 # Shirsendu Roy

## Forum Replies Created

Viewing 5 posts - 1 through 5 (of 40 total)
• Author
Posts
• in reply to: Geometry- Circles #75706 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.

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

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

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.

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.

Viewing 5 posts - 1 through 5 (of 40 total)