Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 WSMO Speed Round Problems/Problem 9"
(→‎Solution 2)
 
(3 intermediate revisions by 3 users not shown)
Line 2: Line 2:
 
Bobby is going to throw 20 darts at the dartboard shown below. It is formed by 4 concentric circles, with radii of <math>1,3,5,</math> and <math>7</math>, 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 <math>\frac{m}{n}\pi,</math> where <math>\gcd{(m,n)}=1</math>. Find <math>m+n</math>.
 
Bobby is going to throw 20 darts at the dartboard shown below. It is formed by 4 concentric circles, with radii of <math>1,3,5,</math> and <math>7</math>, 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 <math>\frac{m}{n}\pi,</math> where <math>\gcd{(m,n)}=1</math>. Find <math>m+n</math>.
  
−
[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("<math>70</math>",(0,0)); label("<math>-28</math>",(0,2)); label("<math>49</math>",(0,4)); label("<math>-21</math>",(0,6)); label("<math>0</math>",(-6,6)); label("<math>0</math>",(6,6)); label("<math>0</math>",(-6,-6)); label("<math>0</math>",(6,-6)); [/asy]
+
<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 <math>\frac{\pi}{196}</math>. The probability of getting a dart in the <math>-28</math> spot is <math>\frac{8\pi}{196}</math>. The probability of getting a dart in the <math>49</math> spot is <math>\frac{16\pi}{196}</math>. The probability of getting a dart in the <math>-21</math> spot is <math>\frac{24\pi}{196}</math>. This gives a result of <math>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</math>.
 
  
−
-OlympusHero
+
==Solution==
 +
The probability of a dart getting 70 points is  <math>\frac{\pi}{196}</math>, the probability of a dart getting <math>-28</math> points is <math>\frac{8\pi}{196}</math>, the probability of a dart getting <math>49</math> points is <math>\frac{16\pi}{196}</math>, and the probability of a dart getting <math>-21</math> points is <math>\frac{24\pi}{196}</math>. This means that on a random throw, the expected number of points Bobby gets is  <cmath>70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \frac{9}{14}\pi</cmath>. After 20 throws, Bobby is expected to get <cmath>20\cdot\frac{9}{14}\pi=\frac{90}{7}\pi\Longrightarrow90+7=\boxed{97}.</cmath>
 +
~pinkpig

Latest revision as of 09:08, 11 July 2022

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

The probability of a dart getting 70 points is $\frac{\pi}{196}$, the probability of a dart getting $-28$ points is $\frac{8\pi}{196}$, the probability of a dart getting $49$ points is $\frac{16\pi}{196}$, and the probability of a dart getting $-21$ points is $\frac{24\pi}{196}$. This means that on a random throw, the expected number of points Bobby gets is \[70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \frac{9}{14}\pi\]. After 20 throws, Bobby is expected to get \[20\cdot\frac{9}{14}\pi=\frac{90}{7}\pi\Longrightarrow90+7=\boxed{97}.\] ~pinkpig

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_WSMO_Speed_Round_Problems/Problem_9&oldid=175791"