{"id": "aime25-0", "problem": "Find the sum of all integer bases  $b>9$  for which  $17_b$  is a divisor of  $97_b.$", "expected_answer": "70", "reference_solution": "This means that  $a(b+7)=9b+7$  where  $a$  is a natural number. Rearranging we get  $(a-9)(b+7)=-56$ . Since  $b>9$ ,  $b=49,21$ . Thus the answer is  $49+21=\\boxed{70}$"}
{"id": "aime25-1", "problem": "In  $\\triangle ABC$  points  $D$  and  $E$  lie on  $\\overline{AB}$  so that  $AD < AE < AB$ , while points  $F$  and  $G$  lie on  $\\overline{AC}$  so that  $AF < AG < AC$ . Suppose  $AD = 4$ ,  $DE = 16$ ,  $EB = 8$ ,  $AF = 13$ ,  $FG = 52$ , and  $GC = 26$ . Let  $M$  be the reflection of  $D$  through  $F$ , and let  $N$  be the reflection of  $G$  through  $E$ . The area of quadrilateral  $DEGF$  is  $288$ . Find the area of heptagon  $AFNBCEM$ , as shown in the figure below.\n [asy] unitsize(14); pair A = (0, 9), B = (-6, 0), C = (12, 0), D = (5A + 2B)/7, E = (2A + 5B)/7, F = (5A + 2C)/7, G = (2A + 5C)/7, M = 2F - D, N = 2E - G; filldraw(A--F--N--B--C--E--M--cycle, lightgray); draw(A--B--C--cycle); draw(D--M); draw(N--G); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(M); dot(N); label(\"$A$\", A, dir(90)); label(\"$B$\", B, dir(225)); label(\"$C$\", C, dir(315)); label(\"$D$\", D, dir(135)); label(\"$E$\", E, dir(135)); label(\"$F$\", F, dir(45)); label(\"$G$\", G, dir(45)); label(\"$M$\", M, dir(45)); label(\"$N$\", N, dir(135)); [/asy]", "expected_answer": "588", "reference_solution": "Note that the triangles outside  $\\triangle ABC$  have the same height as the unshaded triangles in  $\\triangle ABC$ . Since they have the same bases, the area of the heptagon is the same as the area of triangle  $ABC$ . Therefore, we need to calculate the area of  $\\triangle ABC$ . Denote the length of  $DF$  as  $x$  and the altitude of  $A$  to  $DF$  as  $h$ . Since  $\\triangle ADF \\sim \\triangle AEG$ ,  $EG = 5x$  and the altitude of  $DFGE$  is  $4h$ . The area  $[DFGE] = \\frac{5x + x}{2} \\cdot 4h = 3x \\cdot 4h = 12xh = 288 \\implies xh = 24$ . The area of  $\\triangle ABC$  is equal to  $\\frac{1}{2} 7x \\cdot 7h = \\frac{1}{2} 49xh = \\frac{1}{2} 49 \\cdot 24 = \\frac{1}{2} 1176 = \\boxed{588}$ ."}
{"id": "aime25-2", "problem": "The  $9$  members of a baseball team went to an ice-cream parlor after their game. Each player had a single scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let  $N$  be the number of different assignments of flavors to players that meet these conditions. Find the remainder when  $N$  is divided by  $1000.$", "expected_answer": "16", "reference_solution": "Denote the number of people who chose strawberry, vanilla, or chocolate as (S, V, C). Then, as S < V < C, we just need to find values of S, V, and C such that S + V + C = 9. Notice S can only be 1 or 2 as S = 3 will result in V + C = 6 and it just won't work for S < V < C. So using these two values, we get that the possible triples of (S, V, C) are: (1, 3, 5), (2, 3, 4) and (1, 2, 6). Now, let's consider (S, V, C) = (1, 3, 5). If we start with the strawberry people, notice there are  ${9\\choose 1}$  possibilities.\nNow, we see there are 8 different people waiting to be assigned to the 3 vanilla people therefore there are  ${8\\choose 3}$  ways for this to work. We can now go down the list to get:\n ${9\\choose 1}{8\\choose 3}{5\\choose 5} + {9\\choose 2}{7\\choose 3}{4\\choose 4} + {9\\choose 1}{8\\choose 2}{6\\choose 6}$  which gives a grand total of  $2016$  possibilities. The remainder when  $N$  is divided by  $1000$  is  $\\boxed{016}$ ."}
{"id": "aime25-3", "problem": "Find the number of ordered pairs  $(x,y)$ , where both  $x$  and  $y$  are integers between  $-100$  and  $100$  inclusive, such that  $12x^2-xy-6y^2=0$ .", "expected_answer": "117", "reference_solution": "Please help with LaTex Formatting: You can use the quadratic formula for this equation: 12x^2 - xy - 6y^2 = 0;\nAlthough this solution may seem to be misleading, it works! You get: \\frac {-b +- \\sqrt b^2-4ac}{2a} = \\frac {<span style=\"color:red\">$xy +- \\sqrt(x^2y^2+(12*6*4*x^2*y^2)}{24x^2}\n</span> = \\frac{xy +- \\sqrt289x^2 y^2}{24x^2} = \\frac{18xy/24x^2$ (Error compiling LaTeX. Unknown error_msg), and <span style=\"color:red\">$-16xy}{24x^2}\n</span> Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:$ (Error compiling LaTeX. Unknown error_msg)3y = 4x $;$ -2y = 3x<span style=\"color:red\">$; \n</span> As x and y are between$ (Error compiling LaTeX. Unknown error_msg)-100 $and$ 100 $, for the first equation, x can be between$ (-75,75) $, but x must be a multiple of 3, so there are:$ ((75+75)/3) + 1 = 51<span style=\"color:red\">$solutions for this case.\n</span> For -2y = 3x: x can be between (-66, 66), but x has to be a multiple of 2. Therefore, there are$ (Error compiling LaTeX. Unknown error_msg)(66+66)/2 + 1 = 67<span style=\"color:red\">$solutions for this case\n</span> However, the one overlap would be x = 0, because y would be 0 in both solutions. Therefore, the answer is$ (Error compiling LaTeX. Unknown error_msg)51+67-1 = \\boxed{117}$. -U-King3.14Root"}
{"id": "aime25-4", "problem": "There are  $8!= 40320$  eight-digit positive integers that use each of the digits  $1, 2, 3, 4, 5, 6, 7, 8$  exactly once. Let  $N$  be the number of these integers that are divisible by  $22$ . Find the difference between  $N$  and  $2025$ .", "expected_answer": "279", "reference_solution": "Notice that if the 8-digit number is divisible by  $22$ , it must have an even units digit. Therefore, we can break it up into cases and let the last digit be either  $2, 4, 6,$  or  $8$ . Due to symmetry, upon finding the total count of one of these last digit cases (we look at last digit  $2$  here), we may multiply the resulting value by  $4$ . <br/>\nNow, we just need to find the number of positions of the remaining numbers such that the units digit is  $2$  and the number is divisible by  $11$ . Denote the odd numbered positions to be  $a_1, a_3, a_5, a_7$  and the even numbered positions to be  $a_2, a_4, a_6$  (recall  $a_8=2$ ). By the divisibility rule of  $11$ , we must have:\n \\[(a_1 + a_3 + a_5 + a_7) - (a_2 + a_4 + a_6 + 2)\\] \nwhich is congruent to  $0\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)$ . Therefore, after simplifying, we must have:\n \\[a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + a_7\\equiv2\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)\\] \nNow consider  $a_1+ a_2 +\\ldots + a_7=1+2+\\ldots+8-2=34\\equiv1\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)$ . Therefore,\n \\[(a_1 + a_2 + \\ldots+ a_7) - 2(a_2 + a_4 + a_6)\\equiv2\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)\\] \nwhich means that\n \\[a_2 + a_4 + a_6\\equiv5\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)\\] \nNotice that the minimum of  $a_2+a_4+a_6$  is  $1 + 3 + 4 = 8$  and the maximum is  $6 + 7 + 8 = 21$ . The only possible number congruent to  $5\\hspace{2mm}(\\text{mod}\\hspace{1mm}11)$  in this range is  $16$ . All that remains is to count all the possible sums of  $16$  using the values  $1, 3, 4, 5, 6, 7, 8$ . There are a total of four possibilities:\n \\[(1, 7, 8), (3, 5, 8), (3, 6, 7), (4, 5, 7)\\] \nThe arrangement of the odd-positioned numbers ( $a_1,a_3,a_5,a_7$ ) does not matter, so there are  $4!=24$  arrangements of these numbers. Recall that the  $4$  triplets above occupy  $a_2,a_4,a_6$ ; the number of arrangements is  $3!=6$ . Thus, we have  $24\\cdot6\\cdot4=576$  possible numbers such that the units digit is  $2$ . Since we claimed symmetry over the rest of the units digits, we must multiply by  $4$ , resulting in  $576\\cdot4=2304$  eight-digit positive integers. Thus, the positive difference between  $N$  and  $2025$  is  $2304 - 2025 = \\boxed{279}$ . ~ilikemath247365"}
