2021 WSMO Speed Round Problems/Problem 9

Problem

Bobby is going to throw 20 darts at the dartboard shown below. It is formed by 4 concentric circles, with radii of $1,3,5,$ and $7$ , with the largest circle being inscribed in a square. Each point on the dartboard has an equally likely chance of being hit by a dart, and Bobby is guaranteed to hit the dartboard. Each region is labeled with its point value (the number of points Bobby will get if he hits that region). The expected number of points Bobby will get after throwing the 20 darts can be expressed as $\frac{m}{n}\pi,$ where $\gcd{(m,n)}=1$ . Find $m+n$ .

$[asy] size(5cm); filldraw(shift(-7,-7)*((0,0)--(14,0)--(14,14)--(0,14)--cycle),gray); filldraw(circle((0,0),7),white); filldraw(circle((0,0),5),white); filldraw(circle((0,0),3),white); filldraw(circle((0,0),1),white); label("$70$",(0,0)); label("$-28$",(0,2)); label("$49$",(0,4)); label("$-21$",(0,6)); label("$0$",(-6,6)); label("$0$",(6,6)); label("$0$",(-6,-6)); label("$0$",(6,-6)); [/asy]$

Solution 1

The probability of getting a dart in the 70 spot is $\frac{\pi}{196}$ . The probability of getting a dart in the $-28$ spot is $\frac{8\pi}{196}$ . The probability of getting a dart in the $49$ spot is $\frac{16\pi}{196}$ . The probability of getting a dart in the $-21$ spot is $\frac{24\pi}{196}$ . This gives a result of $70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \boxed{\frac{9}{14}}\pi$ .

~OlympusHero

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2021 WSMO Speed Round Problems/Problem 9

Problem

Solution 1