Search found 312 matches
- Mon Oct 20, 2014 10:11 pm
- Forum: Secondary Level
- Topic: Regular Polygon inscribed in a circle
- Replies: 5
- Views: 7175
Regular Polygon inscribed in a circle
Let $A_{1}A_{2}A_{3}......A_{21}$ be a $21$ sided regular polygon inscribed in a circle with center $O$.How many triangles $A_{i} A_{j} A_{k}$,$1\leq i< j< k\leq 21$,contain the point $O$ in their interior
- Sun Oct 19, 2014 5:52 pm
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 7013
Re: Regional Mathematical Olympiad(India) 1994,P6
Oooo....ps!I could not catch itTahmid wrote:because of $ML$ is the median of a right angle triangle $MDC$ .tanmoy wrote: Tahmid vaia, you wrote:$\therefore \angle MLC=180-2x$.This implies,$\angle MCL= \angle CML$.But how
$ML=\frac{DC}{2}=LC$
$\angle MCL= \angle CML$
- Sat Oct 18, 2014 10:20 pm
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 7013
Re: Regional Mathematical Olympiad(India) 1994,P6
little angle chasing , let , $\angle ACD=\angle ABD=x$ and $\angle BAC=\angle BDC=y$ $\therefore \angle MLC=180-2x$ $\therefore \angle OLM=90-\angle MLC=90-180+2x=2x-90$ $\therefore \angle MKB=180-2x$ $\therefore \angle OKM=90-\angle MKB=90-180+2x=2x-90$ $\therefore \angle OLM=\angle OKM$ $\angle K...
- Sat Oct 18, 2014 9:39 pm
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 7013
Re: Regional Mathematical Olympiad(India) 1994,P6
Not that challenging of a problem.... since k,l are mid points, angle k, l are right angles. So, the quadrilateral is concycic. Angle m is a right angle, so should be angle o. So it's a rectangle. Which is, a parallelogram. Your solution is not correct.Since,$O$ is the center and $K$ is the midpoin...
- Thu Oct 16, 2014 7:37 pm
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 7013
Regional Mathematical Olympiad(India) 1994,P6
Let $AC$ and $BD$ be two chords of a circle with center $O$ such that they intersect at right angles inside the circle at the point $M$.Suppose,$K$ and $L$ are the midpoints of the chords $AB$ and $CD$ respectively.Prove that $OKML$ is a parallelogram
- Sat Oct 11, 2014 11:13 pm
- Forum: Combinatorics
- Topic: A probability problem
- Replies: 4
- Views: 4122
A probability problem
Ten distinct letters are selected at random from the alphabet $\left \{ A,B,C.......Z \right \}$.Find the probability that the selection does not include two consecutive letters.
- Tue Sep 30, 2014 11:17 pm
- Forum: Secondary Level
- Topic: Directed Angle
- Replies: 3
- Views: 4279
Directed Angle
What is directed angle
- Tue Sep 16, 2014 11:24 am
- Forum: Junior Level
- Topic: Digital original wanted
- Replies: 2
- Views: 3213
Re: Digital original wanted
suppose,the three digit number is 100x+10y+z.So,
100x+10y+z+297=100z+10y+x.
Or,99x+297=99z.
Or,99(x+3)=99z.
Or,x+3=z.
Again,100x+10y+z+9=100x+10z+y.
Or,9y+9=9z.
Or,9(y+1)=9z.
Or,y+1=z.So,x+3=y+1.Or,y=x+2
so,x+x+2+x+3=11.
Or,3x+5=11.
Or,3x=6.
Or,x=2.So,y=4,z=5
So,the number is 245
100x+10y+z+297=100z+10y+x.
Or,99x+297=99z.
Or,99(x+3)=99z.
Or,x+3=z.
Again,100x+10y+z+9=100x+10z+y.
Or,9y+9=9z.
Or,9(y+1)=9z.
Or,y+1=z.So,x+3=y+1.Or,y=x+2
so,x+x+2+x+3=11.
Or,3x+5=11.
Or,3x=6.
Or,x=2.So,y=4,z=5
So,the number is 245
- Wed Mar 19, 2014 7:52 pm
- Forum: Primary Level
- Topic: A problem to solve
- Replies: 12
- Views: 38911
Re: A problem to solve
Let the three consecutive numbers are $X,X+1$ and $X+2$.So, $X+X+1+X+2=60$.Or,$3X+3=60$.Or, $3X=57$.Or,$X=19$.So,the numbers are $19,20,21$
- Wed Mar 19, 2014 7:38 pm
- Forum: Primary Level
- Topic: Age Fun
- Replies: 2
- Views: 9206
Re: Age Fun
The age of his father is 25.So his age is 2 years and 1 month.