How many triangles with positive area can be drawn on the

Question

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

Bunuel · Accepted Answer

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)... From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear. We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)}; So the final answer would be; 9C3-8=84-8=76. Answer: B. Similar problems with different solutions: arithmetic-og-question-88380.html 700-question-94644.html tough-problem-88958.html Hope it helps.

dungtd · Answer

Thanks a lot!
I got stuck when trying to understand the set of three points which are collinear to make a triangle.

By the way, I like your avatar. It reminds me about my ancestors in my country.

nonameee · Answer

Bunuel, I get 72. Here's my logic:

For the first vertex you get 9 options.
For the second one 8 options.
For the third 6 options.

9 x 8 x 6

But since the order of vertices is not import, so we have to divide by 3!

So, (9x8x6)/3! = 72

Where's the mistake in my reasoning?

Thank you.

Bunuel · Answer

As I understand in 9*8*6 by the last 6 you try to get rid of collinear points for the first two chosen ones. But it's not always true, consider the following:

* **
** * 
***

If you choose red and blue for the first two dots then for the third one you'll have 7 choices, not 6. So there are cases when you have 9*8*7 and cases when you have 9*8*6. That is why you'll get incorrect answer.

Hope it's clear.

nonameee · Answer

Bunuel, thanks a lot. Yes, I've been trying to get rid of colinear elements. But I haven't considered the case you mentioned.

ajit257 · Answer

Bunuel, is there a quicker way to figure out the collinear points

Bunuel · Answer

Points are collinear if they lie on a straight line. So I think it should be quite obvious that 3 vertical points of three, 3 horizontal points of three and 2 diagonal points of three are collinear. Total of 3+3+2=8 collinear points of three.

ajit257 · Answer

yup ...got it ...made a smal mistake ...Thanks a lot

HKHR · Answer

In a plane if there are n points out of which m points are collinear, then  the number of triangles that can be formed by joining them is nC3 - mC3

Here, n=9 and m=8(as we have 8 sets of points that are collinear(3 horizontal, 3 vertical, and 2 diagonal))

Therefore, the number of triangles is:

9C3 - 8C3 ----> 84 - 56 -----> 28

Where am I going wrong? Can someone help please! Thanks.

Bunuel · Answer

We don't have 8 points which are on one line (collinear). We have 8 3-POINT SETS which are collinear.

HKHR · Answer

Thanks Bunuel.
So what will the updated formula be in this case? Is it possible to use this formula for such type problems. Is there a generalized rule?

Bunuel · Answer

I don't think you need any special formulas for such kind of questions. Check similar questions to practice: if-4-points-are-indicated-on-a-line-and-5-points-are-132677.html abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-d-86284.html abcde-is-a-regular-pentagon-with-f-at-its-center-how-many-133328.html the-sides-bc-ca-ab-of-triangle-abc-have-3-4-5-interior-109690.html gmat-diagnostic-test-question-79373.html m03-71107.html how-many-triangles-with-positive-area-can-be-drawn-on-the-98236.html right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html how-many-circles-can-be-drawn-on-the-line-segment-128149.html how-many-triangles-can-be-inscribed-in-the-heptagon-pictured-154277.html Hope it helps.

Jaisri · Answer

Totally missed out on the collinearity of points and solved for the answer to be 9C3=84. Thanks Bunuel for throwing lights on excluding collinear points! And thanks again for the additional questions posted.

zachowdhury · Answer

I don't get the collinear thing. why you deduct 8 from 84?

Bunuel · Answer

If 3 points are collinear (are on the same line) they cannot form a triangle. We have 8 THREE-POINT SETS which are collinear, so 8 triangles cannot be formed.

Hope it's clear.

Magdak · Answer

OMG. This question was so tough to understand. Thank you Bunuel.

manpreetsingh86 · Answer

inequalities 1≤x≤3 and 1≤y≤3 ; represents 9 points on the xy plane. now out of these 9 points total no. of triangles that can be formed =9C3=84

if we put all these points in the xy plane we will see that it will form a square of 3x3. also, point (1,1), (1,2),(1,3) forms the straight line, thus if three points selected lies on this line, then it will not result into a triangle. therefore we will have to subtract all such cases which are:

3 (3 rows) +3 (3 columns) +2 (two diagonals)

therefore total no. of triangles are 84-8=76

siddyj94 · Answer

Bunuel  Not able to understand the elimination as well as how did we figure out the collinear points.

Can you please help!

Bunuel · Answer

If 3 points are collinear (are on the same line) they cannot form a triangle. We have 8 THREE-POINT SETS which are collinear:

3 horizontal: {(1,1), (2,1), (3,1)}; {(1,2) (2,2) (3,2)}; {(1,3) (2,3) (3,3)}.

3 vertical: {(1,1), (1,2), (1,3)}; {(2,1) (2,2) (2,3)}; {(3,1) (3,2) (3,3)}.

2 diagonal {(1,1) (2,2) (3,3)}; {(1,3) (2,2) (3,1)}.

How many triangles with positive area can be drawn on the

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How many triangles with positive area can be drawn on the

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