Probability of Hitting Center of a Square Target - POTW #318 (Jun 13, 2018)

[anemone]

A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge.

Express your answer in the exact form.

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[anemone]

Congratulations to the following members for their correct solution:):

1. MarkFL
2. castor28
3, kaliprasad

Solution from MarkFL:

Spoiler

WLOG, I used a square 4 units in area centered at the origin. I found the area bounded by the positive $y$-axis, the line \(\displaystyle y=x\) and the parabola having its focus at the origin and its directrix at \(\displaystyle y=1\). We find this parabola from:

\(\displaystyle x^2+y^2=(y-1)^2\)

\(\displaystyle x^2+y^2=y^2-2y+1\)

\(\displaystyle y=\frac{1-x^2}{2}\)

We find the first quadrant intersection of the line and the parabola from:

\(\displaystyle x=\frac{1-x^2}{2}\)

\(\displaystyle x^2+2x-1=0\)

\(\displaystyle x=\sqrt{2}-1\)

Thus, the area $A$ of the region in question is:

\(\displaystyle A=8\int_0^{\sqrt{2}-1}\frac{1-x^2}{2}-x\,dx\)

\(\displaystyle A=4\int_0^{\sqrt{2}-1}1-2x-x^2\,dx\)

Dividing by the area of the square, we find the portion is:

\(\displaystyle A=\int_0^{\sqrt{2}-1}1-2x-x^2\,dx=\frac{1}{3}(4\sqrt{2}-5)\)

And so, the probability in question is given by:

\(\displaystyle P(X)=\frac{1}{3}(4\sqrt{2}-5)\approx0.2189514164974602\)