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OneTwoThreeFour
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Compare these two questions for a moment:
2. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
Answer: b
1. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to x axis. The x and y coordinated of P,Q & R are to be integers that satisfy the inequalities -4<=x<=5 & 6<=y<=16.
How many different triangles wth these properties are possible?
A. 110
B. 1100
C. 9900
D. 10000
E. 12100
Answer: C
Now my question is, can we use OG's method to solve the first problem on the second one? I know how to solve the second problem using combinatorics, which is the most straightforward and efficient approach. But, I am having a bit of difficulty trying to solve it using OG's way from the first problem. Or, check out Mitch's way at the link:
https://www.beatthegmat.com/how-many-tri ... 13159.html
Solving the second problem using OG's method from the first problem:
A triangle must have 3 points; the first point can have a total of 9 choices. (3x3. Since each point must have an integer value, we can have a total of 9 dots on a xy plane.) The second point can have a total of 8 possibilities, since it cannot be the exact same point as the first one. So far, we have 9x8. So for the third point, we should have 7 possibilities. Therefore the answer should be 9 x 8 x 7 -8= 496. (We subtract 8, since the 3 points cannot be in a line.) Of course, this is wrong since 496 > 76. So why is (9x8x7)-8 wrong? Is there a correct way to solve it using OG's way on the second problem?
Thanks!
2. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
Answer: b
1. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to x axis. The x and y coordinated of P,Q & R are to be integers that satisfy the inequalities -4<=x<=5 & 6<=y<=16.
How many different triangles wth these properties are possible?
A. 110
B. 1100
C. 9900
D. 10000
E. 12100
Answer: C
Now my question is, can we use OG's method to solve the first problem on the second one? I know how to solve the second problem using combinatorics, which is the most straightforward and efficient approach. But, I am having a bit of difficulty trying to solve it using OG's way from the first problem. Or, check out Mitch's way at the link:
https://www.beatthegmat.com/how-many-tri ... 13159.html
Solving the second problem using OG's method from the first problem:
A triangle must have 3 points; the first point can have a total of 9 choices. (3x3. Since each point must have an integer value, we can have a total of 9 dots on a xy plane.) The second point can have a total of 8 possibilities, since it cannot be the exact same point as the first one. So far, we have 9x8. So for the third point, we should have 7 possibilities. Therefore the answer should be 9 x 8 x 7 -8= 496. (We subtract 8, since the 3 points cannot be in a line.) Of course, this is wrong since 496 > 76. So why is (9x8x7)-8 wrong? Is there a correct way to solve it using OG's way on the second problem?
Thanks!
