $CD$ , there passes a line parallel to $CM$ and in...
Mon Jun 24, 2013 8:49 pm
Forum: Combinatorics
Topic: rotating a colored square
Replies: 5
Views: 2490

### rotating a colored square

A $n$ x $n$ square (where n is an odd positive integer) is divided into $n^2$ unit squares . Each unit square is colored either black or white . Probability of a square being black or white is equal . The square is rotated $90^o$ with respect to the center . If a white square overlaps on a square wh...
Sat Jun 22, 2013 10:19 pm
Forum: Secondary Level
Topic: Prove there is none
Replies: 2
Views: 1371

### Re: Prove there is none

$13^a = (3.4+1)^a = (3.4)^a +\binom{a}{1}(3.4)^{a-1} + ......... + \binom{a}{a-1}(3.4) + 1$ $\equiv \binom{a}{a-1} (3.4) + 1 \not\equiv -1(mod 4^2)$
$\therefore 13^a \not\equiv 15 (mod 4^2)$
Sun May 05, 2013 2:43 pm
Forum: Junior Level
Topic: Brilliant problem
Replies: 1
Views: 1356

### Re: Brilliant problem

$\displaystyle (r_1^2+r_1+1)(r_2^2+r_2+1)(r_3^2+r_3+1)=\frac{r_1^3-1}{r_1-1}.\frac{r_1^3-1}{r_3-1}.\frac{r_3^3-1}{r_3-1}$
put value of $x^3$ and use Vieta's formula .
Tue Apr 09, 2013 1:09 pm
Forum: Higher Secondary Level
Topic: Where is my incenter [self-made]
Replies: 5
Views: 2532

In $ODPE$ , $\angle ODP=\angle OEP=90^o$ , $ODPE$ is cyclic quadrilateral. $OM.MP=DM.ME=XM.MY$ , therefore $O,Y,P,X$ are concyclic. $OX=OY \Rightarrow \angle OXY=\angle OYX\Rightarrow \angle OPY=\angle OXY$ ; let $OP$ intersects $\omega$ at $J$. $JP$ bisects $\angle XPY$. on the other hand , $\angl... Sat Dec 22, 2012 8:18 pm Forum: Geometry Topic: CMO 2012 Replies: 8 Views: 2212 ### Re: CMO 2012 let ,$c_2$intersects$AB$at$X$.$\angle PAD=\angle AEP$,$\angle PAC=\angle PBA$;$\angle PAD-\angle PAC=\angle AEP-\angle PBA\Rightarrow \frac{1}{2}\angle B=(\angle AXP)-(\angle AXP-\angle BPX) \Rightarrow \angle BPX= \frac{1}{2}\angle BAD$intersects$c_1$at$D_1$.$\angle EAD_1=180^o-A-\...
Sat Nov 17, 2012 3:59 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 43397

### Re: Secondary and Higher Secondary Marathon

Sophie Jermain kills it. i couldn't get that ..... i did problem 6 with factorising which results $1,-1$. i think this is general way. solution to problem 7 $\angle BAM=\angle CAM \Rightarrow \angle BCM=\angle CBM$ , $BM=CM$.so $C$ point lies on the $\omega$ circle . now , $XD.DY=BD.DC=AD.DM$ \$\the...
Sat Nov 10, 2012 10:39 am
Forum: News / Announcements
Topic: Active users for marathon
Replies: 23
Views: 6891

### Re: Active users for marathon

I am not confident but i'll visit and try those problems 