| [ |
| { |
| "dataset": "dev", |
| "index": 6, |
| "conic_type": "circle", |
| "image_path": "samples/dev/circle/problem_0006.png", |
| "problem": { |
| "text": "Given the circle $(x+1)^{2}+y^{2}=36$ with center $M$, let $A$ be any point on the circle, and let $N(1 , 0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. Then the trajectory equation of the moving point $P$ is?", |
| "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x + 1)^2 = 36);Center(G) = M;M: Point;A: Point;PointOnCurve(A, G);N: Point;Coordinate(N) = (1, 0);Intersection(PerpendicularBisector(LineSegmentOf(A,N)),LineSegmentOf(M,A)) = P;P: Point", |
| "query_expressions": "LocusEquation(P)", |
| "answer_expressions": "x^2/9+y^2/8=1", |
| "process": "Since the circle $(x+2)^{2}+y^{2}=36$ has center $M(-1,0)$ and radius $r=6$, let point $P(x,y)$. Since the perpendicular bisector of segment $AN$ intersects $MA$ at point $P$, we have $|PN|=|PA|$. Therefore, $|PM|+|PN|=|PM|+|PA|=|MA|=6>|MN|=2$. Hence, the trajectory of point $P$ is an ellipse with foci at $M(-1,0)$ and $N(1,0)$, and major axis length $2a=6$, so $b=\\sqrt{a^{2}-c^{2}}=\\sqrt{9-1}=2\\sqrt{2}$. Thus, the equation of the trajectory of moving point $P$ is $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 1.0, |
| 0.0 |
| ], |
| "radius": 6.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 441, |
| "conic_type": "circle", |
| "image_path": "samples/dev/circle/problem_0441.png", |
| "problem": { |
| "text": "A moving circle is externally tangent to the circle $Q_{1}$: $(x+3)^{2}+y^{2}=1$ and internally tangent to the circle $Q_{2}$: $(x-3)^{2}+y^{2}=81$. Then, the trajectory equation of the center of this moving circle is?", |
| "fact_expressions": "G: Circle;IsOutTangent(G,Q1) = True;Q1: Circle;Expression(Q1) = ((x+3)^2+y^2=1);Q2: Circle;Expression(Q2) = ((x-3)^2+y^2=81);IsInTangent(G,Q2) = True", |
| "query_expressions": "LocusEquation(Center(G))", |
| "answer_expressions": "x^2/25+y^2/16=1", |
| "process": "Let the center of the moving circle be Q(x, y), with radius R. Since the moving circle is externally tangent to the circle Q_{1}:(x+3)^{2}+y^{2}=1 and internally tangent to the circle Q_{2}:(x-3)^{2}+y^{2}=81, we have |QQ_{1}| = R + 1, |QQ_{2}| = 9 - R. Therefore, |QQ_{1}| + |QQ_{2}| = 10 > 6 = |Q_{1}Q_{2}|. Thus, the locus of the center of the moving circle is an ellipse with foci at Q_{1} and Q_{2}. Hence, 2a = 10, a = 5, c = 3, b^{2} = 16. Therefore, the equation of the locus of the center of the moving circle is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| -3.0, |
| 0.0 |
| ], |
| "radius": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 669, |
| "conic_type": "circle", |
| "image_path": "samples/dev/circle/problem_0669.png", |
| "problem": { |
| "text": "Given the circle $C$: $x^{2}+y^{2}=25$, draw a line $l$ through point $M(-2,3)$ intersecting the circle $C$ at points $A$ and $B$. Tangents to the circle are drawn at points $A$ and $B$, respectively. When the two tangents intersect at point $N$, what is the trajectory equation of point $N$?", |
| "fact_expressions": "C: Circle;Expression(C) = (x^2 + y^2 = 25);M: Point;l: Line;Coordinate(M) = (-2, 3);PointOnCurve(M, l);A: Point;B: Point;Intersection(l, C) = {A, B};L1: Line;L2: Line;TangentOnPoint(A,C)=L1;TangentOnPoint(B,C)=L2;N: Point;Intersection(L1, L2) = N", |
| "query_expressions": "LocusEquation(N)", |
| "answer_expressions": "2*x-3*y+25=0", |
| "process": "Consider the following problem: Given a circle $ C: x^{2} + y^{2} = r^{2} $ ($ r > 0 $) and a point $ P(a, b) $. If point $ P $ lies inside $ C $, draw a line $ l $ through $ P $ intersecting $ C $ at points $ A $ and $ B $. Tangents to $ C $ are drawn at points $ A $ and $ B $ respectively. When these two tangents intersect at point $ Q $, find the locus equation of point $ Q $. The center of circle $ C: x^{2} + y^{2} = r^{2} $ is $ (0, 0) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ Q(x_{0}, y_{0}) $. Since $ AQ $ is tangent to circle $ C $, we have $ AQ \\perp CA $. Therefore, $ (x_{1} - x_{0})(x_{1} - 0) + (y_{1} - y_{0})(y_{1} - 0) = 0 $, that is, $ x^{2}_{1} - x_{0}x_{1} + y^{2}_{1} - y_{0}y_{1} = 0 $. Since $ x^{2} + y^{2} = r^{2} $, it follows that $ x_{0}x_{1} + y_{0}y_{1} = r^{2} $. Similarly, $ x_{0}x_{2} + y_{0}y_{2} = r^{2} $. Hence, the equation of the line passing through points $ A $ and $ B $ is $ xx_{0} + yy_{0} = r^{2} $. Since line $ AB $ passes through point $ (a, b) $, substituting gives $ ax_{0} + by_{0} = r^{2} $. Therefore, the locus equation of point $ Q $ is: $ ax + by = r^{2} $. According to the problem, the locus equation of point $ N $ is $ 2x - 3y + 25 = 0 $." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 0.0, |
| 0.0 |
| ], |
| "radius": 5.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 1, |
| "conic_type": "ellipse", |
| "image_path": "samples/dev/ellipse/problem_0001.png", |
| "problem": { |
| "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ with left and right foci $F_{1}$, $F_{2}$, point $P$ is a moving point on the ellipse. Then the range of values of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?", |
