probability 2nd dart closer to 1st than edge of target (2012 AMQ Concours p6)

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probability 2nd dart closer to 1st than edge of target (2012 AMQ Concours p6)

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#1 h Mar 5, 2021, 8:15 PM

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A dart is planted $12$ cm from the center of a circular target $20$ cm in radius. We launch a another dart, which hits the target (each point on the target is as likely as any other to be achieved). What is the probability that the second dart is closer to the first than to the edge of the target?

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#2 h Mar 5, 2021, 8:26 PM • 2 Y

Y by tenebrine, Mango247

This might be wrong, but I got $\frac{9}{25}$ for this problem. Basically the successful region is the region outside the radius $16$ but inside the radius $20$ circle.

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#3 h Mar 5, 2021, 8:45 PM

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@above that's not the correct locus
I suspect it might be an ellipse, but idk.

This post has been edited 1 time. Last edited by natmath, Mar 5, 2021, 8:53 PM

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#4 h Mar 5, 2021, 8:50 PM • 1 Y

Y by tenebrine

oh wait nvm it's one specific point not the circle :fool:

:fool:

that makes it slightly harder

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#5 h Mar 5, 2021, 9:04 PM

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In polar coordinates, the second dart is located at $(r,\theta)$ with $r<20$ .
Position the first dart at $(12,0)$
The distance from the point to the edge is $20-r$
The distance from the point to the first dart is $\sqrt{144+r^2-24r\cos\theta}$
The locus of all points such that the second dart is equidistant to the edge and the first dart is given by
$\sqrt{144+r^2-24r\cos\theta}=20-r$ $144+r^2-24r\cos\theta=400+r^2-40r$ $r(5-3\cos\theta)=32$ $r=\frac{\frac{32}{5}}{1-\frac{3}{5}\cos\theta}$ My physics nerds will recognize this as the formula of an ellipse with its focus at $r=0$ , eccentricity $\frac{3}{5}$ , and semi-latus rectum $\frac{32}{5}$ .
The ellipse has semi-major axis of $\frac{\frac{32}{5}}{1-\frac{9}{25}}=10$ and semi-minor axis of $\sqrt{\frac{32}{5}\cdot 10}=8$ . Hence it's area is $80\pi$

The ratio of the areas is $\boxed{\frac{1}{5}}$

So it did turn out to be an ellipse with its focus at the first dart. We could have possibly guessed this and then figuring out the parameters wouldn't be too hard as its major axis will just be half the diameter of the target.

This post has been edited 3 times. Last edited by natmath, Mar 5, 2021, 9:06 PM

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#6 h Mar 5, 2021, 10:28 PM • 2 Y

Y by natmath, parmenides51

A much easier demonstration that this is an ellipse is as follows: Let the dart be D, the center of the dartboard be O. Now suppose a point P lies on the boundary of the locus. Then it is equidistant from the circle and D. If we let $DP = x$ it follows that P is $x$ away from the circle, hence $OP = 20-x$ . Thus $OP+PD = 20$ . Hence it is an ellipse with foci O,D and major axis $20$ and it is not hard to find its minor axis.

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