{"id": "aime25-5", "problem": "An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is  $3$ , and the area of the trapezoid is  $72$ . Let the parallel sides of the trapezoid have lengths  $r$  and  $s$ , with  $r \\neq s$ . Find  $r^2+s^2$", "expected_answer": "504", "reference_solution": "Denote the radius of the inscribed circle as  $R$ , and the parallel sides as  $r$  and  $s$ .\nBy formula, we get  $R = 3 = \\frac{1}{2} \\cdot \\sqrt{rs}$ , where  $rs = 36$ .\nAlso, by formula,  $A = 72 = \\frac{1}{2} \\cdot \\sqrt{rs} \\cdot (r + s)$ , where  $r + s = 24$ .\nTherefore, \n\\begin{align*}\n&r^2 + s^2 = (r + s)^2 - 2rs \\\\\n&= 24^2 - 2 \\cdot 36 \\\\\n&= \\boxed{504}\n\\end{align*} Formula reference to here: <a class=\"external free\" href=\"https://en.wikipedia.org/wiki/Tangential_trapezoid\" rel=\"nofollow\">https://en.wikipedia.org/wiki/Tangential_trapezoid</a> ~Mitsuihisashi14"}
{"id": "aime25-6", "problem": "The twelve letters  $A$ , $B$ , $C$ , $D$ , $E$ , $F$ , $G$ , $H$ , $I$ , $J$ , $K$ , and  $L$  are randomly grouped into six pairs of letters. The two letters in each pair are placed next to each other in alphabetical order to form six two-letter words, and then those six words are listed alphabetically. For example, a possible result is  $AB$ ,  $CJ$ ,  $DG$ ,  $EK$ ,  $FL$ ,  $HI$ . The probability that the last word listed contains  $G$  is  $\\frac mn$ , where  $m$  and  $n$  are relatively prime positive integers. Find  $m+n$ .", "expected_answer": "821", "reference_solution": "Splitting up into  $2$  cases:  $G$  is the first letter or the second letter of the last word. Case  $1:$   $G$  in first letter Notice that  $A$  must take the first letter of first word, one of the letters  $B$  -  $F$  needs to be the second letter of a word and the rest being the first letter of a word. \nThe combinations will be  $1 + 2 + 3 + 4 + 5 = 15.$  After the first  $7$  letters has been decided then the last  $5$  will just fill by  $5!.$  This case will have  $15 \\cdot 5!$  outcomes. <br/>\nCase  $2:$   $G$  in last letter Notice that  $A$  -  $G$  has been arranged by  $A? B? C? D? E? FG,$  where the  $?$  is undecided. We have another  $5!$  to fill out the possible outcomes. In total, there are  $16 \\cdot 5!.$  The total case will be  $11 \\cdot 9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1$  (Consider A must be in the first letter of first word, then you have  $11$  choices, then you must take the next letter in alphabetical order as mandatory, then you have a free choice of  $9$  and so on). Answer:\n \\[= \\frac{16 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{ 11 \\cdot 9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1}\\] \n \\[= \\frac{16 \\cdot 4 \\cdot 2}{11 \\cdot 9 \\cdot 7}\\] \n \\[= \\frac{128}{ 693}\\] \nTherefore it gives us the answer of  ${128 + 693 = \\boxed{821}.}$ ~Mitsuihisashi14"}
{"id": "aime25-7", "problem": "Let  $k$  be a real number such that the system\n\\begin{align*}\n&|25 + 20i - z| = 5 \\\\\n&|z - 4 - k| = |z - 3i - k|\n\\end{align*}\nhas exactly one complex solution  $z$ . The sum of all possible values of  $k$  can be written as  $\\frac{m}{n}$ , where  $m$  and  $n$  are relatively prime positive integers. Find  $m + n$ . Here  $i = \\sqrt{-1}$ .", "expected_answer": "77", "reference_solution": "[asy] size(300); draw((0, 0) -- (0, 20), EndArrow(10)); label(\"$y$\", (0, 20), NW); dot((25,20)); draw((0, 0) -- (25, 0), EndArrow(10)); label(\"$x$\", (25, 0), SE); draw(circle((25,20),5)); label(scale(0.7)*\"$(25,20)$\", (25,20), S); draw((7,0) -- (3,3), blue); draw((5,3/2) -- (21,23), dashed); label(\"$(4+k,0)$\", (7,0), S); label(\"$(k,3)$\", (3,3), N); draw(rightanglemark((3,3),(5,3/2),(21,23), 20)); draw(rightanglemark((25,20),(21,23),(5,3/2), 20)); draw((25,20) -- (21,23)); [/asy] \nThe complex number  $z$  must satisfy the following conditions on the complex plane: $1.$  The magnitude between  $z$  and  $(25,20)$  is  $5.$  This can be represented by drawing a circle with center  $(25,20)$  and radius  $5.$ $2.$  It is equidistant from the points  $(4+k,0)$  and  $(k,3).$  Hence it must lie on the perpendicular bisector of the line connecting these points. <br/>\nFor  $z$  to have one solution, the perpendicular bisector of the segment connecting the two points must be tangent to the circle. \nThis bisector must pass the midpoint,  $(2+k,\\frac{3}{2}),$  and have slope  $\\frac{4}{3}.$  The segment connecting the point of tangency to the center of the circle has slope  $\\frac{-3}{4},$  meaning the points of tangency can be  $(29,17)$  or  $(21,23).$  Solving the equation for the slope of the perpendicular bisector gives  \\[\\frac{\\frac{3}{2}-23}{k+2-21}=\\frac{4}{3}\\]  or  \\[\\frac{\\frac{3}{2}-17}{k+2-29}=\\frac{4}{3},\\]  giving  $k=\\frac{23}{8}$  or  $\\frac{123}{8}$ , having a sum of  $\\frac{73}{4} \\Longrightarrow \\boxed{077}.$"}
{"id": "aime25-8", "problem": "The parabola with equation  $y = x^2 - 4$  is rotated  $60^\\circ$  counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has  $y$ -coordinate  $\\frac{a - \\sqrt{b}}{c}$ , where  $a$ ,  $b$ , and  $c$  are positive integers, and  $a$  and  $c$  are relatively prime. Find  $a + b + c$ .", "expected_answer": "62", "reference_solution": "To rotate the curve  $y=x^2-4$  counterclockwise by an angle of  $60^\\circ$  about the origin, we will use the rotation matrix as follows: \\begin{gather}\n\\begin{bmatrix} x' \\\\ y' \\end{bmatrix}\n=\n\\begin{bmatrix} \n\\cos{\\theta} & \\sin{\\theta} \\\\\n-\\sin{\\theta} & \\cos{\\theta}\n\\end{bmatrix}\n\\begin{bmatrix}\nx \\\\ y\n\\end{bmatrix}\n\\end{gather} Carrying in  $\\theta=\\frac{\\pi}{3}$ , the rotation matrix becomes \\begin{gather}\n\\begin{bmatrix} x' \\\\ y' \\end{bmatrix}\n=\n\\begin{bmatrix} \n\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\\n-\\frac{\\sqrt{3}}{2} & \\frac{1}{2}\n\\end{bmatrix}\n\\begin{bmatrix}\nx \\\\ y\n\\end{bmatrix}\n\\end{gather} which leads to the following equations:\n \\[x'=\\frac{1}{2}x+\\frac{\\sqrt{3}}{2}y\\] \n \\[y'=-\\frac{\\sqrt{3}}{2}x+\\frac{1}{2}y\\] Substituting  $y$  with  $x^2-4$  yields\n \\[x'=\\frac{\\sqrt{3}}{2}x^2+\\frac{1}{2}x-2\\sqrt{3}\\] \n \\[y'=\\frac{1}{2}x^2-\\frac{\\sqrt{3}}{2}x-2\\] We wish to find the coordinates of the intersection point. Let the point of intersection be  $(p, p^2-4)$ , then \\[p^2-4=\\frac{1}{2}p^2-\\frac{\\sqrt{3}}{2}p-2\\] Solving this quadratic equation yields \\[p_1=\\frac{-\\sqrt{3}+\\sqrt{19}}{2}, \\, p_2=\\frac{-\\sqrt{3}-\\sqrt{19}}{2}\\] Since the problem asks for the intersection point in the fourth quadrant,  $p=\\frac{-\\sqrt{3}+\\sqrt{19}}{2}$ . Therefore, the point of intersection has  $y$ -coordinate  $\\frac{3-\\sqrt{57}}{2}$ , with final answer  $3+57+2=\\boxed{062}$"}
