A VLM answers every question at temperature 0 with reasoning enabled. Its answer is compared with the official ground truth by the verifier described below; matches go to GeomVerse_RS_think and GeomVerse_RS_nothink, non-matches go to GeomVerse_rejected with the ground truth preserved.

The question field is byte-identical to the prompt the model actually received, including any appended instruction, so the stored training text and the inference-time input cannot drift apart.

How a match is decided

Every rule below is stated with its exact constants and a runnable reference implementation, so any accept/reject decision in this dataset can be recomputed. All snippets assume:

def levenshtein(a, b):
    if a == b: return 0
    if not a: return len(b)
    if not b: return len(a)
    prev = list(range(len(b) + 1))
    for i, ca in enumerate(a, 1):
        cur = [i] + [0] * len(b)
        for j, cb in enumerate(b, 1):
            cur[j] = min(prev[j] + 1, cur[j - 1] + 1, prev[j - 1] + (ca != cb))
        prev = cur
    return prev[len(b)]

`anls` — Average Normalized Levenshtein Similarity

The official DocVQA/InfoVQA metric. Both strings are lowercased and stripped, the normalized Levenshtein distance NL = lev(pred, gt) / max(len(pred), len(gt)) is computed, and the score is 1 - NL when NL < 0.5, otherwise 0. A pair is accepted when the score is above zero and both guards below pass.

def anls_pair(pred, gt, tau=0.5):
    p, q = pred.lower().strip(), gt.lower().strip()
    if not p and not q: return 1.0
    m = max(len(p), len(q))
    if m == 0: return 0.0
    nl = levenshtein(p, q) / m
    return (1.0 - nl) if nl < tau else 0.0

Multiple choice. When the question enumerates its own lettered options, the only correct target is the letter, so a single-letter reference is also matched against the first standalone letter in the prediction — C, (C), The answer is C and **Answer: C** all match C.

Two guards that override a positive score

Numeric guard. ANLS was designed for OCR text and is far too permissive on numbers: 99 degrees vs 80 degrees differs by two characters out of ten and scores 0.8. When both sides state numbers and those numbers disagree, the pair is rejected regardless of string similarity. Thousand separators are removed first, so 1,329.87 and 1329.87 are the same number.

_DIGITS = re.compile(r"\d+(?:\.\d+)?")
_THOUS  = re.compile(r"(?<=\d),(?=\d\d\d\b)")

def num_conflict(a, b):
    a, b = _THOUS.sub("", a or ""), _THOUS.sub("", b or "")
    na, nb = _DIGITS.findall(a), _DIGITS.findall(b)
    if not na or not nb:
        return False
    fa, fb = [float(x) for x in na], [float(x) for x in nb]
    if len(fa) == 1 and len(fb) == 1:
        return abs(fa[0] - fb[0]) > 1e-6 * max(1.0, abs(fb[0]))
    return set(na) != set(nb)

Edit budget. ANLS's fixed tau=0.5 lets a 7-character answer differ by 3 characters, which accepts SBOXCAR against a reference of boxcar at a score of 0.857. That threshold is meant for evaluating a model, where OCR noise is expected; selecting TRAINING data needs precision, because an accepted row is shipped as ground truth. The tolerated number of character edits scales with length, and is measured after stripping case, punctuation, whitespace and a leading article — otherwise stop. vs stop would burn the whole budget on a period.

reference length	edits allowed
≤ 8	0 (must be exact)
9–16	1
17–32	2
> 32	length / 12

_ART = re.compile(r"^(?:a|an|the)\s+")

def budget_norm(x):
    return re.sub(r"[^a-z0-9]", "", _ART.sub("", (x or "").strip().lower()))

def edit_budget(gt):
    n = len(gt or "")
    return 0 if n <= 8 else 1 if n <= 16 else 2 if n <= 32 else max(3, n // 12)

def within_budget(pred, gt):
    p, g = budget_norm(pred), budget_norm(gt)
    return p == g or levenshtein(p, g) <= edit_budget(g)

Shared pre-processing (applies to every verifier below)

Both sides are expanded into a small set of candidate strings and the best-scoring pair decides. The prediction contributes: the raw string; its first line; its first paragraph; the text after the last Answer: marker; the span between Answer: and Rationale:; the stated final answer of a worked solution (recognised from \boxed{…}, Answer:, the final answer is …, the answer is …, so the answer is …, 故选X); and a markdown-stripped variant. The reference contributes the same forms.

When the reference is a wrapper around the real answer (The answer is Red), the raw string is replaced by the extracted answer rather than kept beside it: the wrapper text is shared by every wrong answer too, so The answer is Blue would otherwise score 0.78 against it.

import re

def final_answer(gt):
    """Stated answer of a worked solution, or None."""
    pats = [r"\\boxed\{([^{}]{1,80})\}",
            r"(?is)\banswer\s*[:：]\s*(.{1,80}?)\s*$",
            r"(?is)\b(?:the\s+)?final\s+answer\s+is\s*[:：]?\s*(.{1,80}?)\s*[.。]?\s*$",
            r"(?is)\b(?:the\s+)?answer\s+is\s*[:：]?\s*(.{1,80}?)\s*[.。]?\s*$",
            r"(?s)故\s*选\s*[:：]?\s*\(?([A-H])\)?"]
    for p in pats:
        ms = list(re.finditer(p, (gt or "").strip()))
        if ms:
            v = re.sub(r"\s+", " ", ms[-1].group(1).strip().strip("$ \t.。，,;:"))
            if v and len(v) <= 80:
                return v
    return None

Second pass: model-as-judge

String verifiers cannot judge free-form answers at all, and they under-credit answers that are correct but differently phrased. Every item the verifier rejected is therefore re-examined by the same VLM in a text-only pass that sees the question, the reference and the candidate — no image, no re-generation, it only grades text that was already produced.

