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 UniGeo_RS_think and UniGeo_RS_nothink, non-matches go to UniGeo_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)]

`latex` — mathematical expression recognition

For handwritten-formula recognition (CROHME, HME100K) and geometry answers, both sides are normalised as LaTeX before comparison, so two expressions that render identically compare equal.

ALIASES = [(r"\\to\b", r"\\rightarrow"), (r"\\limits\b", ""),
           (r"\\operatorname\s*\{([^}]*)\}", r"\1"), (r"\\mathrm\s*\{([^}]*)\}", r"\1"),
           (r"\\text\s*\{([^}]*)\}", r"\1"),
           (r"\\(left|right|,|;|!|:|quad|qquad)\b", ""), (r"\\cdot\b", r"\\times"),
           (r"\\[lr]?vert\b", "|"), (r"\\[dt]frac\b", r"\\frac"), (r"\\(?=\s|$)", "")]

def latex_norm(x):
    x = re.sub(r"^\$+|\$+$", "", (x or "").strip()).strip()
    x = re.sub(r"^\\\[|\\\]$", "", x).strip()
    x = re.sub(r"^```(?:latex)?|```$", "", x).strip()
    for a, b in ALIASES:
        x = re.sub(a, b, x)
    return re.sub(r"[\s{}]", "", x).lower()

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`	Solve the geometry problem shown in the image. Problem: 如图,在△ABC中,DE∥BC,AD:AB=1:3,若△ADE的面积等于2,则△ABC的面积等于() English translation: As shown in the figure, in △ABC, DE ∥ BC, AD:AB=1:3, if the area of △ADE is equal to 2, then the area of △ABC is equal to () Answer with the correct option letter and value. (A) 6 (B) 8 (C) 12 (D) 18 Answer with the option's letter only, without any other content.
`answer`	D
`org_answer`	D

think (359 chars)

The problem involves similar triangles. Since DE is parallel to BC, triangle ADE is similar to triangle ABC. The ratio of corresponding sides is AD:AB = 1:3. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides. So area(ADE)/area(ABC) = (1/3)² = 1/9. Given area(ADE) = 2, then area(ABC) = 2 × 9 = 18.

The answer is D.

Example 2

field	value
`question`	Solve the geometry problem shown in the image. Problem: 如图,⊙O的直径AB=10cm,CD⊥AB,OE=3cm,则CD的长()cm． English translation: As shown in the figure, the diameter of ⊙O AB = 10, CD ⊥ AB, OE = 3, then the length of CD (). Answer with the correct option letter and value. (A) 4 (B) 6 (C) 8 (D) 10 Answer with the option's letter only, without any other content.
`answer`	C
`org_answer`	C

think (554 chars)

The user wants the answer as a single option letter.

Let me work through the geometry. Circle O has diameter AB = 10, so radius = 5. CD is a chord perpendicular to AB at point E. OE = 3.

Since CD ⊥ AB and AB is a diameter, the perpendicular from the center to a chord bisects the chord. So E is the midpoint of CD, meaning CE = ED, and CD = 2·CE.

Connect OC. OC is a radius = 5. In right triangle OEC, OE = 3, OC = 5, so CE = sqrt(5² - 3²) = sqrt(25 - 9) = sqrt(16) = 4.

Thus CD = 2·4 = 8.

So the answer is C.

The user wants only the option letter.

Example 3

field	value
`question`	Solve the geometry problem shown in the image. Problem: 如图所示,AB∥CD,∠E=27°,∠C=52°,则∠EAB的度数为() English translation: As shown in the figure, AB ∥ CD, ∠E = 27, ∠C = 52, then the degree of ∠EAB is () Answer with the correct option letter and value. (A) 25° (B) 63° (C) 79° (D) 101° Answer with the option's letter only, without any other content.
`answer`	C
`org_answer`	C

think (10840 chars)

The user wants the answer as a single letter. Let me work through the geometry.

We have AB ∥ CD. There's a point E, with lines EC and EA. Looking at the figure, C is bottom left, D is bottom right, so CD is a horizontal line. A is above, B is to the right, so AB is a horizontal line parallel to CD. E is at the top.

The line from C goes up to E (segment CE). The line from A goes up to E (segment AE). So we have a configuration where CE intersects... let's look. C is at bottom left, E at top. The segment CE passes near A? In the figure, CE is a long line from C to E, and A is to the right of that line, with AE going up to E.

Let me set up coordinates. Let CD be horizontal. Let's place C at origin (0,0), D to the right along positive x-axis. AB is parallel to CD, so AB is horizontal at some height h. A is at some point, B to the right.

Angle C = 52° is angle ECD, i.e., the angle between CE and CD. Since CD points right (positive x), and CE goes up-right to E, the angle of CE above horizontal is 52°.

So line CE makes 52° with horizontal.

Now E is on this line. Let's say E = (x_E, y_E) with y_E > 0, and the line from C has slope tan(52°).

A is at height h (same as B). A is to the right of line CE? In figure, A is to the right of the CE line. The segment AE goes from A up to E.

Angle E = 27° is angle AEC, the angle at E between EA and EC.

