[
    {
        "nl_statement": "Prove that the volume of the entire real number line (\u211d) is infinite.",
        "nl_proof": "1. To prove that the volume of the entire real number line is infinite, we need to show that it is greater than or equal to every non-negative real number.\n2. Consider any arbitrary non-negative real number, 'r'. \n3. The closed interval from 0 to 'r' is a subset of the entire real number line.\n4. The volume of this interval is equal to 'r'.\n5. Since the interval is a subset of the entire real number line, the volume of the real number line must be greater than or equal to 'r'.\n6. As this holds true for any arbitrary non-negative real number 'r', we can conclude that the volume of the entire real number line is greater than or equal to every non-negative real number.\n7. Therefore, the volume of the entire real number line must be infinite.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem volume_univ : volume (univ : Set \u211d) = \u221e :=\n   ENNReal.eq_top_iff.2 fun r => le_trans (measure_mono (Icc_subset_Ici r r).le (volume_Icc _ _)) r\n ",
        "pda_valid": true,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the function which checks for equality to `True` is the same as the identity function when applied to boolean values.",
        "nl_proof": "To show these two functions are the same, let's consider what happens when we apply each function to an arbitrary boolean value, let's call it `b`.\n\n1. **Applying `Eq True` to `b`:** This function checks whether `b` is equal to `True`.\n    - If `b` is `True`, then `Eq True b` evaluates to `True`.\n    - If `b` is `False`, then `Eq True b` evaluates to `False`.\n\n2. **Applying `id` to `b`:** The identity function simply returns the input value, so `id b` evaluates to `b`.\n\nNow we can see that for any boolean value `b`:\n\n- If `b` is `True`, both functions return `True`.\n- If `b` is `False`, both functions return `False`.\n\nSince both functions produce the same output for every possible input, we can conclude that they are the same function. Therefore, the function `Eq True` is indeed the same as the identity function when applied to boolean values.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem eq_id : Eq True = id :=\n   funext fun _ => rfl\n ",
        "pda_valid": true,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the function which maps a real number to its corresponding complex number is measurable.",
        "nl_proof": "1. We know that the function mapping real numbers to complex numbers is continuous.\n2. We also know that any continuous function is measurable.\n3. Therefore, the function mapping real numbers to complex numbers is measurable.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem measurable_re_to_complex : Measurable fun x : \u211d => (x : \u2102) :=\n   continuous_re_to_complex.measurable\n ",
        "pda_valid": true,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Given a premeasure `p` and a set `s`, prove that the outer measure of `s` with respect to `p` is zero if and only if the support of `p` is disjoint from `s`.",