The judge is instructed to compare against the reference and never to solve the question itself, so an answer is not promoted merely because the model believes it is true. It answers CORRECT, WRONG, or UNSURE; numbers must agree, and UNSURE is reserved for references that are themselves unusable or ambiguous. UNSURE is never counted as correct — only a CORRECT verdict moves a row into the accepted splits. WRONG and UNSURE both remain in the rejected split, so an unresolved case is never silently promoted to a pass.

Calibration: run over items the string verifier had already accepted, the judge agreed on 59/60. Spot-checking its overturns shows it discriminates rather than rubber-stamps — it keeps \frac{L}{n} rejected against a reference of \frac{L}{n-1}, and returns UNSURE when the model argues no finite answer exists against a numeric reference.

What `answer` holds

In the accepted splits answer is the model's own verified output — the text that passed the verifier — so it agrees with the reasoning in think. The original reference annotation is kept beside it as org_answer. Train on answer; use org_answer when you need the source label.

The two differ more often than one would guess: the verifier accepts on meaning, not on characters, so casing, spacing and formatting routinely differ (bridgestone vs Bridgestone, ( n x m ) vs (n x m)). Storing the reference in the answer slot would pair a model's reasoning with someone else's wording.

In the rejected split the roles are different: answer is the official ground truth and wrong_vlm is what the model said instead.

Schema

image — HF Image() — renders directly in the dataset viewer
qa — list of {question, answer, org_answer, think}
clean_meta — per-row provenance and the cleaning policy that admitted it
image_sha256 — content hash of the image

Examples

Three rows taken straight from this split. Images are the original files as stored in the dataset, and long fields are shown in full.

Example 1

field	value
`question`	If the yellow shape is a rectangle where a semi-circle has been removed from one side of it and the perimeter of the yellow shape is 74, compute the perimeter of the blue parallelogram. Assume $\pi=3.14$. Round computations to 2 decimal places. End your solution with the final answer inside \boxed{}.
`answer`	Setting up the yellow shape: The rectangle has sides $a$ and $10$. The semi-circle is cut from the side of length 10, so that side is replaced by the semi-circular arc. - Arc length = $\frac{\pi \cdot d}{2} = \frac{3.14 \cdot 10}{2} = 15.7$ - Perimeter = $a + a + 10 + 15.7 = 2a + 25.7$ Solving for a: $$2a + 25.7 = 74 \implies 2a = 48.3 \implies a = 24.15$$ Blue parallelogram: Its sides are 12 and $a = 24.15$. Perimeter = $2 \cdot (12 + 24.15) = 2 \cdot 36.15 = 72.3$ \boxed{72.3}
`org_answer`	The diameter of the semi-circle in the yellow shape is equal to the side of the rectangle with length 10 so the shape has two sides with equal but unknown lengths, one side with length 10, and one semi-circle arc with diameter 10. So the perimeter is $2 * UnknownSide + 10 + \frac{10 * \pi}{2}$. So $2 * UnknownSide + 10 + \frac{10 * 3.14}{2} = 74$. So $2 * UnknownSide = 74 - 10 - \frac{10 * 3.14}{2} = 74 - 10 - \frac{31.4}{2} = 74 - 10 - 15.7 = 48.3$. Therefore, the length of the side marked with "$a$" is $\frac{48.3}{2} = 24.15$. The lengths of the two sides of the blue parallelogram are 24.15 and 12, so the perimeter of the blue parallelogram is $2 * (24.15 + 12) = 2 * 36.15 = 72.3$. Therefore the final answer is 72.3.

think (5001 chars)

Let me try understand the figure. There's a yellow rectangle with a semicircle removed from one side. The rectangle has sides: one side is labeled 10 (the top-right side), and the side shared with the blue parallelogram is labeled a.

The yellow shape: a rectangle with a semicircle removed from one side. Looking at the figure, the semicircle is removed from the bottom side (the side parallel to the side of length 10? or the side of length a?).

Let me set up the rectangle's vertices. The rectangle's vertices are approximately: top vertex (1355, 735) [the corner with the right-angle mark], right vertex (2245, 1030) [right-angle mark], bottom vertex (around 1000, 2280?), left vertex (60, 1975). Hmm.

Actually, the rectangle's sides: from top vertex (1355,735) to right vertex (2245,1030) is the side labeled 10. From right vertex (2245,1030) to bottom vertex (around 1000, 2280) is the long side. From bottom vertex to left vertex (60,1975) is the side where the semicircle is removed (the bottom side). From left vertex (60,1975) to top vertex (1355,735) is the side labeled a (shared with the parallelogram).

So the rectangle's sides: a (left side, shared with parallelogram) and 10 (top side). The semicircle is removed from the bottom side, which is parallel to the top side of length 10? Wait, the bottom side is parallel to the top side (length 10). Hmm, but the semicircle in the figure looks large, with a diameter spanning the bottom side.

Let me compute lengths: top side from (1355,735) to (2245,1030): length = sqrt(890^2 + 295^2) = sqrt(792100 + 87025) = sqrt(879125) ≈ 937.6. Labeled 10.

Left side from (60,1975) to (1355,735): length = sqrt(1295^2 + 1240^2) = sqrt(1677025 + 1537600) = sqrt(3214625) ≈ 1792.9. This is a.

Bottom side from (60,1975) to bottom vertex (around 1000, 2280): hmm, the bottom vertex is where the semicircle ends. Actually, the semicircle is removed from the bottom side, so the bottom side's endpoints are the left vertex (60,1975) and the bottom-right vertex. The bottom-right vertex should be left vertex + (right - top) vector = (60+890, 1975+295) = (950, 2270). Yes, matches the figure (the bottom tip is around (1000, 2280)).

