🧵 Untitled Thread
Anonymous at Thu, 6 Jun 2024 02:09:14 UTC No. 16212863
Is /sci/ smart enough to solve this? If you pick three random numbers in a given range, what is the probability that a triangle with side lengths of the numbers you picked is within one degree of being a right triangle (meaning that it has an angle in the 89-91 degree range)?
Anonymous at Thu, 6 Jun 2024 03:32:22 UTC No. 16212994
>>16212863
50%
either it is or it isn't
Anonymous at Thu, 6 Jun 2024 04:54:24 UTC No. 16213090
>>16212994
nice!
Anonymous at Thu, 6 Jun 2024 05:18:50 UTC No. 16213101
>>16212863
sin(2deg)
Anonymous at Thu, 6 Jun 2024 05:46:31 UTC No. 16213124
>>16213101
2*sin2(1°)
Anonymous at Fri, 7 Jun 2024 02:17:26 UTC No. 16218660
>>16213101
>>16213124
>▶
source?
🗑️ Anonymous at Fri, 7 Jun 2024 02:28:16 UTC No. 16218683
>>16218660
Calculating the Probability of Picking Three Numbers for a Right Triangle:
To calculate the probability of randomly selecting three numbers within a given range and forming a triangle that is within one degree of being a right triangle, we need to consider the conditions for a right triangle. In a right triangle, one angle measures 90 degrees. Therefore, for our scenario, we are looking for triangles where one angle falls within the 89-91 degree range.
Step 1: Determine the Range of Possible Numbers:
Let’s assume we are picking three random numbers from a given range, say from 1 to 100. The first number can be any integer from 1 to 100, the second number can also be any integer from 1 to 100, and the third number can also be any integer from 1 to 100.
Step 2: Establishing Conditions for Right Triangle Formation:
For a triangle to be close to a right triangle, one angle must fall within the range of 89-91 degrees. This means that the sum of two sides squared should be approximately equal to the square of the third side.
Step 3: Calculating Probabilities:
To find the probability that three randomly chosen numbers will form a triangle close to a right triangle, we need to determine how many combinations satisfy our conditions out of all possible combinations.
We can use geometric principles and trigonometry to calculate this probability. By considering different combinations of numbers and their corresponding angles, we can determine which sets of numbers would result in triangles with an angle close to 90 degrees.
By calculating these probabilities for various ranges and number selections, we can arrive at an estimate of the overall likelihood of randomly selecting three numbers that form a triangle within one degree of being a right triangle.
Answer:
The probability that three random numbers picked in a given range will form a triangle within one degree of being a right triangle is approximately [insert calculated probability here].
Anonymous at Fri, 7 Jun 2024 02:43:19 UTC No. 16218700
Certainly! Let’s break this down step by step.
Choosing Three Random Numbers: You mentioned picking three random numbers. Let’s denote these numbers as (a), (b), and (c).
Triangle Inequality: For a triangle to exist, the sum of any two side lengths must be greater than the third side length. Mathematically, this can be expressed as: [a + b > c] [a + c > b] [b + c > a]
Right Triangle Condition: A right triangle has one angle measuring 90 degrees. This means that the sum of the other two angles must be 90 degrees. Therefore, we have: [a^2 + b^2 = c^2]
Probability Calculation: We want to find the probability that the randomly chosen side lengths form a triangle within one degree of being a right triangle. To do this, we need to consider the possible combinations of side lengths.
Case 2: Order Doesn’t Matter (With Repeats): If we allow repeated side lengths, we also consider cases like 3-3-5, 5-5-7, and 5-5-9. Each of these can be picked in 3 ways. So, (N(A)) in this case is (3! \cdot 3 + 5).
Total Number of Combinations: There are a total of (N = 5 \cdot 4 \cdot 3) ways to choose three side lengths without restrictions.
Probability: The probability of forming a triangle within one degree of being a right triangle is: [P(A) = \frac{N(A)}{N}]
Calculations:
For Case 1: (P(A) = \frac{3! \cdot 3}{5 \cdot 4 \cdot 3} = 0.3)
For Case 2: (P(A) = \frac{3! \cdot 3 + 5}{5 \cdot 4 \cdot 3} = 0.184)
Final Answer: The overall probability is approximately 0.3 (when order matters) or 0.184 (when order doesn’t matter). Since the question doesn’t specify whether order matters, both answers are valid1. Remember that we assumed the side lengths are chosen randomly and independently.
Feel free to ask if you need further clarification or have additional questions!
Anonymous at Fri, 7 Jun 2024 03:03:17 UTC No. 16218737
>>16218700
Um... anon... Anime posters are great, but saying it is different depending on whether order matters makes no sense. Take a triangle of sides 2, 4, 5. It doesn't matter whether they are in the order 2, 4, 5 or the order 4, 2, 5 or any other order.. any combination of those three numbers will produce an identical triangle.
🗑️ Anonymous at Fri, 7 Jun 2024 03:37:14 UTC No. 16218779
>>16218737
its chatgpt slop. im pretty sure >>16213124 is the answer
Anonymous at Fri, 7 Jun 2024 04:58:02 UTC No. 16218866
>>16212863
The random coordinate case seems easier. Three (x,y) pairs are picked where x,y 0 - 10 in R.
Every line can be rotated such that any third point pick on axis with the line ends will be a right triangle within tolerance. This gives the probability depending on resolution as 2•(angular resolution sweep) / 10^2.
The next thing to consider is when this line is anti rotated back to the original randomized position. Those correct bands are now rotated at some diagonal on the random domain, ranging in length from 0 to 10rt(2 and maintaining the same sweep under the new orientation.
I think this one is actually solvable.
.
The side length version is tough because you don't always generate valid triangles and trig function aren't simple for determining pass condition. I think the latter issue could be resolved, but I don't know how to approach the former.
The three picked sides can be on coordinates (l1,0), (0,l2) which somewhat transforms the problem into coordinate version. It gives sweet spots for p3. but then the other combinations l1,l3 and l2,l3 also have to be factored.
Anonymous at Fri, 7 Jun 2024 12:49:48 UTC No. 16219528
We can represent the probability space using 2 tangent circles of radius x and 2 circles of radius 2x drawn with their center points on the ends of the line between the center points of the two circles. Then we can draw a circle with chord length 2x and a central angle of 89 degrees treating the line between the centers of the two original circles as the chord. And we can repeat the process with a central angle of 91 degrees.
The difference of the area of the disc segments divided by half the area outside the two tangent circles but inside the area of the two circles of radius 2x is the probability.
I leave determining this probability as an exercise for the reader.
Anonymous at Fri, 7 Jun 2024 21:02:36 UTC No. 16220586
>>16219528
whoa...