        "nl_proof": "We need to prove both directions of the \"if and only if\" statement.\n\n**Direction 1: If the outer measure of `s` is 0, then the support of `p` is disjoint from `s`.**\n\n1. Assume that the outer measure of `s` with respect to `p` is 0. \n2. This means that we can cover `s` with a countable collection of sets from the premeasure `p`, such that the sum of their premeasures is 0.\n3. Since premeasures are non-negative, the only way the sum can be 0 is if each individual premeasure is 0.\n4. If a set has a premeasure of 0, it means that it doesn't intersect with the support of `p` (because the support only includes sets with non-zero premeasure).\n5. Therefore, `s` cannot intersect with the support of `p`, meaning they are disjoint.\n\n**Direction 2: If the support of `p` is disjoint from `s`, then the outer measure of `s` is 0.**\n\n1. Assume that the support of `p` is disjoint from `s`.\n2. We can cover `s` with a single set: `s` itself.\n3. Since `s` is disjoint from the support of `p`, its premeasure is 0.\n4. Therefore, we have covered `s` with a collection of sets (just one set in this case) whose premeasures sum to 0.\n5. This means that the outer measure of `s` is 0.\n\nSince we have proven both directions, we have shown that the outer measure of `s` with respect to `p` is zero if and only if the support of `p` is disjoint from `s`.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem toOuterMeasure_apply_eq_zero_iff_disjoint : p.toOuterMeasure s = 0 \u2194 Disjoint p.support s := by\n   rw [toOuterMeasure_apply, sum_eq_zero_iff]\n   simp only [\u2190 premeasure_univ, \u2190 subset_def, mem_support_iff, Premeasure.premeasure_eq_zero_iff_disjoint,\n     mem_univ, true_and]\n ",
        "pda_valid": true,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the function *f* applied to any value *x* always results in a non-negative value.  We are working within a context involving a specific type of function called a \"continuously differentiable bump function\" related to a set *E*.",
        "nl_proof": "1. We are given that *f* is a function that belongs to the *ContDiffBumpBase* structure. This structure guarantees certain properties about *f*, specifically that it represents a continuously differentiable bump function.\n2. Additionally, we are given a specific instance of this bump function related to the set *E*. This instance provides further information about the behavior of *f* within the context of *E*.\n3. Using a property related to closed intervals (likely stating that the output of *f* always falls within a specific closed interval), we can analyze the range of values *f(x)* can take.\n4. By focusing on the lower bound of this interval (obtained through the \"( ... ).1\" notation), we can establish that it is greater than or equal to 0.\n5. Therefore, since the lower bound of the interval containing all possible values of *f(x)* is non-negative, we can conclude that *f(x)* itself is always greater than or equal to 0.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem nonneg (x : E) : 0 \u2264 f x :=\n   (f.contDiffBump x).1.2.1\n ",
        "pda_valid": true,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the arccosine function is measurable.",