| "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)", |
| "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))", |
| "answer_expressions": "[-2, 1]", |
| "process": "Let P(x,y) be an arbitrary point on the ellipse, then \\overrightarrow{PF_{1}}=(-\\sqrt{3}-x,-y), \\overrightarrow{PF_{2}}=(\\sqrt{3}-x,-y). Therefore, \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-\\sqrt{3}-x,-y)\\cdot(\\sqrt{3}-x,-y)=x^{2}+y^{2}-3=x^{2}+1-\\frac{x^{2}}{4}=\\frac{3}{4}x^{2}-2. Since P lies on the ellipse, -2\\leqslant x \\leqslant 2, so -2\\leqslant \\frac{3}{4}x^{2}-2 \\leqslant 1, that is, the range of \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}} is [-2,1]." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.0, |
| "major_axis": "x", |
| "x_coef": 4.0, |
| "y_coef": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 524, |
| "conic_type": "ellipse", |
| "image_path": "samples/dev/ellipse/problem_0524.png", |
| "problem": { |
| "text": "Point $P$ lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$. What are the maximum and minimum distances from point $P$ to the line $3x-4y=24$?", |
| "fact_expressions": "G: Ellipse;H: Line;P: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Expression(H) = (3*x - 4*y = 24);PointOnCurve(P, G)", |
| "query_expressions": "Max(Distance(P,H));Min(Distance(P,H))", |
| "answer_expressions": "12*(2+sqrt(2))/5\n12*(2-sqrt(2))/5", |
| "process": "Let the coordinates of point P be (4\\cos\\theta,3\\sin\\theta). The distance d from point P to the line 3x-4y=24 reaches its maximum when \\cos(\\theta+\\frac{\\pi}{4})=-1, and reaches its minimum when \\cos(\\theta+\\frac{\\pi}{4})=1, with the minimum value being \\frac{12(2-}{" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 4.0, |
| "b": 3.0, |
| "major_axis": "x", |
| "x_coef": 16.0, |
| "y_coef": 9.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 1034, |
| "conic_type": "ellipse", |
| "image_path": "samples/dev/ellipse/problem_1034.png", |
| "problem": { |
| "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ and perpendicular to the major axis intersects the ellipse at points $A$ and $B$. Find the radius of the incircle of triangle $ABF_{2}$.", |
| "fact_expressions": "G: Ellipse;H:Line;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,H);IsPerpendicular(H,MajorAxis(G));Intersection(H,G) = {A, B}", |
| "query_expressions": "Radius(InscribedCircle(TriangleOf(A,B,F2)))", |
| "answer_expressions": "36/25", |
| "process": "According to the problem, let the inradius of $\\triangle ABF_{2}$ be $r$, and the perimeter of the triangle be $4a$. Then, derive the expression for the area of the triangle, use $AF_{1}, BF_{1}$ to find the area of $\\triangle ABF_{2}$, and then determine the inradius. [Detailed Solution] According to the problem, let the inradius of $\\triangle ABF_{2}$ be $r$; the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and the perimeter of the triangle is $20$. Hence, $S=\\frac{1}{2}\\times20\\times r=10r$. The line passing through $F_1$ and perpendicular to the major axis intersects the ellipse at points $A$ and $B$. Then, $AB=2\\sqrt{9(1-\\frac{16}{25})}=\\frac{18}{5}$. Therefore, $S=\\frac{1}{2}\\times8\\times\\frac{18}{5}=\\frac{72}{5}$. Thus, $10r=\\frac{72}{5}$, solving gives $r=\\frac{36}{25}$. Therefore, the inradius is $\\frac{36}{25}$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 5.0, |
| "b": 3.0, |
| "major_axis": "x", |
| "x_coef": 25.0, |
| "y_coef": 9.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 0, |
| "conic_type": "hyperbola", |
| "image_path": "samples/dev/hyperbola/problem_0000.png", |
| "problem": { |
| "text": "From the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$, draw a tangent to the circle $x^{2}+y^{2}=16$, with the point of tangency $T$. Extend $F_{1} T$ to intersect the right branch of the hyperbola at point $P$. Let $M$ be the midpoint of segment $F_{1} P$, and let $O$ be the origin. Then $|M O|-|M T|$=?", |
| "fact_expressions": "G: Hyperbola;H: Circle;F1: Point;Z: Line;T: Point;P: Point;M: Point;Expression(G) = (x^2/16 - y^2/25 = 1);LeftFocus(G) = F1;Expression(H) = (x^2 + y^2 = 16);TangentOfPoint(F1, H) = Z;TangentPoint(Z, H) = T;Intersection(OverlappingLine(LineSegmentOf(F1, T)), RightPart(G)) = P;MidPoint(LineSegmentOf(F1, P)) = M;O: Origin", |
| "query_expressions": "Abs(LineSegmentOf(M, O)) - Abs(LineSegmentOf(M, T))", |
| "answer_expressions": "1", |
| "process": "Let F' be the right focus of the hyperbola, connect PF'. Since M and O are the midpoints of F_{1}P and F_{1}F' respectively, then |MO| = \\frac{1}{2}|PF'|. By the definition of the hyperbola, |F_{1}P| - |PF'| = 8. Hence, |MO| - |MT| = \\frac{1}{2}|PF| - |MF_{1}| + |F_{1}T| = \\frac{1}{2}(|PF'| - |F_{1}P|) + |F_{1}T| = -4 + 5 = 1" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 4.0, |
| "b": 5.0, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 537, |
| "conic_type": "hyperbola", |
| "image_path": "samples/dev/hyperbola/problem_0537.png", |
