AMC10 2025 B

$\frac{350}{20}$ is $17.5$. We round down to find the maximum he can make (he cannot brew half of a cup), so the answer is $\boxed{\text{(B) }17}$.

By a modulo $10$ argument, the ones digits repeat with period $10$ in the following order: \[1,4,9,6,5,6,9,4,1,0\] The sum of the numbers can be verified to be $45$. There are $202$ periods that occur from $1$ to $2025$, and there are five extra numbers, those being $1,4,9,6,5$, corresponding to $2021,2022,2023,2024,2025$. The sum of these numbers is $25$. Hence, the total is \[202\cdot 45+25=9090+25=\boxed{\textbf{(D)}~9115}\]

If we find the sum for each rows, I'll only do the first 5, we find that row 1 has sum $10$, row 2 has sum $11$, row 3 has sum $22$, row 4 has sum $44$, and row 5 has sum $88$. we can see that excluding the first row each row has sum $11*2^{n-2}$ where $n$ is the row. This means the 11th row will have sum $11*2^9=11*512=5632$. The sum of the digits of $5632$ is $16$, so the answer is $\boxed{\textbf{(D) }16}$. Notice that the sum of each row is the slightly different than the normal Pascal's triangle, so the sum of the numbers is similar. Here's a proof of why each number in the normal Pascal's triangle is twice the previous number. To prove that the sum of each row is $2$ times the previous row, use the equation ${(1+1)}^n$ with the normal Pascal's triangle and use the binomial theorem from there.

By definition of bases, $\underline{a}~\underline{b}$ base seven is equal to $7a+b$, and $\underline{b}~\underline{a}$ base nine is equal to $9b+a$. Therefore, we must have $7a+b=9b+a$, or $6a=8b$, or $3a=4b$. But in base seven, we must have $a,b<7$. Testing cases yields $a=4,b=3$ as the only solution. Their sum is $\boxed{\textbf{(A)}~7}$. The first equation comes from the following idea. In base $10$, a two-digit number can be represented as $10$ times the tens digit plus the units digit, or $10^1 \cdot a + 10^0 \cdot b$. If we insert the base numbers into this expression for $\underline{a}~\underline{b}$ and $\underline{b}~\underline{a}$, we get $7^1 \cdot a + b = 9^1 \cdot b + a$. The rest of the solution is above.

Let us extend $\overline{AO}$ to meet $\odot{O}$ at $D$. Since $m\angle{ABC}=130$°, by the Inscribed Angle Theorem, the measure of major arc $\widehat{AC}=260$°. $\overline{AD}$ is a diameter of $\odot{O}$, so the measure of top arc $\widehat{AD}=180$, so $m\widehat{CD}=260-180=80$° and $m\angle{CAO}=\boxed{\textbf{(C) }40\textbf{°}}$.

We solve this by finding the areas of trapezoids. The line is $y = \frac{1}{3}x + 1$. Total Area ($A_{\text{total}}$) of Lower Region: The region from $x=0$ to $x=2$ is a trapezoid. \[A_{\text{total}} = \frac{1}{2}(y(0) + y(2)) \times (2 - 0) = \frac{1}{2}\left(1 + \frac{5}{3}\right)(2) = \frac{8}{3}\] Half Area ($A_{\text{half}}$): The area of the region from $x=0$ to $x=a$ must be $\frac{1}{2}A_{\text{total}}$. \[A_{\text{half}} = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}\] Solve for $a$: This half-region is also a trapezoid. \[A_{\text{half}} = \frac{1}{2}(y(0) + y(a)) \times (a - 0) = \frac{4}{3}\] \[\frac{1}{2}\left(1 + \left(\frac{1}{3}a + 1\right)\right)a = \frac{4}{3}\] \[\frac{1}{2}\left(2 + \frac{a}{3}\right)a = \frac{4}{3}\] \[a + \frac{a^2}{6} = \frac{4}{3}\] Multiplying by 6 gives the quadratic equation: \[a^2 + 6a - 8 = 0\] \[a = \frac{-6 \pm \sqrt{6^2 - 4(1)(-8)}}{2} = \frac{-6 \pm \sqrt{68}}{2} = \frac{-6 \pm 2\sqrt{17}}{2} = -3 \pm \sqrt{17}\] Since $a$ must be positive, $a = \sqrt{17} - 3$. \[s + t = 17 + 3 = 20\] The correct option is $C$. Note: We can find the area of the trapezoid by integrating. We have the area is $\int_{0}^{2} \frac{1}{3}x + 1 \ dx = \frac{1}{6}x^2 + x \bigg|_{x=0}^{2} = \frac{8}{3}$. But this sort of destroys the purpose of this simple solution without calculus. With calculus, the solution below would be the most efficient.