We need angle EAB, the angle at A between AE and AB. AB points right (positive x direction).

Let me think about the direction of lines. At point E, we have two rays: EC going down-left to C, and EA going down to A (down and slightly left or right?). In figure, A is below E and slightly left? E is at top, A is below and a bit left of E? Looking at figure, E is at top, A is below E but the line EA looks nearly vertical, maybe slightly left. Hard to say.

Let's use angles with horizontal.

Ray EC direction: from E to C is down-left. The line CE makes 52° with horizontal (CD). So from C, direction to E is 52° above positive x. Thus from E, direction to C is 52° + 180° = 232° (i.e., pointing down-left). In standard angle measure, that's 180+52 = 232°.

Ray EA direction: from E to A. We need to find this.

Angle between EC and EA at E is 27°.

Now consider line AB horizontal. At A, ray AB points to positive x (0°). Ray AE points from A to E, which is up (and some horizontal direction).

Let me denote the direction of ray AE (from A to E) as angle θ measured from positive x-axis. Since E is above A, θ is between 0 and 180.

The direction of ray EA (from E to A) is θ + 180 (mod 360).

The direction of ray EC (from E to C) is 232°.

The angle between EA and EC is 27°. So |θ + 180 - 232| = 27 or 360 - that = 27. So |θ - 52| = 27 (mod considerations). So θ + 180 - 232 = θ - 52. Set equal to ±27.

Case 1: θ - 52 = 27 → θ = 79°.
Case 2: θ - 52 = -27 → θ = 25°.

Now angle EAB is the angle between ray AE (direction θ) and ray AB (direction 0°). That's |θ - 0| = θ (if θ between 0 and 180). So angle EAB = θ = 79° or 25°.

Hmm both are options (A) 25 and (C) 79. Need to determine which based on figure.

Let's look at figure. In the figure, line CE is steep (52° from horizontal, that's fairly steep but not vertical). Line EA appears steeper than CE? In figure, EA looks almost vertical, maybe slightly tilted. CE is tilted left as it goes up (from C bottom-left to E top, E is to the right of C, so CE tilts right as it goes up). Wait C is bottom left, E is top middle. So CE goes up and right. Yes 52° from horizontal means it goes up-right.

In figure, E is at top, A is below E and to the left of E? Let's look: E label at top, the line from E goes down to A which is below and slightly left? Actually in the image, E is at top, A is below it, the segment EA looks like it tilts: top at E, bottom at A, A appears slightly left of E? Hard.

Let's think about positions. Let's place C=(0,0). Line CE at 52°, so E = (t cos52, t sin52) for some t. cos52≈0.6157, sin52≈0.788. So E ≈ (0.616 t, 0.788 t).

A is at height h = 0.788 t * (something less than 1) since A is below E. Let's say A = (x_A, h). In figure, A's x-coordinate: A appears to be to the right of C and maybe around the middle. The line CE at height h has x = h / tan52 = h / 1.2796 ≈ 0.781 h. So the point on CE at height h is at x ≈ 0.781 h. In figure, A is to the right of the CE line? Look: the CE line passes through the figure; A is to the right of it. Yes in the figure, the long line from C to E is on the left, and A is to the right of that line. So x_A > 0.781 h.

Now E's x-coordinate is 0.616 t. h = y_A. Let's pick t such that E's y is, say, 10. Then h maybe 4. Then E = (6.16, 10). The CE line at height 4 has x = 4/1.2796 = 3.13. A is to the right, say x_A = 4. So A = (4,4). Then vector from A to E = (6.16-4, 10-4) = (2.16, 6). That's up and right, angle θ = atan2(6, 2.16) = atan(2.78) ≈ 70°. Hmm.

Vector from E to A = (-2.16, -6), angle = atan2(-6,-2.16) = 180+70 = 250°. Vector from E to C = (-6.16,-10), angle = atan2(-10,-6.16)=180+58.4=238.4° (since 52° line, from E to C is 180+52=232? wait compute: direction from C to E is 52°, so from E to C is 52+180=232°. But my numbers: E-C = (6.16,10), angle = atan2(10,6.16)=58.4°, not 52. Because I used t=10 giving E=(6.16,10) but that's not on a 52° line from origin? Check: 10/6.16 = 1.623, tan52=1.279. Not matching. I messed: if line is 52°, then y/x = tan52 =1.279, so if y=10, x=10/1.279=7.82. So E=(7.82,10). Then at height 4, x=4/1.279=3.13. A to right, say x_A=4. Then A=(4,4). Vector A to E = (3.82,6), angle=atan2(6,3.82)=57.5°. Vector E to A = 57.5+180=237.5°. Vector E to C = 232°. Difference = 5.5°, not 27.

I need to choose x_A to satisfy angle condition. Let's solve generally. Let h=4, E=(7.82,10). A=(x_A,4). Vector E->A = (x_A-7.82, -6). Vector E->C = (-7.82,-10). The angle between them is 27°.

We want angle EAB = θ where θ = angle of vector A->E = (7.82 - x_A, 6). Since 6>0, θ in (0,180). 7.82 - x_A could be positive or negative.