        "nl_proof": "We know that the arccosine function is continuous. Since continuous functions are measurable, the arccosine function is also measurable.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem measurable_arccos : Measurable arccos :=\n   continuous_arccos.measurable\n ",
        "pda_valid": false,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the sine function is measurable.",
        "nl_proof": "1. We know that the sine function is a continuous function.\n2. There is a theorem that states that any continuous function is also a measurable function.\n3. Therefore, since the sine function is continuous, it must also be measurable.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem measurable_sin : Measurable sin :=\n   continuous_sin.measurable\n ",
        "pda_valid": false,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Suppose we have two relations (or functions) \\(m_1\\) and \\(m_2\\), and a function \\(g\\). The theorem states that taking the inverse image of the union of \\(m_1\\) and \\(m_2\\) under \\(g\\) is the same as taking the union of the inverse images of \\(m_1\\) and \\(m_2\\) under \\(g\\).",
        "nl_proof": "1. The proof relies on a previously established property (likely the `gc_comap_map` lemma) that connects inverse images, unions, and functions.\n2. This property is applied specifically to the left-hand side of the equation, showing that taking the inverse image of the union (the `comap` of the `sup`) can be decomposed.\n3. By applying this property, we directly arrive at the conclusion that taking the inverse image of the union of \\(m_1\\) and \\(m_2\\) under \\(g\\) is equivalent to taking the union of the inverse images of \\(m_1\\) and \\(m_2\\) under \\(g\\).\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem comap_sup : (m\u2081 \u2294 m\u2082).comap g = m\u2081.comap g \u2294 m\u2082.comap g :=\n   (gc_comap_map g).l_sup\n ",
        "pda_valid": false,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that the sine function is measurable.",
        "nl_proof": "1. We know that the sine function is continuous.\n2. It is a known property that continuous functions are measurable.\n3. Therefore, since the sine function is continuous, it is measurable.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem measurable_sin : Measurable sin :=\n   continuous_sin.measurable\n ",
        "pda_valid": false,
        "source": "lean_basic"
    },
    {
        "nl_statement": "Prove that applying `.toFun` to a function `f` is the same as directly applying `f` using the notation `\u21d1f`.",
        "nl_proof": "The statement is true by definition.  The `.toFun` operation on a function `f` is defined to be the same as applying `f` directly. Therefore, there is nothing to prove, as the two sides are immediately equivalent.\n\nTranslate the statement and proof in natural language to lean4:",
        "fl_statementproof": "theorem toFun_eq_coe : f.toFun = \u21d1f :=\n   rfl\n ",
        "pda_valid": false,
        "source": "lean_basic"
    },
    {
        "nl_statement": "For some complex number $\\omega$ with $|\\omega| = 2,$ there is some real $\\lambda > 1$ such that $\\omega,$ $\\omega^2,$ and $\\lambda \\omega$ form an equilateral triangle in the complex plane.  Find $\\lambda.$",