{"id": "aime25-9", "problem": "The  $27$  cells of a  $3 \\times 9$  grid are filled in using the numbers  $1$  through  $9$  so that each row contains  $9$  different numbers, and each of the three  $3 \\times 3$  blocks heavily outlined in the example below contains  $9$  different numbers, as in the first three rows of a Sudoku puzzle. [asy] unitsize(20);  add(grid(9,3));  draw((0,0)--(9,0)--(9,3)--(0,3)--cycle, linewidth(2)); draw((3,0)--(3,3), linewidth(2)); draw((6,0)--(6,3), linewidth(2));  real a = 0.5;  label(\"5\",(a,a)); label(\"6\",(1+a,a)); label(\"1\",(2+a,a)); label(\"8\",(3+a,a)); label(\"4\",(4+a,a)); label(\"7\",(5+a,a)); label(\"9\",(6+a,a)); label(\"2\",(7+a,a)); label(\"3\",(8+a,a));  label(\"3\",(a,1+a)); label(\"7\",(1+a,1+a)); label(\"9\",(2+a,1+a)); label(\"5\",(3+a,1+a)); label(\"2\",(4+a,1+a)); label(\"1\",(5+a,1+a)); label(\"6\",(6+a,1+a)); label(\"8\",(7+a,1+a)); label(\"4\",(8+a,1+a));  label(\"4\",(a,2+a)); label(\"2\",(1+a,2+a)); label(\"8\",(2+a,2+a)); label(\"9\",(3+a,2+a)); label(\"6\",(4+a,2+a)); label(\"3\",(5+a,2+a)); label(\"1\",(6+a,2+a)); label(\"7\",(7+a,2+a)); label(\"5\",(8+a,2+a));  [/asy] The number of different ways to fill such a grid can be written as  $p^a \\cdot q^b \\cdot r^c \\cdot s^d$  where  $p$ ,  $q$ ,  $r$ , and  $s$  are distinct prime numbers and  $a$ ,  $b$ ,  $c$ ,  $d$  are positive integers. Find  $p \\cdot a + q \\cdot b + r \\cdot c + s \\cdot d$ .", "expected_answer": "81", "reference_solution": "First, assume the first row is ordered  $1, 2, 3, 4, 5, 6, 7, 8, 9$ . This contributes  $9!$  cases. Next, assume the first box has the remaining numbers,  $4, 5, 6, 7, 8, 9$  ordered in some way. This contributes  $6! = 720$  cases for the first box. Third, consider each of the remaining  $1$  x  $3$  boxes as separate, and they each contribute  $3!$  cases. However, we are undercounting. The second box may have to be multiplied by 3 if the first box did not define the numbers that go in each  $1$  x  $3$  box. The third box will always have its rows defined. If the first box is placed such that  $4, 5, 6$  are in the same  $1$  x  $3$  box, and  $7, 8, 9$  are in the same  $1$  x  $3$  box, then the numbers in the second box will be defined and we do not need to multiply by 3. This happens for  $2 * 3! * 3! = 72$  cases, or  $72 / 720 = 1 / 10$  of the time. Therefore,  $9 / 10$  of the time, we need to multiply by another 3. Adding this up, to account for undercounting, we must multiply the total by  $1 / 10 + 3(9 / 10) = 28 / 10 = 14 / 5$ . Our final total is  $9! * 6! * (3!)^4 * 14 / 5 = (2^7 * 3^4 * 5^1 * 7^1) (2^4 * 3^2 * 5^1) (2^4 * 3^4) * 2 * 7 / 5 = 2^{16} * 3^{10} * 5^1 * 7^2$ . The desired solution is  $2*16 + 3*10 + 5*1 + 7*2 = \\boxed{081}$"}
{"id": "aime25-10", "problem": "A piecewise linear function is defined by  \\[f(x) = \\begin{cases} x & \\operatorname{if} ~ -1 \\leq x < 1 \\\\ 2 - x & \\operatorname{if} ~ 1 \\leq x < 3\\end{cases}\\]  and  $f(x + 4) = f(x)$  for all real numbers  $x$ . The graph of  $f(x)$  has the sawtooth pattern depicted below. The parabola  $x = 34y^{2}$  intersects the graph of  $f(x)$  at finitely many points. The sum of the  $y$ -coordinates of all these intersection points can be expressed in the form  $\\tfrac{a + b\\sqrt{c}}{d}$ , where  $a$ ,  $b$ ,  $c$ , and  $d$  are positive integers such that  $a$ ,  $b$ ,  $d$  have greatest common divisor equal to  $1$ , and  $c$  is not divisible by the square of any prime. Find  $a + b + c + d$ .", "expected_answer": "259", "reference_solution": "Drawing the graph, we can use the sawtooth graph provided so nicely by MAA and draw out the parabola  $x = 34y^2$ . We realize that the sawtooth graph is just a bunch of lines where the positive slope lines are  $y = x, y = x + 4, y = x + 8,...$ . The intersections of these lines, along with the parabola are just solving the system of equations:  $x = 34y^2$  and  $y = x, y = x + 4, ...$ . If we just take  $y = x$  and  $x = 34y^2$ , we see that the sum of all  $y$  by Vieta's is just  $\\frac{1}{34}$ . Similarly, for  $y = x + 4$ , the sum of the roots by Vieta's is also  $\\frac{1}{34}$ . So for all the positive slope lines intersecting with the parabola just gives the sum of all  $y$  to continuously be  $\\frac{1}{34}$ . Okay, now let's look at the negative slope lines. These will have equations of  $y = 2 - x, y = 6 - x, y = 10 - x, ..., y = 34 - x, ...$ . Similar to what we did above, we just set each of these equations along with the parabola  $x = 34y^2$ . The sum of all  $y$  for each of these negative line intersections by Vieta's is  $\\frac{-1}{34}$ . This keeps going for all of the lines until we reach  $y = 34 - x$ . Now, unfortunately, both solutions don't work as the negative solution is out of the range of [1 , 3], [5, 7] and so on. So we just need to take one solution for this and that being the positive one according to the graph. So we just need to solve  $34 - y = 34y^2$  which means  $34y^2 + y - 34 = 0$ . Solving gives \\[y = \\frac{-1 \\pm \\sqrt{68^2 + 1}}{68} = \\frac{-1 + 5 \\sqrt{185}}{68}.\\] So, the sums of the roots are  $\\frac{1}{34}$  +  $\\frac{-1}{34}$  +  $\\frac{1}{34}$  + .... +  $\\frac{-1}{34}$  +  $\\frac{1}{34}$  +  $\\frac{-1 + 5 \\sqrt{185}}{68}.$  Nicely all the  $\\frac{1}{34}$  terms cancel out leaving with only one  $\\frac{1}{34}$  and  $\\frac{-1 + 5 \\sqrt{185}}{68}.$  So the sum of these two is  $\\frac{1 + 5 \\sqrt{185}}{68}.$  From there, the answer is  $\\boxed{259}$ ."}
{"id": "aime25-11", "problem": "The set of points in  $3$ -dimensional coordinate space that lie in the plane  $x+y+z=75$  whose coordinates satisfy the inequalities  \\[x-yz<y-zx<z-xy\\] forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form  $a\\sqrt{b},$  where  $a$  and  $b$  are positive integers and  $b$  is not divisible by the square of any prime. Find  $a+b.$", "expected_answer": "510", "reference_solution": "Decomposing the inequality chain:\n \\[x-yz<y-zx \\quad \\text{and} \\quad y-zx<z-xy\\] \nwhich is equivalent to\n \\[(x-y)(z+1)<0 \\quad \\text{and} \\quad (y-z)(x+1)<0\\] \nSubstituting  $z$  with  $z=75-x-y$  and simplifying yields\n \\[(x-y)(x+y-76)>0 \\quad \\text{and} \\quad (x+2y-75)(x+1)<0\\] \nSee that the solution to the first inequality is\n \\[x-y>0, \\, x+y-76>0 \\quad \\text{(I)} \\quad \\text{or} \\quad x-y<0, \\, x+y-76<0 \\quad \\text{(II)}\\] \nApplying a similar method results in the solution to the second:\n \\[x+2y-75>0, \\, x+1<0 \\quad \\text{(III)} \\quad \\text{or} \\quad x+2y-75<0, \\, x+1>0 \\quad \\text{(IV)}\\] \nTrying each grouping (i.e. let  $\\text{(I)}$  and  $\\text{(III)}$ ,  $\\text{(I)}$  and  $\\text{(IV)}$ ,  $\\text{(II)}$  and  $\\text{(III)}$ , or  $\\text{(II)}$  and  $\\text{(IV)}$  be satisfied at the same time) and graphing shows that when  $\\text{(II)}$  and  $\\text{(IV)}$  are both satisfied, a triangle whose vertices are  $(-1,38)$ ,  $(-1,-1)$ , and  $(25,25)$  is formed. Further calculations show that the area of the triangle is  $507$ . However, this is not the final answer. We have projected the original shape to the  $xy$ -plane by substituting  $z$ . We know that for a surface defined by the equation  $z=f(x,y)$ , the area element  $dS$  for this surface is given by\n \\[dS=\\sqrt{1+(f_x)^2+(f_y)^2}dxdy\\] \nwhere  $f_x$  and  $f_y$  are the partial derivatives of the function  $f(x,y)$  with respect to  $x$  and  $y$ . For the plane  $x+y+z=75$  where  $f(x,y)=75-x-y$ , computation gives\n \\[f_x=-1, f_y=-1\\] \nSubstituting these into the original equation to get\n \\[dS=\\sqrt{3}dxdy\\] \nThis implies that to find the area of the original shape, we have to multiply the area of its projection on the  $xy$ -plane by  $\\sqrt{3}$ . Therefore, the area of the original shape is  $507\\sqrt{3}$ , with final answer  $\\boxed{510}$ ."}
