README: example (AIME 2025 II #15, Quark vs Qwen3.5-9B), Ollama in Usage, Theo in the intro

README.md

@@ -24,7 +24,7 @@ our curated physics, maths and coding mix.
 It outperforms the base model on a range of benchmarks, and reaches
 graduate-level science scores comparable to much larger open models.
 
-Quark is the first in a series of small, powerful scientific reasoning models. You can use Quark locally or natively on [Theo Platform](https://firstprinciples.com/sign-up).
+Quark is the first in a series of small, powerful scientific reasoning models. You can run it locally (see [Usage](#usage)) or on [Theo Platform](https://firstprinciples.com/sign-up), a research platform for complex problems, where you can use it alongside other frontier models, check its results with Python, Mathematica and Sage, and keep each session as a research graph you can revise one step at a time.
 
 ## Results
 
@@ -63,9 +63,7 @@ Both models were measured by us with the same setup.
 
 Quark was fine-tuned on our curated mix of about 200,000 physics, science, maths and coding problems,
 each with a worked solution written by [Qwen3.8-Flash-Next](https://huggingface.co/Qwen/Qwen3.8-Flash-Next).
-Solutions were checked for correctness, and problems overlapping the benchmarks above were removed. Most
-examples include the full reasoning; a fifth are direct answers, so the model keeps both thinking and
-non-thinking modes.
+Solutions were checked for correctness, and problems overlapping the benchmarks above were removed.
 