        "nl_proof": "Note that $\\omega,$ $\\omega^2,$ and $\\lambda \\omega$ form an equilateral triangle if and only if 1, $\\omega,$ and $\\lambda$ form an equilateral triangle.\n\nGiven 1 and $\\lambda > 1,$ there are two complex numbers $\\omega$ such that 1, $\\omega,$ and $\\lambda$ form an equilateral triangle.  Both complex numbers $\\omega$ have the same magnitude, so assume that the imaginary part of $\\omega$ is positive.\n\n[asy]\nunitsize (0.6 cm);\n\npair L, W;\n\nL = (5,0);\nW = 1 + 4*dir(60);\n\ndraw((-1,0)--(6,0));\ndraw((0,-1)--(0,4));\ndraw((1,0)--W--L);\n\nlabel(\"$1$\", (1,0), S);\nlabel(\"$\\lambda$\", L, S);\nlabel(\"$\\omega$\", W, N);\n[/asy]\n\nThen the side length of the equilateral triangle is $\\lambda - 1,$ so\n\\begin{align*}\n\\omega &= 1 + e^{\\pi i/3} (\\lambda - 1) \\\\\n&= 1 + \\left( \\frac{1}{2} + \\frac{\\sqrt{3}}{2} i \\right) (\\lambda - 1) \\\\\n&= \\frac{\\lambda + 1}{2} + \\frac{(\\lambda - 1) \\sqrt{3}}{2} i.\n\\end{align*}Hence,\n\\begin{align*}\n|\\omega|^2 &= \\left( \\frac{\\lambda + 1}{2} \\right)^2 + \\left( \\frac{(\\lambda - 1) \\sqrt{3}}{2} \\right)^2 \\\\\n&= \\frac{\\lambda^2 + 2 \\lambda + 1}{4} + \\frac{3 \\lambda^2 - 6 \\lambda + 3}{4} \\\\\n&= \\frac{4 \\lambda^2 - 4 \\lambda + 4}{4} = \\lambda^2 - \\lambda + 1.\n\\end{align*}But $|\\omega|^2 = 2^2 = 4,$ so $\\lambda^2 - \\lambda + 1 = 4,$ or\n\\[\\lambda^2 - \\lambda - 3 = 0.\\]By the quadratic formula,\n\\[\\lambda = \\frac{1 \\pm \\sqrt{13}}{2}.\\]Since $\\lambda > 1,$\n\\[\\lambda = \\boxed{\\frac{1 + \\sqrt{13}}{2}}.\\]\nThe answer is \\frac{1 + \\sqrt{13}}{2}\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem omega_two_norm : \u2016(2 : \u2102)\u2016 = 2 :=\n   rfl\n ",
        "pda_valid": true,
        "source": "math"
    },
    {
        "nl_statement": "The equation $\\sin^2 x + \\sin^2 2x + \\sin^2 3x + \\sin^2 4x = 2$ can be reduced to the equivalent equation\n\\[\\cos ax \\cos bx \\cos cx = 0,\\]for some positive integers $a,$ $b,$ and $c.$  Find $a + b + c.$",
        "nl_proof": "From the double angle formula,\n\\[\\frac{1 - \\cos 2x}{2} + \\frac{1 - \\cos 4x}{2} + \\frac{1 - \\cos 6x}{2} + \\frac{1 - \\cos 8x}{2} = 2,\\]so $\\cos 2x + \\cos 4x + \\cos 6x + \\cos 8x = 0.$  Then by sum-to-product,\n\\[\\cos 2x + \\cos 8x = 2 \\cos 5x \\cos 3x\\]and\n\\[\\cos 4x + \\cos 6x = 2 \\cos 5x \\cos x,\\]so\n\\[2 \\cos 5x \\cos 3x + 2 \\cos 5x \\cos x= 0,\\]or $\\cos 5x (\\cos x + \\cos 3x) = 0.$\n\nAgain by sum-to-product, $\\cos x + \\cos 3x = 2 \\cos 2x \\cos x,$ so this reduces to\n\\[\\cos x \\cos 2x \\cos 5x = 0.\\]Thus, $a + b + c = 1 + 2 + 5 = \\boxed{8}.$\nThe answer is 8\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem add_coe_sin_pow_two : (1 : \u2124) + (2 : \u2124) + (3 : \u2124) + (4 : \u2124) = 8 := by\n   simpa using @sin_pow_two_add_cos_mul_cos 2 4\n ",
        "pda_valid": true,
        "source": "math"
    },
    {
        "nl_statement": "A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $ \\frac{r-\\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t.$",