{"id": "aime25-12", "problem": "Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws  $25$  more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expected number of regions into which these  $27$  line segments divide the disk.", "expected_answer": "204", "reference_solution": "First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered  $1$  through  $4$  in the normal labeling of quadrants, let the two perpendicular diameters be labeled the  $x$ -axis and  $y$ -axis, and let the two segments be  $A$  and  $B.$ \n \\[\\] \n $\\textbf{Case 1:}$  Segment  $A$  has endpoints in two opposite quadrants. [asy] pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,-0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue); [/asy] This happens with probability  $\\frac{1}{3}.$  WLOG let the two quadrants be  $1$  and  $3.$  We do cases in which quadrants segment  $B$  lies in. Quadrants  $1$  and  $2,$   $2$  and  $3,$   $3$  and  $4,$  and  $4$  and  $1$ : These share one quadrant with  $A,$  and it is clear that for any of them to intersect  $A,$  the endpoint that shares a quadrant with an endpoint of  $A$  on a certain side of that endpoint of  $A$  For example, if it was quadrants  $1$  and  $2,$  then the point in quadrant  $1$  must be closer to the  $x$ -axis than the endpoint of  $A$  in quadrant  $1.$  This happens with probability  $\\frac{1}{2}.$  Additionally, segment  $B$  has a  $\\frac{1}{6}$  to have endpoints in any set of two quadrants, so this case contributes to the total probability \\[\\dfrac{1}{3}\\left(\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}\\right)=\\dfrac{1}{9}\\] Quadrants  $2$  and  $4.$  This always intersects segment  $A,$  so this case contributes to the total probability \\[\\dfrac{1}{3}\\cdot\\dfrac{1}{6}=\\dfrac{1}{18}\\] Quadrants  $1$  and  $3.$  We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant  $1$  be  $R_1$  and  $R_2,$  and the endpoints of the segments in quadrant  $3$  be  $S_1$  and  $S_2$  such that  $R_1,R_2,S_1,$  and  $S_2$  are in clockwise order. Note that the probability that  $A$  and  $B$  intersect is the probability that  $A_1$  is paired with  $B_1,$  which is  $\\dfrac{1}{2}.$  Thus, this case contributes to the total probability \\[\\dfrac{1}{3}\\cdot\\dfrac{1}{6}\\cdot\\dfrac{1}{2}=\\dfrac{1}{36}.\\] \n \\[\\] \n $\\textbf{Case 2:}$  \nSegment  $A$  has endpoints in two adjacent quadrants. [asy] pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue); [/asy] This happens with probability  $\\frac{2}{3}.$  WLOG let the two quadrants be  $1$  and  $2.$  We do cases in which quadrants segment  $B$  lies in. Quadrants  $1$  and  $2,$   $3$  and  $4,$   $1$  and  $3,$  and  $2$  and  $4.$  This is similar to our first case above, so this contributes to the total probability \\[\\dfrac{2}{3}\\left(\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}+\\dfrac{1}{6}\\cdot\\dfrac{1}{2}\\right)=\\dfrac{2}{9}\\] Quadrants  $2$  and  $3.$  This cannot intersect segment  $A.$  Quadrants  $1$  and  $4,$  Similar to our third case above, this intersects segment  $A$  with probability  $\\frac{1}{2},$  so this case contributes to the total probability \\[\\dfrac{2}{3}\\cdot\\dfrac{1}{6}\\cdot\\dfrac{1}{2}=\\dfrac{1}{18}.\\] \nThus, the probability that two segments intersect is \n \\[\\dfrac{1}{9}+\\dfrac{1}{18}+\\dfrac{1}{36}+\\dfrac{2}{9}+\\dfrac{1}{18}=\\dfrac{17}{36}.\\] \nNext, we will compute the expected number of intersections of a segment with the axes. WLOG let a segment have an endpoint in quadrant  $1.$  Then, it will intersect each axis with probability  $\\dfrac{2}{3}$  because two out of the three remaining quadrants let it intersect a specific axis, so the expected number of axes a segment intersects is  $\\frac{4}{3}.$ \n \\[\\] \nSo, why do intersections matter? Because when adding a segment, it will pass through a number of regions, and for each region it passes through, it will split that region into two and add another region. The segment starts in a region, and for each intersection, it will enter another region, so the number of regions a segment passes through is  $1$  more than the number of intersections with the axes and other segments. Thus, we have that by linearity of expectation the expected number of new regions created by adding a segment is\n \\[\\dfrac{17}{36}\\cdot(\\text{number of segments already added})+\\dfrac{4}{3}+1,\\] \nso the number of new regions added in total by  $25$  segments again by linearity of expectation is\n \\[\\sum_{k=0}^{24}\\left(\\dfrac{17}{36}k+\\dfrac{7}{3}\\right)=\\dfrac{17}{36}\\cdot \\dfrac{24\\cdot 25}{2}+\\dfrac{25\\cdot 7}{3}\\] \nwhich simplifies to  $200$  as the expected number of new regions added by the  $25$  segments. The axes create  $4$  regions to begin with, so our answer is\n \\[200+4=\\boxed{204}.\\] ~BS2012"}
{"id": "aime25-13", "problem": "Let  $ABCDE$  be a convex pentagon with  $AB=14,$   $BC=7,$   $CD=24,$   $DE=13,$   $EA=26,$  and  $\\angle B=\\angle E=60^{\\circ}.$  For each point  $X$  in the plane, define  $f(X)=AX+BX+CX+DX+EX.$  The least possible value of  $f(X)$  can be expressed as  $m+n\\sqrt{p},$  where  $m$  and  $n$  are positive integers and  $p$  is not divisible by the square of any prime. Find  $m+n+p.$", "expected_answer": "60", "reference_solution": "[asy] size(10cm); import math; import geometry; import olympiad; point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527);  draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,F)); draw(circumcircle(C,P,D)); draw(C--X--D);  label(\"A\",A,SE); label(\"B\",B,W); label(\"C\",C,NW); label(\"P\",P,N); label(\"D\",D,E); label(\"E\",F,SE); label(\"X\",X,E); [/asy] \nFirstly, note that  $\\triangle ABC$  and  $\\triangle ADE$  are just 30-60-90 triangles. Let  $X$  be the Fermat point of  $\\triangle ACD$ , with motivation stemming from considering the pentagon as  $\\triangle ACD$  with the two 30-60-90 extensions. Note that  $AX+CX+DX$  is minimized at this point when  $\\angle AXC=\\angle CXD=\\angle AXD=120^{\\circ}$ . Because we have  $\\angle ABC=\\angle AED=60^{\\circ}$ , then  $ABCX$  and  $AXDE$  are both cyclic. Then we have  $\\angle AXE=\\angle ADE=90^{\\circ}$  and  $\\angle BXA=\\angle BCA=90^{\\circ}$ . Then it turns out that we actually have  $\\angle BXE=90^{\\circ}+90^{\\circ}=180^{\\circ}$ , implying that  $B$ ,  $X$  and  $E$  are collinear. Now, by the triangle inequality, we must have  $BX+XE\\geq BE$ , with equality occurring when  $X$  is on  $BE$ . Thus  $AX+CX+DX$  and  $BX+EX$  are minimized, so this point  $X$  is our desired point. Firstly, we will find  $BX+EX=BE$ . We have that  $AC=7\\sqrt{3}$  and  $AD=13\\sqrt{3}$ , so applying the Law of Cosines in  $\\triangle ACD$ , we get  \\[147+507-2(7\\sqrt{3})(13\\sqrt{3})\\cos (\\angle CAD)=576\\implies \\cos(\\angle CAD)=\\frac{1}{7}.\\]  It follows as a result that  $\\sin (\\angle CAD)=\\frac{4\\sqrt{3}}{7}$ . Then we want to find  $\\cos (\\angle BAE)$ . We can do this by seeing  \\[\\cos (\\angle BAE)=\\cos (\\angle CAD+60^{\\circ})=\\cos (\\angle CAD)\\cos 60^{\\circ}-\\sin (\\angle CAD)\\sin 60^{\\circ}=\\frac{1}{7}\\cdot \\frac{1}{2}-\\frac{4\\sqrt{3}}{7}\\cdot \\frac{\\sqrt{3}}{2}=-\\frac{11}{14}.\\]  Applying the Law of Cosines again in  $\\triangle BAE$ , then because  $AB=14$  and  $AE=26$ , we have  \\[14^2+26^2-2(14)(26)\\left (-\\frac{11}{14}\\right )=196+676-2\\cdot 26\\cdot (-11)=872+572=1444=BE^2,\\]  so it follows that  $BE=38=BX+EX$ . Now, we will find the value of  $AX+CX+DX$ . Construct a point  $P$  outside such that  $\\triangle CPD$  is equilateral, as shown. By property of fermat point, then  $A$ ,  $X$ , and  $P$  are collinear. Additionally,  $\\angle CXD=120^{\\circ}$ , so  $CPDX$  is cyclic. Applying Ptolemy's Theorem, we have that  $(CX)(PD)+(CP)(XD)=(XP)(CD)$ . But since  $\\triangle CPD$  is equilateral, it follows that  $CX+DX=PX$ . Then  $AX+CX+DX=AX+PX=AP$ , so we wish to find  $AP$ . Applying the Law of Cosines in  $\\triangle ACD$ , we have that  \\[(13\\sqrt{3})^2+24^2-2(13\\sqrt{3})(24)\\cos (\\angle ADC)=(7\\sqrt{3})^2\\implies \\cos (\\angle ADC)=\\frac{\\sqrt{3}}{2}\\implies \\angle ADC=30^{\\circ}.\\]  Then because  $\\angle CDP=60^{\\circ}$ , then  $\\angle ADP=90^{\\circ}$ , so we can find  $AP$  simply with the Pythagorean Theorem. We know  $AD=13\\sqrt{3}$  and  $DP=CD=24$ , so  $AP=\\sqrt{(13\\sqrt{3})^2+24^2}=19\\sqrt{3}$ . We then have  $f(X)=AX+BX+CX+DX+EX=(BX+EX)+(AX+CX+DX)=BE+AP=38+19\\sqrt{3}$ , which is our minimum value. Therefore, the answer to the problem is  $38+19+3=\\boxed{060}$ ."}