Let Frances's location be $F$, the locked gate be $L$, the box be $B$, the unlocked gate on the west be $U$, and the unlocked gate on the east be $V$. We are given that $FL = 15$, $LB = x$, $BV$ = 9, and $LU = 8$. Because $FL$ is vertical and the fence is horizontal, $m \angle FLU = m \angle FLV = 90^\circ$. If Frances takes the western path, she must first travel the length of $FV$, and then $x + 8$ meters. By the Pythagorean theorem, $FV = \sqrt{8^2 + 15^2} = 17$. The remaining distance from Francis and the box of chocolates after she walks that distance is $8 + x$ meters. Thus, the distance Francis travels if she takes the western path is $25 + x$. If Frances takes the eastern path, she first travels $\sqrt{(x + 9)^2 + 15^2} = \sqrt{(x + 9)^2 + 225}$ meters, and then $9$ more meters. Her total distance on the eastern path is $\sqrt{(x + 9)^2 + 225} + 9$ meters. Now, we set the two distances equal to each other: \[x + 25 = \sqrt{(x + 9)^2 + 225} + 9\] \[x + 16 = \sqrt{(x + 9)^2 + 225}\] \[x^2 + 32x + 256 = (x + 9)^2 + 225\] \[x^2 + 32x + 256 = x^2 + 18x + 81 + 225\] \[14x = 50\] \[x = \frac{50}{14} = \frac{25}{7}.\] This corresponds with the answer $\boxed{\textbf{(C)}\hspace{3pt}3\frac{4}{7}}$.

The problem is essentially telling us that $\underline A~\underline B~\underline B.\underline B~\underline A$ is divisible by $36$, so it is divisible by $9$. Then, by the divisibility by $9$ condition, \[A+B+B+B+A=2A+3B\equiv 0\pmod{9}\] This means that $A$ must be divisible by $3$, or otherwise $2A+3B$ would not be divisible by $3$ and thus $9$. Since $A\ne0$, we must have one of $A=3,6,9$. But $A$ must be even or else the entire number would not be even and thus would not be divisible by $4$. Hence $A=6$. Then, $2\cdot 6+3B\equiv 0\pmod{9}$, so $4+B\equiv 0\pmod{3}$, and thus $B\equiv 2\pmod{3}$. This yields $B=2,5,8$, of which $B=5$ is the only number that allows $\underline A~\underline B~\underline B.\underline B~\underline A$ to be divisible by $4$. Thus the answer is $6+5=\boxed{\textbf{(C)}~11}$.

We have: \[\begin{cases} x+y+z \ge 2, \\ -x+y+z \le 2, \\ x-y+z \le 2, \\ x+y-z \le 2. \end{cases}\] Subtract the second inequality from the first: \[(x+y+z)-(-x+y+z) \ge 2-2 \quad\Rightarrow\quad 2x \ge 0 \Rightarrow x\ge 0.\]By symmetry in $x,y,z,$ we also obtain \[y\ge 0,\qquad z\ge 0.\] WLOG & by symmetry, let \[0 \le x \le y \le z.\] up to permutation. From the second inequality of we have $y+z \le 2+x.$ Using $x\le y$, we get \[y+z \le 2+y \quad\Rightarrow\quad z \le 2.\] Hence, $0 \le x \le y \le z \le 2.$ Divide into casework on $x:$ Case 1: $x=0$ Then the first two inequalities become \[y+z \ge 2,\qquad y+z \le 2 \Rightarrow y+z=2\]And because we have $0\le y\le z\le 2, (y,z)=(0,2), (1,1).$ Case 2: $x=1$ \[1+y+z \ge 2 \Rightarrow y+z\ge 1,\qquad -1+y+z \le 2 \Rightarrow y+z \le 3.\]$1\le y\le z\le 2$, so $(y,z)=(1,1)\ \text{or}\ (1,2).$ Case 3: $x=2$ \[2+y+z \ge 2 \qquad -2+y+z \le 2 \Rightarrow y+z \le 4.\]$2\le y\le z\le 2$ so the only combination $(y,z)=(2,2).$ Thus up to permutation we have $(0,0,2),\ (0,1,1),\ (1,1,1),\ (1,1,2),\ (2,2,2)$ Now count distinct ordered triples for each set of three numbers under symmetry: $3+3+3+1+1=\boxed{11}.$