        "nl_proof": "Place the cube in space so that $A$ is at the origin, and the three vertices adjacent to $A$ are $(10,0,0),$ $(0,10,0),$ and $(0,0,10).$  Let the equation of the plane be\n\\[ax + by + cz + d = 0,\\]where $a^2 + b^2 + c^2 = 1.$  Then, the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d.$\n\n[asy]\nimport three;\n\n// calculate intersection of line and plane\n// p = point on line\n// d = direction of line\n// q = point in plane\n// n = normal to plane\ntriple lineintersectplan(triple p, triple d, triple q, triple n)\n{\n  return (p + dot(n,q - p)/dot(n,d)*d);\n}\n\n// projection of point A onto plane BCD\ntriple projectionofpointontoplane(triple A, triple B, triple C, triple D)\n{\n  return lineintersectplan(A, cross(B - D, C - D), B, cross(B - D, C - D));\n}\n\nsize(250);\ncurrentprojection = perspective(6,3,2);\n\ntriple A, B, C, D, X, Y, Z, P, Q, R, T;\ntriple[] S;\nreal a, b, c, d;\n\nA = (0,0,0);\nB = (10,0,0);\nC = (0,10,0);\nD = (0,0,10);\na = 0.471548;\nb = 0.571548;\nc = 0.671548;\nd = 5.28452;\nX = (-d/a,0,0);\nY = (0,-d/b,0);\nZ = (0,0,-d/c);\nP = projectionofpointontoplane(B, X, Y, Z);\nQ = projectionofpointontoplane(C, X, Y, Z);\nR = projectionofpointontoplane(D, X, Y, Z);\nT = projectionofpointontoplane(A, X, Y, Z);\nS[1] = -0.5*X + 2*Y - 0.5*Z;\nS[2] = 2*X - 0.5*Y - 0.5*Z;\nS[3] = S[2] + 0.5*cross((a,b,c),S[1] - S[2]);\nS[4] = S[1] + S[3] - S[2];\n\ndraw(surface(S[1]--S[2]--S[3]--S[4]--cycle),paleyellow,nolight);\ndraw(S[1]--S[2]--S[3]--S[4]--cycle);\ndraw(A--B);\ndraw(A--C);\ndraw(A--D);\ndraw(B--P,dashed);\ndraw(C--Q,dashed);\ndraw(D--R,dashed);\ndraw(A--T,dashed);\n\ndot(\"$(0,0,0)$\", A, NE);\ndot(\"$(10,0,0)$\", B, NW);\ndot(\"$(0,10,0)$\", C, NE);\ndot(\"$(0,0,10)$\", D,  N);\ndot(P);\ndot(Q);\ndot(R);\ndot(T);\n[/asy]\n\nSo, by looking at the three vertices, we have $10a+d=10,$ $10b+d=11,$ and $10c+d=12.$  Then $10a = 10 - d,$ $10b = 11 - d,$ and $10c = 12 - d,$ so\n\\[(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100.\\]Solving for $d,$ we find\n\\[d = 11 \\pm 7 \\sqrt{\\frac{2}{3}}.\\]Note that the distance from the origin to the plane is $d,$ which must be less than 10, so\n\\[d = 11 - 7 \\sqrt{\\frac{2}{3}} = \\frac{33 - \\sqrt{294}}{3}.\\]The final answer is $33+294+3=\\boxed{330}$.\nThe answer is 330\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem mod_cast_three_three_zero : (10 : \u2124) % (1000 : \u2124) = 330 := by\n   simp\n ",
        "pda_valid": true,
        "source": "math"
    },
    {
        "nl_statement": "A certain regular tetrahedron has three of its vertices at the points $(0,1,2),$ $(4,2,1),$ and $(3,1,5).$  Find the coordinates of the fourth vertex, given that they are also all integers.",
        "nl_proof": "The side length of the regular tetrahedron is the distance between $(0,1,2)$ and $(4,2,1),$ which is\n\\[\\sqrt{(0 - 4)^2 + (1 - 2)^2 + (2 - 1)^2} = \\sqrt{18} = 3 \\sqrt{2}.\\]So if $(x,y,z)$ is the fourth vertex, with integer coordinates, then\n\\begin{align*}\nx^2 + (y - 1)^2 + (z - 2)^2 &= 18, \\\\\n(x - 4)^2 + (y - 2)^2 + (z - 1)^2 &= 18, \\\\\n(x - 3)^2 + (y - 1)^2 + (z - 5)^2 &= 18.\n\\end{align*}Subtracting the first and third equations, we get $6x + 6z - 30 = 0$, so $x + z = 5,$ which means $z = 5 - x.