So the bottom side length equals the top side = 10 (in actual units). The semicircle's diameter is the bottom side = 10? But in the figure the semicircle looks large relative to the side... The semicircle's diameter spans the entire bottom side. In the figure, the bottom side length in pixels is sqrt(890^2+295^2) ≈ 937.6, and the semicircle's diameter looks about that. The semicircle's radius in the figure is about 470 pixels, which is half of 937. Yes, the semicircle's diameter equals the bottom side.

So the yellow shape's perimeter: the rectangle's perimeter minus the bottom side (since it's removed) plus the semicircle's arc. Wait, the semicircle is removed from the bottom side, so the boundary consists of: left side (a), top side (10), right side (a), and the semicircle's arc (diameter = bottom side = 10). The bottom side itself is not part of the boundary (replaced by the arc).

Perimeter = a + 10 + a + (π * d / 2) = 2a + 10 + (3.14 * 10 / 2) = 2a + 10 + 15.7 = 2a + 25.7.

Set equal to 74: 2a = 48.3, a = 24.15.

Hmm, but wait: the side labeled 10 is the top side; the bottom side is also 10. The semicircle's diameter = 10, arc length = π*10/2 = 15.7.

So perimeter = a + a + 10 + 15.7 = 2a + 25.7 = 74 → 2a = 48.3 → a = 24.15.

Now the blue parallelogram: sides are 12 and a? The parallelogram's sides: left side labeled 12 (vertical side from (60,1265) to (60,1975)), and the side shared with the rectangle is a. The parallelogram's perimeter = 2*(12 + a) = 2*(12 + 24.15) = 2*36.15 = 72.3.

So the answer is 72.3.

But let me double-check the rounding: a = 24.15, perimeter = 2*(12+24.15) = 72.3.

Hmm, but maybe the semicircle's diameter is the side of length a? Let me look at the figure: the semicircle is at the bottom, and its diameter appears to be along the bottom side (the side from left vertex to bottom vertex), which is parallel to the side labeled 10. In the figure, the bottom side's pixel length ≈ 937.6, same as the top side. The semicircle's diameter spans from (60,1975) to (950,2270)? The drawn semicircle: the arc goes from around (60,1975) to around (1000,2280), bulging inward. Yes, the diameter is the bottom side.

Thus the yellow perimeter = a (left) + 10 (top) + a (right) + semicircle arc (diameter 10) = 2a + 10 + 15.7.

Wait, but the right side length equals the left side = a. Yes, rectangle.

So 2a + 25.7 = 74 → a = 24.15.

Parallelogram perimeter = 2*(12 + a) = 2*36.15 = 72.3.

Rounded to 2 decimal places: 72.3.

Thus the final answer \boxed{72.3}.

Let me double-check the arc: semicircle with diameter 10: arc length = π r = 3.14 * 5 = 15.7. Yes.

So the answer is 72.3.

But wait: the problem says "round the computation to 2 decimal places." a = (74 - 10 - 15.7)/2 = 48.3/2 = 24.15. Parallelogram perimeter = 2*(12+24.15) = 72.3.

Thus the final: 72.3.

Example 2

field	value
`question`	If the area of the ABCD parallelogram is 108, the ADEF shape is a combination of a rectangle and a semi-circle, the perimeter of the ADEF shape is 42, the ABHIJ shape is a rectangle where an equilateral triangle has been removed from one side of it and the area of the ABHIJ shape is 120, compute the degree of the BAD angle. Assume $\pi=3.14$. Round computations to 2 decimal places. End your solution with the final answer inside \boxed{}.
`answer`	Step 1 – Find AD (from shape ADEF): The semicircle has diameter 6, so its arc length = π·r = 3.14·3 = 9.42. Perimeter = top + two sides + arc: 6 + 2·AD + 9.42 = 42 → 2·AD = 26.58 → AD = 13.29. Step 2 – Find AB (from shape ABHIJ): The rectangle has side BH = 11, so the opposite side (where the triangle was cut) is also 11. Equilateral triangle area = (√3/4)·11² ≈ 0.433·121 = 52.39. Rectangle area = 120 + 52.39 = 172.39 → AB = 172.39 / 11 = 15.67. Step 3 – Find angle BAD (from parallelogram ABCD): Area = AB · AD · sin(BAD) → 108 = 15.67 · 13.29 · sin(BAD) sin(BAD) = 108 / 208.25 ≈ 0.52 BAD = arcsin(0.52) ≈ 31.33° \boxed{31.33}
`org_answer`	The perimeter of the ADEF shape is 42 and the length of the DE side is 6, so $2 * OtherSide + 6 + \frac{6 * 3.14}{2} = 42$. So $2 * OtherSide = 42 - 6 - \frac{6 * 3.14}{2} = 42 - 6 - \frac{18.84}{2} = 42 - 6 - 9.42 = 26.58$. Therefore, the length of the AD side is $\frac{26.58}{2} = 13.29$. The area of the ABHIJ shape is 120 and the length of the BH side is 11, so $OtherSide * 11 - \frac{\sqrt{3}}{4} * 11^2 = 120$, so $OtherSide * 11 = 120 + \frac{\sqrt{3}}{4} * 11^2 = 120 + \frac{1.73}{4} * 121 = 120 + 0.43 * 121 = 120 + 52.03 = 172.03$. Therefore, the length of the AB side is $\frac{172.03}{11} = 15.64$. The lengths of the AB and the AD sides of the ABCD parallelogram are 15.64 and 13.29 and the area is 108 so the sine of the BAD angle is $\frac{108}{15.64 * 13.29} = 0.52$ and so the angle in degrees is $\arcsin(0.52) = 31.33$. Therefore the final answer is 31.33.

think (13156 chars)

Let me work through this problem carefully.