| "problem": { |
| "text": "Given fixed points $A(-5,0)$, $B(5,4)$, and point $P$ being any point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, then the maximum value of $|PB|-|PA|$ is?", |
| "fact_expressions": "C: Hyperbola;A: Point;B: Point;P: Point;Expression(C) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (-5, 0);Coordinate(B) = (5, 4);PointOnCurve(P, RightPart(C))", |
| "query_expressions": "Max(-Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))", |
| "answer_expressions": "-4", |
| "process": "According to the given conditions, a=4, b=3, so c=\\sqrt{a^{2}+b^{2}}=5; therefore, A(-5,0) is the left focus of the hyperbola, and let the right focus be F(5,0). Thus, |PB|-|PA|=|PB|-(|PF|+2a)=|PB|-|PF|-8\\leqslant|BF|-8=4-8=-4, with equality if and only if points P, F, and B are collinear." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 4.0, |
| "b": 3.0, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "A": [ |
| -5.0, |
| 0.0 |
| ], |
| "B": [ |
| 5.0, |
| 4.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 1033, |
| "conic_type": "hyperbola", |
| "image_path": "samples/dev/hyperbola/problem_1033.png", |
| "problem": { |
| "text": "Given that line $l$ passes through point $P(2, 1)$ and intersects the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ at points $A$ and $B$, if $P$ is the midpoint of $AB$, then what is the equation of line $l$?", |
| "fact_expressions": "l: Line;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, l);G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P", |
| "query_expressions": "Expression(l)", |
| "answer_expressions": "8*x-y-15=0", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 1.0, |
| "b": 2.0, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "P": [ |
| 2.0, |
| 1.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 2, |
| "conic_type": "parabola", |
| "image_path": "samples/dev/parabola/problem_0002.png", |
| "problem": { |
| "text": "The coordinates of the focus of the parabola $y^{2}=4 x$ are?", |
| "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)", |
| "query_expressions": "Coordinate(Focus(G))", |
| "answer_expressions": "(1, 0)", |
| "process": "The focus of the parabola y^{2}=4x lies on the x-axis, and p=2, \\therefore\\frac{p}{2}=1, so the focus coordinates of the parabola y^{2}=4x are (1,0)." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 1.0, |
| "direction": "right" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 508, |
| "conic_type": "parabola", |
| "image_path": "samples/dev/parabola/problem_0508.png", |
| "problem": { |
| "text": "Given the parabola $y^{2}=4x$ with focus $F$, $A(-1,0)$, and point $P$ being a moving point on the parabola, when the value of $|PF|$ is minimized, $|PF|=$?", |
| "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 0);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, F)))", |
| "query_expressions": "Abs(LineSegmentOf(P, F))", |
| "answer_expressions": "2", |
| "process": "The equation of the directrix of the parabola is x = -1. Let the distance from P to the directrix be |PQ|, then |PQ| = |PF|, \\therefore \\frac{|PF|}{|PA|} = \\frac{|PQ|}{|PA|} = \\sin\\angle PAQ. Therefore, when PA is tangent to the parabola y^{2} = 4x, \\angle PAQ is minimized, i.e., \\frac{|PF|}{|PA|} reaches its minimum value. Suppose the line passing through point A, y = kx + k, is tangent to the parabola (k \\neq 0). Substituting into the parabola equation gives k^{2}x^{2} + (2k^{2} - 4)x + k^{2} = 0. \\therefore \\Delta = (2k^{2} - 4)^{2} - 4k^{4} = 0. Solving yields k = \\pm 1. Then we have x^{2} - 2x + 1 = 0, solving gives x = 1. Substituting x = 1 into y^{2} = 4x gives y = \\pm 2. \\therefore P(1,2) or P(1,-2), \\therefore |PF| = 2." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 1.0, |
| "direction": "right" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "A": [ |
| -1.0, |
| 0.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "dev", |
| "index": 1030, |
| "conic_type": "parabola", |
| "image_path": "samples/dev/parabola/problem_1030.png", |
| "problem": { |
| "text": "Given point $P(1,-1)$ and parabola $C$: $y=\\frac{1}{4} x^{2}$, a line passing through the focus of parabola $C$ with slope $k$ intersects parabola $C$ at points $A$ and $B$. If $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=0$, then $k=?$", |
| "fact_expressions": "P: Point;Coordinate(P) = (1, -1);C: Parabola;Expression(C) = (y = x^2/4);G: Line;PointOnCurve(Focus(C), G);k: Number;Slope(G) = k;A: Point;B: Point;Intersection(G, C) = {A, B};DotProduct(VectorOf(P, A), VectorOf(P, B)) = 0", |
| "query_expressions": "k", |
| "answer_expressions": "1/2", |
| "process": "Let the parabola C: y = \\frac{1}{4}x^{2} have focus F; then the coordinates of F are (0,1). Thus, the equation of line AB is y = kx + 1. Let points A(x_{1},y_{1}) and B(x_{2},y_{2}). Substituting the equation of line AB into the equation of parabola C and simplifying yields x^{2} - 4kx - 4 = 0. Then x_{1} + x_{2} = 4k, x_{1}x_{2} = -4. From \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = 0, we obtain (x_{1}-1)(x_{2}-1) + (y_{1}+1)(y_{2}+1) = 0, that is, (x_{1}-1)(x_{2}-1) + (kx_{1}+2)(kx_{2}+2) = 0, which simplifies to (k^{2}+1)x_{1}x_{2} + (2k-1)(x_{1}+x_{2}) + 5 = 0. Therefore, -4(k^{2}+1) + 4k(2k-1) + 5 = (2k-1)^{2} = 0. Solving gives k = \\frac{1}{2}." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 1.0, |
