MHB What is the Probability of Hitting the Center of a Square Target?

[Ackbach]

Here is this week's POTW:

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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.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to MarkFL for his correct, more general solution, which follows.

Spoiler

Let's generalize and say the dartboard is an $n$-gon. Consider an n-gon whose sides are 2 units in length. This regular polygon may be deconstructed into $n$ isosceles triangles. We take one of these triangles, bisect it and orient it in the Cartesian plane as in the diagram:

https://www.physicsforums.com/attachments/395

We have:

$\displaystyle \theta=\frac{\pi}{n}$ where $\displaystyle 2<n\in\mathbb{N}$

$\displaystyle h=\cot(\theta)$

The curve in red represents the locus of points equidistant from the upper vertex and the base. We know this curve must be parabolic by the definition of a parabola, and we may find the equation representing it as follows:

$\displaystyle x^2+(y-h)^2=y^2$

$\displaystyle x^2+y^2-2hy+h^2=y^2$

$\displaystyle y=\frac{x^2+h^2}{2h}$

The side of the triangle drawn in green coincides with the line given by:

$\displaystyle y=-hx+h$

Next, we determine the quadrant I x-coordinate where the parabola and the line intersect:

$\displaystyle \frac{x^2+h^2}{2h}=-hx+h$

$\displaystyle x^2+2h^2x-h^2=0$

Taking the positive root, we find by the quadratic formula:

$\displaystyle x_a=\frac{-2h^2+\sqrt{4h^4+4h^2}}{2}=-h^2+h\sqrt{h^2+1}=h(\sqrt{h^2+1}-h)$

Now, to find the area of the region of the triangle above the parabola, we use:

$\displaystyle \int_0^{x_a}(-hx+h)-\left(\frac{x^2+h^2}{2h} \right)\,dx=\frac{1}{2h}\int_0^{x_a} h^2-2h^2x-x^2\,dx$

Dividing this by the area of the triangle $\displaystyle \frac{h}{2}$, we obtain:

$\displaystyle P=\frac{1}{h^2}\int_0^{x_a} h^2-2h^2x-x^2\,dx=\int_0^{x_a} 1-2x-\frac{1}{h^2}x^2\,dx=$

$\displaystyle \left[x-x^2-\frac{1}{3h^2}x^3 \right]_0^{x_a}=x_a-x_a^2-\frac{1}{3h^2}x_a^3$

Recall, we have:

$\displaystyle x_a=h\left(\sqrt{h^2+1}-h \right)=\cot(\theta)\left(\sqrt{\cot^2(\theta )+1}-\cot(\theta) \right)=\cot(\theta)(\csc(\theta)-\cot(\theta))=$

$\displaystyle \cot(\theta)\frac{1-\cos(\theta)}{\sin(\theta)}=\cos(\theta)\frac{1-\cos(\theta)}{1-\cos^2(\theta)}=\frac{\cos(\theta)}{\cos(\theta)+1}$

$\displaystyle x_a^2=\frac{\cos^2(\theta)}{(\cos(\theta)+1)^2}$

$\displaystyle \frac{1}{3h^2}x_a^3=\frac{1}{3}\tan^2(\theta)\frac{\cos^3(\theta)}{(\cos(\theta)+1)^3}=\frac{\sin^2(\theta)\cos(\theta)}{3(\cos(\theta)+1)^3}=\frac{ \cos(\theta)(1-\cos(\theta))}{3(\cos(\theta)+1)^2}$

Thus, we have:

$\displaystyle x_a-x_a^2-\frac{1}{3h^2}x_a^3=$

$\displaystyle \frac{3\cos(\theta)(\cos(\theta)+1)-3\cos^2(\theta)-\cos(\theta)(1-\cos(\theta))}{3(\cos(\theta)+1)^2}=$

$\displaystyle \frac{\cos(\theta)(\cos(\theta)+2)}{3(\cos(\theta)+1)^2}$

And since $\displaystyle \theta=\frac{\pi}{n}$ we have:

$\displaystyle P(n)=\frac{\cos(\frac{\pi}{n})(\cos(\frac{\pi}{n})+2)}{3(\cos(\frac{\pi}{n})+1)^2}$

Now, since the problem gave $n = 4$, we find:

$\displaystyle P(4)=\frac{\cos(\frac{\pi}{4})(\cos(\frac{\pi}{4})+2)}{3(\cos(\frac{\pi}{4})+1)^2}=\frac{4\sqrt{2}-5}{3}\approx0.21895141649746$

I also include my solution which is a bit more ho-hum, but also more direct.

Spoiler

Let the square be centered at the origin, with side length $2$. We may restrict our attention, by symmetry, to the first quadrant, and even more specifically, to the region between the line $y=x$ and the $x$-axis in the first quadrant. The curve defining points equidistant from the origin and the side at $x=1$ is given by $\sqrt{x^2+y^2}=1-x$, or
$$x=\frac{1-y^2}{2}.$$
The area of the triangle in question is $1/2$. To find the area of the region closer to the origin than the curve above, we perform an iterated integral. We must set $(1-y^2)/2=y$ to find the point of intersection of the line $y=x$ and the curve; this occurs at $y=\sqrt{2}-1$. So the iterated integral is
$$\int_{0}^{\sqrt{2}-1}\int_{y}^{(1-y^2)/2}dx \, dy=\frac16 [ 4\sqrt{2}-5].$$
Dividing this by the $1/2$, which is the area of the entire triangle, yields that the probability is
$$\frac13 [4\sqrt{2}-5].$$