We earlier derived θ = 79 or 25 from the angle-between-rays equation, but that derivation assumed the direction of EC exactly 232°, which holds. Let's recompute: direction of E->C is 232°. Direction of E->A = θ+180 (mod 360). The angle between = 27. So (θ+180) - 232 = ±27 mod 360. θ -52 = ±27 → θ = 79 or 25. Good, independent of coordinates. So θ is either 79 or 25.

Now θ = angle of A->E. If θ=79, then A->E points up-right (since 79° in first quadrant, both components positive, so 7.82 - x_A >0 → x_A < 7.82). If θ=25, A->E points up-right but more horizontal, also x_A<7.82 but more so: tan25=0.466, so 6/(7.82-x_A)=0.466 → 7.82-x_A=12.87 → x_A=-5.05, which is left of C, not matching figure (A is to the right of C). For θ=79: tan79=5.14, 6/(7.82-x_A)=5.14 → 7.82-x_A=1.167 → x_A=6.65. That's to the right of C and to the right of CE line at height 4 (which is 3.13). Good, matches figure: A is right of CE line. Also x_A=6.65 <7.82 so A is left of E's x, meaning E is to the right of A, so EA tilts right as it goes up. In figure, does EA tilt right as it goes up? Look: E at top, A below. In the image, E appears to be slightly to the right of A? The line EA in the figure looks like it goes from A (lower) up to E (upper) tilting to the right? Hmm the figure shows E at top with the line going down-left to A? Let's look carefully. The image: E at top center. Below it, the line goes down to A. A is labeled below the line's lower end. The line EA appears to tilt: top end (E) is to the right of bottom end (A)? Or left? In many such figures, EA is drawn nearly vertical with E slightly right. Hard.

Let's look at the visual: The segment from E to A: E is at the top, the line goes down and slightly left to A? Actually I see in the image the line EA is slanted such that E is at top-right of that segment and A at bottom-left? The label A is at the bottom of the segment, and the segment's top is E. The slant: I'd guess E is to the right of A? Hmm.

Let's look at the overall shape. C bottom-left, line CE goes up-right to E. A is somewhere middle, line AB goes right. The line EA connects A to E. In the figure, the angle at E between EC and EA is small (27°), and EC goes down-left steeply, EA goes down (steeply). For the angle between them to be 27°, EA should be rotated 27° from EC. EC direction (from E) is 232° (down-left). Rotating by 27° clockwise (toward down) gives 232-27=205°? or counterclockwise 232+27=259°. Direction 259° is down and slightly left (since 270 is straight down, 259 is down-left a bit). Direction 205° is down-left more. Hmm wait we need E->A direction. We had E->A = θ+180. If θ=79, E->A=259°. If θ=25, E->A=205°.

259°: cos259≈-0.19, sin259≈-0.98 → down and slightly left. So A is below E and slightly left of E. That means E is to the right of A. So EA tilts right going up.

205°: cos205≈-0.906, sin205≈-0.423 → down-left significantly. So A is below and left of E, more left. Then EA tilts right going up but more horizontal? Wait if A is left of E, then going from A to E you go right, so A->E has positive x, consistent with θ in first quadrant (25°). Yes.

Now which matches figure? In figure, the horizontal distance between E and A looks small (EA nearly vertical). 259° gives horizontal component -0.19 (small), vertical -0.98 (large) → nearly vertical. 205° gives horizontal -0.906, vertical -0.423 → more horizontal, slanted. In the figure, EA looks fairly steep (nearly vertical), so 259° i.e., θ=79° matches better.

Also check position of A relative to CE line. For θ=79 we found x_A=6.65 with E x=7.82, CE line at height 4 is 3.13, so A (6.65) is well right of CE line, matches figure where A is to the right of the long CE line. For θ=25, x_A=-5.05 which is left of C, contradicts figure. So definitely θ=79.

Thus angle EAB = 79°.

Hence answer C.

Let's double check with a known solution approach: Extend EA to meet CD at point F, or draw a parallel line. Common method: Since AB ∥ CD, draw through E a line parallel? Or use the fact that angle EAB is exterior angle.

Alternative: Let line CE intersect AB at point G (since CE goes from C up to E, and AB is horizontal between, CE will cross AB at some point G). In figure, does CE cross AB? AB is at height h from A to B (rightwards). CE at height h is at x=3.13 (in my numbers). A is at x=6.65, B further right. So the point G=(3.13,4) is to the left of A, so not on segment AB but on its extension leftwards. Hmm.

Better method: Use the "M" or zigzag angle theorem. Actually consider the line through E parallel to AB and CD. Hmm.

Let's just trust coordinate result: 79°.

Let's verify angle EAB directly with coordinates: A=(6.65,4), B to the right say (10,4). Ray AB direction 0°. Ray AE direction = E-A = (7.82-6.65, 10-4)=(1.17,6), angle=atan2(6,1.17)=79°. Yes angle EAB=79°.

Thus answer C.

The user wants only the option letter. So output "C".

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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UniGeo — UniGeo_RS_think

How the data was produced