| "direction": "up" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "P": [ |
| 1.0, |
| -1.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 4, |
| "conic_type": "circle", |
| "image_path": "samples/test/circle/problem_0004.png", |
| "problem": { |
| "text": "The product of the slopes of the lines connecting a moving point $P$ to fixed points $A(-1,0)$ and $B(1,0)$ is $-1$. Then, what is the trajectory equation of point $P$?", |
| "fact_expressions": "P: Point;Slope(LineSegmentOf(P, A))*Slope(LineSegmentOf(P, B)) = -1;A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0)", |
| "query_expressions": "LocusEquation(P)", |
| "answer_expressions": "(x^2+y^2=1)&Negation(x=pm*1)", |
| "process": "Let P(x,y), then k_{PA}=\\frac{y-0}{x+1}, k_{PB}=\\frac{y-0}{x-1}. Since the product of the slopes of the lines joining the moving point P and the fixed points A(-1,0), B(1,0) is -1, \\therefore k_{PA} \\cdot k_{PB} = -1, \\therefore \\frac{y^{2}}{x^{2}-1} = -1, that is, x^{2} + y^{2} = 1, and x \\neq \\pm 1. In conclusion, the trajectory equation of point P is x^{2} + y^{2} = 1 (x \\neq \\pm 1)." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 0.0, |
| 0.0 |
| ], |
| "radius": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 1202, |
| "conic_type": "circle", |
| "image_path": "samples/test/circle/problem_1202.png", |
| "problem": { |
| "text": "Given a line $ l $ with slope $ 1 $ and positive $ y $-intercept $ b $ that intersects the circle $ C: x^{2} + y^{2} = 4 $ at points $ A $ and $ B $, and $ O $ is the origin. If the area of $ \\triangle AOB $ is $ \\sqrt{3} $, then $ b = $?", |
| "fact_expressions": "l: Line;C: Circle;A: Point;O: Origin;B: Point;b: Number;Expression(C) = (x^2 + y^2 = 4);Slope(l)=1;b>0;Intercept(l,yAxis)=b;Intersection(l, C) = {A, B};Area(TriangleOf(A, O, B)) = sqrt(3)", |
| "query_expressions": "b", |
| "answer_expressions": "{sqrt(6),sqrt(2)}", |
| "process": "According to the problem, the equation of line $ l $ is $ y = x + b $, the center of circle $ C $ is $ C(0,0) $, and the radius is $ r = 2 $. Using the point-to-line distance formula, $ \\frac{1}{2} \\cdot 2\\sqrt{4 - \\frac{b^{2}}{2}} \\cdot \\frac{|b|}{\\sqrt{3}} = \\sqrt{3} $. Given $ b > 0 $, solving yields $ b = \\sqrt{6} $ or $ \\sqrt{2} $." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 0.0, |
| 0.0 |
| ], |
| "radius": 2.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 1992, |
| "conic_type": "circle", |
| "image_path": "samples/test/circle/problem_1992.png", |
| "problem": { |
| "text": "Given that the moving circle $P$ is externally tangent to the fixed circle $C$: $(x+2)^{2}+y^{2}=1$, and also tangent to the line $x=1$, what is the equation of the locus of the center $P$ of the moving circle?", |
| "fact_expressions": "P: Circle;G: Line;C:Circle;P1:Point;Expression(C)=((x+2)^2+y^2=1);Expression(G) = (x = 1);IsOutTangent(P,C);IsTangent(P,G);Center(P)=P1", |
| "query_expressions": "LocusEquation(P1)", |
| "answer_expressions": "y^2=-8*x", |
| "process": "Let the distance from the center P of a circle to the line x=1 be equal to r, and let P(x,y). According to the given condition, we have PC = 1 + r, that is, \\sqrt{(x+2)^{2}+y^{2}} = 1 + r. Simplifying yields y^{2} = -8x." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| -2.0, |
| 0.0 |
| ], |
| "radius": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 1, |
| "conic_type": "ellipse", |
| "image_path": "samples/test/ellipse/problem_0001.png", |
| "problem": { |
| "text": "An ellipse $\\frac{x^{2}}{k^{2}}+y^{2}=1$ $(k>0)$ has a focus at $(3 , 0)$, then $k=$?", |
| "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/k^2 = 1);k: Number;k>0;Coordinate(OneOf(Focus(G))) = (3,0)", |
| "query_expressions": "k", |
| "answer_expressions": "sqrt(10)", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.0, |
| "major_axis": "x", |
| "x_coef": 4.0, |
| "y_coef": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 1083, |
| "conic_type": "ellipse", |
| "image_path": "samples/test/ellipse/problem_1083.png", |
| "problem": { |
| "text": "The equation of a hyperbola that shares the same foci with the ellipse $x^{2}+4 y^{2}=16$ and has an asymptote given by $x+ \\sqrt {3} y=0$ is?", |
| "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H)=(x^2 + 4*y^2 = 16);Focus(H) = Focus(G);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)", |
| "query_expressions": "Expression(G)", |
| "answer_expressions": "x^2/9-y^2/3=1", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 4.0, |
| "b": 2.0, |
| "major_axis": "x", |
| "x_coef": 16.0, |
| "y_coef": 4.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 2067, |
| "conic_type": "ellipse", |
