Problem Solving

Sorry to restart an old thread, but I am still having trouble fathoming the solution to this problem. From my understanding, -4<=x <= 5 and 6<=y<=16. Therefore we are assuming that the height and width of the triangle are within the ranges of 11 and 10. This I understand. However, when actually using combinatorics to determine the number of possibilities of P, shouldn't it be 9 *10 rather than 10 * 11? By having 11 * 10 instead of 9 *10, we are assuming that it is possible for P to be either at (-4, 16) or (5,6) on the xy graph. However these two points are irrational since that would imply that either point Q (height) or point R (width) would be beyond the limits of 16 and 5. Thus, I believe the solution should be 9 *10 for P, 10 for Q (Since Q and R can't be equal) and 9 for R, (Since R and P can't be equal.) So the answer should be 9 * 10 * 10 * 9 = 8,100.

Please tell me where my logic is wrong. Thanks!

Never mind, I forgot to realize the possibility of flipping the right triangle within the range given in the question. The solution now makes sense. Thanks!

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

Mitch,

Can you please help me solve the above problem using the technique outlined below.

Thanks

GMATGuruNY wrote:When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.

We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.

Point P:
x value: -4≤x≤5, giving us 10 choices.

y value: 6≤y≤16, giving us 11 choices.

Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:

(number of choices for x)*(number of choices for y)=10*11=110 choices for P.

Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.

y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 11-1=10 choices for Q's y value.

(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.

Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.

x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 10-1=9 choices for R's x value.

(number of choices for x)*(number of choices for y)=9*1=9 choices for R.

So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:

(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.

The correct answer is C.

Hope this helps!

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

Mitch,

Can you please help me solve the above problem using the technique outlined below.

Thanks

This too is a combinations question, but I would handle it differently.

In the other triangle problem, we had to count the number of ways we could form a right triangle whose base would be parallel to the x axis. The coordinates of each vertex affected the coordinates of the other two vertices: P had to have the same y coordinate as R and the same x coordinate as Q. So we had to determine how many choices we had for each vertex separately.

In the problem above, we have no such restrictions. We can form any kind of triangle we want. So there is no need to determine how many choices we have for each vertex separately.

A triangle is a combination of 3 points. Since 1≤x≤3 and 1≤y≤3, we have 9 points from which to choose:



The number of combinations of 3 that can formed from 9 choices = 9C3 = 84.

But some of these combinations will result in a straight line. Using the 9 points, 8 lines can be formed:



Thus, we need to subtract these 8 lines from our total:

Total possible triangles = 84-8 = 76.

The correct answer is B.

Mitch,

Thanks for the solution. It was brief and nicely explained. However, I have a couple of questions.

1) Just for my own curiosity and to have a back up technique, how can we solve the below problem without using the combinations (9c3) formula - meaning determine how many choices we have for each vertex separately.

2) Also, is it safe to say that in a case with restrictions one can count choices for each vertex separately and in a case without restrictions one can use the combinations formula?

Thanks again

GMATGuruNY wrote:

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

Mitch,

Can you please help me solve the above problem using the technique outlined below.

Thanks

This too is a combinations question, but I would handle it differently.

In the other triangle problem, we had to count the number of ways we could form a right triangle whose base would be parallel to the x axis. The coordinates of each vertex affected the coordinates of the other two vertices: P had to have the same y coordinate as R and the same x coordinate as Q. So we had to determine how many choices we had for each vertex separately.

In the problem above, we have no such restrictions. We can form any kind of triangle we want. So there is no need to determine how many choices we have for each vertex separately.

A triangle is a combination of 3 points. Since 1≤x≤3 and 1≤y≤3, we have 9 points from which to choose:



The number of combinations of 3 that can formed from 9 choices = 9C3 = 84.

But some of these combinations will result in a straight line. Using the 9 points, 8 lines can be formed:



Thus, we need to subtract these 8 lines from our total:

Total possible triangles = 84-8 = 76.

The correct answer is B.

gmat1978 wrote:Mitch,

Thanks for the solution. It was brief and nicely explained. However, I have a couple of questions.

1) Just for my own curiosity and to have a back up technique, how can we solve the below problem without using the combinations (9c3) formula - meaning determine how many choices we have for each vertex separately.

2) Also, is it safe to say that in a case with restrictions one can count choices for each vertex separately and in a case without restrictions one can use the combinations formula?

Thanks again

GMATGuruNY wrote:

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

Mitch,

Can you please help me solve the above problem using the technique outlined below.

Thanks

This too is a combinations question, but I would handle it differently.