$  Subtracting the first and second equation, we get $8x + 2y - 2z - 16 = 0,$ so\n\\[y = z - 4x + 8 = (5 - x) - 4x + 8 = 13 - 5x.\\]Substituting into the first equation, we get\n\\[x^2 + (12 - 5x)^2 + (3 - x)^2 = 18.\\]This simplifies to $27x^2 - 126x + 135 = 0,$ which factors as $9(x - 3)(3x - 5) = 0.$  Since $x$ is an integer, $x = 3.$  Then $y = -2$ and $z = 2.$  Thus, the fourth vertex is $\\boxed{(3,-2,2)}.$\nThe answer is (3,-2,2)\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem fourth_vert : (0 : \u2124) = -4 * (3 : \u2124) := by norm_num\n ",
        "pda_valid": true,
        "source": "math"
    },
    {
        "nl_statement": "One line is parameterized by\n\\[\\begin{pmatrix} -1 + s \\\\ 3 - ks \\\\ 1 + ks \\end{pmatrix}.\\]Another line is parameterized by\n\\[\\begin{pmatrix} t/2 \\\\ 1 + t \\\\ 2 - t \\end{pmatrix}.\\]If the lines are coplanar (i.e. there is a plane that contains both lines), then find $k.$",
        "nl_proof": "First, we check if the two lines can intersect.  For the two lines to intersect, we must have\n\\begin{align*}\n-1 + s &= \\frac{t}{2}, \\\\\n3 - ks &= 1 + t, \\\\\n1 + ks &= 2 - t.\n\\end{align*}Adding the second equation and third equation, we get $4 = 3,$ contradiction.  Thus, the two lines cannot intersect.\n\nSo for the two lines to be coplanar, the only other possibility is that they are parallel.  For the two lines to be parallel, their direction vectors must be proportional.  The direction vectors of the lines are $\\begin{pmatrix} 1 \\\\ -k \\\\ k \\end{pmatrix}$ and $\\begin{pmatrix} 1/2 \\\\ 1 \\\\ -1 \\end{pmatrix},$ respectively.  These vectors are proportional when\n\\[2 = -k.\\]Hence, $k = \\boxed{-2}.$\nThe answer is -2\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem coplanar_lines_two_three : (k : \u211d) = -2 := by\n   rw [\u2190 two_add_two_eq_four, \u2190 sub_eq_iff_eq_add, \u2190 neg_eq_iff_eq_neg, \u2190 sub_eq_iff_eq_add,\n     sub_sub_sub_cancel_right, \u2190 neg_eq_iff_eq_neg, \u2190 neg_eq_iff_eq_neg]\n   exact (not_and_or.mpr (not_intersecting_lines_two_three k s t).symm).resolve_left (by norm_num)\n ",
        "pda_valid": true,
        "source": "math"
    },
    {
        "nl_statement": "Compute $\\begin{pmatrix} 2 & 3 \\\\ 7 & -1 \\end{pmatrix} \\begin{pmatrix} 1 & -5 \\\\ 0 & 4 \\end{pmatrix}.$",
        "nl_proof": "We have that\n\\[\\begin{pmatrix} 2 & 3 \\\\ 7 & -1 \\end{pmatrix} \\begin{pmatrix} 1 & -5 \\\\ 0 & 4 \\end{pmatrix} = \\begin{pmatrix} (2)(1) + (3)(0) & (2)(-5) + (3)(4) \\\\ (7)(1) + (-1)(0) & (7)(-5) + (-1)(4) \\end{pmatrix} = \\boxed{\\begin{pmatrix} 2 & 2 \\\\ 7 & -39 \\end{pmatrix}}.\\]\nThe answer is \\begin{pmatrix} 2 & 2 \\\\ 7 & -39 \\end{pmatrix}\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem two_three_seven_neg_one_one_neg_five_zero_four : (2 : \u2124) * (1 : \u2124) + 3 * 0 = 2 :=\n   rfl\n ",
        "pda_valid": false,
        "source": "math"
    },
    {
        "nl_statement": "The curve $y = \\sin x$ cuts the line whose equation is $y = \\sin 70^\\circ$ into segments having the successive ratios\n\\[\\dots p : q : p : q \\dots\\]with $p < q.$  Compute the ordered pair of relatively prime positive integers $(p,q).$",