We have parallelogram ABCD with area 108. The shape ADEF is a combination of a rectangle and a semicircle. The perimeter of ADEF is 42. The shape ABHIJ is a rectangle with an equilateral triangle removed from one side, and its area is 120.

We need to find the angle BAD.

Let me set up the figure. Points: E is top-left, D is top-right of the pink shape, F is bottom-left, A is bottom-right. The pink shape ADEF: rectangle with a semicircle on the bottom (FA side? Actually the semicircle is on the bottom, between F and A). The top edge ED = 6. So the rectangle has width ED = 6 (horizontal), and vertical sides EF and DA. The semicircle is on the bottom, with diameter FA = 6 (same as the top width). Wait, the rectangle's width is 6, so FA = 6, and the semicircle's diameter is FA = 6.

Perimeter of ADEF: the shape consists of the rectangle's top edge ED, the two vertical sides EF and DA, and the semicircular arc on the bottom (diameter FA). The perimeter includes ED (6), EF, DA, and the semicircle arc (π * r = π * 3 = 9.42). The bottom edge FA is not part of the perimeter because it's replaced by the semicircle.

So perimeter = ED + EF + DA + arc = 6 + 2*h + 9.42 = 42, where h = EF = DA.

Thus 2h = 42 - 6 - 9.42 = 26.58, h = 13.29.

So AD = 13.29.

Now parallelogram ABCD: vertices A, B, C, D. Sides: AB, BC, CD, DA. Area = base * height. We know AD = 13.29. Hmm, but we need the angle BAD.

In parallelogram ABCD, the sides are AB and AD (adjacent sides at vertex A). Area = AB * AD * sin(angle DAB).

We need AB. From the green shape ABHIJ: a rectangle with an equilateral triangle removed from one side. The rectangle is probably ABHI? Let's see: the shape's vertices are A, B, H, I, J. The rectangle is A B H ... hmm.

Looking at the figure: B is top, H is right, I is bottom, J is the inner notch. The rectangle is probably A-B-H-? Actually, the rectangle's vertices are A, B, H, and some point, with the equilateral triangle removed from side ... The shape ABHIJ: rectangle ABH? Let me think: rectangle with vertices A, B, H, X, where X is the fourth corner; the equilateral triangle is removed from side ... The notch at J suggests the triangle removed is A-J-I? Hmm.

Actually, typical problem: "ABHIJ is a rectangle with an equilateral triangle removed from one side." The rectangle is probably ABH? Let me think: the rectangle's sides: AB and BH (length 11 given on BH). The rectangle is A B H I? But then the triangle removed is from side ... The shape's boundary: A to B, B to H, H to I, I to J, J to A. So the rectangle is A B H I? Wait, rectangle A-B-H-I would have sides AB, BH, HI, IA. But the shape's boundary goes H to I directly? In the figure, H connects to I (the long edge on the right side going down to I). And I connects to J, J connects to A. So the rectangle is A B H I? Hmm, but then the removed triangle is A I J? That is, the rectangle's side IA is replaced by two sides IJ and JA, forming an inward notch (equilateral triangle removed from side IA). So the rectangle is ABHI with sides AB and BH=11, and the equilateral triangle removed is on side IA (which equals AB). Wait, rectangle ABHI: vertices in order A, B, H, I. Sides: AB, BH, HI, IA. Opposite sides: AB = HI, BH = IA. So IA = BH = 11. The equilateral triangle removed from side IA has side 11, area = (sqrt(3)/4)*121 ≈ 0.433*121 = 52.39.

Area of shape = rectangle area - triangle area = AB * 11 - 52.39 = 120. So AB * 11 = 172.39, AB = 15.67.

Hmm, but wait: the right angle marks are at B and H. At B, the angle between BC? Actually, the right angle at B is between AB and BH? The gray square at B is between the orange parallelogram's side and the green shape. The right angle at H is between BH and HI. So the rectangle's corners are at B and H, consistent with rectangle ABHI having right angles at B and H.

So AB = (120 + 52.39)/11 = 172.39/11 = 15.67.

Now parallelogram ABCD: sides AB = 15.67 and AD = 13.29. Area = AB * AD * sin(angle BAD) = 108.

sin(angle BAD) = 108 / (15.67 * 13.29) = 108 / 208.25 = 0.5186.

Angle = arcsin(0.5186) ≈ 31.24 degrees.

Let me compute more precisely with rounding to 2 decimals at each step as instructed.

Step 1: semicircle arc: diameter = 6, radius 3, arc length = π * r = 3.14 * 3 = 9.42.

Perimeter: 6 + 2*AD + 9.42 = 42 => 2*AD = 26.58 => AD = 13.29.

Step 2: equilateral triangle side = 11 (since BH = 11, and the removed triangle is on the opposite side IA which equals BH = 11). Area = (sqrt(3)/4) * 11^2. sqrt(3) ≈ 1.732. 0.433 * 121 = 52.393 ≈ 52.39.

Rectangle area = 120 + 52.39 = 172.39. AB = 172.39 / 11 = 15.6718... ≈ 15.67.

Step 3: sin(BAD) = 108 / (15.67 * 13.29). 15.67*13.29 = 208.2543 ≈ 208.25. 108/208.25 = 0.518607... ≈ 0.52.

arcsin(0.52) ≈ 31.33 degrees? Let me compute: sin 31° = 0.5150, sin 32° = 0.5299. Interpolate: 0.52 is (0.52-0.5150)/(0.5299-0.5150)=0.005/0.0149=0.3356 => 31.34°.

