https://github.com/bgreenwell/concentric
Random triangles on concentric circles
https://github.com/bgreenwell/concentric
Last synced: about 2 months ago
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Random triangles on concentric circles
- Host: GitHub
- URL: https://github.com/bgreenwell/concentric
- Owner: bgreenwell
- Created: 2019-01-17T15:19:36.000Z (over 6 years ago)
- Default Branch: master
- Last Pushed: 2019-01-17T15:39:59.000Z (over 6 years ago)
- Last Synced: 2025-02-02T00:48:28.912Z (4 months ago)
- Language: R
- Size: 3.15 MB
- Stars: 0
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.Rmd
Awesome Lists containing this project
README
---
output: github_document
---
```{r setup, include = FALSE}
knitr::opts_chunk$set(
cache = TRUE,
collapse = TRUE,
comment = "#>"
)
# Required packages
library(ggplot2)
# Function to generate three concentric circles over a grid of points
genCircles <- function(n = 500L, radii = c(1, 2, 5)) {
angles <- seq(from = 0, to = 2 * pi, length = n)
names(radii) <- 1:3
plyr::ldply(radii, .fun = function(r) {
data.frame(x = r * cos(angles),
y = r * sin(angles))
}, .id = "circle")
}
circles <- genCircles()
circle1 <- circles[circles$circle == 1, ]
circle2 <- circles[circles$circle == 2, ]
circle3 <- circles[circles$circle == 3, ]
# Function to simulate data frame of random triangles
simTri <- function(n = 1L, radii = c(1, 2, 3)) {
plyr::rdply(n, {
angles <- runif(3, min = 0, max = 2 * pi)
data.frame(x = radii * cos(angles),
y = radii * sin(angles))
}, .id = "triangle")
}
# Function to determine if a triangle is obtuse
isObtuse <- function(triangle) {
x <- triangle$x
y <- triangle$y
# Lengths
aa <- sqrt((x[1]-x[2])^2 + (y[1]-y[2])^2)
bb <- sqrt((x[1]-x[3])^2 + (y[1]-y[3])^2)
cc <- sqrt((x[2]-x[3])^2 + (y[2]-y[3])^2)
# Test if each triangle is obtuse
if (aa^2 + bb^2 < cc^2 | aa^2 + cc^2 < bb^2 | bb^2 + cc^2 < aa^2) {
TRUE
} else {
FALSE
}
}
```
# concentric
What is the probability that a _random triangle_ is obtuse? The answer, depends on how you define a random trinagle. In the most general case, the answer is $1/4$. To obtain this answer, one can use the fact that, for any triangle $T$, there exists some circle $C$ such that the vertices of $T$ all lie on its circumference. Hence, the problem is equivalent to finding the probability that the points randomly distirbuted on the circumference of a circle form an obtuse triangle.
A more challenging problem, inspired by the question above, is the following. Concider three concenctric circles $C_1$, $C_2$, and $C_3$ with radii $r_1$, $r_2$, and $r_3$, respectively, and that $r_1 < r_2 < r3$. What is the probability that three randomly selected points, one on each circle, forms an obtuse triagle? In other words, what is the probability that a random triangle is obtuse if each vertex must lie on the circumference of a different concentric circle? Some simulated examples (with $r_1 = 1$, $r_2 = 2$, and $r_3 = 5$) are illustrated in the figure below.
```{r examples, echo=FALSE, fig.width=6, fig.asp=1, out.width="70%", fig.align="center"}
# Simulate some random triangles
set.seed(101)
triangles <- simTri(n = 16, radii = c(1, 2, 5))
# Plot random triangles
ggplot(triangles, aes(x = x, y = y, fill = as.factor(triangle))) +
geom_polygon() +
theme_light() +
facet_wrap( ~ triangle, nrow = 4) +
annotate("path", x = circle1$x, y = circle1$y) +
annotate("path", x = circle2$x, y = circle2$y) +
annotate("path", x = circle3$x, y = circle3$y) +
geom_hline(aes(yintercept = 0), alpha = 0.25) +
geom_vline(aes(xintercept = 0), alpha = 0.25) +
theme(legend.position = "none",
strip.background = element_blank(),
strip.text = element_blank()) +
xlab("") +
ylab("")
```
```{r proportion}
# Simulate some random triangles
set.seed(101)
triangles <- simTri(n = 1000L, radii = c(1, 1, 5))
# What proportion are obtuse?
mean(plyr::daply(triangles, .variables = "triangle", isObtuse))
(4*5 - 1) / (4 * 5)
```
```{r proportion-plot}
props <- sapply(1:10, FUN = function(x) {
triangles <- simTri(n = 1000, radii = c(1, 1, x))
mean(plyr::daply(triangles, .variables = "triangle", isObtuse))
})
plot(1:10, props, type = "b", pch = 19,
xlab = expression(r[3]), ylab = "Probability")
```