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Algebra
22 Feb 2007 15:41:38 IST
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vishak p
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Joined: 9 Feb 2007
Posts: 1347
22 Feb 2007 15:51:40 IST
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please help !
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22 Feb 2007 16:43:13 IST
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Hi
By using Pick's theorem
Area of a triangle (or any polygon) whose vertices
are integer co ordinates is
are integer co ordinates is
A = I+B/2-1, where I is the number of integer co ordinates inside the polygon, and B is the number of integer co ordinates on the boundary.
So what's B? The number of integer co ordinates exactly on the line from
(x1,y1) to (x2,y2), including one of its endpoints, will be the GCD of
(x1-x2) and (y1-y2). Add these and you'll have B.
So my formula for I, given three vertices (x1,y1), (x2,y2), and
(x3,y3) rounded to integers, would be
I = A + 1 - [gcd(x1-x2, y1-y2) + gcd(x2-x3, y2-y3)
+ gcd(x3-x1, y3-y1)/2
(x1,y1) to (x2,y2), including one of its endpoints, will be the GCD of
(x1-x2) and (y1-y2). Add these and you'll have B.
So my formula for I, given three vertices (x1,y1), (x2,y2), and
(x3,y3) rounded to integers, would be
I = A + 1 - [gcd(x1-x2, y1-y2) + gcd(x2-x3, y2-y3)
+ gcd(x3-x1, y3-y1)/2
The points given are (0,0) , (21,0) , (0,21)
There fore
I = 1/2 * 21*21 + 1 - ( gcd( -21 ,0) +gcd (21,-21) + gcd(0,-21) )/2
= (441 +2 - (21+21+21) )/3
= (443 - 63)/2
= 380/2 = 190.
Hence the number of Integer co ordinates inside (0,0) , (21,0) , (0,21) are 190.
Cheers!
22 Feb 2007 17:21:35 IST
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The correct answer is 190.Here is another method Vish0001.
For 0<n<21 ,the line x=n contains 20-n integral points inside the triangle.x=1 contains the integral points (1,1),(1,2)....(1,19) See that (1,20)lies on the edge of the triangle and not inside the triangle.Similarly x=2 contains the points (2,1),(2,2)....(2,18) and similarly for x=3,x=4etc
So total no. of points =19+18+17+.....+1=19*20/2=190
Cheers!!!
22 Feb 2007 17:48:37 IST
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it is difficult to draw so i am telling u steps draw it on ur copy
take x &y axis as two vertices of the triangle. x-axis intercept is BC and y-axis
intercept is AB and AC is the line which is along the hypotenuse of the triangle.(point B is origin).
let us take general point P inside the triangle whose coordinates are (h,k).
the eqn of line Ac is x+y=21
we know that
1- point A & P lie on the same side of BC
2- point B & P lie ---------------------------- CA
3- point C &P ----------------------------------AB
so using the condition of points lie on same side ,we get
1- h-21<0 so h
(-
,21)
(-,21)2- h+k-21<0
3- k-21<0 so k (-,21)
(-,21)but h&k cannot be -ve in 1st quad.
so h (0,21) & k (0,21)
(0,21) & k (0,21) h&k can be 1,2,3-------------------20
and furthur by permutations no. of coordinates will be 190.
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