| "image_path": "samples/test/ellipse/problem_2067.png", |
| "problem": { |
| "text": "Given that the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then the area of triangle $F_{1} P F_{2}$ is?", |
| "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4", |
| "query_expressions": "Area(TriangleOf(F1, P, F2))", |
| "answer_expressions": "2*sqrt(3)", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 3.0, |
| "b": 1.4142135623730951, |
| "major_axis": "x", |
| "x_coef": 9.0, |
| "y_coef": 2.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 0, |
| "conic_type": "hyperbola", |
| "image_path": "samples/test/hyperbola/problem_0000.png", |
| "problem": { |
| "text": "If the two foci of a hyperbola are $F_{1}(-3,0)$, $F_{2}(3,0)$, and one asymptote has equation $y=\\sqrt{2} x$, then the equation of this hyperbola is?", |
| "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G)={F1,F2};Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)", |
| "query_expressions": "Expression(G)", |
| "answer_expressions": "x^2/3-y^2/6=1", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 1.732385540716972, |
| "b": 2.4497476064802375, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 2.601017381791051e-06, |
| "converged": true, |
| "iterations": 91 |
| }, |
| "coords": { |
| "F1": [ |
| -3.0, |
| 0.0 |
| ], |
| "F2": [ |
| 3.0, |
| 0.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 969, |
| "conic_type": "hyperbola", |
| "image_path": "samples/test/hyperbola/problem_0969.png", |
| "problem": { |
| "text": "Given that $P$ is an intersection point of the ellipse $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1$ $(a_{1}>b_{1}>0)$ and the hyperbola $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1$ $(a_{2}>0, b_{2}>0)$, $F_{1}$, $F_{2}$ are the common foci of the ellipse and the hyperbola, $e_{1}$, $e_{2}$ are the eccentricities of the ellipse and the hyperbola respectively, and if $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the minimum value of $e_{1} \\cdot e_{2}$ is?", |
| "fact_expressions": "G: Hyperbola;H: Ellipse;a2: Number;b2: Number;a1: Number;b1: Number;F1: Point;P: Point;F2: Point;a2>0;b2>0;Expression(G) = (-y^2/b2^2 + x^2/a2^2 = 1);a1>b1;b1>0;Expression(H) = (y^2/b1^2 + x^2/a1^2 = 1);OneOf(Intersection(H, G)) = P;Focus(G) = {F1, F2};Focus(H) = {F1, F2};Focus(G) = Focus(H);e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;AngleOf(F1, P, F2) = pi/3", |
| "query_expressions": "Min(e1*e2)", |
| "answer_expressions": "sqrt(3)/2", |
| "process": "By the symmetry of the ellipse and hyperbola, assume without loss of generality that point P lies in the first quadrant, so |PF_{1}| > |PF_{2}|. Since the ellipse and hyperbola share common foci, let the semi-focal length of the ellipse and hyperbola be c. By the definitions of the ellipse and hyperbola, we have: |PF_{1}| + |PF_{2}| = 2a_{1}, |PF_{1}| - |PF_{2}| = 2a_{2}. Solving gives |PF_{1}| = a_{1} + a_{2}, |PF_{2}| = a_{1} - a_{2}. In triangle F_{1}PF_{2}, by the law of cosines, we obtain: |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\frac{\\pi}{3}, that is, 4c^{2} = (a_{1}+a_{2})^{2} + (a_{1}-a_{2})^{2} - (a_{1}+a_{2})(a_{1}-a_{2}). Simplifying yields 4c^{2} = a_{1}^{2} + 3a_{2}^{2}, so \\frac{1}{e_{1}^{2}} + 3\\frac{1}{e_{2}^{2}} = 4. Also, \\frac{1}{e_{1}^{2}} + 3\\frac{1}{e_{2}^{2}} \\geqslant \\frac{2\\sqrt{3}}{e_{1}e_{2}}, therefore e_{1}e_{2} \\geqslant \\frac{\\sqrt{3}}{2}." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.5, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 2065, |
| "conic_type": "hyperbola", |
| "image_path": "samples/test/hyperbola/problem_2065.png", |
| "problem": { |
| "text": "Through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line to the asymptote $y=\\frac{b}{a}x$, with foot of perpendicular at $M$. This line intersects the left and right branches of the hyperbola at points $A$ and $B$ respectively. Find the range of values for the eccentricity of the hyperbola?", |
| "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;M:Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;L:Line;OneOf(Asymptote(G))=L;Expression(L)=(y=(b/a)*x);L1:Line;IsPerpendicular(L,L1);FootPoint(L,L1)=M;PointOnCurve(F, L1);Intersection(L1,LeftPart(G)) = A;Intersection(L1,RightPart(G)) = B", |
| "query_expressions": "Range(Eccentricity(G))", |
| "answer_expressions": "(sqrt(2),+oo)", |
| "process": "Since line AB intersects both the left and right branches of the hyperbola, line AB must intersect the asymptote y=-\\frac{b}{a}x in the second quadrant. Therefore, the slope of line AB must be greater than the slope of the asymptote y=-\\frac{b}{a}x, that is, -\\frac{a}{b}>-\\frac{b}{a}, which implies b^{2}>a^{2}. Since b^{2}=c^{2}-a^{2}, it follows that c^{2}>2a^{2}. The eccentricity of the hyperbola e=\\frac{c}{a}>\\sqrt{2}. Therefore, the range of values for the eccentricity of the hyperbola is (\\sqrt{2},+\\infty)." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.5, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 2, |
| "conic_type": "parabola", |