 ## Usage
 
@@ -76,6 +74,12 @@ vllm serve FirstPrinciplesLabs/quark \
   --max-model-len 135168 --reasoning-parser qwen3
 ```
 
+Or run it locally with [Ollama](https://ollama.com/FirstPrinciples/quark) (BF16 weights, with the recommended sampling settings and a 131,072-token context built in):
+
+```bash
+ollama run firstprinciples/quark
+```
+
 Recommended sampling (from the Qwen3.5 model card):
 
 | | Thinking (default) | Non-thinking |
@@ -85,6 +89,246 @@ Recommended sampling (from the Qwen3.5 model card):
 | presence penalty | 1.5 | 2.0 |
 | max output tokens | up to 131,072 | 32,768 |
 
+## Example
+
+A hard problem from AIME 2025 (II, problem 15), in thinking mode with the recommended sampling settings:
+
+> Let \\(f(x)=\dfrac{(x-18)(x-72)(x-98)(x-k)}{x}\\).
+> There exist exactly three positive real values of \\(k\\) such that \\(f\\) has a minimum at exactly two real values of \\(x\\). Find the sum of these three values of \\(k\\).
+
+| | Qwen3.5-9B | **Quark** |
+|---|---|---|
+| Correct, out of 16 attempts | 1 | **10** |
+| Most common answer | 188 ✗ (14 of 16 attempts) | **240 ✓** |
+| Approach | Finds no way in and guesses that k must be one of the existing roots: 18 + 72 + 98 = 188 | Turns "a minimum at two points" into an equation for k and solves it |
+
+**How Quark solves it** (summarised from its answer):
+
+1. If the minimum value \\(m\\) is reached at two points \\(u\\) and \\(v\\), the numerator minus \\(mx\\) must be a perfect square:
+$$
+(x-18)(x-72)(x-98)(x-k)-mx=(x-u)^2(x-v)^2 .
+$$
+2. Matching coefficients and eliminating \\(u\\) and \\(v\\) leaves one equation in \\(k\\):
+$$
+(k^2-376k-5120)^2 = 8128512\,k .
+$$
+3. It factors as \\((k-8)(k-32)(k-200)(k-512)=0\\), and checking the sign before squaring rules out \\(k=512\\).
+4. So \\(k = 8, 32, 200\\), and the answer is **240**.
+
+Both models' full written answers (each follows several thousand tokens of reasoning, not shown):
+
+<details>
+<summary><b>Quark</b>: 240 (correct)</summary>
+
+
+We interpret “ \\(f\\) has a minimum at exactly two real values of \\(x\\)” as meaning that the minimum value is attained at exactly two distinct \\(x\\)-values (the intended interpretation; otherwise \\(f\\) has no global minimum on all real \\(x\neq0\\)).
+
+Let
+
+$$
+P(x)=(x-18)(x-72)(x-98)(x-k).
+$$
+
+First expand \\(P(x)\\). We have
+
+$$
+18+72+98=188,
+$$
+
+$$
+18\cdot72+18\cdot98+72\cdot98=10116,
+$$
+
+$$
+18\cdot72\cdot98=127008.
+$$
+
+Thus
+
+$$
+P(x)=x^4-(k+188)x^3+(10116+188k)x^2-(127008+10116k)x+127008k.
+$$
+
+Suppose the minimum value is \\(m\\), attained at two distinct \\(x\\)-values \\(u\neq v\\). Then \\(f(x)-m\\) has double zeros at \\(u\\) and \\(v\\). Since
+
+$$
+x(f(x)-m)=P(x)-mx,
+$$
+
+we must have
+
+$$
+P(x)-mx=(x-u)^2(x-v)^2.
+$$
+
+Now expand the right-hand side:
+
+$$
+(x-u)^2(x-v)^2
+=
+x^4-2(u+v)x^3+\bigl((u+v)^2+2uv\bigr)x^2-2uv(u+v)x+u^2v^2.
+$$
+
+Matching coefficients gives
+
+$$
+2(u+v)=k+188, \tag{1}
+$$
+
+$$
+(u+v)^2+2uv=10116+188k, \tag{2}
+$$
+
+$$
+u^2v^2=127008k. \tag{3}
+$$
+
+Let
+
+$$
+r=u+v,\qquad s=uv.
+$$
+
+Since \\(u,v\gt 0\\), we have \\(r\gt 0\\) and \\(s\gt 0\\). From (1),
+
+$$
+r=\frac{k+188}{2}.
+$$
+
+From (3),
+
+$$
+s=\sqrt{127008k}.
+$$
+
+Substitute into (2):
+
+$$
+\left(\frac{k+188}{2}\right)^2+2\sqrt{127008k}=10116+188k.
+$$
+
+Simplify:
+
+$$
+\frac{k^2}{4}+94k+8836+2\sqrt{127008k}=10116+188k,
+$$
+
+$$
+\frac{k^2}{4}-94k-1280=-2\sqrt{127008k}.
+$$
+
+Multiplying by \\(4\\),
+
+$$
+k^2-376k-5120=-8\sqrt{127008k}. \tag{4}
+$$
+
+Now square both sides:
+
+$$
+(k^2-376k-5120)^2=64\cdot127008\,k.
+$$
+
+Since \\(64\cdot127008=8128512\\),
+
+$$
+(k^2-376k-5120)^2-8128512k=0.
+$$
+
+This factors as
+
+$$
+(k-8)(k-32)(k-200)(k-512)=0.
+$$
+
+But equation (4) requires the left-hand side \\(k^2-376k-5120\\) to be negative. Checking the candidates:
+
+$$
+\begin{aligned}
+k=8&:\quad 64-3008-5120=-8064,\\
+k=32&:\quad 1024-12032-5120=-16128,\\
+k=200&:\quad 40000-75200-5120=-40320,\\
+k=512&:\quad 262144-192512-5120=64512.
+\end{aligned}
+$$
+
+Thus \\(k=512\\) is extraneous, while
+
+$$
+k=8,\;32,\;200
+$$
+
+are valid.
+
+Their sum is
+
+$$
+8+32+200=240.
+$$
+
+$$
+\boxed{240}
+$$
+
+
+</details>
+
+<details>
+<summary><b>Qwen3.5-9B</b>: 188 (incorrect)</summary>
+
+
+The function is given by \\(f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}\\).
+
+We are looking for the number of local minima.
+
+Let \\(P(x) = (x-18)(x-72)(x-98)(x-k)\\). Then \\(f(x) = P(x)/x\\).
+
+The critical points of \\(f(x)\\) are given by \\(f'(x) = 0\\), which is equivalent to \\(x P'(x) - P(x) = 0\\). Let \\(Q(x) = x P'(x) - P(x)\\).
+
+\\(Q(x)\\) is a polynomial of degree 4.
+
+We analyzed the signs of \\(Q(x)\\) at the roots of \\(P(x)\\), denoted \\(r_1, r_2, r_3, r_4\\) (sorted).
+
+We found that \\(Q(r_i)\\) alternates in sign: \\(Q(r_1) \lt  0\\), \\(Q(r_2) \gt  0\\), \\(Q(r_3) \lt  0\\), \\(Q(r_4) \gt  0\\).
+
+Also \\(Q(0) \lt  0\\).
+
+Using the Intermediate Value Theorem, \\(Q(x)\\) has roots in the intervals \\((-\infty, 0)\\), \\((r_1, r_2)\\), \\((r_2, r_3)\\), and \\((r_3, r_4)\\).
+
+Let these roots be \\(c_1, c_2, c_3, c_4\\).
+
+\\(c_1\\) is in \\((-\infty, 0)\\), and corresponds to a local maximum (since \\(f \to -\infty\\) at both ends of this interval).
+
+\\(c_2\\) is in \\((r_1, r_2)\\), and corresponds to a local minimum (since \\(f\\) changes from 0 to negative to 0).
+
+\\(c_3\\) is in \\((r_2, r_3)\\), and corresponds to a local maximum (since \\(f\\) changes from 0 to positive to 0).
+
+\\(c_4\\) is in \\((r_3, r_4)\\), and corresponds to a local minimum (since \\(f\\) changes from 0 to negative to 0).
+
+Thus, for generic \\(k\\) (distinct roots), \\(f(x)\\) has exactly two local minima.
+
+The problem states there are exactly three values of \\(k\\). This suggests that for these values, the condition holds, or perhaps the interpretation of "minimum" or the number of extrema changes. However, based on standard interpretations of such problems where the set of solutions is finite and associated with the roots of the polynomial, the values of \\(k\\) are likely the roots of \\(P(x)\\) where the polynomial has a multiple root.
+
+If \\(k\\) coincides with one of the existing roots \\(18, 72, 98\\), the polynomial \\(P(x)\\) has a root of multiplicity at least 2.
+
+Specifically, if \\(k=18\\), \\(r_1=18\\) is a double root. If \\(k=72\\), \\(r_2=72\\) is a double root. If \\(k=98\\), \\(r_3=98\\) is a double root.
+
+These values are natural candidates.
+
+If \\(k\\) is distinct from 18, 72, 98, and positive, the number of minima is consistently found to be 2.
+
+Given the constraint "exactly three positive real values", it implies that the condition holds specifically for \\(k=18, 72, 98\\).
+
+(Note: There might be a subtle nuance regarding the definition of minimum or global minimum, or perhaps my analysis of generic \\(k\\) having 2 minima implies the condition holds for all \\(k\\), but the problem structure implies a finite set. In competitions, such "exactly three" questions often point to the roots).
+
+Assuming the values are \\(18, 72, 98\\).
+
+Sum \\(= 18 + 72 + 98 = 188\\).
+
+\\(\boxed{188}\\)
+
+</details>
+
 ## Limitations
 
 - **Needs a long output budget.** At a 32K-token limit about 20% of GPQA answers are cut off and GPQA drops