        "nl_proof": "The graph of $y = \\sin x$ intersects the line $y = \\sin 70^\\circ$ at points of the form $(70^\\circ + 360^\\circ n, \\sin 70^\\circ)$ and $(110^\\circ + 360^\\circ n, \\sin 70^\\circ),$ where $n$ is an integer.\n\n[asy]\nunitsize(1.2 cm);\n\nreal func (real x) {\n  return(sin(x));\n}\n\ndraw(graph(func,-2*pi,2*pi),red);\ndraw((-2*pi,Sin(70))--(2*pi,Sin(70)),blue);\ndraw((-2*pi,0)--(2*pi,0));\ndraw((0,-1)--(0,1));\ndraw((70*pi/180,0)--(70*pi/180,Sin(70)),dashed);\ndraw((110*pi/180,0)--(110*pi/180,Sin(70)),dashed);\ndraw((-290*pi/180,0)--(-290*pi/180,Sin(70)),dashed);\ndraw((-250*pi/180,0)--(-250*pi/180,Sin(70)),dashed);\n\nlabel(\"$70^\\circ$\", (70*pi/180,0), S, fontsize(10));\nlabel(\"$110^\\circ$\", (110*pi/180,0), S, fontsize(10));\nlabel(\"$-290^\\circ$\", (-290*pi/180 - 0.1,0), S, fontsize(10));\nlabel(\"$-250^\\circ$\", (-250*pi/180 + 0.1,0), S, fontsize(10));\n[/asy]\n\nThe ratio of the lengths is then\n\\[\\frac{110 - 70}{70 + 250} = \\frac{40}{320} = \\frac{1}{8},\\]so $(p,q) = \\boxed{(1,8)}.$\nThe answer is (1,8)\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem sin_70_ratios : (1 : \u2124) / 8 = 1 / 8 := by simp\n ",
        "pda_valid": false,
        "source": "math"
    },
    {
        "nl_statement": "Compute $\\begin{pmatrix} 2 & - 1 \\\\ - 3 & 4 \\end{pmatrix} \\begin{pmatrix} 3 \\\\ - 1 \\end{pmatrix}.$",
        "nl_proof": "We have that\n\\[\\begin{pmatrix} 2 & - 1 \\\\ - 3 & 4 \\end{pmatrix} \\begin{pmatrix} 3 \\\\ - 1 \\end{pmatrix} = \\begin{pmatrix} (2)(3) + (-1)(-1) \\\\ (-3)(3) + (4)(-1) \\end{pmatrix} = \\boxed{\\begin{pmatrix} 7 \\\\ -13 \\end{pmatrix}}.\\]\nThe answer is \\begin{pmatrix} 7 \\\\ -13 \\end{pmatrix}\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem two_neg_one_neg_three_four_three_neg_one : (2 : \u2124) * 3 + (-1 : \u2124) * (-1) = 7 := by\n   simp\n ",
        "pda_valid": false,
        "source": "math"
    },
    {
        "nl_statement": "A square with side length 1 is rotated about one vertex by an angle of $\\alpha,$ where $0^\\circ < \\alpha < 90^\\circ$ and $\\cos \\alpha = \\frac{4}{5}.$  Find the area of the shaded region that is common to both squares.\n\n[asy]\nunitsize(3 cm);\n\npair A, B, C, D, Bp, Cp, Dp, P;\n\nA = (0,0);\nB = (-1,0);\nC = (-1,-1);\nD = (0,-1);\nBp = rotate(aCos(4/5))*(B);\nCp = rotate(aCos(4/5))*(C);\nDp = rotate(aCos(4/5))*(D);\nP = extension(C,D,Bp,Cp);\n\nfill(A--Bp--P--D--cycle,gray(0.7));\ndraw(A--B---C--D--cycle);\ndraw(A--Bp--Cp--Dp--cycle);\n\nlabel(\"$\\alpha$\", A + (-0.25,-0.1));\n[/asy]",
        "nl_proof": "Let the squares be $ABCD$ and $AB'C'D',$ as shown.  Let $P$ be the intersection of $\\overline{CD}$ and $\\overline{B'C'}.$\n\n[asy]\nunitsize(3 cm);\n\npair A, B, C, D, Bp, Cp, Dp, P;\n\nA = (0,0);\nB = (-1,0);\nC = (-1,-1);\nD = (0,-1);\nBp = rotate(aCos(4/5))*(B);\nCp = rotate(aCos(4/5))*(C);\nDp = rotate(aCos(4/5))*(D);\nP = extension(C,D,Bp,Cp);\n\nfill(A--Bp--P--D--cycle,gray(0.7));\ndraw(A--B---C--D--cycle);\ndraw(A--Bp--Cp--Dp--cycle);\ndraw(A--P);\n\nlabel(\"$\\alpha$\", A + (-0.25,-0.1));\nlabel(\"$A$\", A, NE);\nlabel(\"$B$\", B, NW);\nlabel(\"$C$\", C, SW);\nlabel(\"$D$\", D, SE);\nlabel(\"$B'$\", Bp, W);\nlabel(\"$C'$\", Cp, S);\nlabel(\"$D'$\", Dp, E);\nlabel(\"$P$\", P, SW);\n[/asy]\n\nThen $\\angle B'AD = 90^\\circ - \\alpha,$ and by symmetry, $\\angle B'AP = \\angle DAP = \\frac{90^\\circ - \\alpha}{2} = 45^\\circ - \\frac{\\alpha}{2}.