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Difference between revisions of "2016 AIME I Problems/Problem 13"

(Created page with "Notice that we don't really care about what the <math>x</math>-coordinate of the frog is. So let's let <math>f(y)</math> denote the expected number of times Freddy will jump a...")
 
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==Solution==
 
Notice that we don't really care about what the <math>x</math>-coordinate of the frog is. So let's let <math>f(y)</math> denote the expected number of times Freddy will jump at a <math>y</math> coordinate of <math>y</math> until he reaches the line <math>y = 24</math>. So therefore we want to find <math>f(21)</math>.  
 
Notice that we don't really care about what the <math>x</math>-coordinate of the frog is. So let's let <math>f(y)</math> denote the expected number of times Freddy will jump at a <math>y</math> coordinate of <math>y</math> until he reaches the line <math>y = 24</math>. So therefore we want to find <math>f(21)</math>.  
  
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<cmath>f(k) = z - k - 2k^2 </cmath>
 
<cmath>f(k) = z - k - 2k^2 </cmath>
 
Now, we also have that <math>f(24) = 0</math> so that gives us <math>f(24) = z - 24 - 2 \cdot 576 = 0</math> so <math>z = 1176</math>. So now we know <math>f(k) = 1176 - k - 2k^2</math> so plugging in <math>k = 21</math> we get <math>f(21) = 1176 - 21 - 882 = \boxed{273}</math>
 
Now, we also have that <math>f(24) = 0</math> so that gives us <math>f(24) = z - 24 - 2 \cdot 576 = 0</math> so <math>z = 1176</math>. So now we know <math>f(k) = 1176 - k - 2k^2</math> so plugging in <math>k = 21</math> we get <math>f(21) = 1176 - 21 - 882 = \boxed{273}</math>
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== See also ==
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{{AIME box|year=2015|n=I|before=First Problem|num-a=2}}
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{{MAA Notice}}

Revision as of 16:00, 4 March 2016

Solution

Notice that we don't really care about what the $x$-coordinate of the frog is. So let's let $f(y)$ denote the expected number of times Freddy will jump at a $y$ coordinate of $y$ until he reaches the line $y = 24$. So therefore we want to find $f(21)$.

So we have $f(24) = 0$. Suppose Freddy is at $y = n$. He has a $\frac{1}{2}$ probability of moving horizontally, $\frac{1}{4}$ chance of moving up and $\frac{1}{4}$ chance of moving down. So we have \[f(n) = 1 + \frac{1}{2} f(n) + \frac{1}{4} f(n-1) + \frac{1}{4} f(n+1)\] So we get the recursion $2f(n) - f(n-1) - f(n+1) = 4$. Rearranging we see $f(n) - f(n-1) = f(n+1) - f(n) + 4$. That means the difference between consecutive terms goes down by $4$ each time. So for convenience let's let $z = f(0)$ and $d = f(1) - f(0)$. So that means \[f(k) = z + d + (d-4) + (d-8) + \cdots + (d - 4(k-1)) = z + kd - 2k(k-1)\] Yes, this is a quadratic. Now, notice that since there is a boundary, we have to give special care to $f(0)$. We have $f(0) = 1 + \frac{2}{3}f(0) + \frac{1}{3} f(1)$ so this turns into $\frac{1}{3}z = 1 + \frac{1}{3}(z+d)$ and this results in $d = -3$. So now we know \[f(k) = z - k - 2k^2\] Now, we also have that $f(24) = 0$ so that gives us $f(24) = z - 24 - 2 \cdot 576 = 0$ so $z = 1176$. So now we know $f(k) = 1176 - k - 2k^2$ so plugging in $k = 21$ we get $f(21) = 1176 - 21 - 882 = \boxed{273}$

See also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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