Observe that both $f(n)$ and $g(n)$ can be factored using RZT. By trying out a few values, we get that $n = -1$ is a root of $f(n)$. Then, we can use synthetic division (regular polynomial division works, too) to get the other roots of $f(n)$, which are $n = 4$ and $n = 2$. Now we factor $g(n)$. By inspection, $n = -1$ also works on $g(n)$, and so synthetic division and factoring the remaining quadratic gets us the roots $n = 3$ and $n = 4.$ We can now express $\frac{f(n)}{g(n)}$ as: \[\frac{(n + 1)(n - 4)(n - 2)}{(n + 1)(n - 4)(n - 3)}.\] We can cancel the top and bottom to get: \[\frac{(n - 2)}{(n - 3)}\] Notice that the only values for which this expression is an integer are $n = 4$ and $n = 2$, because they both cause the denominator to have a magnitude of $1$. However, $n = 4$ is an extraneous solution as plugging it back into the original expression would make it evaluate to $\frac{0}{0}$. Our answer is then $\boxed{\textbf{(A)}\hspace{3pt} 2}$.

On Monday, the first tutor has 6 choices for kids to be assigned to, the second has 5, and so on. Hence, there are $6!$ possible ways to assign the tutors to the kids. On Tuesday, there are $\binom{6}{2}$ ways to choose the kid-tutor pairs that remain the same from Monday. For the other 4 kids, they need to be assigned to the other tutors in such a way that none of those kids are assigned to the same tutor that they were assigned to on Monday. There are $9$ ways to do this. Thus, our final answer is $\frac{\binom{6}{2} \cdot 9}{6!}$ or $\boxed{\textbf{(B)}\hspace{3pt}\frac{3}{16}}$. Remark: We can get $9$ either by listing all the possible configurations or by rounding $\frac{4!}{e}$. This is called a derangement and is notated as $!n$. It can be calculated with the explicit formula $!n = n!\sum _{i=0}^{n}\frac{(-1)^{i}}{i!}$

To solve this problem, we can dissect the different shapes into equilateral triangles. We can keep figure $T$ the same, and then divide the rhombus $R$ into two equilateral triangles (as a result of one of the angles equaling $60^\circ$), with one circle in each that is the incircle to its triangle. We can do the same for the hexagon $H$, dividing it into six equilateral triangles with one circle in each, and each circle is tangent to the triangle that encloses it. We can see that $H = R$, since the proportion of the shaded circles to the unshaded portions in the rhombus and the hexagon is the same. Both are equal to an equilateral triangle with a circle inscribed in it; therefore they have an equal ratio. We can use this to estimate that the ratio of the circle to the triangle is the largest for the first figure, as we can obviously see this through intuition and observation. Furthermore, there is no answer choice for $T < H = R$, and it is evidently not $H = R = T$. Hence, the answer to the question is as follows: $\boxed{\textbf{(C)}\ H = R < T}.$

Without loss of generality, let $\triangle ABC$ have side-lengths $AB=2, BC=2\sqrt{3},$ and $AC=4.$ Let $D$ be the foot of the perpendicular from $B$ to $\overline{AC}, \ E$ be the midpoint of $\overline{AB},$ and $F$ be the intersection of $\overline{CE}$ and $\overline{BD}.$ Note that $\triangle ADB$ and $\triangle BDC$ are both $30^\circ{-}60^\circ{-}90^\circ$ triangles. From the side-length ratio, we get $AD=1$ and $DC=3.$ We obtain the following diagram: From here, we will proceed with mass points. Throughout this solution, we will use $W_P$ to denote the weight of point $P.$ Let $W_C=1.$ Since $3AD=DC,$ it follows that $W_A=3$ and $W_D=W_C+W_A=4.$ Since $AE=EB$ and $W_A=3,$ it follows that $W_B=3.$ Now we focus on $\overline{BD}:$ Since $W_B=3$ and $W_D=4,$ we have $\frac{DF}{FB}=\frac xy=\frac34.$ Therefore, the answer is \[\frac{x}{x+y}=\frac{3}{3+4}=\boxed{\textbf{(A) } \dfrac{3}{7}}.\]