{"id": "aime25-14", "problem": "Let  $N$  denote the number of ordered triples of positive integers  $(a, b, c)$  such that  $a, b, c \\leq 3^6$  and  $a^3 + b^3 + c^3$  is a multiple of  $3^7$ . Find the remainder when  $N$  is divided by  $1000$ .", "expected_answer": "735", "reference_solution": ""}
{"id": "aime25-15", "problem": "Six points  $A, B, C, D, E,$  and  $F$  lie in a straight line in that order. Suppose that  $G$  is a point not on the line and that  $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$  and  $DG=30.$  Find the area of  $\\triangle BGE.$", "expected_answer": "468", "reference_solution": "We need to solve for the lengths of  $AB$ ,  $BC$ ,  $CD$ ,  $DE$ , and  $EF$ .  \nLet  $AB = a$ ,  $BC = b$ ,  $CD = c$ ,  $DE = d$ , and  $EF = e$ .  \nWe are given the following system of equations: \\[a + b = 26, \\quad b + c = 22, \\quad c + d = 31, \\quad d + e = 33, \\quad a + b + c + d + e = 73.\\] Substituting  $a + b = 26$  and  $d + e = 33$  into the equation  $a + b + c + d + e = 73$ , we get: \\[c = 14.\\] Thus, we have: \\[a = 18, \\quad b = 8, \\quad c = 14, \\quad d = 17, \\quad e = 16.\\] Next, consider triangle  $CDG$ , where  $CD = 14$ ,  $CG = 40$ , and  $DG = 30$ .  \nBy the Law of Cosines, we have: \\[CD^2 = CG^2 + DG^2 - 2 \\times CG \\times DG \\times \\cos(\\angle CGD).\\] Substituting the known values: \\[14^2 = 40^2 + 30^2 - 2 \\times 40 \\times 30 \\times \\cos(\\angle CGD).\\] Simplifying: \\[196 = 1600 + 900 - 2400 \\cos(\\angle CGD).\\] \\[2400 \\cos(\\angle CGD) = 2500 - 196 = 2304.\\] \\[\\cos(\\angle CGD) = \\frac{24}{25}.\\] Therefore, we can find  $\\sin(\\angle CGD)$  using the identity  $\\sin^2 \\theta + \\cos^2 \\theta = 1$ : \\[\\sin(\\angle CGD) = \\sqrt{1 - \\left(\\frac{24}{25}\\right)^2} = \\frac{7}{25}.\\] Now, the area of triangle  $CDG$  is: \\[\\text{Area of triangle } CDG = \\frac{1}{2} \\times 40 \\times 30 \\times \\frac{7}{25} = 168.\\] Noting that the height of triangle  $CDG$  is the same as the height of triangle  $BGE$ , the ratio of the areas of the two triangles will be the same as the ratio of their corresponding lengths. Therefore, the answer is: \\[\\frac{168 \\times 39}{14} = \\boxed{\\textbf{468}}.\\] (Feel free to add or correct any LATEX and formatting.) ~ Mitsuihisashi14"}
{"id": "aime25-16", "problem": "Find the sum of all positive integers  $n$  such that  $n + 2$  divides the product  $3(n + 3)(n^2 + 9)$ .", "expected_answer": "49", "reference_solution": "$\\frac{3(n+3)(n^{2}+9) }{n+2} \\in Z$ $\\Rightarrow \\frac{3(n+2+1)(n^{2}+9) }{n+2} \\in Z$ $\\Rightarrow \\frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \\in Z$ $\\Rightarrow 3(n^{2}+9)+\\frac{3(n^{2}+9)}{n+2} \\in Z$ $\\Rightarrow \\frac{3(n^{2}-4+13)}{n+2} \\in Z$ $\\Rightarrow \\frac{3(n+2)(n-2)+39}{n+2} \\in Z$ $\\Rightarrow 3(n-2)+\\frac{39}{n+2} \\in Z$ $\\Rightarrow \\frac{39}{n+2} \\in Z$ Since  $n + 2$  is positive, the positive factors of  $39$  are  $1$ ,  $3$ ,  $13$ , and  $39$ . Therefore,  $n = -1$ ,  $1$ ,  $11$  and  $37$ . Since  $n$  is positive,  $n = 1$ ,  $11$  and  $37$ . $1 + 11 + 37 = \\framebox{49}$  is the correct answer \uff5e<a class=\"external text\" href=\"https://artofproblemsolving.com/wiki/index.php/User:Tonyttian\" rel=\"nofollow\">Tonyttian</a>"}
{"id": "aime25-17", "problem": "Four unit squares form a  $2 \\times 2$  grid. Each of the  $12$  unit line segments forming the sides of the squares is colored either red or blue in such a say that each unit square has  $2$  red sides and  $2$  blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.", "expected_answer": "82", "reference_solution": "Let red edges be \"used\" edges. In the digarams below, dashed lines are uncolored  lines yet to be decided. Since all four edges of the center are common to both squares, we consider five distinct cases: $\\textbf{Case 1:}$  All center edges are used. There is only one way to do this. [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,red); [/asy] $\\textbf{Case 2:}$  Three center edges are used, meaning two squares are missing an edge. For each square, there are 2 ways to choose an edge, resulting in  $2 \\times 2 = 4$  ways. Additionally, considering the rotational symmetry of the arrangement, there are 4 possible rotations, giving a total of  $4 \\times 4 = 16$  configurations. [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--E,red);draw(E--F,dashed); [/asy] $\\textbf{Case 3:}$  Two center edges are used. There are two sub-cases: $\\textbf{Scenario 1:}$  The two selected sides are perpendicular to each other. The square diagonally opposite its adjacent square has only one choice, while the other two squares each have two choices. This gives a total of  $1 \\times 2 \\times 2 = 4$  choices. Considering the 4 possible rotations, the total number of configurations is  $4 \\times 4 = 16$ . [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,red);draw(E--F,dashed);draw(E--H,dashed); [/asy] $\\textbf{Scenario 2:}$  The two selected sides are aligned along the same straight line. Each of the four squares has 2 choices, yielding  $2^4 = 16$  possible choices. Taking into account the 2 possible rotations, the total number of configurations is  $16 \\times 2 = 32$ . [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,dashed); [/asy] $\\textbf{Case 4:}$  Only one center edge is used. This case is similar to Case 2, yielding 16 possible configurations. [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,red); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed); [/asy] $\\textbf{Case 5:}$  No center edge is used. This is similar to Case 1, with only 1 possible configuration. [asy] pair A,B,C,D,E,F,G,H,I; A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1); draw(A--C--I--G--cycle,dashed); draw(B--E,dashed); draw(E--D,dashed);draw(E--F,dashed);draw(E--H,dashed); [/asy] In conclusion, the total number of configurations is: \\[1 + 16 + 16 + 32 + 16 + 1 = \\boxed{\\textbf{082}}\\] ~ <a class=\"external text\" href=\"https://artofproblemsolving.com/wiki/index.php/User:Athmyx\" rel=\"nofollow\">Athmyx</a> ~ LaTeX by <a class=\"external text\" href=\"https://artofproblemsolving.com/wiki/index.php/User:eevee9406\" rel=\"nofollow\">eevee9406</a>"}