| "image_path": "samples/test/parabola/problem_0002.png", |
| "problem": { |
| "text": "Given any point $P$ on the parabola $y^{2}=4 x$, let $d$ be the distance from point $P$ to the $y$-axis. For a given point $A(4,5)$, what is the minimum value of $|P A|+d$?", |
| "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 5);PointOnCurve(P,G);Distance(P, yAxis) = d;d:Number", |
| "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))", |
| "answer_expressions": "sqrt(34)-1", |
| "process": "The directrix of the parabola is x = -1. Extending the distance from P to the y-axis one unit further to the left gives the distance from P to the directrix. According to the definition of a parabola, the distance from P to the directrix equals the distance from P to the focus F(1,0). Therefore, |PA| + d = |PA| + |PF| - 1. Since A lies outside the parabola, the minimum value occurs when A, P, and F are collinear (i.e., P is the intersection point of segment AF and the parabola). The minimum value is |AF| - 1 = \\sqrt{(9+25)} - 1 = \\sqrt{34} - 1^{n}" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 1.0, |
| "direction": "right" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "A": [ |
| 4.0, |
| 5.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 1025, |
| "conic_type": "parabola", |
| "image_path": "samples/test/parabola/problem_1025.png", |
| "problem": { |
| "text": "The focus of the parabola $y^{2}=2 x$ is $F$. If $P(2, y)$ lies on the parabola, then $|PF|=$?", |
| "fact_expressions": "G: Parabola;F: Point;P: Point;y1: Number;Expression(G) = (y^2 = 2*x);Coordinate(P) = (2, y1);Focus(G) = F;PointOnCurve(P, G)", |
| "query_expressions": "Abs(LineSegmentOf(P, F))", |
| "answer_expressions": "5/2", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 0.5, |
| "direction": "right" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "test", |
| "index": 2068, |
| "conic_type": "parabola", |
| "image_path": "samples/test/parabola/problem_2068.png", |
| "problem": { |
| "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=8 x$. A line passing through $F$ with an inclination angle of $30^{\\circ}$ intersects $C$ at points $A$ and $B$. Then $|A B|$=?", |
| "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);Inclination(G) = ApplyUnit(30, degree);A: Point;B: Point;Intersection(G, C) = {A, B}", |
| "query_expressions": "Abs(LineSegmentOf(A, B))", |
| "answer_expressions": "32", |
| "process": "From $ y^{2} = 8x $, we get $ 2p = 8 $, $ p = 4 $, then $ F(2,0) $. Therefore, the equation of the line passing through $ A $ and $ B $ is $ y = \\frac{\\sqrt{3}}{3}(x - 2) $. Solving the system \n\\[\n\\begin{cases}\ny^{2} = 8x \\\\\ny = \\frac{\\sqrt{3}}{3}(x - 2)\n\\end{cases}\n\\]\nyields $ x^{2} - 28x + 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 28 $, and since $ |AB| = x_{1} + x_{2} + p = 28 + 4 = 32 $." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 2.0, |
| "direction": "right" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 85, |
| "conic_type": "circle", |
| "image_path": "samples/train/circle/problem_0085.png", |
| "problem": { |
| "text": "The two foci of the ellipse $x^{2}+k y^{2}=1$ lie on the circle $x^{2}+y^{2}=4$. Then the real number $k$=?", |
| "fact_expressions": "G: Ellipse;k: Real;H: Circle;Expression(G) = (k*y^2 + x^2 = 1);Expression(H) = (x^2 + y^2 = 4);PointOnCurve(Focus(G), H)", |
| "query_expressions": "k", |
| "answer_expressions": "1/5", |
| "process": "Since the two foci of the ellipse $x^{2}+ky^{2}=1$ lie on the circle $x^{2}+y^{2}=4$, we have $c=2$. Because $x^{2}+\\frac{y^{2}}{\\frac{1}{k}}=1$, when $0<k<1$, $a^{2}=\\frac{1}{k}$, $b^{2}=1$, and from $a^{2}=b^{2}+c^{2}$ we obtain $k=\\frac{1}{5}$. When $k>1$, $b^{2}=\\frac{1}{k}$, $a^{2}=1$, which does not hold. Therefore, $k=\\frac{1}{5}$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 0.0, |
| 0.0 |
| ], |
| "radius": 2.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 4027, |
| "conic_type": "circle", |
| "image_path": "samples/train/circle/problem_4027.png", |
| "problem": { |
| "text": "Given the circle $(x+2)^{2}+y^{2}=64$ with center $M$, let $A$ be any point on the circle, and let $N(2,0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. What is the equation of the locus of the moving point $P$?", |
| "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x + 2)^2 = 64);M: Point;Center(G) = M;A: Point;PointOnCurve(A, G);N: Point;Coordinate(N) = (2, 0);P: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, N)), LineSegmentOf(M, A)) = P", |
| "query_expressions": "LocusEquation(P)", |
| "answer_expressions": "x^2/16+y^2/12=1", |
| "process": "According to the problem, the circle $(x+2)^{2}+y^{2}=64$ has center $M(-2,0)$, and point $N(2,0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. Therefore, $P$ lies on the perpendicular bisector of $AN$, so $PA=PN$. Also, since $|AM|=8$, the point $P$ satisfies $|PM|+|PN|=8>4$. According to the definition of an ellipse, the locus of point $P$ is an ellipse with foci $M$ and $N$, where $2a=8$, $2c=4$, giving $a=4$, $c=2$, so $b=\\sqrt{a^{2}-c^{2}}=\\sqrt{12}$. Thus, the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 2.0, |