We will arrange the nine athletes in a line, with the first three being in the blue group, the next three being in the red group, and the last three being in the green group. Suppose the three tallest athletes are named $A, B, C$, in some order. We have $3!$ ways to choose which group each of these athletes can be in. We then have $3 \cdot 3 \cdot 3$ ways to determine which of the three positions in that group they can take. From here, we must distribute the remaining six athletes in the remaining six spaces, which can be done in $6!$ ways. There are therefore $6 \cdot 27 \cdot 6!$ favorable outcomes. There are also $9!$ total ways to arrange the athletes with no restrictions. Our answer is $\frac{6 \cdot 27 \cdot 6!}{9!} = \frac{6 \cdot 27}{9 \cdot 8 \cdot 7} = \boxed{\textbf{(C) }\frac{9}{28}}.$

Through partial sum decomposition we obtain \[\frac{1}{k^3 + 6k^2 + 8k} = \frac{1}{8} \left( \frac{1}{k} - \frac{2}{k+2} + \frac{1}{k+4} \right),\] so one can write the expression as 18∑k=1∞(1k−2k+2+1k+4)=18∑k=1∞(1k−1k+2−1k+2+1k+4)=18∑k=1∞(1k−1k+2)+18∑k=1∞(1k+4−1k+2)=18∑k=1∞(1k−1k+2)−18∑k=1∞(1k+2−1k+4)=18∑k=1∞(1k−1k+2)−18∑k=3∞(1k−1k+2)=18∑k=12(1k−1k+2) Simplifying the sum and multiplying results in $\frac{11}{96}$, in which we finally have $11 + 96 =$ $\boxed{\textbf{(D)}~107}$ Note: For more info on Partial Sum Decompositions; AoPS Intermediate Algebra Chapter 17.2 (Telescoping) gives a brief introduction to them, which is sufficient enough to solve this problem.

Model the circle as the 6-cycle $C_6$ with labeled positions $1,2,3,4,5,6$ in circular order. We count proper colorings using $R,G,B$, each appearing exactly twice. There are two structural possibilities for how the two copies of each color are placed: {Case 1: All three colors occupy opposite positions.} Opposite positions in $C_6$ are the three pairs $(1,4)$, $(2,5)$, $(3,6)$. If each color occupies one opposite pair, then the colors must alternate around the circle, yielding a pattern of the form $\{A,B,C,A,B,C\}$ for some ordering $(A,B,C)$ of $(R,G,B)$. The number of such colorings is $3!=6$. {Case 2: Exactly one color occupies opposite positions.} Choose the color that is opposite: $3$ choices. Choose which opposite pair it occupies: $3$ choices, one of $(1,4)$, $(2,5)$, $(3,6)$, for a total of $3\cdot 3=9$ placements. Removing those two opposite positions leaves two disjoint adjacent pairs among the remaining four positions. For example, if the opposite color is at $(1,4)$, the remaining positions $\{2,3,5,6\}$ form the two adjacent pairs $(2,3)$ and $(5,6)$. The other two colors must each appear twice, and no adjacent positions may match; therefore each pair must be colored with different colors, and to achieve two copies of each color overall, the two pairs must be complementary. $(2,3)$ is $(G,B)$ or $(B,G)$, and then $(5,6)$ in the opposite ordering. This gives $2$ valid assignments for the four positions per fixed opposite placement. Hence, this case contributes $9\cdot 2=18$. Summing the $2$ cases $6+18=24$ $(D)\ 24$.

Since the mean of the first three terms is $2025$, their sum is $2025 * 3 = 6075$. Then, incorporating the fourth term makes the mean $2025-1=2024$, so their sum must be $2024 * 4 = 8096$, implying that the 4th term is $8096-6075=2021$ Doing the same for the 5th term, $2023 * 5= 10115$, 5th term is $10115-8096=2019$ Algebraically, we can model this pattern for the $k$th term = x as $(k-1)(2029-k) + x = (k)(2028-k)$ $2029k-2029-k^2+k+x=2028k-k^2$ $x=2029-2k$ So the maximum $k\le n$ for which $x$ is positive is 1014, giving our answer $n=\boxed{\textbf{(B) }1014}$

