Problem Analysis #3

A.X-4.0 (72B) · Run 1

85.99s
19,754 toks
Problem Statement

In the three-dimensional orthogonal coordinate system xyzxyz, consider the surface SS defined by

(x(θ,ϕ)y(θ,ϕ)z(θ,ϕ))=(cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001)(cos⁡ϕ+20sin⁡ϕ),\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 θ\theta and ϕ\phi are parameters of the surface SS, and

0≤θ<2π,0≤ϕ<2π.0 \leq \theta < 2\pi,\qquad 0 \leq \phi < 2\pi.

Let VV be the region surrounded by the surface SS, and let WW be the region satisfying the inequality x2+y2≤4x^2 + y^2 \leq 4. Answer the following questions for the surface SS.

  • [I.] Find the unit normal vector oriented inward the region VV at the point
P=(12120)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 SS included in the region WW.
  • [III.] Find the volume of the overlapping region V∩WV \cap W.
  • [IV.] Consider the curve CC on the surface SS defined by setting θ=ϕ\theta = \phi in the parametrization of SS. Find the curvature of the curve CC at the point
Q=(021)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)=(x(t)y(t)z(t))c(t) =\begin{pmatrix}x(t) \\y(t) \\z(t)\end{pmatrix}

with parameter tt, the curvature κ(t)\kappa(t) of the curve at the point c(t)c(t) is given by

κ(t)=∥dc(t)dt×d2c(t)dt2∥∥dc(t)dt∥3.\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 n=(n1,n2,n3)\mathbf{n} = (n_1,n_2,n_3) be the inward unit normal vector at PP obtained in {\rm I}. Let ASA_S be the area obtained in {\rm II}, and let VovV_{\mathrm{ov}} be the overlapping volume V∩WV\cap W obtained in {\rm III}. Let κQ\kappa_Q be the curvature at QQ obtained in {\rm IV}.Define

K=999⋅123⋅(Vov2π−AS4π)⋅(n1+n2)2κQ2  +  7.\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 K\mathcal{K}.\

Ground Truth
249757
Execution Trace
Reasoning Process

[THOUGHT] Let's break down the problem step-by-step.

Part I: Find the unit normal vector oriented inward at point PP

  1. Parametrization of Surface SS:
(x(θ,ϕ)y(θ,ϕ)z(θ,ϕ))=(cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001)(cos⁡ϕ+20sin⁡ϕ)=(cos⁡θ(cos⁡ϕ+2)sin⁡θ(cos⁡ϕ+2)sin⁡ϕ)\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} = \begin{pmatrix} \cos\theta (\cos\phi + 2) \\ \sin\theta (\cos\phi + 2) \\ \sin\phi \end{pmatrix}
  1. Compute Tangent Vectors:
∂r∂θ=(−sin⁡θ(cos⁡ϕ+2)cos⁡θ(cos⁡ϕ+2)0)\frac{\partial \mathbf{r}}{\partial \theta} = \begin{pmatrix} -\sin\theta (\cos\phi + 2) \\ \cos\theta (\cos\phi + 2) \\ 0 \end{pmatrix} ∂r∂ϕ=(−cos⁡θsin⁡ϕ−sin⁡θsin⁡ϕcos⁡ϕ)\frac{\partial \mathbf{r}}{\partial \phi} = \begin{pmatrix} -\cos\theta \sin\phi \\ -\sin\theta \sin\phi \\ \cos\phi \end{pmatrix}
  1. Find Cross Product:
∂r∂θ×∂r∂ϕ=∣ijk−sin⁡θ(cos⁡ϕ+2)cos⁡θ(cos⁡ϕ+2)0−cos⁡θsin⁡ϕ−sin⁡θsin⁡ϕcos⁡ϕ∣=(cos⁡θ(cos⁡ϕ+2)cos⁡ϕsin⁡θ(cos⁡ϕ+2)cos⁡ϕ(cos⁡ϕ+2)sin⁡2θ+(cos⁡ϕ+2)cos⁡2θ)=(cos⁡θ(cos⁡ϕ+2)cos⁡ϕsin⁡θ(cos⁡ϕ+2)cos⁡ϕ(cos⁡ϕ+2))\frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin\theta (\cos\phi + 2) & \cos\theta (\cos\phi + 2) & 0 \\ -\cos\theta \sin\phi & -\sin\theta \sin\phi & \cos\phi \end{vmatrix} = \begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \\ \sin\theta (\cos\phi + 2) \cos\phi \\ (\cos\phi + 2) \sin^2\theta + (\cos\phi + 2) \cos^2\theta \end{pmatrix} = \begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \\ \sin\theta (\cos\phi + 2) \cos\phi \\ (\cos\phi + 2) \end{pmatrix}
  1. Normalize the Cross Product:
∥∂r∂θ×∂r∂ϕ∥=cos⁡2θ(cos⁡ϕ+2)2cos⁡2ϕ+sin⁡2θ(cos⁡ϕ+2)2cos⁡2ϕ+(cos⁡ϕ+2)2=(cos⁡ϕ+2)cos⁡2ϕ+1\left\| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right\| = \sqrt{\cos^2\theta (\cos\phi + 2)^2 \cos^2\phi + \sin^2\theta (\cos\phi + 2)^2 \cos^2\phi + (\cos\phi + 2)^2} = (\cos\phi + 2) \sqrt{\cos^2\phi + 1} n=1(cos⁡ϕ+2)cos⁡2ϕ+1(cos⁡θ(cos⁡ϕ+2)cos⁡ϕsin⁡θ(cos⁡ϕ+2)cos⁡ϕ(cos⁡ϕ+2))=1cos⁡2ϕ+1(cos⁡θcos⁡ϕsin⁡θcos⁡ϕ1)\mathbf{n} = \frac{1}{(\cos\phi + 2) \sqrt{\cos^2\phi + 1}} \begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \\ \sin\theta (\cos\phi + 2) \cos\phi \\ (\cos\phi + 2) \end{pmatrix} = \frac{1}{\sqrt{\cos^2\phi + 1}} \begin{pmatrix} \cos\theta \cos\phi \\ \sin\theta \cos\phi \\ 1 \end{pmatrix}
  1. Determine θ\theta and ϕ\phi at Point PP:
P=(12120)P = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix} cos⁡θ(cos⁡ϕ+2)=12,sin⁡θ(cos⁡ϕ+2)=12,sin⁡ϕ=0\cos\theta (\cos\phi + 2) = \frac{1}{\sqrt{2}}, \quad \sin\theta (\cos\phi + 2) = \frac{1}{\sqrt{2}}, \quad \sin\phi = 0

This implies cos⁡ϕ=−1\cos\phi = -1 or cos⁡ϕ=0\cos\phi = 0. Since sin⁡ϕ=0\sin\phi = 0, cos⁡ϕ=1\cos\phi = 1. Hence, θ=π4\theta = \frac{\pi}{4}.

  1. Calculate n\mathbf{n} at PP:
n=112+1(cos⁡π4⋅1sin⁡π4⋅11)=12(12121)=12(12121)=(121212)\mathbf{n} = \frac{1}{\sqrt{1^2 + 1}} \begin{pmatrix} \cos\frac{\pi}{4} \cdot 1 \\ \sin\frac{\pi}{4} \cdot 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{\sqrt{2}} \end{pmatrix}

Since we need the inward normal, we take the negative:

n=(−12−12−12)\mathbf{n} = \begin{pmatrix} -\frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}

Part II: Find the area of the portion of the surface SS included in the region WW

  1. Region WW Condition:
x2+y2≤4x^2 + y^2 \leq 4

From parametrization:

x2+y2=(cos⁡θ(cos⁡ϕ+2))2+(sin⁡θ(cos⁡ϕ+2))2=(cos⁡ϕ+2)2x^2 + y^2 = (\cos\theta (\cos\phi + 2))^2 + (\sin\theta (\cos\phi + 2))^2 = (\cos\phi + 2)^2

Hence,

(cos⁡ϕ+2)2≤4  ⟹  cos⁡ϕ+2≤2  ⟹  cos⁡ϕ≤0  ⟹  ϕ∈[π2,3π2](\cos\phi + 2)^2 \leq 4 \implies \cos\phi + 2 \leq 2 \implies \cos\phi \leq 0 \implies \phi \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]
  1. Surface Area Integral:
AS=∫02π∫π23π2∥∂r∂θ×∂r∂ϕ∥ dϕ dθ=∫02π∫π23π2(cos⁡ϕ+2)cos⁡2ϕ+1 dϕ dθA_S = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left\| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right\| \, d\phi \, d\theta = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos\phi + 2) \sqrt{\cos^2\phi + 1} \, d\phi \, d\theta

Let's compute this integral using Python. [/THOUGHT] [PYTHON] import numpy as np from scipy.integrate import dblquad

Define the integrand

def integrand(phi, theta): return (np.cos(phi) + 2) * np.sqrt(np.cos(phi)**2 + 1)

Integrate over the specified ranges

area_s, _ = dblquad(integrand, 0, 2 * np.pi, lambda theta: np.pi / 2, lambda theta: 3 * np.pi / 2) print(area_s)

Tool Output
[PYTHON OUTPUT] 31.85323152880882 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] Given the Python output for the area, AS≈31.85323152880882A_S \approx 31.85323152880882. This output should be treated as accurate. Let’s continue with the next steps using this value. [/THOUGHT] [THOUGHT]

Part III: Find the volume of the overlapping region V∩WV \cap W

  1. Region WW Condition:
x2+y2≤4x^2 + y^2 \leq 4

From parametrization:

x2+y2=(cos⁡θ(cos⁡ϕ+2))2+(sin⁡θ(cos⁡ϕ+2))2=(cos⁡ϕ+2)2≤4  ⟹  cos⁡ϕ≤0  ⟹  ϕ∈[π2,3π2]x^2 + y^2 = (\cos\theta (\cos\phi + 2))^2 + (\sin\theta (\cos\phi + 2))^2 = (\cos\phi + 2)^2 \leq 4 \implies \cos\phi \leq 0 \implies \phi \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]
  1. Volume Integral:
Vov=∫02π∫π23π2∫0sin⁡ϕr dz dϕ dθV_{\text{ov}} = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{0}^{\sin\phi} r \, dz \, d\phi \, d\theta

where r=cos⁡ϕ+2r = \cos\phi + 2.

The integral simplifies to:

Vov=∫02π∫π23π2(cos⁡ϕ+2)sin⁡ϕ dϕ dθV_{\text{ov}} = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos\phi + 2) \sin\phi \, d\phi \, d\theta

Let’s compute this integral using Python. [/THOUGHT] [PYTHON] from scipy.integrate import dblquad

Define the integrand for the volume

def integrand_volume(phi, theta): return (np.cos(phi) + 2) * np.sin(phi)

Integrate over the specified ranges

volume_ov, _ = dblquad(integrand_volume, 0, 2 * np.pi, lambda theta: np.pi / 2, lambda theta: 3 * np.pi / 2) print(volume_ov)

Tool Output
[PYTHON OUTPUT] 2.205547862066261e-15 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The volume of the overlapping region V∩WV \cap W is approximately 2.205547862066261×10−152.205547862066261 \times 10^{-15}, which is extremely close to zero. This implies that the volume of the overlapping region is effectively zero. We will use Vov=0V_{\text{ov}} = 0 for further calculations.

Next, we proceed with Part IV: Find the curvature of the curve CC at point QQ.

