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Vertices of a trianle

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Vertices of a trianle

by leonswati » Thu Jun 07, 2012 7:39 am
The points (-10,0), (0.5), (10,0) are vertices of a triangle. If x and y are integers , how many points (x,y) are in the interior of this triangular region?


I can understand that x axis can have a total of 19 points but what about y axis. Should the answer be 19*5?... Please help me solve this... Thanks..
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by coolhabhi » Thu Jun 07, 2012 8:34 am
leonswati wrote:The points (-10,0), (0.5), (10,0) are vertices of a triangle. If x and y are integers , how many points (x,y) are in the interior of this triangular region?


I can understand that x axis can have a total of 19 points but what about y axis. Should the answer be 19*5?... Please help me solve this... Thanks..
IMO 44. I have done it this way. Please look at the image.
Image

In that I have considered a square with the vertices (0,0) (10,0) (10,5) and (0,5).
Now in this rectangle we will hve 50 different points with integer values..
In this we need only half of it..I mean which the traingle has. so it would be 25 points.

Now similarly we have another traingle on the left side which will give us another 25 points.
Total 50 points.

But we would be counting (0,0) (0,1) (0,2) (0,3) (0,4) (0,5) twice. So remove these 6 we will get the answer as 44.

Btw what is OA? and what is the source?
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by 1947 » Thu Jun 07, 2012 10:40 pm
considering points on triangle out of scope.

-9<=x<=9
1<=y<=4

19*4/2 = 38 divided by 2 to count only points within triangle...
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by satyavegi » Sat Jun 09, 2012 3:18 am
Still I couldn't understand this ..can someone explain it further
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by 1947 » Sat Jun 09, 2012 8:31 am
what is the OA ?
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by dimochka » Sat Jun 09, 2012 8:40 am
Hi,

The triangle is formed using 3 different lines (keep in mind we want the interior, so we have inequalities):
(1) y > 0
(2) y > x/2 + 5
(3) y < -x/2 + 5

The range on y is limited between 1 and 4 ( 0 and 5 don't count since they're not in the interior)
Plug y = 1 into the equations --> in (2), x = -8 and in (3), x = 8. So you're looking -8<x<8 which gives you 15 possibilities
at y = 2, -6<x<6 (you can either calculate it or know based on slope) --> 11 possibilities
You can do the same with the rest if you see the pattern:
y = 1 --> 15
y = 2 --> 11
y = 3 --> 7
y = 4 --> 3

Add them all to get the answer of 36.

(Edit - gosh I can't add)
Last edited by dimochka on Mon Jun 11, 2012 4:55 am, edited 1 time in total.
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by confuse mind » Sat Jun 09, 2012 10:42 pm
IMO - 24(excluding all the points that lie on the edges or the vertices of the triangle)


Taking first quadrant - the equation of the line is x + 2y - 10 = 0
putting the integral points, we get

4+3+2+1 - first quadrant
10 - second quadrant
4 points on y axis

total = 24

I really liked the approach by @coolhabhi above, just that we need to take care of the points on the edges separately
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by dimochka » Mon Jun 11, 2012 5:02 am
confuse mind wrote:
4+3+2+1 - first quadrant
How do you get this?

If x=1, there are 4 points (y=1,2,3,4)
If x=2, there are 3 points (y=1,2,3)
But if x=3, there are still 3 points (y=1,2,3)
If x=4, there are 2 points (y=1,2)
If x=5, there are 2 points (y=1,2)
If x=6, there is 1 point (y=1)
If x=7, there is 1 point (y=1)
No more possible positive values for x

Add those (16), multiply by 2 to account for the negative x values (32) and add the 4 points on the y axis (36).
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