A point is moving randomly inside an equilateral triangle

Question

#1. A point is moving randomly inside an equilateral triangle whose side length is 3 cm. What is the probability that its distance to any corner is greater than 1?


#2. Two persons agree to meet at between 2 PM to 4 PM, but each of them will wait 30 minutes for the late comer. What is the probability that they will meet?

duttsit · Answer

#1. 
If we try to find the area of all these points, it will form an equilateral triangle similar to given triangle.

to find this probability we need to find ratio of (area of equilateral triangle that is made whose vertices are 1 cm from given triangle) to (given triangle).

Side of inscribed triangle = 3 - sqrt(3)

this triangle will be similar to given equilateral triagle. Using similar triangle property: ration of area = raio of sqaure of their sides

therefore, the ratio of area of two equilateral triangles = (3 - sqrt(3)) ^ 2 / (3 ^ 2)

or (12 - 6 sqrt(3)) / 9


#2. 

if first person reaches at any time T, the probability that second person reaches with time period T + 30 is 30/120 or 0.25

lazy_k · Answer

#1: all points within a distance of 1 cm from the corners of the triangle are lying in a 60Â° bow around each corner with radius 1cm. So we have 3 bows = 180Â° -> Area = 1/2 * Pi * r^2 = 1/2 Pi

Area of the triangle: sqrt(3)/4 * a^2 = 9*sqrt(3) / 4

-> Probability that the point is within 1 cm distance P1 = Area (circle) / Area (triangle) = (Pi/2) / (9*sqrt(3) / 4) = (2*Pi) / 9*sqrt(3)
-> Probability that the point is outside this area: 1 - P1 = 1 - (2*Pi) / 9*sqrt(3), which is around 60%

jdtomatito · Answer

Agree with lazy-k in 1.

For 2 I get 30/150 you need to add the half hour the first one to arrive is willing to wait.

jdtomatito · Answer

Agree with lazy-k in 1.

For 2 I get 30/150 you need to add the half hour the first one to arrive is willing to wait.

pavrnd · Answer

Agree with lazy-k in 1

For (2) I get 7/16
I solved it in the following way:
Let us assume that the interval is 0-120 min. (2-4 hrs)
Let x be the time (in minutes) when A arrives in the 0-120 min interval
Let y be the time (in minutes) when B arrives in the 0-120 min interval

Probability of A arriving in a small dx interval around x minutes is dx/120
Probability that A and B will meet is

1) If x is in first 30 min and y is between 0 to x+30 min
2) If x is in 30 to 90 min and y is between x-30 to x+30 min
3) If x is in 90 to 120 min and y is between x-30 to 120 min

i.e. 
1) { intergrate   } : limits x-> 0 to 30
+
2) { intergrate   } : limits x-> 30 to 90
+
3) { intergrate   } : limits x-> 90 to 120

 = 7/16

jdtomatito · Answer

You are absolutely right pavrnd, I really oversimplified the problem.

7/16 in the second question for me too.

vlad33 · Answer

How do you all get this as the area of the triangle (from above)?
"Area of the triangle: sqrt(3)/4 * a^2 = 9*sqrt(3) / 4 "

Where did any of those numbers and the a^2 come from?

Is the base not 3 and the height not 1.5(sqrt(3))

Is the area not 1/2bh = 1/2 * 3 * 1.5(sqrt(3))

Thanks, I just can't see where you all get the numbers from.

A point is moving randomly inside an equilateral triangle

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