Shirsendu Roy

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  • in reply to: Geometry- Circles #75706
    Shirsendu Roy
    Moderator

    from definition of a tangent, orthogonality of points A and B is trivial.

    We are to show that the other two points of intersection A’ and B’ of the circle with diameter AB with the original two circles are such that OA’ and OB’ are tangents where O is the centre of the third circle.

    Let P be the centre of one of the circles whose tangents are drawn OA and OA’. Here triangle OAP and OA’P are congruent by S-S-S test.

    Hence angle OAP = angle OA’P=90 degrees.

    Thus, OA’ is also a tangent which completes the proof.

    in reply to: Combinatorics #75057
    Shirsendu Roy
    Moderator

    For convex polygon K, with vertices A1, A2, A3,….,A20.

    Total number of ways to select three sides in 20 sides of k polygon ={20 \choose 3}

    =\frac{20!}{17!3!}=\frac{20\times19\times18}{6}=1140

    now, number of ways of selection when exactly three sides, are common ({A1A2,A2A3,A3A4} type all selection)=20

    now, number of ways of selection when exactly two sides are common{A1A2,A2A3} then we can’t select A3A4 and A1A20}=20 \times 16=320

    again number of ways of selection when the selected two sides are A1A2 and A3A4(one side gap between them)=20

    Two cases possible for third side

    case1 third side selected at a gap of two from above selected two sides

    {if we take A1A2,A3A4 then we can not choose 3rd side A3A4, A4A5,A20A1, A19A20}

    number of ways of third side section=20-6-1=13

    required ways=20 \times 13=260

    case2 3rd sides is at a gap of one from any selected side=20

    so number of ways=1140-20-320-260-20=520

     

    in reply to: Number Theory #74978
    Shirsendu Roy
    Moderator

    gcd (a,b)=1

    or, there exists u,v integers au+bv=1

    or, (a+b)u+bv=1+bu

    or, (a+b)u+b(v-u)=1

    or, gcd(a+b,b)=1 is first equation

    similar way gcd(a+b,a)=1 is second equation

    from both equations

    gcd(a+b, ab)=1

     

    in reply to: number theory #74715
    Shirsendu Roy
    Moderator

    Let 3^{n-1}+5^{n-1}|3^n+5^n then there exists some positive integer k such that 3^n+5^n=k(3^{n-1}+5^{n-1})

    if k >=5

    k(3^{n-1}+5^{n-1})>=5(3^{n-1}+5^{n-1})=5.3^{n-1}+5.5^{n-1}>3^n+5^n

    then k<=4

    3^n+5^n=3.3^{n-1}+5.5^{n-1}>3(3^{n-1}+5^{n-1})

    then k>=4 that is k=4

    3^n+5^n=4(3^{n-1}+5^{n-1}) which gives

    5^{n-1}=3^{n-1} which becomes impossible if n>1

    then n=1 as we see 2|8.

    in reply to: Combinatorics #74639
    Shirsendu Roy
    Moderator

    There are 4 even numbers and 4 odd numbers

    if an even and odd number is paired then no pair of gcd 2

    which is 4!=24 ways.

    the order of odd numbers can be shuffled that many different ways in each case pairing the first odd number with 2, the second one with 4, and so on.

    if even and odd are not paired then

    two even numbers is paired

    here the pair of even numbers in the set that have gcd not 2 is 4 and 8.

    so pair of 4 and 8 taken.

    choose two odd numbers to pair together in {4 \choose 2}=6 different ways and pair the remaining two odd numbers in 2 different ways with 2 and 6

    which gives (2)(6)=12 different ways

    so two methods give 24+12=36 ways.

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