ABJeIIy

avatar
Nombre de usuarioABJeIIy
Puntuación940
Membership
Stats
Preguntas 236
Respuestas 11

 #1
avatar+940 
0

There are two ways to tackle this problem:

 

Method 1: Counting Valid Triangles

 

Total Points: We have 3 choices for x (1, 2, or 3) and 8 choices for y (3, 4, 5, ..., 10), resulting in a total of 3 * 8 = 24 points.

 

Choosing 3 Points: We need to choose 3 out of these 24 points to form a triangle.

 

The number of ways to do this can be calculated using combinations: nCr = n! / (r! * (n-r)!) where n is the total number of elements (24) and r is the number of elements to choose (3). 24C3 = 24! / (3! * (24 - 3)!) = 2280

 

Not All Triangles are Valid: However, not all combinations of 3 points will form valid triangles.\

 

The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side. We need to check how many of the 2280 combinations violate this rule.

 

Loop through each combination of 3 points.

 

Calculate the distances between all three pairs of points using the distance formula (square root of the sum of squared differences in x and y coordinates).

 

If any of the distances violate the triangle inequality (i.e., the sum of two distances is less than the third distance), discard that combination.

Valid Triangles: After checking all combinations, count the remaining ones that satisfy the triangle inequality. This will be the number of valid triangles.

 

Method 2: Faster Approach (Counting Invalid Triangles)

 

Total Combinations: Same as method 1, there are 2280 total ways to choose 3 points from 24.

 

Invalid Triangles: We can count the number of triangles that violate the triangle inequality and subtract them from the total to get the valid ones. There are two main cases for invalid triangles:

 

All points on a straight line: If all three chosen points fall on the same line (e.g., (1, 3), (2, 3), (3, 3)), they cannot form a triangle.

 

Degenerate Triangle: If two points have the same x-coordinate or the same y-coordinate, and the third point does not lie above the line connecting them, they cannot form a valid triangle. (Imagine a straight line segment connecting the two points with the same coordinate. The third point must be above this line segment for a triangle to form).

 

Count the number of ways to choose 3 points that fall on the same horizontal or vertical line (3 choices for x-coordinate * 8 choices for y-coordinate = 24)

 

Count the number of ways to choose 2 points with the same x-coordinate (3 choices for x-coordinate * 7 choices for y-coordinate for the first point * 6 choices for y-coordinate for the second point, excluding the chosen y-coordinate = 126).

 

Do the same for points with the same y-coordinate.

 

Valid Triangles: The total number of valid triangles is then: Total Combinations - Invalid Triangles 2280 - (24 + 126 + 126) = 2004

 

Both methods will give you the answer: there are 2004​ triangles that can be formed using 3 of the points.

9 abr 2024
 #2
avatar+940 
0

The key to this problem is finding the least common multiple (LCM) of 360, 450, and 540. This represents the smallest number of days that will take for all three planets to complete one full rotation around the sun relative to their starting positions.

 

Here's why:

 

If all planets complete their rotations in multiples of their individual periods (e.g., 360 days for X, 900 days for Y, etc.), they might not be lined up

again.

 

The LCM ensures that after that specific number of days, X will have completed exactly m rotations (for some integer m), Y will have completed exactly n rotations (for some integer n), and Z will have completed exactly o rotations (for some integer o).

 

To find the LCM, we can use various methods. In this case, since the numbers are relatively close, you might be able to find it by inspection.

 

However, a more general approach is to use the Euclidean Algorithm:

 

Find the greatest common divisor (GCD) of the largest two numbers (here, 450 and 540).

 

Divide the LCM of the initial three numbers (which doesn't exist yet) by the GCD you just found. This gives you the LCM of the first two numbers.

 

Now, find the GCD of the LCM you obtained in step 2 and the remaining number (here, 360).

 

The final LCM is the product of the GCDs you found in steps 1 and 3.

 

Following these steps, you'll find the GCD of 450 and 540 to be 90. The LCM of 450 and 540 is then (450 * 540) / 90 = 300.

 

Finally, the GCD of 300 and 360 is 60. Therefore, the LCM of 360, 450, and 540 is (300 * 60) = 18000.

 

However, 18000 is not the minimum positive number of days. Why? Because all three planets might be lined up again at a smaller common multiple. We need to find the least common multiple.

 

Looking closer, we see that 540 (Z's rotation period) is divisible by both 360 (X's period) and 450 (Y's period).

 

This means that every time Z completes a full rotation, X will have completed some integer number of rotations and Y will have completed some other integer number of rotations. In other words, they will all be lined up again after every 540 days.

 

Therefore, the minimum positive number of days before they are on the same line again is 540​ days.

9 abr 2024