Problem Analysis #7

Gemini 3.5 Flash · Run 1

Correct
9.80s
2,071 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.

Let f(x)=2x3−4x2+3f(x)=2x^3-4x^2+3. For 0<t<40<t<4, let CtC_t be the circle passing through (t,f(t))(t,f(t)), having the same tangent there as y=f(x)y=f(x), and having its center on the xx-axis. If the center is (c(t),0)(c(t),0) and the radius is r(t)r(t), find

c(1)+r(1)2.c(1)+r(1)^2.
Ground Truth
4
Recorded model response / solution
Reasoning Process

To find the value of c(1)+r(1)2c(1) + r(1)^2, we proceed step by step:

  1. Evaluate the function and its derivative at x=1x = 1: The given function is:
f(x)=2x3−4x2+3f(x) = 2x^3 - 4x^2 + 3

At x=1x = 1:

f(1)=2(1)3−4(1)2+3=1f(1) = 2(1)^3 - 4(1)^2 + 3 = 1

The derivative of the function is:

f′(x)=6x2−8xf'(x) = 6x^2 - 8x

At x=1x = 1:

f′(1)=6(1)2−8(1)=−2f'(1) = 6(1)^2 - 8(1) = -2
  1. Find the equation of the normal line at x=1x = 1: The point of tangency is P(1,1)P(1, 1). The slope of the tangent line at this point is f′(1)=−2f'(1) = -2. The slope of the normal line is the negative reciprocal of the tangent slope:
mnormal=−1f′(1)=12m_{\text{normal}} = -\frac{1}{f'(1)} = \frac{1}{2}

The equation of the normal line passing through P(1,1)P(1, 1) is:

y−1=12(x−1)y - 1 = \frac{1}{2}(x - 1)
  1. Find the center of the circle Q(c(1),0)Q(c(1), 0): Since the circle is tangent to the curve at PP, its center Q(c(1),0)Q(c(1), 0) must lie on the normal line. Substituting y=0y = 0 into the normal line equation:
0−1=12(c(1)−1)0 - 1 = \frac{1}{2}(c(1) - 1) −2=c(1)−1  ⟹  c(1)=−1-2 = c(1) - 1 \implies c(1) = -1
  1. Find the radius squared r(1)2r(1)^2: The radius r(1)r(1) is the distance between the center Q(−1,0)Q(-1, 0) and the point P(1,1)P(1, 1):
r(1)2=(1−(−1))2+(1−0)2=22+12=5r(1)^2 = (1 - (-1))^2 + (1 - 0)^2 = 2^2 + 1^2 = 5
  1. Calculate c(1)+r(1)2c(1) + r(1)^2:
c(1)+r(1)2=−1+5=4c(1) + r(1)^2 = -1 + 5 = 4

4\boxed{4}

Final Model Output
4

Comparative Analysis (P7) (Average across runs)

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

Avg Time Taken

DeepSeek V4 Pro
20.2
K-EXAONE-236B-A23B
19.9
Gemini 3.5 Flash
9.1
GPT-5.5
7.4
Solar Pro 3
6.1
Claude Opus 4.8
3.7
KT Mi:dm 2.0 Base Instruct
3.3

Avg Token Usage

K-EXAONE-236B-A23B
2,870
Gemini 3.5 Flash
2,161
KT Mi:dm 2.0 Base Instruct
1,866
Solar Pro 3
1,143
DeepSeek V4 Pro
996
Claude Opus 4.8
551
GPT-5.5
523
    Tokyo · Problem 7 · Gemini 3.5 Flash | EntropyMath