Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Guest Apr 14, 2017

#1**+3 **

Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Within the circle section around A, X is not more than 1 unit from A.

The probability sought is equal to the ratio of the area of the circular segment to the rest of the area of the triangle.

The side of the triangle is a. r = 1

\(probability=\frac{\pi r^2}{6}:\\(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi r^2}{6})\)

\(probability=\frac{\pi }{6}:(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi}{6})\) [ \(\times\frac{6}{\pi}\)

\(probability=1:(\frac{a^2\cdot 6 \cdot\sqrt{3}}{4\pi}-1)\)

\(prohability=1:(\frac{a^2\cdot 3 \cdot\sqrt{3}}{2\pi}-1)\)

\(prohablity=1:(0.827a^2-1)\)

!

asinus
Apr 14, 2017

#2**+4 **

I didn't have any idea how to do this until I read asinus's answer!

But after I saw his I wanted to try it.

Here, A is the center of the circle with a radius of 1.

probability that X ≤ 1 unit away from A = \(\frac{\text{area of blue sector}}{\text{area of triangle}}\)

\(\frac{\text{area of blue sector}}{\pi*1^2}=\frac{60^{\circ}}{360^{\circ}} \rightarrow \text{area of blue sector}=\frac{\pi}{6}\)

\(\text{area of triangle}=(2)*(\frac12)*(\frac32)*(\frac{3\sqrt3}{2}) =\frac{9\sqrt3}{4}\)

\(\text{probability}=\frac{\frac{\pi}{6}}{\frac{9\sqrt3}{4}}=\frac{\pi}6\cdot\frac4{9\sqrt3}=\frac{4\pi}{54\sqrt3}=\frac{2\sqrt3\pi}{81}\)

My answer is a little different because I put the area of the whole triangle in the denominator, but asinus put the area of the rest of the triangle, not including the sector, in the denominator. I don't really know for sure which is correct.

hectictar
Apr 14, 2017

#3**+1 **

Yours is correct, hectictar.........( target area of triamgle) / (whole area of triangle) = desired probability

CPhill
Apr 14, 2017

#4**+1 **

The point X meets randomly from above on two surfaces of equal importance:

the circular section\((\leq 1 \ unit\ from\ vertex\ A)\) and the rest of the triangle\((>1 \ unit\ from\ vertex\ A)\).

The probability can only be the quotient of these two surfaces.

asinus
Apr 14, 2017

#5**+2 **

Hmm.. this is my (very limited!) understanding of probability...

\(\text{probability}=\frac{\text{number of ways something can happen}}{\text{total number of possible outcomes}}\)

So... if you have a bag of 3 blue marbles and 8 red marbles,

the probability of drawing a blue marble = \(\frac{3}{11}\) (and not \(\frac38\))

hectictar
Apr 14, 2017

#7**+2 **

Asinus....you have given the * odds *that "x" is no more than 1 unit away from "A," not the probability

To see the difference .....look at the following image :

The* probability* that we choose the green area in the above circle =

[ area that is green] / [ area that is green + area that is not green ] =

[1/4 of circle] / [ 1/4 of circle + 3/4 of circle] =

[1/4 of circle] / [ whole circle] = 1/4

But the * odds* that we choose the green area =

[area that is green] / [area that is not green] =

[1/4 of circle] / [ 3/4 of the circle] = 1/3 = 1 : 3

CPhill
Apr 15, 2017