Difference between revisions of "2021 AMC 12A Problems/Problem 23"

m (added my solution)
m (fixed some formatting)
Line 90: Line 90:
  
 
==Solution 5==
 
==Solution 5==
Imagine an infinite grid of <math>2</math> by <math>2</math> squares such that there is a <math>2</math> by <math>2</math> square centered at <math>(3x, 3y)</math> for all ordered pairs of integers <math>(x, y).</math> It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at <math>(0, 0)</math>. (minus the teleportations) Since counting the complement set is easier, we'll count the number of <math>4</math>-step paths such that Frieda never reaches a corner point. In other words, since the reachable corner points are <math>(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), </math> and <math>(\pm 2, \pm 2),</math> Frieda can only travel along the collection of points included in <math>S</math>, where <math>S</math> is all points on <math>x=0</math> and <math>y=0</math> such that <math>|y|<4</math> and <math>|x|<4</math>, respectively, plus all points on the big square with side length <math>6</math> centered at <math>(0, 0).</math> We then can proceed with casework:
+
Imagine an infinite grid of <math>2</math> by <math>2</math> squares such that there is a <math>2</math> by <math>2</math> square centered at <math>(3x, 3y)</math> for all ordered pairs of integers <math>(x, y).</math> It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at <math>(0, 0)</math>. (minus the teleportations) Since counting the complement set is easier, we'll count the number of <math>4</math>-step paths such that Frieda never reaches a corner point.  
 +
 
 +
In other words, since the reachable corner points are <math>(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1), </math> and <math>(\pm 2, \pm 2),</math> Frieda can only travel along the collection of points included in <math>S</math>, where <math>S</math> is all points on <math>x=0</math> and <math>y=0</math> such that <math>|y|<4</math> and <math>|x|<4</math>, respectively, plus all points on the big square with side length <math>6</math> centered at <math>(0, 0).</math> We then can proceed with casework:
 +
 
 
Case <math>1</math>: Frieda never reaches <math>(0, \pm 3)</math> nor <math>(\pm 3, 0).</math>
 
Case <math>1</math>: Frieda never reaches <math>(0, \pm 3)</math> nor <math>(\pm 3, 0).</math>
When Frieda only moves horizontally or vertically for her four moves, she can do so in <math>2^4 - 4 = 12</math> ways for each case . Thus, <math>12 \cdot 2</math> total paths for the subcase of staying in one direction. (For instance, all length <math>4</math> combinations of <math>F</math> and <math>B</math> except <math>FFFF</math>, <math>BBBB</math>, <math>FFFB</math>, and <math>BBBF</math> for the horizontal direction.) There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are <math>FBUD</math>, <math>FBUU</math>, <math>UDFB</math>, and <math>UDFF.</math> Thus, a total of <math>4 \cdot 4 = 16</math> ways for this subcase.  
+
 
 +
When Frieda only moves horizontally or vertically for her four moves, she can do so in <math>2^4 - 4 = 12</math> ways for each case . Thus, <math>12 \cdot 2</math> total paths for the subcase of staying in one direction. (For instance, all length <math>4</math> combinations of <math>F</math> and <math>B</math> except <math>FFFF</math>, <math>BBBB</math>, <math>FFFB</math>, and <math>BBBF</math> for the horizontal direction.)  
 +
 
 +
There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are <math>FBUD</math>, <math>FBUU</math>, <math>UDFB</math>, and <math>UDFF.</math> Thus, a total of <math>4 \cdot 4 = 16</math> ways for this subcase.  
 +
 
 
Total for Case <math>1</math>: <math>24 + 16 = 40</math>
 
Total for Case <math>1</math>: <math>24 + 16 = 40</math>
 +
 
Case <math>2</math>: Frieda reaches <math>(0, \pm 3)</math> or <math>(\pm 3, 0)</math>.
 
Case <math>2</math>: Frieda reaches <math>(0, \pm 3)</math> or <math>(\pm 3, 0)</math>.
 +
 
Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of <math>4 \cdot 4 = 16</math> paths for this case.
 
Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of <math>4 \cdot 4 = 16</math> paths for this case.
 +
 
Our total number of paths never reaching coroners is thus <math>16+40=56,</math> making for an answer of <cmath>\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.</cmath>
 
Our total number of paths never reaching coroners is thus <math>16+40=56,</math> making for an answer of <cmath>\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.</cmath>
  

Revision as of 15:00, 12 February 2021

Problem

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$

Solution 1 (complementary counting)

We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability.

From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework.

$\textbf{Case 1:}$ She goes back to the center.

Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1

$\textbf{Case 2:}$ She goes to another edge (rightmost).

Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$

Subcase 2: She goes to the center. Now any move works.

$\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2

She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$

So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$

But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$. ~ firebolt360

Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360

Solution 2 (direct counting and probability states)

We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have $1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}$, so the answer is $\boxed{D}$. ~IceWolf10

Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)

Denominator

There are $4^4=256$ ways to make 4 hops without restrictions.


Numerator (Casework)

Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.


(1) Hop #2 (Hops #3 and #4 have no restrictions)

The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has $4\cdot2\cdot4\cdot4=128$ ways.


(2) Hop #3 (Hope #4 has no restriction)

No matter which direction the first hop takes, the second hop must "wrap around".

The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has $4\cdot1\cdot2\cdot4=32$ ways.


(3) Hop #4

Two sub-cases:

(3.1) The second hop "wraps around". It follows that the third hop also "wraps around".

The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has $4\cdot1\cdot1\cdot2=8$ ways.

(3.2) The second hop backs to the center.

The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has $4\cdot1\cdot4\cdot2=32$ ways.

Together, Case (3) has $8+32=40$ ways.


The numerator is $128+32+40=200.$


Probability

$\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.$


This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3


~MRENTHUSIASM

Solution 4

Let $C_n$ be the probability that Frieda is on the central square after n moves, $E_n$ be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and $V_n$ (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in $1$ specific direction out of $4$ total directions from the middle of an edge, so $C_{n+1}=\frac{E_n}{4}$. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in $1$ specific direction from the middle of an edge, so $E_{n+1}=C_n+\frac{E_n}{4}$. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in $2$ specific directions from the middle of an edge, so $V_{n+1}=V_n+\frac{E_n}{2}$. Since Frieda always start from the center, $C_0=1$, $E_0=0$, and $V_0=0$. We use the previous formulas to work out $V_4$ and find it to be $\boxed{\textbf{(D)} ~\frac{25}{32}}$.

-SmileKat32

Solution 5

Imagine an infinite grid of $2$ by $2$ squares such that there is a $2$ by $2$ square centered at $(3x, 3y)$ for all ordered pairs of integers $(x, y).$ It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at $(0, 0)$. (minus the teleportations) Since counting the complement set is easier, we'll count the number of $4$-step paths such that Frieda never reaches a corner point.

In other words, since the reachable corner points are $(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1),$ and $(\pm 2, \pm 2),$ Frieda can only travel along the collection of points included in $S$, where $S$ is all points on $x=0$ and $y=0$ such that $|y|<4$ and $|x|<4$, respectively, plus all points on the big square with side length $6$ centered at $(0, 0).$ We then can proceed with casework:

Case $1$: Frieda never reaches $(0, \pm 3)$ nor $(\pm 3, 0).$

When Frieda only moves horizontally or vertically for her four moves, she can do so in $2^4 - 4 = 12$ ways for each case . Thus, $12 \cdot 2$ total paths for the subcase of staying in one direction. (For instance, all length $4$ combinations of $F$ and $B$ except $FFFF$, $BBBB$, $FFFB$, and $BBBF$ for the horizontal direction.)

There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are $FBUD$, $FBUU$, $UDFB$, and $UDFF.$ Thus, a total of $4 \cdot 4 = 16$ ways for this subcase.

Total for Case $1$: $24 + 16 = 40$

Case $2$: Frieda reaches $(0, \pm 3)$ or $(\pm 3, 0)$.

Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of $4 \cdot 4 = 16$ paths for this case.

Our total number of paths never reaching coroners is thus $16+40=56,$ making for an answer of \[\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.\]

-fidgetboss_4000

Video Solution by OmegaLearn (Using Probability States)

https://youtu.be/V_Sn30N2q50

~ pi_is_3.14

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS