Problem Analysis #3

142.69s

21,249 toks

Problem Statement

In the three-dimensional orthogonal coordinate system $xyz$ , consider the surface $S$ defined by $\begin{pmatrix}x(\theta, \phi) \\y(\theta, \phi) \\z(\theta, \phi)\end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta & 0 \\\sin\theta & \cos\theta & 0 \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}\cos\phi + 2 \\0 \\\sin\phi\end{pmatrix},$ where $\theta$ and $\phi$ are parameters of the surface $S$ , and $0 \leq \theta < 2\pi,\qquad 0 \leq \phi < 2\pi.$ Let $V$ be the region surrounded by the surface $S$ , and let $W$ be the region satisfying the inequality $x^2 + y^2 \leq 4$ . Answer the following questions for the surface $S$ .

[I.] Find the unit normal vector oriented inward the region $V$ at the point $P = \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \\ 0 \end{pmatrix}$ on the surface $S$ .
[II.] Find the area of the portion of the surface $S$ included in the region $W$ .
[III.] Find the volume of the overlapping region $V \cap W$ .
[IV.] Consider the curve $C$ on the surface $S$ defined by setting $\theta = \phi$ in the parametrization of $S$ . Find the curvature of the curve $C$ at the point $Q = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ on the curve $C$ . Note that, in general, given a three-dimensional curve defined by $c(t) =\begin{pmatrix}x(t) \\y(t) \\z(t)\end{pmatrix}$ with parameter $t$ , the curvature $\kappa(t)$ of the curve at the point $c(t)$ is given by $\kappa(t)= \frac{\left\lVert \dfrac{dc(t)}{dt} \times \dfrac{d^2c(t)}{dt^2}\right\rVert} {\left\lVert \dfrac{dc(t)}{dt}\right\rVert^3}.$ Let $\mathbf{n} = (n_1,n_2,n_3)$ be the inward unit normal vector at $P$ obtained in {\rm I}. Let $A_S$ be the area obtained in {\rm II}, and let $V_{\mathrm{ov}}$ be the overlapping volume $V\cap W$ obtained in {\rm III}. Let $\kappa_Q$ be the curvature at $Q$ obtained in {\rm IV}.Define $\mathcal{K}= 999 \cdot 123 \cdot\left(\frac{V_{\mathrm{ov}}}{2\pi}- \frac{A_S}{4\pi}\right)\cdot\frac{(n_1 + n_2)^2}{\kappa_Q^2}\;+\; 7.$ Compute the integer $\mathcal{K}$ .\

Ground Truth

249757

Execution Trace

Reasoning Process

[THOUGHT] The surface is a torus with major radius 2, minor radius 1. For part I, at P corresponding to θ=π/4, φ=π, the inward unit normal is (1/√2, 1/√2, 0), so n1 + n2 = √2, (n1 + n2)^2 = 2. For II, A_S = 4π² - 4π. For III, V_ov = 2π² - 4π/3. Then V_ov/(2π) - A_S/(4π) = 1/3. For IV, verify κ_Q at t=π/2 for c(t) = ((2 + cos t) cos t, (2 + cos t) sin t, sin t) using sympy to compute ||c' × c''|| / ||c'||^3, expecting √41 / 5^{3/2}, so κ_Q^2 = 41/125. Then the expression simplifies to 999 * 250 + 7 = 249757. Now verify κ_Q with code. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t') alpha = 2 + sp.cos(t) cx = alpha * sp.cos(t) cy = alpha * sp.sin(t) cz = sp.sin(t)

c = sp.Matrix([cx, cy, cz]) c1 = sp.diff(c, t) c2 = sp.diff(c1, t)

t_val = sp.pi / 2 c1_val = c1.subs(t, t_val) c2_val = c2.subs(t, t_val)

cross = c1_val.cross(c2_val) norm_cross = sp.sqrt(cross.dot(cross)) norm_c1 = sp.sqrt(c1_val.dot(c1_val)) kappa = norm_cross / norm_c1**3

print(kappa) print(float(kappa2)) # to check 41/125=0.328 print(kappa2.simplify()) [/PYTHON]

Tool Output

[PYTHON OUTPUT] SyntaxError: invalid decimal literal (<string>, line 24) [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] Fix the SyntaxError by removing comments and the float line. Compute kappa exactly and simplify kappa**2. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t') alpha = 2 + sp.cos(t) cx = alpha * sp.cos(t) cy = alpha * sp.sin(t) cz = sp.sin(t)

c = sp.Matrix([cx, cy, cz]) c1 = c.diff(t) c2 = c1.diff(t)

t_val = sp.pi / 2 c1_val = c1.subs(t, t_val) c2_val = c2.subs(t, t_val)

cross = c1_val.cross(c2_val) norm_cross = sp.sqrt(cross.dot(cross)) norm_c1 = sp.sqrt(c1_val.dot(c1_val)) kappa = norm_cross / norm_c1**3

print(kappa) print(kappa.simplify()) print((kappa**2).simplify()) [/PYTHON]

Tool Output

[PYTHON OUTPUT] sqrt(205)/25 sqrt(205)/25 41/125 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The calculations confirm: $\mathbf{n} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$ , so $(n_1 + n_2)^2 = 2$ . $A_S = 4\pi^2 - 4\pi$ . $V_{\mathrm{ov}} = 2\pi^2 - \frac{4\pi}{3}$ . Thus $\frac{V_{\mathrm{ov}}}{2\pi} - \frac{A_S}{4\pi} = \frac{1}{3}$ . $\kappa_Q^2 = \frac{41}{125}$ . Then $\mathcal{K} = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{2}{41/125} + 7 = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{250}{41} + 7$ . Since $123 = 3 \cdot 41$ , this simplifies to $333 \cdot 3 \cdot 250 + 7 = 999 \cdot 250 + 7 = 249750 + 7 = 249757$ . [/THOUGHT] \boxed{249757}

Explanation

The reasoning derives the values for each component of (\mathcal{K}) systematically from the surface parametrization and problem constraints.

For part I, the point (P = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)) corresponds to (\theta = \pi/4), (\phi = \pi). Compute partial derivatives: (\mathbf{r}\theta = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)), (\mathbf{r}\phi = (0, 0, -1)). The cross product (\mathbf{r}\theta \times \mathbf{r}\phi = \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right)) has unit length and points outward for the solid torus (V). Thus, the inward unit normal is (\mathbf{n} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)), so (n_1 + n_2 = \sqrt{2}) and ((n_1 + n_2)^2 = 2).

For part II, the area element is (|( \mathbf{r}\theta \times \mathbf{r}\phi )| = 2 + \cos \phi). Region (W) requires (\sqrt{x^2 + y^2} = 2 + \cos \phi \leq 2), so (\cos \phi \leq 0) or (\phi \in [\pi/2, 3\pi/2]). Thus, (A_S = \int_0^{2\pi} \int_{\pi/2}^{3\pi/2} (2 + \cos \phi) , d\phi , d\theta = 2\pi \left[ 2\phi + \sin \phi \right]_{\pi/2}^{3\pi/2} = 2\pi (2\pi - 2) = 4\pi^2 - 4\pi).

For part III, volume (V_{\mathrm{ov}}) uses cylindrical coordinates: (V_{\mathrm{ov}} = 4\pi \int_1^2 \rho \sqrt{1 - (\rho - 2)^2} , d\rho). Substitute (u = \rho - 2), yielding (\int_{-1}^0 (u + 2) \sqrt{1 - u^2} , du = -\frac{1}{3} + \frac{\pi}{2}). Thus, (V_{\mathrm{ov}} = 4\pi \left( \frac{\pi}{2} - \frac{1}{3} \right) = 2\pi^2 - \frac{4\pi}{3}).

Then, (\frac{V_{\mathrm{ov}}}{2\pi} = \pi - \frac{2}{3}) and (\frac{A_S}{4\pi} = \pi - 1), so (\frac{V_{\mathrm{ov}}}{2\pi} - \frac{A_S}{4\pi} = \frac{1}{3}).

For part IV, curve (C): (\mathbf{c}(t) = \left( (2 + \cos t) \cos t, (2 + \cos t) \sin t, \sin t \right)). At (Q) ((t = \pi/2)), Python computation gives (\kappa_Q = \sqrt{205}/25), so (\kappa_Q^2 = 41/125).

Substitute into (\mathcal{K}): (999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{2}{41/125} + 7 = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{250}{41}). Since (123 = 3 \cdot 41), simplify to (333 \cdot 3 \cdot 250 + 7 = 999 \cdot 250 + 7 = 249750 + 7 = 249757).

Final Model Output

249757

Comparative Analysis (P3) (Average across runs)

Avg Time Taken

Gemini-3-Pro-Preview

343.1

GPT-5.1 (high)

185.0

Deepseek-V3.2

168.5

Grok-4.1-fast

142.7

Solar-Pro-2 (31B)(high)

139.6

Llama-VARCO-8B-Instruct

92.2

A.X-4.0 (72B)

86.0

Claude-Opus-4.5

84.8

HCX-007(high)

31.8

EXAONE-4.0.1-32B (high)

12.4

Avg Token Usage

Solar-Pro-2 (31B)(high)

22878.0

Grok-4.1-fast

21249.0

A.X-4.0 (72B)

19754.0

Claude-Opus-4.5

16732.0

Deepseek-V3.2

15050.0

Gemini-3-Pro-Preview

14578.0

GPT-5.1 (high)

9991.0

HCX-007(high)

9092.0

EXAONE-4.0.1-32B (high)

7591.0

Llama-VARCO-8B-Instruct

6884.0