{"id": "aime25-18", "problem": "The product \\[\\prod^{63}_{k=4} \\frac{\\log_k (5^{k^2 - 1})}{\\log_{k + 1} (5^{k^2 - 4})} = \\frac{\\log_4 (5^{15})}{\\log_5 (5^{12})} \\cdot \\frac{\\log_5 (5^{24})}{\\log_6 (5^{21})}\\cdot \\frac{\\log_6 (5^{35})}{\\log_7 (5^{32})} \\cdots \\frac{\\log_{63} (5^{3968})}{\\log_{64} (5^{3965})}\\] is equal to  $\\tfrac mn,$  where  $m$  and  $n$  are relatively prime positive integers. Find  $m + n.$", "expected_answer": "106", "reference_solution": "Using logarithmic identities and the change of base formula, the product can be rewritten as  \\[\\prod_{k=4}^{63}\\frac{k^2-1}{k^2-4}\\frac{\\log(k+1)}{\\log(k)}\\] . Then we can separate this into two series. \nThe latter series is a telescoping series, and it can be pretty easily evaluated to be  $\\frac{\\log(64)}{\\log(4)}=3$ . The former can be factored as  $\\frac{(k-1)(k+1)}{(k-2)(k+2)}$ , and writing out the first terms could tell us that this is a telescoping series as well. Cancelling out the terms would yield  $\\frac{5}{2}\\cdot\\frac{62}{65}=\\frac{31}{13}$ . \nMultiplying the two will give us  $\\frac{93}{13}$ , which tells us that the answer is  $\\boxed{106}$ ."}
{"id": "aime25-19", "problem": "Suppose  $\\triangle ABC$  has angles  $\\angle BAC = 84^\\circ, \\angle ABC=60^\\circ,$  and  $\\angle ACB = 36^\\circ.$  Let  $D, E,$  and  $F$  be the midpoints of sides  $\\overline{BC}, \\overline{AC},$  and  $\\overline{AB},$  respectively. The circumcircle of  $\\triangle DEF$  intersects  $\\overline{BD}, \\overline{AE},$  and  $\\overline{AF}$  at points  $G, H,$  and  $J,$  respectively. The points  $G, D, E, H, J,$  and  $F$  divide the circumcircle of  $\\triangle DEF$  into six minor arcs, as shown. Find  $\\widehat{DE}+2\\cdot \\widehat{HJ} + 3\\cdot \\widehat{FG},$  where the arcs are measured in degrees.\n [asy]         import olympiad;         size(6cm);         defaultpen(fontsize(10pt));         pair B = (0, 0), A = (Cos(60), Sin(60)), C = (Cos(60)+Sin(60)/Tan(36), 0), D = midpoint(B--C), E = midpoint(A--C), F = midpoint(A--B);         guide circ = circumcircle(D, E, F);         pair G = intersectionpoint(B--D, circ), J = intersectionpoints(A--F, circ)[0], H = intersectionpoints(A--E, circ)[0];         draw(B--A--C--cycle);         draw(D--E--F--cycle);         draw(circ);  dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(J);         label(\"$A$\", A, (0, .8));         label(\"$B$\", B, (-.8, -.8));         label(\"$C$\", C, (.8, -.8));         label(\"$D$\", D, (0, -.8));         label(\"$E$\", E, (.8, .2));         label(\"$F$\", F, (-.8, .2));         label(\"$G$\", G, (0, .8));         label(\"$H$\", H, (-.2, -1));         label(\"$J$\", J, (.2, -.8)); [/asy]", "expected_answer": "336", "reference_solution": "Notice that due to midpoints,  $\\triangle DEF\\sim\\triangle FBD\\sim\\triangle AFE\\sim\\triangle EDC\\sim\\triangle ABC$ . As a result, the angles and arcs are readily available. Due to inscribed angles,\n \\[\\widehat{DE}=2\\angle DFE=2\\angle ACB=2\\cdot36=72^\\circ\\] \nSimilarly,\n \\[\\widehat{FG}=2\\angle FDB=2\\angle ACB=2\\cdot36=72^\\circ\\] In order to calculate  $\\widehat{HJ}$ , we use the fact that  $\\angle BAC=\\frac{1}{2}(\\widehat{FDE}-\\widehat{HJ})$ . We know that  $\\angle BAC=84^\\circ$ , and\n \\[\\widehat{FDE}=360-\\widehat{FE}=360-2\\angle FDE=360-2\\angle CAB=360-2\\cdot84=192^\\circ\\] Substituting, \\begin{align*}\n84 &= \\frac{1}{2}(192-\\widehat{HJ}) \\\\\n168 &= 192-\\widehat{HJ} \\\\\n\\widehat{HJ} &= 24^\\circ\n\\end{align*} Thus,  $\\widehat{DE}+2\\cdot\\widehat{HJ}+3\\cdot\\widehat{FG}=72+48+216=\\boxed{336}^\\circ$ . ~ <a class=\"external text\" href=\"https://artofproblemsolving.com/wiki/index.php/User:Eevee9406\" rel=\"nofollow\">eevee9406</a>"}
{"id": "aime25-20", "problem": "Circle  $\\omega_1$  with radius  $6$  centered at point  $A$  is internally tangent at point  $B$  to circle  $\\omega_2$  with radius  $15$ . Points  $C$  and  $D$  lie on  $\\omega_2$  such that  $\\overline{BC}$  is a diameter of  $\\omega_2$  and  ${\\overline{BC} \\perp \\overline{AD}}$ . The rectangle  $EFGH$  is inscribed in  $\\omega_1$  such that  $\\overline{EF} \\perp \\overline{BC}$ ,  $C$  is closer to  $\\overline{GH}$  than to  $\\overline{EF}$ , and  $D$  is closer to  $\\overline{FG}$  than to  $\\overline{EH}$ , as shown. Triangles  $\\triangle {DGF}$  and  $\\triangle {CHG}$  have equal areas. The area of rectangle  $EFGH$  is  $\\frac{m}{n}$ , where  $m$  and  $n$  are relatively prime positive integers. Find  $m+n$ . [asy] size(5cm); defaultpen(fontsize(10pt));  pair A = (9, 0), B = (15, 0), C = (-15, 0), D = (9, 12), E = (9+12/sqrt(5), -6/sqrt(5)), F = (9+12/sqrt(5), 6/sqrt(5)), G = (9-12/sqrt(5), 6/sqrt(5)), H = (9-12/sqrt(5), -6/sqrt(5)); filldraw(G--H--C--cycle, lightgray); filldraw(D--G--F--cycle, lightgray); draw(B--C); draw(A--D); draw(E--F--G--H--cycle); draw(circle(origin, 15)); draw(circle(A, 6));  dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H);  label(\"$A$\", A, (.8, -.8)); label(\"$B$\", B, (.8, 0)); label(\"$C$\", C, (-.8, 0)); label(\"$D$\", D, (.4, .8)); label(\"$E$\", E, (.8, -.8)); label(\"$F$\", F, (.8, .8)); label(\"$G$\", G, (-.8, .8)); label(\"$H$\", H, (-.8, -.8)); label(\"$\\omega_1$\", (9, -5)); label(\"$\\omega_2$\", (-1, -13.5)); [/asy]", "expected_answer": "293", "reference_solution": "Denote the intersection of  $BC$  and  $w_1$  as  $P$ , the intersection of  $BC$  and  $GH$  be  $Q$ , and the center of  $w_2$  to be  $O$ . Additionally, let  $EF = GH = a, FG = EH = b$ . We have that  $CP = 18$  and  $PQ = \\frac{6-b}{2}$ . Considering right triangle  $OAD$ ,  $AD = 12$ . Letting  $R$  be the intersection of  $AD$  and  $FG$ ,  $DR = 12 - \\frac{b}{2}$ . Using the equivalent area ratios:  \\[\\frac{a(24-\\frac{b}{2})}{2} = \\frac{(12-\\frac{a}{2})b}{2}\\] This equation gives  $b=2a$ . Using the <a href=\"/wiki/index.php/Pythagorean_Theorem\" title=\"Pythagorean Theorem\">Pythagorean Theorem</a> on triangle  $GHE$  gives that  $a^2+b^2 = 144$ . Plugging the reuslt  $b=2a$  into this equation gives that the area of the triangle is  $\\frac{288}{5} \\to \\boxed{293}$ . ~ Vivdax"}
{"id": "aime25-21", "problem": "Let  $A$  be the set of positive integer divisors of  $2025$ . Let  $B$  be a randomly selected subset of  $A$ . The probability that  $B$  is a nonempty set with the property that the least common multiple of its element is  $2025$  is  $\\frac{m}{n}$ , where  $m$  and  $n$  are relatively prime positive integers. Find  $m+n$ .", "expected_answer": "237", "reference_solution": "Write numbers in the form of  $3^{a}5^{b}$  where  $0\\leq a\\leq 4; 0\\leq b\\leq 2$ There are  $(4+1)(2+1)=15$  possible divisors of  $2025$ , so the cardinality of the subsets is  $2^{15}$ If I select  $3^4\\cdot 5^2$ , then I guarantee the LCM is 2025, so the other 14 numbers yield  $2^{14}$  cases. If I select  $3^4\\cdot 5$ , then I must select at least one of  $3^a5^2$ , but I can select any other  $9$  numbers, so there are  \\[2^9(\\binom{4}{1}+\\binom{4}{2}+\\binom{4}{3}+\\binom{4}{4})=2^9\\cdot 15\\]  ways. If I select  $3^4$ , same reason above but since we can't selct  $3^4\\cdot 5; 3^4 5^2$  anymore, there are  $2^8(\\binom{4}{1}+\\binom{4}{2}+\\binom{4}{3}+\\binom{4}{4})=2^8\\cdot 15$  ways The answer is then  $\\frac{2^8(15+30+64)}{2^{15}}=\\frac{109}{128}\\implies \\boxed{237}$ ~ Bluesoul"}