In the other triangle problem, we had to count the number of ways we could form a right triangle whose base would be parallel to the x axis. The coordinates of each vertex affected the coordinates of the other two vertices: P had to have the same y coordinate as R and the same x coordinate as Q. So we had to determine how many choices we had for each vertex separately.

In the problem above, we have no such restrictions. We can form any kind of triangle we want. So there is no need to determine how many choices we have for each vertex separately.

A triangle is a combination of 3 points. Since 1≤x≤3 and 1≤y≤3, we have 9 points from which to choose:



The number of combinations of 3 that can formed from 9 choices = 9C3 = 84.

But some of these combinations will result in a straight line. Using the 9 points, 8 lines can be formed:



Thus, we need to subtract these 8 lines from our total:

Total possible triangles = 84-8 = 76.

The correct answer is B.

We could generate the same result (9C3) by handling the coordinates of each vertex separately. Let's call the vertices A, B, and C.

Vertex A:
Number of choices for x coordinate = 3. (We could use x=1, x=2, or x=3.)
Number of choices for y coordinate = 3. (We could use y=1, y=2, or y=3.)
To combine our choices for x and our choices for y, we multiply: 3*3 = 9.

Vertex B:
Number of choices for x coordinate = 3. (We could use x=1, x=2, or x=3.)
Number of choices for y coordinate = 3. (We could use y=1, y=2, or y=3.)
To combine our choices for x and our choices for y, we multiply: 3*3 = 9.
Since we can't reuse the same combination for x and y that was used for vertex A, we lose one possible combination: 9-1 = 8.

Vertex C:
Number of choices for x coordinate = 3. (We could use x=1, x=2, or x=3.)
Number of choices for y coordinate = 3. (We could use y=1, y=2, or y=3.)
To combine our choices for x and our choices for y, we multiply: 3*3 = 9.
Since we can't reuse the same combinations for x and y that were used for vertices A and B, we lose two possible combinations: 9-1 = 7.

To combine our choices for A, B and C, we multiply the results above: 9*8*7 = 504.

The result above represents the number of ways that we could arrange our choices for A, B and C. But the order of the vertices doesn't matter: ABC will yield the same triangle as BAC. Each duplicate combination (ABC, CAB, BCA) will result in the same triangle. To eliminate the duplicate combinations, we divide by (number of elements being chosen!). Since we are choosing 3 vertices, the result above must be divided by 3!: 504/3! = 84.

(Please note that in the PQR triangle problem no division was necessary because the order of the vertices did matter: the right angle had to be at point P and side PR had to be parallel to the x axis. In this problem, we have no such restrictions.)

Finally, we subtract from the result above the 8 combinations that will yield straight lines: 84-8 = 76.

Hope the explanation above helps. Here's the big idea when dealing with restrictions: start with the most restricted positions. If a boy can't be first, start with the first position. If a girl can't be last, start with the last position. If the right angle has to be at point P, start with point P. Then determine the number of choices for the remaining positions.

Many thanks Mitch. Your explanation is of great help.

Cheers!

Hi,

I understand the solution to this problem. What confuses me is why we did not filter the total number of possible triangles for the side1 - side 3 < side 2 < side 1 + side 3 rule? How can we be sure that the points we have selected will always form a triangle? Am I missing something or was there something to this effect implied in the question?

Thanks,

smk1407 wrote:Hi,

I understand the solution to this problem. What confuses me is why we did not filter the total number of possible triangles for the side1 - side 3 < side 2 < side 1 + side 3 rule? How can we be sure that the points we have selected will always form a triangle? Am I missing something or was there something to this effect implied in the question?

Thanks,

Any combination of 3 points THAT ARE NOT COLLINEAR can serve to form a triangle.
Try it out:
Plot any 3 points that are not collinear.
Connect the points.
The result will be a triangle.

Hi GURY NY

plz explain ur apporoch for below 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

Brilliant
Thank you.

GMATGuruNY wrote:When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.

We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.

Point P:
x value: -4≤x≤5, giving us 10 choices.

y value: 6≤y≤16, giving us 11 choices.

Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:

(number of choices for x)*(number of choices for y)=10*11=110 choices for P.

Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.

y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 11-1=10 choices for Q's y value.

(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.

Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.

x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 10-1=9 choices for R's x value.

(number of choices for x)*(number of choices for y)=9*1=9 choices for R.

So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:

(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.

The correct answer is C.

Hope this helps!

Brent@GMATPrepNow wrote:Here's a 10-page discussion on this question: https://www.beatthegmat.com/how-many-tri ... 28974.html

Cheers,
Brent

Good grief!

Got it, but how can we simply solve this?