$  Then\n\\[B'P = \\tan \\left( 45^\\circ - \\frac{\\alpha}{2} \\right) = \\frac{\\tan 45^\\circ - \\tan \\frac{\\alpha}{2}}{1 + \\tan 45^\\circ \\tan \\frac{\\alpha}{2}} = \\frac{1 - \\tan \\frac{\\alpha}{2}}{1 + \\tan \\frac{\\alpha}{2}}.\\]Since $\\alpha$ is acute,\n\\[\\sin \\alpha = \\sqrt{1 - \\cos^2 \\alpha} = \\sqrt{1 - \\left( \\frac{4}{5} \\right)^2} = \\frac{3}{5},\\]so\n\\[\\tan \\frac{\\alpha}{2} = \\frac{\\sin \\alpha}{1 + \\cos \\alpha} = \\frac{3/5}{1 + 4/5} = \\frac{1}{3}.\\]Then\n\\[BP = \\frac{1 - 1/3}{1 + 1/3} = \\frac{1}{2},\\]so $[AB'P] = \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot 1 = \\frac{1}{4}.$  Also, $[ADP] = \\frac{1}{4},$ so the area of the shaded region is $\\boxed{\\frac{1}{2}}.$\nThe answer is \\frac{1}{2}\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem area_of_half_square_rotated : (1 / 2 : \u211d) = (2 : \u211d) / 4 := by\n   norm_num\n ",
        "pda_valid": false,
        "source": "math"
    },
    {
        "nl_statement": "Find the least positive integer $n$ such that $$\\frac 1{\\sin 45^\\circ\\sin 46^\\circ}+\\frac 1{\\sin 47^\\circ\\sin 48^\\circ}+\\cdots+\\frac 1{\\sin 133^\\circ\\sin 134^\\circ}=\\frac 1{\\sin n^\\circ}.$$",
        "nl_proof": "Each term is of the form $\\frac{1}{\\sin k^\\circ \\sin (k + 1)^\\circ}.$  To deal with this term, we look at $\\sin ((k + 1)^\\circ - k^\\circ).$  From the angle subtraction formula,\n\\[\\sin ((k + 1)^\\circ - k^\\circ) = \\sin (k + 1)^\\circ \\cos k^\\circ - \\cos (k + 1)^\\circ \\sin k^\\circ.\\]Then\n\\begin{align*}\n\\frac{\\sin 1^\\circ}{\\sin k^\\circ \\sin (k + 1)^\\circ} &= \\frac{\\sin ((k + 1)^\\circ - k^\\circ)}{\\sin k^\\circ \\sin (k + 1)^\\circ} \\\\\n&= \\frac{\\sin (k + 1)^\\circ \\cos k^\\circ - \\cos (k + 1)^\\circ \\sin k^\\circ}{\\sin k^\\circ \\sin (k + 1)^\\circ} \\\\\n&= \\frac{\\cos k^\\circ}{\\sin k^\\circ} - \\frac{\\cos (k + 1)^\\circ}{\\sin (k + 1)^\\circ} \\\\\n&= \\cot k^\\circ - \\cot (k + 1)^\\circ.\n\\end{align*}Hence,\n\\[\\frac{1}{\\sin k^\\circ \\sin (k + 1)^\\circ} = \\frac{1}{\\sin 1^\\circ} (\\cot k^\\circ - \\cot (k + 1)^\\circ).\\]Then\n\\begin{align*}\n&\\frac{1}{\\sin 45^\\circ \\sin 46^\\circ} + \\frac{1}{\\sin 47^\\circ \\sin 48^\\circ} + \\dots + \\frac{1}{\\sin 133^\\circ \\sin 134^\\circ} \\\\\n&= \\frac{1}{\\sin 1^\\circ} (\\cot 45^\\circ - \\cot 46^\\circ + \\cot 47^\\circ - \\cot 48^\\circ + \\dots + \\cot 133^\\circ - \\cot 134^\\circ).\n\\end{align*}Since $\\cot (180^\\circ - x) = -\\cot x,$ the sum reduces to\n\\[\\frac{\\cot 45^\\circ - \\cot 90^\\circ}{\\sin 1^\\circ} = \\frac{1}{\\sin 1^\\circ}.\\]Thus, the smallest such positive integer $n$ is $\\boxed{1}.$\nThe answer is 1\n\nTranslate the statement and proof in natural language to lean4:",
        "formal": "theorem find_n_sin_add_sin : \u2203 n : \u2115, (n : \u211d) = 1 \u2227 \u2191n = 1 :=\n   \u27e81, by simp\u27e9\n ",
        "pda_valid": false,
        "source": "math"
    }
]