But if we use 0.5186: sin 31.2°? sin 31° = 0.51504, sin 31.5° = 0.52250. 0.5186 is (0.5186-0.51504)/(0.52250-0.51504)=0.00356/0.00746=0.477 => 31.24°.

Hmm, rounding: the problem says round computations to 2 decimal places. So sin = 0.52, angle = arcsin(0.52) = 31.33°? Let me compute arcsin(0.52) precisely: sin(31.33°)? Let me compute: 31.33° in radians = 0.54683. sin = 0.51999? Let me compute: sin(31°) = 0.515038, cos(31°)=0.857167. Derivative per degree: cos * π/180 = 0.857167*0.0174533 = 0.014960 per degree. Need 0.52 - 0.515038 = 0.004962. 0.004962/0.014960 = 0.3317. So angle ≈ 31.33°.

If we use 0.5186: 0.5186-0.515038=0.003562; /0.014960=0.2381 => 31.24°.

Which is expected? The problem probably expects a specific numeric answer. These auto-generated problems (from a dataset) usually compute with rounding at each step. Let me think about the typical solution text: "The perimeter of ADEF is 42 and one side is 6, so the other side is (42 - 6 - 3.14*3)/2 = 13.29." Then "the area of the equilateral triangle with side 11 is 11*11*sqrt(3)/4 = 52.39." Then "the area of the rectangle is 120+52.39=172.39, so AB = 172.39/11 = 15.67." Then "the area of the parallelogram is AB*AD*sin(BAD) => sin = 108/(15.67*13.29) = 0.52, angle = 31.33."

Hmm, but maybe they compute sin = 108/208.25 = 0.5186 and round to 0.52, then arcsin(0.52) = 31.33. Or they might keep 0.5186 and get 31.24.

Let me compute arcsin(0.5186) more precisely: using more precise values: sin(31.24°)? 31.24° rad = 0.545235. sin = ?

Let me compute sin(31.24°): sin(31°)=0.5150381, cos(31°)=0.8571673. Δ=0.24° = 0.00418879 rad. sin ≈ 0.5150381*cos(0.00418879) + 0.8571673*sin(0.00418879) ≈ 0.5150381*0.9999912 + 0.8571673*0.00418878 ≈ 0.5150336 + 0.0035905 = 0.5186241. So sin(31.24°) ≈ 0.51862, matches 0.5186. So arcsin(0.5186) ≈ 31.24°.

So the answer is either 31.24 or 31.33 depending on rounding.

The instruction "round computations to 2 decimal places" suggests rounding intermediate results to 2 decimals. The sin value 0.5186 rounds to 0.52. Then arcsin(0.52) = 31.33.

But many of these dataset solutions do: "sin(BAD) = 108 / (15.67 * 13.29) = 0.52. Therefore the BAD angle is 31.33 degrees." Hmm.

Let me check the product: 15.67 * 13.29 = 15.67*13 = 203.71, plus 15.67*0.29 = 4.5443, total 208.2543. 108/208.2543 = 0.51860. Rounded to 2 decimals: 0.52.

arcsin(0.52): let me compute precisely. sin(31.33°)? 31.33° = 31° + 0.33°. 0.33° = 0.00575959 rad. sin ≈ 0.5150381*0.9999833 + 0.8571673*0.00575956 ≈ 0.5150295 + 0.0049369 = 0.5199664. Slightly less than 0.52. Need 0.52: difference 0.0000336; per degree 0.01496 => 0.00225°. So arcsin(0.52) ≈ 31.332°. So 31.33.

Thus the final answer is probably 31.33.

But maybe the dataset's answer is 31.24? Let me think about how these problems are generated. This looks like a "geometry" dataset with synthetic problems (probably from "MathVision" or similar). The phrasing "round computations to 2 decimal places" appears in many of these generated problems, and the solutions typically round each intermediate to 2 decimals. The final answer is computed from the rounded intermediates.

Let me reconstruct the likely solution:

- Semicircle diameter = 6 => arc = 3.14*6/2 = 9.42.
- AD = (42 - 6 - 9.42)/2 = 13.29.
- Equilateral triangle side = 11 => area = 11*11*0.4330127/4? They might use sqrt(3)/4 ≈ 0.433. 121*0.433 = 52.393 ≈ 52.39.
- Rectangle area = 120 + 52.39 = 172.39; AB = 172.39/11 = 15.6718 ≈ 15.67.
- Parallelogram: area = AB * AD * sin(BAD) => sin = 108/(15.67*13.29) = 108/208.25 = 0.5186 ≈ 0.52.
- BAD = arcsin(0.52) = 31.33°.

I think the expected answer is 31.33.

But wait: maybe they compute the parallelogram area as base*height with base AD and height = AB*sin? Same thing.

Alternatively, maybe they intend angle BAD such that area = AB*AD*sin(angle). Yes.

Let me double-check the orientation: In parallelogram ABCD, the vertices in order A, B, C, D. Sides AB, BC, CD, DA. Angle at A is between AB and AD. Yes.

So sin(BAD) = area/(AB*AD).

Thus answer ≈ 31.33 (or 31.24). Need to decide.

Given the instruction to round computations to 2 decimals, I'd present sin = 0.52 and angle = 31.33.

Hmm, but maybe they don't round the sin before taking arcsin; they might compute arcsin(0.5186) = 31.24 and round to 2 decimals. The phrase "round computations to 2 decimal places" is ambiguous.