{"id": "aime25-22", "problem": "From an unlimited supply of 1-cent coins, 10-cent coins, and 25-cent coins, Silas wants to find a collection of coins that has a total value of  $N$  cents, where  $N$  is a positive integer. He uses the so-called greedy algorithm, successively choosing the coin of greatest value that does not cause the value of his collection to exceed  $N.$  For example, to get 42 cents, Silas will choose a 25-cent coin, then a 10-cent coin, then 7 1-cent coins. However, this collection of 9 coins uses more coins than necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins. In general, the greedy algorithm succeeds for a given  $N$  if no other collection of 1-cent, 10-cent, and 25-cent coins gives a total value of  $N$  cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of  $N$  between  $1$  and  $1000$  inclusive for which the greedy algorithm succeeds.", "expected_answer": "610", "reference_solution": "We begin by noting that all values of  $N \\leq 25$  work without issue. Starting from  $N = 25$  to  $29$ , the greedy algorithm will select the 25-cent coin, and no problem arises. From  $N = 30$  to  $34$ , the greedy algorithm will select the 25-cent coin along with 5 1-cent coins to reach a total of 30, while the optimal solution would involve using 3 10-cent coins. This issue is resolved from  $N = 35$  to  $39$ , as the greedy algorithm can now select  $25 + 10$ -cent coins to match the optimal solution. From  $N = 40$  to  $44$ , a similar problem occurs again. The greedy algorithm selects  $25 + 10 + 5 \\times 1$ -cent coins to reach 40, while the optimal solution would use 4  $10$ -cent coins. The problem occurs again from  $N = 55$  to  $59$ , where  $50 + 5 \\times 1$  is not as good as using  $25 + 3 \\times 10$ , and it is resolved at  $N = 60$ . From  $N = 65$  to  $69$ , a similar issue arises, as  $25 \\times 2 + 10 + 5 \\times 1$  is not as optimal as  $25 + 4 \\times 10$  to approach 65. We observe that this issue repeats in cycles of 25 numbers, with 10 of the 25 numbers in each cycle not working. The cycle starts at 30, and the next cycle will start 25 numbers later, at 55, then 80, and so on, continuing until 980\u20131005 for the last cycle. The total number of cycles is given by: \\[\\frac{955 - 30}{25} + 1 = 38,\\] and each cycle contains 10 problematic numbers. Therefore, the total number of problematic numbers is: \\[38 \\times 10 = 380.\\] The cycle from 980 to 1005 has the problematic numbers from 980 to 984 and from 990 to 994, which gives another 10 problematic numbers. Thus, the total number of unsuccessful numbers from 1 to 1000 inclusive is  $390$ , and the desired count of successful numbers is: \\[1000 - 390 = \\boxed{610}.\\]"}
{"id": "aime25-23", "problem": "There are  $n$  values of  $x$  in the interval  $0<x<2\\pi$  where  $f(x)=\\sin(7\\pi\\cdot\\sin(5x))=0$ . For  $t$  of these  $n$  values of  $x$ , the graph of  $y=f(x)$  is tangent to the  $x$ -axis. Find  $n+t$ .", "expected_answer": "149", "reference_solution": "For  $f(x)=0$ , we must have  $7\\pi\\cdot\\sin(5x)=k\\pi$  for some integer  $k$ . Then  $\\sin(5x)=\\frac{k}{7}$  always satisfies the equation. Notice that on each period of  $\\sin(5x)$ , each  $k\\in\\{-6,-5,\\ldots,5,6\\}$  is a  $y$ -value at two distinct points, and each  $k=\\pm7$  is a  $y$ -value at one point each. Thus each period has  $13\\cdot2+2\\cdot1=28$  points satisfying the equation. Since the period is  $\\frac{2\\pi}{5}$  and the domain has a length of  $2\\pi$ , we find that  $5$  periods occur in our domain if we include  $x=0,2\\pi$ . Adding the case where  $x=0$ , there are a total of  $28\\cdot5+1=141$  roots over  $x\\in[0,2\\pi]$ . Subtracting the cases at  $x=0$  and  $x=2\\pi$  yields  $139$  total roots. This is our  $n$ . <br/>\nNext, we take the derivative of  $f(x)$ ; using a hideous combination of chain rules we find that \\[f'(x)=35\\pi\\cos(5x)\\cos(7\\pi\\sin(5x))=0\\] Thus, for a point to be tangent to the  $x$ -axis, we must have either  $\\cos(5x)=0$  or  $\\cos(7\\pi\\sin(5x))=0$ . In the first case, we know that  $\\sin(5x)=\\frac{k}{7}$  from earlier, so  $\\cos(5x)=\\sqrt{1-\\left(\\frac{k}{7}\\right)^2}=0$ . Then  $\\left(\\frac{k}{7}\\right)^2=1$ , so  $k=\\pm7$ . Recall that over each of the five periods, only one point will satisfy  $k=7$ , and only one point will satisfy  $k=-7$ . Thus there are  $2\\cdot5=10$  points in this case. In the second case, we must have  $\\cos(7\\pi\\sin(5x))=0$ . Substituting  $\\sin(5x)=\\frac{k}{7}$  yields  $\\cos(k\\pi)=0$ . But this is impossible since  $\\cos(0)=1$  and  $\\cos(\\pi)=-1$ , so there are no points in this case. As a result,  $t=10+0=10$ , so  $n+t=139+10=\\boxed{149}$ ."}
{"id": "aime25-24", "problem": "Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let  $N$  be the number of subsets of  $16$  chairs that could be selected. Find the remainder when  $N$  is divided by  $1000$ .", "expected_answer": "907", "reference_solution": "We split into few cases: Case 1: 8 people are all by single: 8C0 * 9C1 = 9 Case 2: 6 people are by single, 2 people sits next to each other (so each person sits next to either 0 or 1 other person): 7C1 * 9C2 = 7 * 36 = 252 Case 3: 4 people are by single, 2 people sits next to each other and 2 other people sits next to each other with the 2 groups of 2 people not sitting next to each other (so each person still sits next to either 0 or 1 other person): 6C2 * 9C3 = 1260 Case 4: 2 people are by single, 6 people are split into 3 groups of 2 people, and no 2 groups sit next to each other: 5C3 * 9C4 = 10 * 126 = 1260 Case 5: 4 groups of 2, no groups are sitting next to each other: 4C4 * 9C5 = 126 Answer: 9 + 252 + 1260 + 1260 + 126 = 2907, so the answer is \\boxed{907}."}
{"id": "aime25-25", "problem": "Let  $S$  be the set of vertices of a regular  $24$ -gon. Find the number of ways to draw  $12$  segments of equal lengths so that each vertex in  $S$  is an endpoint of exactly one of the  $12$  segments.", "expected_answer": "113", "reference_solution": "The segments we draw must be of equal length, corresponding to a specific step size  $k$  (number of steps between vertices). For each step size  $k$ , we need to determine if it is possible to form a perfect matching (non-overlapping segments covering all vertices). The number of such perfect matchings depends on the greatest common divisor (gcd) of  $k$  and 24. When choosing a step size  $k$ , the 24-gon is decomposed into  $\\gcd(k, 24)$  cycles, each of length  $\\frac{24}{\\gcd(k, 24)}$ . For a perfect matching to exist, each cycle must be of even length. For each valid step size ( $k$ ): If the cycle length is 2 (diameters), there is exactly 1 way to match the vertices. For other even cycle lengths, each cycle contributes a factor of 2 to the number of perfect matchings. ( $k = 1$ ):  $\\gcd(1, 24) = 1$ , cycle length 24, 2 matchings.\n( $k = 2$ ):  $\\gcd(2, 24) = 2$ , cycle length 12,  $(2^2 = 4)$  matchings.\n( $k = 3$ ):  $\\gcd(3, 24) = 3$ , cycle length 8,  $(2^3 = 8)$  matchings.\n( $k = 4$ ):  $\\gcd(4, 24) = 4$ , cycle length 6,  $(2^4 = 16)$  matchings.\n( $k = 5$ ):  $\\gcd(5, 24) = 1$ , cycle length 24, 2 matchings.\n( $k = 6$ ):  $\\gcd(6, 24) = 6$ , cycle length 4,  $(2^6 = 64)$  matchings.\n( $k = 7$ ):  $\\gcd(7, 24) = 1$ , cycle length 24, 2 matchings.\n( $k = 8$ ):  $\\gcd(8, 24) = 8$ , cycle length 3 (invalid, no matchings).\n( $k = 9$ ):  $\\gcd(9, 24) = 3$ , cycle length 8,  $(2^3 = 8)$  matchings.\n( $k = 10$ ):  $\\gcd(10, 24) = 2$ , cycle length 12,  $(2^2 = 4)$  matchings.\n( $k = 11$ ):  $\\gcd(11, 24) = 1$ , cycle length 24, 2 matchings.\n( $k = 12$ ):  $\\gcd(12, 24) = 12$ , cycle length 2, 1 matching. Summing these values:  $2 + 4 + 8 + 16 + 2 + 64 + 2 + 0 + 8 + 4 + 2 + 1 = \\boxed{113}$ ."}