All terms from the 1st to the 3rd will equal 1, because the value inside the floor function will be greater than or equal to $\sqrt{1} = 1$ but less than $\sqrt{4} = 2$. Following this logic, all terms from the 4th to the 8th will equal 2, all terms from the 9th to the 15th will equal 3, and all terms from number $n^2$ to $(n + 1)^2 - 1$ will equal $\sqrt{n}$. Thus, there will be 3 terms equal to 1, 5 terms equal to 2, and so on. Writing out a few terms of the total sum, we have: \[1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + 44 \cdot 89\] The expression above can be written in summation format as: \[\sum_{n=1}^{44} n(2n + 1).\] We can expand this to the following: \[\sum_{n=1}^{44} 2n^2 + \sum_{n=1}^{44} n\] Using the sum of integers formulas, this becomes \[\frac{2 \cdot 44 \cdot 45 \cdot 89}{6} + \frac{44 \cdot 45}{2}\] The units digit of that expression evaluates to 0. However, we excluded $\lfloor \sqrt{2025} \rfloor = 45$ from our sum because it appears only once in the sequence, so we have to add it back into our expression. Our final answer is $\boxed{\textbf{(D)}\hspace{3pt}5}$.

Extend the edges pointing downwards to converge at a point $A$ to form a square pyramid. Consider 3 square pyramids, the large one formed by the top vertices of the original figure and $A$, the middle one formed by the medians running through the sides of the original figure and point $A$, and the smaller one formed by the bottom vertices of the original figure and point $A$. Note that all pyramids are similar since they are all essentially scaled by a certain factor. The median length is $\frac{3+1}{2}=2$ Using side length to volume ratios, find that the volumes must have ratios $1:8:27$ Then, you get that the ratio of the volume thus filled to the volume that we must fill is equivalent to $8-1:27-8 = 7:19$. Thus, it will take $\frac{19}{7}$ more time to fill the remaining volume giving us an answer of $\frac{19}{7} * 35 = \boxed{\textbf{(D) }95}$

Let the radius of the large semicircles be $r$, and let the diameter of the inner be $D$. Draw lines as follows: By the Pythagorean Theorem, $r^2+(1-r)^2=(2r)^2$, so $2r^2+2r-1=0$. Hence $r=\frac{\sqrt{3}-1}{2}$ (we take the positive solution). Now apply the Pythagorean Theorem on the quadrilateral’s altitude down to get $(1-2r)^2+1=(2r+D)^2$. Solving yields $2r+D=\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$. Then $D=\sqrt{6}-\sqrt{2}-\sqrt{3}+1=(\sqrt{3}-1)(\sqrt{2}-1)$. Hence the answer is $3-1+2-1=\boxed{\textbf{(A) } 3}$.

Denote $1=\text{red}$, $2=\text{blue}$, $3=\text{yellow}$. We need $1\to 2\to 3\to 1$. WLOG place $1$ in the center $(0,0)$, $2$ on the left edge $(-1,0)$, $3$ on the top-left corner $(-1,1)$. \[\begin{bmatrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] $3$ must see $1$, so the top edge $(0,1)$ must also have $1$. Then, $1$ must see $2$, so the top-right corner $(1,1)$ becomes $2$, which must see $3$, so the right edge $(1,0)$ must have $3$. \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix}\] Now this bottom-right corner can vary either as $1$, $2$ or $3$. Cases on $(1,-1)$: If $1$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}\] but the 3 needs a 1 and does not have it, so there are $0$ cases. If $2$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ b & a & 2 \end{bmatrix}\] if $a=1$, $b$ can be $1$ or $3$. If $a=2$, $b=3$, but the 3 needs a 1 and can't get it. If $a=3$, $b=1,2$, so there are $4$ cases in total. If $3$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 2 & 1 & 3 \end{bmatrix}\] Since the 2 in the bottom left corner does not have a 3 nearby, there are $0$ cases. WLOG the center fixes a factor of $3$, so the answer is $4\cdot 3=\boxed{\textbf{(C) } 12}$.

