| { |
| "key": "aimo-validation-aime", |
| "title": "AI-MO AIMO validation AIME", |
| "provider": "huggingface", |
| "reference": "AI-MO/aimo-validation-aime", |
| "purpose": [ |
| "held-out validation", |
| "AIME-style difficulty checks" |
| ], |
| "answer_type": "integer answer in [0, 99999]", |
| "local_artifact": null, |
| "notes": "Use as a shared eval set across all lanes.", |
| "split": "train", |
| "rows": 90, |
| "columns": [ |
| "id", |
| "problem", |
| "solution", |
| "answer", |
| "url" |
| ], |
| "sample": [ |
| { |
| "id": 0, |
| "problem": "Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$", |
| "solution": "Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.\nNote that\n\\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\\\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \\end{alignat*}\nso the slope of $R(x)$ is $\\frac{106-108}{20-16}=-\\frac12.$\nIt follows that the equation of $R(x)$ is \\[R(x)=-\\frac12x+c\\] for some constant $c,$ and we wish to find $R(0)=c.$\nWe substitute $x=20$ into this equation to get $106=-\\frac12\\cdot20+c,$ from which $c=\\boxed{116}.$\n~MRENTHUSIASM\nLet\n\\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*}\nfor some constants $a,b,c$ and $d.$\nWe are given that\n\\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*}\nand we wish to find \\[P(0)+Q(0)=b+d.\\]\nWe need to cancel $a$ and $c.$ Since $\\operatorname{lcm}(16,20)=80,$ we subtract $4\\cdot[(3)+(4)]$ from $5\\cdot[(1)+(2)]$ to get \\[b+d=5\\cdot(54+54)-4\\cdot(53+53)=\\boxed{116}.\\]\n~MRENTHUSIASM\nLike Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us\n\\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*}\nWe can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$\nNext, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \\cdot (-2) + (b+d) = 108,$ so therefore $b+d = \\boxed{116}.$\n~jessiewang28\nLet\n\\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*}\nBy substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get:\n\\begin{align*} 2(16)^2 + 16a + b &= 54, \\\\ 2(20)^2 + 20a + b &= 53. \\end{align*}\nHence, $a = -72.25$ and $b = 698.$\nSimilarly,\n\\begin{align*} -2(16)^2 + 16c + d &= 54, \\\\ -2(20)^2 + 20c + d &= 53. \\end{align*}\nHence, $c = 71.75$ and $d = -582.$\nNotice that $b = P(0)$ and $d = Q(0).$\nTherefore \\[P(0) + Q(0) = 698 + (-582) = \\boxed{116}.\\]\n~Littlemouse\nAdd the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$. This equation must also pass through the two points $(16,54)$ and $(20,53)$.\nLet $m=a+c$ and $n=b+d$. We then have two equations:\n\\begin{align*} 108&=16m+n, \\\\ 106&=20m+n. \\end{align*}\nWe are trying to solve for $n=P(0)$. Using elimination:\n\\begin{align*} 540&=80m+5n, \\\\ 424&=80m+4n. \\end{align*}\nSubtracting both equations, we find that $n=\\boxed{116}$.\n~eevee9406", |
| "answer": "116", |
| "url": "https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1" |
| }, |
| { |
| "id": 1, |
| "problem": "Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$.", |
| "solution": "This solution refers to the Diagram section.\nWe let $\\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:\n\nBecause the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \\sqrt{560}$. Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \\implies x = 4$.\nWe have $AO_A = BD$ because of the rectangle, so $\\sqrt{11^2-r^2} = \\sqrt{13^2-r^2}-4$.\nSquaring, we have $121-r^2 = 169-r^2 + 16 - 8 \\cdot \\sqrt{169-r^2}$.\nSubtracting, we get $8 \\cdot \\sqrt{169-r^2} = 64 \\implies \\sqrt{169-r^2} = 8 \\implies 169-r^2 = 64 \\implies r^2 = 105$.\nWe also notice that since we had $\\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$.\nWe take a cross-section that contains $A$ and $C$, which contains these two spheres but not the third, as shown below:\n\nWe have $CO_C = \\sqrt{19^2-r^2} = \\sqrt{361 - 105} = \\sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \\sqrt{30^2 - 12^2} = \\sqrt{900-144} = \\sqrt{756}$. Therefore, $O_AE^2 = AC^2 = \\boxed{756}$.\n~KingRavi\nLet the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.\nAccording to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$\nThe desired value is $(11+19)^2-(16-4)^2=\\boxed{756}.$\n~bluesoul\nDenote by $r$ the radius of three congruent circles formed by the cutting plane.\nDenote by $O_A$, $O_B$, $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$, $B$, $C$, respectively.\nBecause three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$, $O_A O_C = 11 + 19 = 30$.\nWe have $O_A A^2 = 11^2 - r^2$, $O_B B^2 = 13^2 - r^2$, $O_C C^2 = 19^2 - r^2$.\nBecause $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\\angle O_A A B = \\angle O_B BA = 90^\\circ$.\nHence,\n\\begin{align*} O_B B - O_A A & = \\sqrt{O_A O_B^2 - AB^2} \\\\ & = 4 . \\hspace{1cm} (1) \\end{align*}\nRecall that\n\\begin{align*} O_B B^2 - O_A A^2 & = \\left( 13^2 - r^2 \\right) - \\left( 11^2 - r^2 \\right) \\\\ & = 48 . \\hspace{1cm} (2) \\end{align*}\nHence, taking $\\frac{(2)}{(1)}$, we get\n\\[ O_B B + O_A A = 12 . \\hspace{1cm} (3) \\]\nSolving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$.\nThus, $r^2 = 11^2 - O_A A^2 = 105$.\nThus, $O_C C = \\sqrt{19^2 - r^2} = 16$.\nBecause $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\\angle O_A A C = \\angle O_C CA = 90^\\circ$.\nTherefore,\n\\begin{align*} AC^2 & = O_A O_C^2 - \\left( O_C C - O_A A \\right)^2 \\\\ & = \\boxed{756}. \\end{align*}\n$\\textbf{FINAL NOTE:}$ In our solution, we do not use the condition that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem. \n$\\textbf{MORE FINAL NOTE:}$ the above note is incorrect because that condition was used at the start when claiming $O_AO_B=24$. Perhaps the note is referring to spheres $B$ and $C$.\n~Steven Chen (www.professorcheneeu.com)\n~anonymous (minor edits)", |
| "answer": "756", |
| "url": "https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10" |
| } |
| ] |
| } |
|
|