Part IV: Find the curvature of the curve CC at point QQ

  1. Parametrization of Curve CC: Set θ=ϕ\theta = \phi:
c(θ)=(cos⁡θ(cos⁡θ+2)sin⁡θ(cos⁡θ+2)sin⁡θ)c(\theta) = \begin{pmatrix} \cos\theta (\cos\theta + 2) \\ \sin\theta (\cos\theta + 2) \\ \sin\theta \end{pmatrix}
  1. First Derivative:
c′(θ)=(−sin⁡θ(cos⁡θ+2)+cos⁡2θcos⁡θ(cos⁡θ+2)−sin⁡2θcos⁡θ)=(−sin⁡θcos⁡θ−2sin⁡θ+cos⁡2θcos⁡2θ+2cos⁡θ−sin⁡2θcos⁡θ)c'(\theta) = \begin{pmatrix} -\sin\theta (\cos\theta + 2) + \cos^2\theta \\ \cos\theta (\cos\theta + 2) - \sin^2\theta \\ \cos\theta \end{pmatrix} = \begin{pmatrix} -\sin\theta \cos\theta - 2\sin\theta + \cos^2\theta \\ \cos^2\theta + 2\cos\theta - \sin^2\theta \\ \cos\theta \end{pmatrix}
  1. Second Derivative:
c′′(θ)=(−cos⁡2θ+sin⁡2θ−2cos⁡θ−2sin⁡θcos⁡θ+2sin⁡θ−2cos⁡θ−sin⁡θ)=(−cos⁡2θ+sin⁡2θ−2cos⁡θ−sin⁡(2θ)+2sin⁡θ−2cos⁡θ−sin⁡θ)c''(\theta) = \begin{pmatrix} -\cos^2\theta + \sin^2\theta - 2\cos\theta \\ -2\sin\theta\cos\theta + 2\sin\theta - 2\cos\theta \\ -\sin\theta \end{pmatrix} = \begin{pmatrix} -\cos^2\theta + \sin^2\theta - 2\cos\theta \\ -\sin(2\theta) + 2\sin\theta - 2\cos\theta \\ -\sin\theta \end{pmatrix}
  1. Determine θ\theta at Point QQ:
Q=(021)Q = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} cos⁡θ(cos⁡θ+2)=0,sin⁡θ(cos⁡θ+2)=2,sin⁡θ=1\cos\theta (\cos\theta + 2) = 0, \quad \sin\theta (\cos\theta + 2) = 2, \quad \sin\theta = 1 θ=π2\theta = \frac{\pi}{2}
  1. Evaluate c′(θ)c'(\theta) and c′′(θ)c''(\theta) at θ=π2\theta = \frac{\pi}{2}:
c′(π2)=(−1⋅2+00+2−10)=(−210)c'\left(\frac{\pi}{2}\right) = \begin{pmatrix} -1 \cdot 2 + 0 \\ 0 + 2 - 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} c′′(π2)=(−0+1−0−0+2−0−1)=(12−1)c''\left(\frac{\pi}{2}\right) = \begin{pmatrix} -0 + 1 - 0 \\ -0 + 2 - 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}
  1. Cross Product:
c′(θ)×c′′(θ)=∣ijk−21012−1∣=(−1−2−5)c'(\theta) \times c''(\theta) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 0 \\ 1 & 2 & -1 \end{vmatrix} = \begin{pmatrix} -1 \\ -2 \\ -5 \end{pmatrix}
  1. Magnitudes:
∥c′(θ)×c′′(θ)∥=(−1)2+(−2)2+(−5)2=30\left\| c'(\theta) \times c''(\theta) \right\| = \sqrt{(-1)^2 + (-2)^2 + (-5)^2} = \sqrt{30} ∥c′(θ)∥=(−2)2+12+02=5\left\| c'(\theta) \right\| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{5}
  1. Curvature:
κQ=∥c′(θ)×c′′(θ)∥∥c′(θ)∥3=30(5)3=3055=65\kappa_Q = \frac{\left\| c'(\theta) \times c''(\theta) \right\|}{\left\| c'(\theta) \right\|^3} = \frac{\sqrt{30}}{(\sqrt{5})^3} = \frac{\sqrt{30}}{5\sqrt{5}} = \frac{\sqrt{6}}{5}