Let me look at similar problems from this dataset: I recall problems like "the area of the parallelogram is X, compute the angle" with solutions like "sin = ... = 0.5, angle = 30". Usually they compute sin to 2 decimals and then take arcsin.

I think the expected final answer is 31.33.

Let me double-check the perimeter of ADEF: shape ADEF: rectangle A D E F? Order: A (bottom-right), D (top-right), E (top-left), F (bottom-left). Rectangle with top ED=6, sides EF and DA vertical, bottom FA. Semicircle attached on bottom (diameter FA). Perimeter = ED + DE? Let me list boundary: from E to D (top, 6), D to A (right side, h), A to F via semicircle arc (9.42), F to E (left side, h). Total = 6 + 2h + 9.42 = 42 => h = 13.29. Yes.

So AD = 13.29.

Now AB: rectangle ABHI? Wait, the shape is ABHIJ: order A, B, H, I, J. Rectangle corners: A, B, H, and the fourth corner would be I? But the notch J is inside. The rectangle is A B H I with side IA removed and replaced by I-J-A (equilateral triangle A I J removed). For the triangle to be equilateral with side IA = 11, we need AJ = JI = 11. In the figure, J is inside, plausible.

Right angles at B (between AB and BH) and H (between BH and HI). Good.

So rectangle area = AB * BH = AB * 11. Triangle area = (sqrt3/4)*11^2 ≈ 52.39. Shape area = 11*AB - 52.39 = 120 => AB = 172.39/11 = 15.6718 ≈ 15.67.

Now sin(BAD) = 108/(15.67*13.29) = 0.5186.

Angle = 31.24° (if using 0.5186) or 31.33° (if using 0.52).

Hmm.

Let me think about the source: this is probably from the "GeoQA" or "SynthGeo" dataset? The style with colored shapes and labels and "round computations to 2 decimal places" matches a known dataset of generated geometry problems (I've seen similar: "If the area of the ABCD parallelogram is ..., compute the degree of the BAD angle"). I recall the solutions in that dataset do step-by-step with rounding to 2 decimals at each step, and the final answer is like "31.33".

Actually, I recall a similar problem: "the area of the parallelogram is 100, ... compute the angle" with solution "sin = 0.5, angle = 30". They compute sin as a rounded 2-decimal number and then arcsin.

Given the ambiguity, maybe I should compute arcsin(0.52) = 31.33 and present that.

But let me double-check arcsin(0.52) in degrees: sin 31.33° ≈ 0.51997, sin 31.34° ≈ 0.52012? Let me compute: per degree 0.01496. 31.33 gave 0.519966; add 0.000034 to reach 0.52 => 0.0023° => 31.3323°. So 31.33.

Thus final: \boxed{31.33}.

Alternatively, if they keep more precision: 31.24.

Let me see which is more standard: In many of these generated solutions I've seen, they write e.g. "sin(BAD) = 108 / (15.67 * 13.29) = 0.52. Therefore, the BAD angle is 31.33 degrees." I think that's the pattern: they round sin to 2 decimals then take arcsin.

Actually, I've seen solutions like: "the sine of the angle is 0.52, so the angle is 31.33 degrees." Yes.

I'll go with 31.33.

Let me double-check the triangle area: sqrt(3)/4 = 0.4330127019. *121 = 52.39453693 ≈ 52.39. Good.

120 + 52.39 = 172.39. /11 = 15.671818... ≈ 15.67.

Product 15.67*13.29: 15.67*13.29 = 15.67*13 + 15.67*0.29 = 203.71 + 4.5443 = 208.2543 ≈ 208.25.

108/208.25 = 0.518607... ≈ 0.52.

arcsin(0.52) = 31.33°.

Thus final answer 31.33.

I'll present the solution steps.

One more sanity check: could the removed equilateral triangle be on side AB instead? The notch J is between I and A, and the triangle removed is from side IA? The shape's boundary includes J as a reflex vertex. The rectangle's side that is "removed" is the one not present in the boundary: the boundary has A-B, B-H, H-I, I-J, J-A. The rectangle's fourth side would be I-A, which is replaced by I-J and J-A. So the triangle removed is A-I-J with side IA. IA = BH = 11 (opposite sides of rectangle). Yes.

Thus AB = 15.67.

All consistent.

Therefore angle BAD ≈ 31.33 degrees.

I'll write the final answer as \boxed{31.33}.

Example 3

field	value
`question`	If the length of the BC side is 10 and the arc length of the ABC sector is 10.28, compute the degree of the CBA angle. Assume $\pi=3.14$. Round computations to 2 decimal places. End your solution with the final answer inside \boxed{}.
`answer`	The sector is centered at B, so BC is the radius: r = 10. The arc from C to A has length 10.28. The angle at the center relates to the arc by: angle = 360 · (arc length / (2·π·r)) Compute the circumference factor: 2·π·r = 2 · 3.14 · 10 = 62.8 Compute the fraction (rounded to 2 dp): 10.28 / 62.8 = 0.16 Compute the angle: 360 · 0.16 = 57.6 Therefore the CBA angle is 57.6 degrees. \boxed{57.6}
`org_answer`	The BC radius of the ABC sector is 10 and the arc length is 10.28. So the CBA angle can be computed as $\frac{ArcLength}{2 \pi r} * 360 = \frac{10.28}{2 \pi * 10} * 360 = \frac{10.28}{62.8} * 360 = 0.16 * 360 = 57.6$. Therefore the final answer is 57.6.

think (9991 chars)

The user wants me to find the angle CBA of a sector.

Let me look at the image. There's a sector with center at B, and the arc goes from C to A. The sides BC and BA are radii. The arc is CA (the arc of the sector ABC).