Let the number be represented as $\overline{a_1a_2a_3a_4a_5a_6a_7}.$ We first seek to find the number of seven-digit integers with $a_1+a_2+a_3+a_4+a_5+a_6+a_7 = 61.$ We would love to do stars and bars here, but we have the condition that $0 \leq a_n \leq 9.$ Therefore, we define $a_n' = 9-a_n \implies a_n = 9-a_n'.$ This is a bijective operation, so the total number of integers doesn't change when we use these digits instead. Substituting in gives that ∑k=17(9−ak′)=61,a1′+a2′+a3′+a4′+a5′+a6′+a7′=2. Since the $a_n'$ are nonnegative integers and the condition $a_n' \leq 9$ doesn't matter with a sum less than 9, we can do stars and bars to figure out the number of permutations of digits that satisfy the constraint: ${{2+6}\choose {2}} = {8\choose 2} = 28.$ Now, we need to find how many of these integers are divisible by $11.$ We can use the divisibility by $11$ rule, which says that $(a_1+a_3+a_5+a_7) - (a_2+a_4+a_6) = 11k$ for some integer $k.$ Notice that $k \geq 1$ because, otherwise, $a_2+a_4+a_6 > 27$ which is the maximum sum of three digits. In addition, $k \leq 1$ because, otherwise $a_1+a_3+a_5+a_7 > 36$ which is the maximum sum of four digits. Therefore $k = 1.$ Then we have that $(a_1+a_3+a_5+a_7) - (a_2+a_4+a_6) = 11,$ so $a_1+a_3+a_5+a_7 = 36$ and $a_2+a_4+a_6 = 25.$ The first equation can only result in $a_1=a_3=a_5=a_7 = 9.$ The second equation can be rewritten in a similar manner as before to get that (9−a2′)+(9−a4′)+(9−a6′)=25,a2′+a4′+a6′=2. Again, with stars and bars, there exists ${{2+2}\choose {2}} = {4\choose 2} = 6$ ordered triples of $(a_2, a_4, a_6)$. Therefore, our answer is \[\frac{6}{28} = \boxed{\text{(A) } \frac{3}{14}}.\]

Let's say $n$ is one of the numbers that got written twice in the same box. Suppose it is at column $x$ and row $y$. We will write an expression for $n$ in terms of $x$ and $y$ in two ways: from Horace's perspective and Vera's perspective. From Horace's perspective, $n = (y-1)(141) + x$. This is because there are $y-1$ full rows before $n$, and we then need $x$ more numbers to reach $n$. Similarly, Vera will say $n = (x-1)(91) + y$. We now have the Diophantine equation \[(y-1)(141) + x = (x-1)(91)+y\] \[141y-141+x = 91x-91+y\] \[140y=90x+50\] \[14y = 9x + 5\] One such solution is, of course, $x=y=1$. We find further solutions by adding $14$ to $x$ and $9$ to $y$. For example, the second solution is $(15,10)$. This continues until $(141,91)$ is reached. There are $\boxed{11}$ ordered pairs in this list.

We will solve this using states. Let $P_n$ be the probability of reaching $4$, given that you start from $n$. We want to find $P_0$. Of course, $P_4 = 1$. We also know that \[P_3 = \frac{1}{4}P_4 + \frac{1}{4}P_2\] \[P_2 = \frac{1}{4}P_3 + \frac{1}{4}P_1\] \[P_1 = \frac{1}{4}P_2 + \frac{1}{4}P_0\] \[P_0 = \frac{1}{2}P_1.\] Solving the system, we find that $P_1 = \frac{2}{97}$ and $P_0 = \boxed{\frac{1}{97}}.$

The square can be placed with $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$. Then $P=\left(0,\frac{8}{5}\right)$, $Q=\left(\frac{10}{3},4\right).$ Straight line reflection is equivalent to passing through. The straight line from $P$ to $Q$ is $y=\frac{18}{25}x+\frac{8}{5}$ $25y=18x+40\equiv 4\pmod{18}$ Notice that $25\times 2=7\times 2\equiv -4\pmod{18}$, so $25\times(-2)\equiv 4\pmod{18}$, $y=-2\equiv 16\pmod{18}$, then $x=20$, the line passed through $(20,16)$. From $\left(0,\frac{8}{5}\right)$ to $(20,16)$, $x$ coordinate increases $20$ whole units, $y$ coordinate increases $15$ whole units. Excluding the case where $PQ$ passes through a lattice point at the final grid point, the segment $PQ$ crosses integer grid lines a total of $(20-1)+(15-1)=33$ times. Each time the line crosses an integer grid line, it corresponds to one reflection; every four reflections, the path completes a full rotation within the square. $33\equiv 1\pmod{4}$, after one reflection off side $CD$, the line $PQ$ will pass through point $\boxed{B}$.