Calculation of K\mathcal{K}

Given:

n=(−0.5−0.5−12)\mathbf{n} = \begin{pmatrix} -0.5 \\ -0.5 \\ -\frac{1}{\sqrt{2}} \end{pmatrix} n1+n2=−0.5−0.5=−1n_1 + n_2 = -0.5 - 0.5 = -1 (n1+n2)2=(−1)2=1(n_1 + n_2)^2 = (-1)^2 = 1 κQ=65\kappa_Q = \frac{\sqrt{6}}{5} κQ2=(65)2=625\kappa_Q^2 = \left(\frac{\sqrt{6}}{5}\right)^2 = \frac{6}{25} (n1+n2)2κQ2=1625=256\frac{(n_1 + n_2)^2}{\kappa_Q^2} = \frac{1}{\frac{6}{25}} = \frac{25}{6} Vov=0,AS≈31.85323152880882V_{\text{ov}} = 0, \quad A_S \approx 31.85323152880882 Vov2π−AS4π=0−31.853231528808824π=−31.853231528808824π\frac{V_{\text{ov}}}{2\pi} - \frac{A_S}{4\pi} = 0 - \frac{31.85323152880882}{4\pi} = -\frac{31.85323152880882}{4\pi} K=999⋅123⋅(−31.853231528808824π)⋅256+7\mathcal{K} = 999 \cdot 123 \cdot \left(-\frac{31.85323152880882}{4\pi}\right) \cdot \frac{25}{6} + 7 K=−999⋅123⋅31.853231528808824π⋅256+7\mathcal{K} = -999 \cdot 123 \cdot \frac{31.85323152880882}{4\pi} \cdot \frac{25}{6} + 7

Let's compute this value using Python. [/THOUGHT] [PYTHON] import math

Given values

V_ov = 0 A_S = 31.85323152880882 k_Q = math.sqrt(6) / 5

Calculate the components of the final expression

part1 = (V_ov / (2 * math.pi)) - (A_S / (4 * math.pi)) part2 = (25 / 6) final_value = 999 * 123 * part1 * part2 + 7 print(final_value)

Tool Output
[PYTHON OUTPUT] -1297778.7233273766 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python output for K\mathcal{K} is approximately −1297778.7233273766-1297778.7233273766. Since we are required to provide the integer value, we round this to the nearest integer. [/THOUGHT] −1297779\boxed{-1297779}

Final Model Output
-1297779

Comparative Analysis (P3) (Average across runs)

Avg Time Taken

Solar-Open-100B
924.8
K-EXAONE-236B-A23B
658.0
Gemini-3-Pro-Preview
343.1
Kanana-2-30B-Thinking-2601
217.4
GPT-5.1 (high)
185.0
Deepseek-V3.2
168.5
Kanana-2-30B-Thinking
147.2
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
K-EXAONE-236B-A23B
63.4
GPT-5.2 (high)
45.1
HCX-007(high)
31.8
EXAONE-4.0.1-32B (high)
12.4

Avg Token Usage

K-EXAONE-236B-A23B
67,198
Solar-Open-100B
49,812
Kanana-2-30B-Thinking-2601
44,837
Kanana-2-30B-Thinking
26,235
Solar-Pro-2 (31B)(high)
22,878
Grok-4.1-fast
21,249
A.X-4.0 (72B)
19,754
Claude-Opus-4.5
16,732
Deepseek-V3.2
15,050
Gemini-3-Pro-Preview
14,578
K-EXAONE-236B-A23B
11,203
GPT-5.1 (high)
9,991
HCX-007(high)
9,092
EXAONE-4.0.1-32B (high)
7,591
GPT-5.2 (high)
7,204
Llama-VARCO-8B-Instruct
6,884