Qwen3-Next-REAP-25B-A3B-Thinking has the following specifications:

Type: Causal Language Models
Number of Parameters: 25B in total and 3B activated
Hidden Dimension: 2048
Number of Layers: 48
Hybrid Layout: 12 * (3 * (Gated DeltaNet -> MoE) -> 1 * (Gated Attention -> MoE))
Gated Attention:
Number of Attention Heads: 16 for Q and 2 for KV
Head Dimension: 256
Rotary Position Embedding Dimension: 64
Gated DeltaNet:
**Number of Linear Attention Heads: 32 for V and 16 for QK
**Head Dimension: 128
Mixture of Experts:
**Number of Experts: 160 (uniformly pruned from 512)
**Number of Activated Experts: 10
**Number of Shared Experts: 1
Context Length: 262,144 natively and extensible up to 1,010,000 tokens
Compression Method: REAP (Router-weighted Expert Activation Pruning)
Compression Ratio: 68.75% expert pruning
Specialized: Math, Physics, Control Engineering, Scientific Writing

Live test video: https://www.bilibili.com/video/BV1ndrRBZEm5/

Solve math problem with IQ1_M

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build : b7738-d98b54812 model : Qwen3-Next-REAP-25B-A3B-Thinking-IQ1_M.gguf modalities : text

available commands: /exit or Ctrl+C stop or exit /regen regenerate the last response /clear clear the chat history /read add a text file

Determine all real numbers t with 0 < t < 1 for which the following statement is true: If, in an acute triangle, the sides with lengthsa, b, and c are opposite the angles with measures α, β, and γ, respectively, as usual, and a/(b + c) = t, then α ≤ β and α ≤ γ

To determine all real numbers $ t \in (0, 1) $ such that the implication holds:

If in an acute triangle, the sides $ a, b, c $ (opposite angles $ \alpha, \beta, \gamma $ respectively) satisfy $ \frac{a}{b + c} = t $, then $ \alpha \leq \beta $ and $ \alpha \leq \gamma $,

we analyze the geometric and algebraic constraints imposed by the triangle being acute and the side-angle correspondence.

Key Observations

In any triangle, the smallest side corresponds to the smallest angle. Thus, the condition $ \alpha \leq \beta $ and $ \alpha \leq \gamma $ is equivalent to $ a \leq b $ and $ a \leq c $, i.e., $ a $ is the smallest side.
For a triangle with sides $ a, b, c $, and $ \frac{a}{b + c} = t $, the triangle inequality requires $ a < b + c $, so $ t < 1 $.
The triangle is acute if all angles are acute, which implies: $$ a^2 < b^2 + c^2,\quad b^2 < a^2 + c^2,\quad c^2 < a^2 + b^2. $$

We are looking for values of $ t $ such that every acute triangle with $ \frac{a}{b + c} = t $ has $ a $ as the smallest side (i.e., $ a \leq b $ and $ a \leq c $).

Critical Analysis

Let’s consider the case where $ a $ is not the smallest side. Suppose there exists a triangle with $ \frac{a}{b + c} = t $ and $ b < a $ (since $ b $ is smaller than $ a $, and $ c > a $, due to triangle inequality). Then:

Let $ b = ka $, $ c = ma $, with $ 0 < k < 1 $ and $ m > 1 $.
Triangle inequality: $ a < b + c \Rightarrow 1 < k + m $.
Acute condition: $ c^2 < a^2 + b^2 \Rightarrow m^2 < 1 + k^2 $.
The value of $ t = \frac{a}{b + c} = \frac{1}{k + m} $.

We now find the minimum possible value of $ t $ for which such a triangle exists with $ a $ not the smallest side (i.e., $ b < a $).

From the acute condition $ m < \sqrt{1 + k^2} $, and triangle inequality $ k + m > 2 $, we solve for the minimal $ t $ such that these constraints are satisfied.

The function $ f(k) = k + \sqrt{1 + k^2} $ is increasing in $ k $, and achieves its maximum at $ k = 1 $, where: $f(1) = 1 + \sqrt{2}, \quad t = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1 \approx 0.414.$

Thus, for $ t > \sqrt{2} - 1 $, there exists an acute triangle with $ \frac{a}{b + c} = t $ and $ a $ not the smallest side (e.g., $ a = 1 $, $ b = 0.95 $, $ c = 1.27 $ gives $ t \approx 0.454 > \sqrt{2} - 1 $).

For $ t < \sqrt{2} - 1 $, it is impossible to have a triangle with $ \frac{a}{b + c} = t $ and $ a $ not the smallest side. Why?

If $ b < a $, then $ m < \sqrt{1 + k^2} $, and $ k + m > 2 $, but for $ k < 1 $, the maximum value of $ k + m $ is less than $ 1 + \sqrt{2} $, so $ t > \sqrt{2} - 1 $, contradicting $ t < \sqrt{2} - 1 $.
Hence, for any acute triangle with $ t < \sqrt{2} - 1 $, $ a $ must be the smallest side.

Conclusion

For $ t < \sqrt{2} - 1 $, every acute triangle with $ \frac{a}{b + c} = t $ has $ a $ as the smallest side ⇒ the implication holds.
For $ t \geq \sqrt{2} - 1 $, there exist acute triangles where $ \frac{a}{b + c} = t $ and $ a $ is not the smallest side ⇒ the implication fails.

Final Answer

$\boxed{(0,\ \sqrt{2} - 1)}$

[ Prompt: 199.7 t/s | Generation: 38.6 t/s ]

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