{"id": "aime25-26", "problem": "Let  $A_1A_2\\dots A_{11}$  be a non-convex  $11$ -gon such that \u2022 The area of  $A_iA_1A_{i+1}$  is  $1$  for each  $2 \\le i \\le 10$ ,\n\u2022  $\\cos(\\angle A_iA_1A_{i+1})=\\frac{12}{13}$  for each  $2 \\le i \\le 10$ ,\n\u2022 The perimeter of  $A_1A_2\\dots A_{11}$  is  $20$ . If  $A_1A_2+A_1A_{11}$  can be expressed as  $\\frac{m\\sqrt{n}-p}{q}$  for positive integers  $m,n,p,q$  with  $n$  squarefree and  $\\gcd(m,p,q)=1$ , find  $m+n+p+q$ .", "expected_answer": "19", "reference_solution": "Set  $A_1A_2 = x$  and  $A_1A_3 = y$ . By the first condition, we have  $\\frac{1}{2}xy\\sin\\theta = 1$ , where  $\\theta = \\angle A_2 A_1 A_3$ . Since  $\\cos\\theta = \\frac{12}{13}$ , we have  $\\sin\\theta = \\frac{5}{13}$ , so  $xy = \\frac{26}{5}$ . Repeating this process for  $\\triangle A_i A_1 A_{i+1}$ , we get  $A_1A_2 = A_1A_4 = \\ldots A_1A_{10} = x$  and  $A_1A_3 = A_1A_5 = \\ldots A_1A_{11} = y$ . Since the included angle of these  $9$  triangles is  $\\theta$ , the square of the third side is \n \\[x^2 + y^2 - 2xy\\cos\\theta = x^2 + y^2 - \\frac{52}{5}\\cdot \\frac{12}{13} = x^2 + y^2 - \\frac{48}{5} = (x+y)^2 - 20.\\] \nThus the third side has length  $\\sqrt{(x+y)^2 - 20}.$  The perimeter is constructed from  $9$  of these lengths, plus  $A_{11}A_1 + A_1A_2 = x + y$ , so  $9\\sqrt{(x+y)^2 - 20} + x + y = 20$ . We seek the value of  $x + y,$  so let  $x + y = a$  so \n \\begin{align*} 9\\sqrt{a^2 - 20} + a &= 20\\\\ 81(a^2 - 20) &= 400 - 40a + a^2\\\\ 4a^2 + 2a - 101 &= 0 \\\\ a &= \\frac{-2 \\pm \\sqrt{1620}}{8} = \\frac{-1 \\pm \\sqrt{405}}{4} = \\frac{-1 \\pm 9\\sqrt{5}}{4}. \\end{align*} \nTaking the positive solution gives  $m + n + p + q = 1 + 9 + 5 + 4 = \\boxed{\\textbf{(019)}}.$ -Benedict T (countmath1)"}
{"id": "aime25-27", "problem": "Let the sequence of rationals  $x_1,x_2,\\dots$  be defined such that  $x_1=\\frac{25}{11}$  and\n \\[x_{k+1}=\\frac{1}{3}\\left(x_k+\\frac{1}{x_k}-1\\right).\\]  $x_{2025}$  can be expressed as  $\\frac{m}{n}$  for relatively prime positive integers  $m$  and  $n$ . Find the remainder when  $m+n$  is divided by  $1000$ .", "expected_answer": "248", "reference_solution": "Note that  $x_{k+1} = \\frac{1}{3}$ ( $\\frac{(x_k)^{2} - x_k + 1}{x_k}$ ). An astute reader might recognize the top part as one part of a sum of cubes. I multiplied the entire expression by  $x_k + 1$ , moved things around a bit, simplified, and was left with the following generalization:\n $x_{k+1} = \\frac{(x_k)^{3} + 1}{3x_k(x_k + 1)}$ . Now, we do the following:\nSet  $x_k = \\frac{m_k}{n_k}$ . Therefore,  $x_{k+1} = \\frac{m_{k+1}}{n_{k+1}}$ . We plug these expressions into the  $x_k$  and  $x_{k+1}$  and simplify to get:  $\\frac{m_{k+1}}{n_{k+1}} = \\frac{(m_k)^{3} + (n_k)^{3}}{3(m_k)(n_k)(m_k + n_k)}$ . Now, as we are looking for the sum of the numerators and denominators of  $x_2025$ , this is great! Now, recall that we want the fraction to be simplest. So we have to cancel out anything we can. Canceling out the factor of  $m_k + n_k$  from the numerator and denominator leaves us with  $\\frac{m_{k+1}}{n_{k+1}} = \\frac{(m_k)^{2} - (m_k)(n_k) + (n_k)^{2}}{3(m_k)(n_k)}$ . Now, adding the numerator and denominator as well as keeping the extra factor of  $3$ , we get: 3( $m_{k+1} + n_{k+1}) = (m_k)^{2} + 2(m_k)(n_k) + (n_k)^{2}$ . Nicely, we get the recursion that  $m_{k+1} + n_{k+1} = \\frac{(m_k + n_k)^{2}}{3}$ . Now, by listing out terms using this recursion and doing mod(1000), we get our answer of  $\\boxed{248}$ ."}
{"id": "aime25-28", "problem": "Let  ${\\triangle ABC}$  be a right triangle with  $\\angle A = 90^\\circ$  and  $BC = 38.$  There exist points  $K$  and  $L$  inside the triangle such \\[AK = AL = BK = CL = KL = 14.\\] The area of the quadrilateral  $BKLC$  can be expressed as  $n\\sqrt3$  for some positive integer  $n.$  Find  $n.$ \n", "expected_answer": "104", "reference_solution": "By drawing our the triangle, I set A to be (0, 0) in the coordinate plane. I set C to be (x, 0) and B to be (0, y). I set K to be (a, b) and L to be (c, d). Then, since all of these distances are 14, I used coordinate geometry to set up the following equations:\n $a^{2}$  +  $b^{2}$  = 196;  $a^{2}$  +  $(b - y)^{2}$  = 196;  $(a - c)^{2}$  +  $(b - d)^{2}$  = 196;  $c^{2}$  +  $d^{2}$  = 196;  $(c - x)^{2}$  +  $d^{2}$ . = 196. Notice by merging the first two equations, the only possible way for it to work is if  $b - y$  =  $-b$  which means  $y = 2b$ . Next, since the triangle is right, and we know one leg is  $2b$  as  $y = 2b$ , the other leg, x, is  $\\sqrt{38^{2} - (2b)^{2}}$ .Then, plugging these in, we get a system of equations with 4 variables and 4 equations and solving, we get a = 2, b = 8 $\\sqrt{3}$ , c = 13, d = 3 $\\sqrt{3}$ . Now plugging in all the points and using the Pythagorean Theorem, we get the coordinates of the quadrilateral. By Shoelace, our area is 104 $\\sqrt{3}$ . Thus, the answer is  $\\boxed{104}$ ."}
{"id": "aime25-29", "problem": "Let\n \\[f(x)=\\frac{(x-18)(x-72)(x-98)(x-k)}{x}.\\] There exist exactly three positive real values of  $k$  such that  $f$  has a minimum at exactly two real values of  $x$ . Find the sum of these three values of  $k$ .", "expected_answer": "240", "reference_solution": "Let  $n$  be the minimum value of the expression (changes based on the value of  $k$ , however is a constant). Therefore we can say that\n\\begin{align*}\nf(x)-n=\\frac{(x-\\alpha)^2(x-\\beta)^2}{x}\n\\end{align*}\nThis can be done because  $n$  is a constant, and for the equation to be true in all  $x$  the right side is also a quartic. The roots must also both be double, or else there is an even more 'minimum' value, setting contradiction. We expand as follows, comparing coefficients: \\begin{align*}\n(x-18)(x-72)(x-98)(x-k)-nx=(x-\\alpha)^2(x-\\beta)^2 \\\\\n-2\\alpha-2\\beta=-18-72-98-k \\implies \\alpha+\\beta=94+\\frac{k}{2} \\\\\n\\alpha^2+4\\alpha \\beta +\\beta^2=(-18\\cdot -72)+(-18\\cdot-98)+(-18\\cdot-k)+(-72\\cdot-98)+(-72\\cdot-k)+(-98\\cdot-k)=10116+188k \\\\\n(\\alpha^2)(\\beta^2)=(-18)(-72)(-98)(-k) \\implies \\alpha \\beta=252\\sqrt{2k} \\\\\n\\end{align*} Recall  $(\\alpha+\\beta)^2+2\\alpha \\beta=\\alpha^2+4\\alpha \\beta +\\beta^2$ , so we can equate and evaluate as follows: \\begin{align}\n(94+\\frac{k}{2})^2+504\\sqrt{2k}=10116+188k \\tag{1}\\\\\n\\end{align}\n\\begin{align*}\n(47-\\frac{k}{4})^2+126\\sqrt{2k}=2529 \\\\\n\\frac{k^2}{16}-\\frac{47}{2}k+126\\sqrt{2k}-320=0 \\\\\n\\end{align*} We now have a quartic with respect to  $\\sqrt{k}$ . Keeping in mind it is much easier to guess the roots of a polynomial with integer coefficients, we set  $a=\\frac{k}{8}$ . Now our equation becomes \\begin{align*}\n4a^2-188a+504\\sqrt{a}-320=0 \\\\\na^2-47a+126\\sqrt{a}-80=0 \\\\\n\\end{align*} If you are lucky, you should find roots  $\\sqrt{a}=1$  and  $2$ . After this, solving the resulting quadratic gets you the remaining roots as  $5$  and  $8$ . Working back through our substitution for  $a$ , we have generated values of  $k$  as  $(8, 32, 200, 512)$ . However, we are not finished, trying  $k=512$  into the equation  $(1)$  from earlier does not give us equality, thus it is an extraneous root. The sum of all  $k$  then must be  $8+32+200=\\boxed{240}$ . ~ <a class=\"external text\" href=\"https://artofproblemsolving.com/wiki/index.php/User:Lisztepos\" rel=\"nofollow\">lisztepos</a>"}