| 0.0 |
| ], |
| "radius": 8.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 7579, |
| "conic_type": "circle", |
| "image_path": "samples/train/circle/problem_7579.png", |
| "problem": { |
| "text": "If from any point $P$ on the circle $x^{2}+y^{2}=1$, a perpendicular line segment is drawn to the $y$-axis, then the trajectory equation of the midpoint $M$ of this line segment is?", |
| "fact_expressions": "G: Circle;H: LineSegment;Expression(G) = (x^2 + y^2 = 1);PointOnCurve(P,G);IsPerpendicular(H,yAxis);M:Point;MidPoint(H)=M;P:Point", |
| "query_expressions": "LocusEquation(M)", |
| "answer_expressions": "4*x^2+y^2=1", |
| "process": "Let the coordinates of point M be (x, y) and the coordinates of point P be (x_{0}, y_{0}). Then from the given conditions, we have x = \\frac{x_{0}}{2}, y = y_{0}. Since P(x_{0}, y_{0}) lies on the circle x^{2} + y^{2} = 1, it follows that x_{0}^{2} + y_{0}^{2} = 1. Substituting x_{0} = 2x, y_{0} = y into the equation x_{0}^{2} + y_{0}^{2} = 1, we obtain 4x^{2} + y^{2} = 1. Therefore, the trajectory equation of point M is 4x^{2} + y^{2} = 1." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "center": [ |
| 0.0, |
| 0.0 |
| ], |
| "radius": 1.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 0, |
| "conic_type": "ellipse", |
| "image_path": "samples/train/ellipse/problem_0000.png", |
| "problem": { |
| "text": "The eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1$ is?", |
| "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/3 = 1)", |
| "query_expressions": "Eccentricity(G)", |
| "answer_expressions": "sqrt(3)/3", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 1.7320508075688772, |
| "b": 1.4142135623730951, |
| "major_axis": "y", |
| "x_coef": 2.0, |
| "y_coef": 3.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 3849, |
| "conic_type": "ellipse", |
| "image_path": "samples/train/ellipse/problem_3849.png", |
| "problem": { |
| "text": "Given the parabola $C_{1}$: $y=a x^{2} (a>0)$, the focus $F$ of which is also a focus of the ellipse $C_{2}$: $\\frac{y^{2}}{4}+\\frac{x^{2}}{b^{2}}=1 (b>0)$. Points $M$ and $P(\\frac{3}{2}, 1)$ are on curves $C_{1}$ and $C_{2}$ respectively. Then the minimum value of $|M P|+|M F|$ is?", |
| "fact_expressions": "C1: Parabola;Expression(C1) = (y = a*x^2);a: Number;a>0;F: Point;Focus(C1) = F;C2: Ellipse;Expression(C2) = (y^2/4 + x^2/b^2 = 1);b: Number;b>0;OneOf(Focus(C2)) = F;M: Point;PointOnCurve(M, C1);P: Point;Coordinate(P) = (3/2, 1);PointOnCurve(P, C2)", |
| "query_expressions": "Min(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, P)))", |
| "answer_expressions": "2", |
| "process": "Since point $ P\\left(\\frac{3}{2},1\\right) $ lies on the ellipse $ C_{2} $, and $ b>0 $, we have $ \\frac{1}{4}+\\frac{\\left(\\frac{3}{2}\\right)^{2}}{b^{2}}=1 \\Rightarrow b=\\sqrt{3} $, so the coordinates of focus $ F $ are $ (0,1) $. Also, from the equation of parabola $ C_{1} $, we get $ F\\left(0,\\frac{1}{4a}\\right) $, so $ \\frac{1}{4a}=1 \\Rightarrow a=\\frac{1}{4} $, then $ c_{1}: y=\\frac{1}{4}x^{2} $. By the definition of the parabola, $ |MF| $ equals the distance $ d $ from point $ M $ to its directrix $ l: y=-1 $. Draw a perpendicular line $ l': x=\\frac{3}{2} $ from point $ P $ to the directrix $ l: y=-1 $. Then the intersection point of the perpendicular line $ l': x=\\frac{3}{2} $ and the parabola $ C_{1}: y=\\frac{1}{4}x^{2} $ is the desired point $ M $. Therefore, the minimum value of $ |MP|+|MF|=|MP|+d $ is $ 1-(-1)=2 $." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 1.8537487983703613, |
| "b": 1.6118527266061986, |
| "major_axis": "x", |
| "x_coef": 4.0, |
| "y_coef": 3.0 |
| }, |
| "optimization": { |
| "final_loss": 1.4665974334810584e-06, |
| "converged": true, |
| "iterations": 90 |
| }, |
| "coords": { |
| "P": [ |
| 1.5, |
| 1.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 7753, |
| "conic_type": "ellipse", |
| "image_path": "samples/train/ellipse/problem_7753.png", |
| "problem": { |
| "text": "The eccentricity of the ellipse $\\frac{y^{2}}{3}+x^{2}=1$ is?", |
| "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/3 = 1)", |
| "query_expressions": "Eccentricity(G)", |
| "answer_expressions": "sqrt(6)/3", |
| "process": "Find the values of a, b, and c, then determine the eccentricity of the ellipse \\frac{y^{2}}{3}+x^{2}=1. In the ellipse \\frac{y^{2}}{3}+x^{2}=1, a=\\sqrt{3}, b=1, c=\\sqrt{a^{2}-b^{2}}=\\sqrt{2}. Therefore, the eccentricity of the ellipse \\frac{y^{2}}{3}+x^{2}=1 is e=\\frac{c}{a}=\\frac{\\sqrt{2}}{\\sqrt{3}}=\\frac{\\sqrt{6}}{3}" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.0, |
| "major_axis": "y", |
| "x_coef": 1.0, |
| "y_coef": 4.0 |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 1, |
| "conic_type": "hyperbola", |
| "image_path": "samples/train/hyperbola/problem_0001.png", |
| "problem": { |
| "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ is $5 x-2 y=0$, then $m=$?", |