Wait, the problem says "arc length of the ABC sector is 10.28". The sector is named ABC, with center at B (since the angle we want is CBA, the vertex is B). So the radii are BC and BA, and the arc is from C to A.

Given BC = 10, so radius r = 10.

Arc length formula: arc length = (angle/360) * 2 * pi * r.

So 10.28 = (angle/360) * 2 * 3.14 * 10 = (angle/360) * 62.8.

Thus angle/360 = 10.28 / 62.8 = 0.16369... Let me compute: 10.28 / 62.8 = 0.163694...

Round to 2 decimal places? The problem says "Round computations to 2 decimal places." Hmm. Let's think. 10.28 / 62.8 = 0.1636942675. Rounded to 2 decimal places = 0.16.

Then angle = 0.16 * 360 = 57.6.

Hmm but maybe they want us to round intermediate computations to 2 decimal places. Let's see typical problems of this type. Usually they compute the fraction then multiply.

Let me compute more precisely: 10.28 / 62.8 = 0.16369426751592357. If we round to 2 decimals, 0.16. Then 0.16 * 360 = 57.6.

Alternatively, maybe they intend the arc length formula as arc = (theta/360)*2*pi*r, and they expect theta = arc * 360 / (2*pi*r) = 10.28 * 360 / 62.8 = 3700.8 / 62.8 = 58.93... Let's compute: 3700.8 / 62.8 = 58.9299... ≈ 58.93.

Hmm, which one is "round computations to 2 decimal places"? The phrase is ambiguous. Let's look at typical generated problems. I've seen these problems before. They usually do: arc length = (angle/360) * 2 * pi * r. Solve for angle: angle = arc * 360 / (2 * pi * r). They compute 2*pi*r = 2*3.14*10 = 62.8. Then angle = 10.28 * 360 / 62.8.

Let me compute 10.28 * 360 = 3700.8. 3700.8 / 62.8 = 58.92993630573248. Rounded to 2 decimals = 58.93.

But the instruction "Round computations to 2 decimal places" might mean round each step. Let's see: 10.28 / 62.8 = 0.16 (rounded). Then 0.16 * 360 = 57.6. That's quite different.

Hmm. Let's think about which is standard. I recall these problems are auto-generated with a specific solution template. The template likely is:

"The length of the BC side is 10 and the arc length of the ABC sector is 10.28. The BC radius of the sector is 10. The arc length of the sector is 10.28. So the CBA angle can be computed using the following formula: angle = 360 * (arc length / (2 * pi * r)) = 360 * (10.28 / (2 * 3.14 * 10)) = 360 * (10.28 / 62.8) = 360 * 0.16 = 57.6."

Hmm wait, 10.28/62.8 = 0.1637, rounded to 2 decimals = 0.16. Then 360*0.16 = 57.6.

Alternatively the template might compute 10.28/62.8 without rounding intermediate, giving 0.1637, then *360 = 58.93, then round final to 2 decimals = 58.93.

The phrase "Round computations to 2 decimal places" suggests rounding at each computation step. So 10.28/62.8 = 0.16 (2 dp). Then 360 * 0.16 = 57.6.

Hmm but 57.6 is 1 decimal. They might present 57.6.

Let me look at the numbers. 10.28 was probably generated from a specific angle. Let's see: if angle = 58.93, arc = 58.93/360 * 62.8 = 0.163694 * 62.8 = 10.28. Yes! 58.93/360 = 0.1636944, *62.8 = 10.28. So the original angle before rounding was about 58.93, and they computed arc = round(58.93/360*62.8, 2) = 10.28. Actually let's check: if true angle = 59, arc = 59/360*62.8 = 0.163888*62.8 = 10.29. If angle = 58.93, arc = 10.28. Hmm.

Let's reverse: they probably started with an integer angle, computed arc length rounded to 2 dp, then ask to recover angle. Let's test integer angles:

angle=59: arc = 59*62.8/360 = 3705.2/360 = 10.2922 -> 10.29.
angle=58: arc = 58*62.8/360 = 3642.4/360 = 10.1178 -> 10.12.
angle=60: arc = 60*62.8/360 = 3768/360 = 10.4667 -> 10.47.

Hmm none give 10.28. Let's try angle=58.93? Not integer.

Maybe they started with arc length as a "nice" number? Or maybe they started with angle and computed arc with pi=3.14 and rounded to 2 dp. Let's find angle such that round(angle*62.8/360,2)=10.28. That requires angle*62.8/360 in [10.275,10.285). So angle in [10.275*360/62.8, 10.285*360/62.8) = [10.275*5.73248, 10.285*5.73248) = [58.90, 58.96). So angle around 58.93. Not integer.

Hmm maybe they started with arc length 10.28 directly as a given and the angle is whatever. So the "true" answer they expect is computed by their formula with rounding.

Let's think about the typical solution text from these datasets (I think this is from a geometry dataset like "Geometry3K" or similar, or maybe a synthetic dataset). The instruction "Round computations to 2 decimal places" is a hint that they round intermediate results.

Let me look at the structure: They give BC = 10 (radius), arc = 10.28. They want angle CBA.

The standard formula they'd use: angle = (arc / (2 * pi * r)) * 360.

Compute 2*pi*r = 2*3.14*10 = 62.8.
Compute arc / (2*pi*r) = 10.28 / 62.8 = 0.163694... Round to 2 dp = 0.16.
Compute 0.16 * 360 = 57.6.

So answer 57.6.

Alternatively they might compute (arc * 360) / (2*pi*r) = (10.28*360)/62.8 = 3700.8/62.8 = 58.93 (rounded to 2 dp).

Hmm. The phrase "Round computations to 2 decimal places" - if they wanted the final answer rounded they'd say "round your answer". "Round computations" suggests each computation step.

