Problem Analysis #5

Gemini 3.5 Flash · Run 1

Correct
11.84s
3,022 toks

API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.

Problem Statement

Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.

For

f(x)=∫01∣t−x∣1+t2 dt(0≤x≤1),f(x)=\int_0^1\frac{|t-x|}{1+t^2}\,dt\qquad(0\le x\le1),

let α\alpha be the number satisfying 0<α<π/40<\alpha<\pi/4 and f′(tan⁡α)=0f'(\tan\alpha)=0. Let u=tan⁡αu=\tan\alpha. The value of uu is irrational, so instead compute the integer

(u+1u)2.\left(u+\frac1u\right)^2.
Ground Truth
8
Recorded model response / solution
Reasoning Process

To find the value of (u+1u)2\left(u + \frac{1}{u}\right)^2, we first express the function f(x)f(x) for 0≤x≤10 \le x \le 1 by splitting the integral at t=xt = x:

f(x)=∫0xx−t1+t2 dt+∫x1t−x1+t2 dtf(x) = \int_0^x \frac{x-t}{1+t^2} \, dt + \int_x^1 \frac{t-x}{1+t^2} \, dt

Using the Leibniz rule to differentiate f(x)f(x) with respect to xx, we get:

f′(x)=∫0x11+t2 dt−∫x111+t2 dtf'(x) = \int_0^x \frac{1}{1+t^2} \, dt - \int_x^1 \frac{1}{1+t^2} \, dt

Evaluating these integrals:

f′(x)=arctan⁡(x)−(arctan⁡(1)−arctan⁡(x))=2arctan⁡(x)−π4f'(x) = \arctan(x) - \left( \arctan(1) - \arctan(x) \right) = 2\arctan(x) - \frac{\pi}{4}

We are given that f′(tan⁡α)=0f'(\tan\alpha) = 0 for 0 < $$\alpha$$ < $$\frac{\pi}{4}$$$. Substituting x = \tan\alpha$:

2arctan⁡(tan⁡α)−π4=0  ⟹  2α=π4  ⟹  α=π82\arctan(\tan\alpha) - \frac{\pi}{4} = 0 \implies 2\alpha = \frac{\pi}{4} \implies \alpha = \frac{\pi}{8}

Thus, u=tan⁡α=tan⁡(π8)u = \tan\alpha = \tan\left(\frac{\pi}{8}\right).

Using the double-angle formula for tangent:

tan⁡(2⋅π8)=2u1−u2  ⟹  1=2u1−u2  ⟹  u2+2u−1=0\tan\left(2 \cdot \frac{\pi}{8}\right) = \frac{2u}{1-u^2} \implies 1 = \frac{2u}{1-u^2} \implies u^2 + 2u - 1 = 0

Dividing by uu (since u≠0u \neq 0):

u+2−1u=0  ⟹  u−1u=−2u + 2 - \frac{1}{u} = 0 \implies u - \frac{1}{u} = -2

We want to find the value of (u+1u)2\left(u + \frac{1}{u}\right)^2. Using the algebraic identity:

(u+1u)2=(u−1u)2+4\left(u + \frac{1}{u}\right)^2 = \left(u - \frac{1}{u}\right)^2 + 4

Substituting u−1u=−2u - \frac{1}{u} = -2:

(u+1u)2=(−2)2+4=4+4=8\left(u + \frac{1}{u}\right)^2 = (-2)^2 + 4 = 4 + 4 = 8

\boxed{8}

Final Model Output
8

Comparative Analysis (P5) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

K-EXAONE-236B-A23B
24.1
DeepSeek V4 Pro
18.2
Gemini 3.5 Flash
11.7
GPT-5.5
8.0
Solar Pro 3
6.9
Claude Opus 4.8
3.5
KT Mi:dm 2.0 Base Instruct
3.2

Avg Token Usage

K-EXAONE-236B-A23B
3597.7
Gemini 3.5 Flash
2820.0
KT Mi:dm 2.0 Base Instruct
1642.3
Solar Pro 3
1309.0
DeepSeek V4 Pro
1304.0
Claude Opus 4.8
590.0
GPT-5.5
559.3
    Tokyo · Problem 5 · Gemini 3.5 Flash | EntropyMath