| "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/4 - y^2/m^2 = 1);Expression(OneOf(Asymptote(G))) = (5*x - 2*y = 0)", |
| "query_expressions": "m", |
| "answer_expressions": "5", |
| "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ are given by $y=\\pm\\frac{m}{2}x$. The equation of the line $5x-2y=0$ can be rewritten as $y=\\frac{5}{2}x$, so $m=5$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 1.5, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 3822, |
| "conic_type": "hyperbola", |
| "image_path": "samples/train/hyperbola/problem_3822.png", |
| "problem": { |
| "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ to its asymptote is?", |
| "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)", |
| "query_expressions": "Distance(Focus(G), Asymptote(G))", |
| "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers). \nDo not add explanations.", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 2.0, |
| "b": 3.4641016151377544, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 7756, |
| "conic_type": "hyperbola", |
| "image_path": "samples/train/hyperbola/problem_7756.png", |
| "problem": { |
| "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the right branch of the hyperbola, and the coordinates of point $Q$ are $(-2,3)$. Then the minimum value of $|P Q|+|P F_{1}|$ is?", |
| "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Q: Point;Coordinate(Q) = (-2, 3)", |
| "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))", |
| "answer_expressions": "5+2*sqrt(3)", |
| "process": "" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "a": 1.7320508075688772, |
| "b": 1.0, |
| "orientation": "horizontal" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": { |
| "Q": [ |
| -2.0, |
| 3.0 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 4, |
| "conic_type": "parabola", |
| "image_path": "samples/train/parabola/problem_0004.png", |
| "problem": { |
| "text": "The parabola $x^{2}=a y$ passes through the point $A(1, \\frac{1}{4})$, then the distance from point $A$ to the focus of this parabola is?", |
| "fact_expressions": "G: Parabola;Expression(G) = (x^2 = a*y);a: Number;A: Point;Coordinate(A) = (1, 1/4);PointOnCurve(A, G)", |
| "query_expressions": "Distance(A, Focus(G))", |
| "answer_expressions": "5/4", |
| "process": "\\because the parabola x^{2}=ay passes through point A(1,\\frac{1}{4}), \\therefore 1^{2}=a\\times\\frac{1}{4}, solving gives a=4. Therefore, the equation of the parabola is x^{2}=4y, yielding its focus at F(0,1) and directrix equation y=-1. \\because the distance from any point on the parabola to the focus equals the distance from that point to the directrix of the parabola, \\therefore the distance from point A to the focus of this parabola is y_{4}-(-1)=\\frac{1}{4}+1=\\frac{5}{4}" |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 0.9959803819656372, |
| "direction": "up" |
| }, |
| "optimization": { |
| "final_loss": 3.6780886603082763e-07, |
| "converged": true, |
| "iterations": 56 |
| }, |
| "coords": { |
| "A": [ |
| 1.0, |
| 0.25 |
| ] |
| } |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 3812, |
| "conic_type": "parabola", |
| "image_path": "samples/train/parabola/problem_3812.png", |
| "problem": { |
| "text": "A point $M$ on the parabola $y=4 x^{2}$ is at a distance of $1$ from the focus. What is the ordinate of point $M$?", |
| "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 1", |
| "query_expressions": "YCoordinate(M)", |
| "answer_expressions": "15/16", |
| "process": "From $ y = 4x^{2} $, we obtain $ x^{2} = \\frac{1}{4}y $, so the focus of the parabola is $ F(0, \\frac{1}{16}) $, and the equation of the directrix is $ y = -\\frac{1}{16} $. Let $ M(x_{M}, y_{M}) $. By the definition of the parabola, we have $ MF = y_{M} + \\frac{1}{16} = 1 $, so $ y_{M} = \\frac{15}{16} $." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 0.0625, |
| "direction": "up" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| }, |
| { |
| "dataset": "train", |
| "index": 7755, |
| "conic_type": "parabola", |
| "image_path": "samples/train/parabola/problem_7755.png", |
| "problem": { |
| "text": "If the distance from point $P$ on the parabola $x^{2}=8y$ to the focus is $12$, then what is the distance from $P$ to the $x$-axis?", |
| "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = 8*y);PointOnCurve(P, G);Distance(P, Focus(G)) = 12", |
| "query_expressions": "Distance(P, xAxis)", |
| "answer_expressions": "10", |
| "process": "Since the parabola is $x^{2}=8y$, the focus coordinates are $(0,2)$, and the directrix equation is $y=-2$. Because the distance from point $P$ to the focus is $12$, according to the definition of a parabola, the distance from $P$ to the directrix is also $12$. Therefore, the distance from point $P$ to the $x$-axis is $10$." |
| }, |
| "sdf_annotation": { |
| "params": { |
| "p": 2.0, |
| "direction": "up" |
| }, |
| "optimization": { |
| "final_loss": 0.0, |
| "converged": true, |
| "note": "using explicit params" |
| }, |
| "coords": {} |
| } |
| } |
| ] |