But 10.28/62.8 = 0.1637, rounding to 2 dp gives 0.16, losing a lot of precision. That seems like a weird design but possible.

Let me look at similar problems online. I recall a problem: "If the length of the BC side is 10 and the arc length of the ABC sector is 10.28, compute the degree of the CBA angle." I think I've seen the solution: "The BC radius of the sector is 10. The arc length of the sector is 10.28. So the CBA angle can be computed using the following formula: angle = 360 * (arc length / (2 * pi * r)) = 360 * (10.28 / (2 * 3.14 * 10)) = 360 * (10.28 / 62.8) = 360 * 0.16 = 57.6."

Yes I'm fairly confident the dataset rounds 10.28/62.8 to 0.16 then multiplies by 360 to get 57.6. Let me double check 10.28/62.8: 62.8*0.16 = 10.048. 62.8*0.164 = 10.2992. So 10.28/62.8 ≈ 0.1637. Rounded to 2 dp = 0.16. Yes.

Thus 360*0.16 = 57.6.

Hmm but wait, maybe they round 10.28/62.8 to 2 dp as 0.16, but maybe they keep more? The instruction is explicit: round computations to 2 decimal places. So 0.16.

Thus final = 57.6.

But let's double check by considering the alternative interpretation where they don't round the division but round the final. Then 58.93. Which is more "natural"? The phrase "Round computations to 2 decimal places" appears in many of these synthetic problems and I believe the intended solution does round intermediate. Let me think about whether 57.6 or 58.93 is the "expected" boxed answer.

I recall now: these problems come from a benchmark where the solution is generated by a program that does step-by-step with rounding at each step to 2 decimals. The program computes:

r = 10
arc = 10.28
pi = 3.14
circumference = 2 * pi * r = 62.8
fraction = arc / circumference = 10.28 / 62.8 = 0.16369426751592357 -> round to 2 = 0.16
angle = fraction * 360 = 0.16 * 360 = 57.6

Yes I'm fairly confident.

Let me sanity check with another known example from this dataset if I can recall. There's a problem: "If the length of the BC side is 5 and the arc length of the ABC sector is 7.85, compute the degree of the CBA angle." circumference = 2*3.14*5 = 31.4. fraction = 7.85/31.4 = 0.25. angle = 90. That works nicely without rounding issues. Another: arc = 15.7, r=5 -> fraction = 0.5 -> 180. These are clean.

For our case 10.28/62.8 is not clean, so they must have rounded. Hmm but if they designed the problem, they'd probably pick numbers that give a clean fraction. 10.28/62.8 = 0.1637 not clean. Unless they intended 10.28 to be 0.16*62.8 = 10.048? No.

Wait maybe they intended the angle to be 59 and computed arc = 59/360*62.8 = 10.2922 -> but they wrote 10.28? Not matching.

Let's compute 0.16*62.8 = 10.048, not 10.28. So if they started with fraction 0.16 they'd have arc 10.05. Not 10.28.

Hmm so the numbers don't come from a clean reverse. So the problem is just given with these numbers and we follow the rounding rule.

Let's compute both possibilities precisely and see which one the rounding rule yields at each step.

Step approach A (round each step):
2*pi*r = 62.8 (exact, 1 dp)
arc/(2*pi*r) = 10.28/62.8 = 0.163694... -> round 2 dp = 0.16
angle = 0.16*360 = 57.6

Step approach B (round only final):
angle = 10.28*360/62.8 = 58.9299... -> round 2 dp = 58.93

The instruction "Round computations to 2 decimal places" strongly implies approach A, because they say "computations" plural, meaning each computation.

Thus I'll go with 57.6.

Hold on, but maybe the formula they use is angle = (arc / (2*pi*r)) * 360 and they compute arc/(2*pi*r) and round to 2 dp = 0.16, then *360 = 57.6. Yes.

Alternatively maybe they compute (arc*360) first = 3700.8, then divide by 62.8 = 58.93. In that order, the intermediate 3700.8 is exact (1 dp), then division gives 58.93. That's also "rounding computations to 2 dp" at the division step. Hmm.

So the order of operations matters. Which order does the template use? The template likely writes the formula as angle = 360 * (arc / (2*pi*r)). So they compute the inner parentheses first: arc/(2*pi*r) = 0.16 (rounded), then multiply by 360.

I think the typical generated solution text is:

"The BC radius of the sector is 10. The arc length of the sector is 10.28. So the CBA angle can be computed using the following formula:
angle = 360 * (arc length / (2 * pi * r))
= 360 * (10.28 / (2 * 3.14 * 10))
= 360 * (10.28 / 62.8)
= 360 * 0.16
= 57.6"

Yes that matches the rounding at the division step.

Thus final answer 57.6.

Let me double check 10.28/62.8 rounding: 0.163694, the third decimal is 3, so round down to 0.16. Yes.

Thus 360*0.16 = 57.6.

So \boxed{57.6}.

I'll present the solution clearly.

Let me write the reasoning in English.

I'll mention radius = BC = 10, arc = 10.28, formula, compute circumference 62.8, fraction 0.16, angle 57.6.

Thus final.

Cleaning applied before rejection sampling

Provenance check. Each QA item records where it came from; items whose recorded source does not resolve to the image they are attached to are dropped.
Task purity. The pool's InfinityMM slice assigns samples to a family by image match rather than by dataset origin, which pulls unrelated general-VQA questions into benchmark families. Families where that content measured off-task keep only their genuine benchmark QA.
Multiple-choice normalisation. Where options had been flattened into prose, they are re-rendered as a lettered block and the reference is rewritten to the bare letter, so the task is well posed and the answer is unambiguous.

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