MathTeacher

An agentic-RAG architecture system that replicates a mathematical professor

Usage

• Copy the .env.model file and rename it to .env
• Obtain GOOGLE_API_KEY and TAVILY_API_KEY and enter in the .env file
• Create and activate your virtual python environment

  python3 -m venv .venv
  source .venv/bin/activate

• Install all the dependencies

  pip install -r requirements.txt

• Run all the cells in data/prepare.ipynb to generate the vector database
• Run the application

  streamlit run src/main.py

Agent Graph Architecure

In this context, green nodes represent components powered by a large language model (LLM), while orange nodes indicate components that utilize external tools.

In the above diagram, KB denotes a Knowledge Base.

Input Guard Rails

This node uses a LLM to determine weather:

• The question is clearly related to mathematics (e.g., algebra, calculus, geometry, etc.)
• The input is phrased respectfully and contains no offensive or inappropriate content
• The input is free from prompt injection or attempts to manipulate the system
• The input is concise, relevant, and meaningful for tutoring or problem-solving
• The input does not request answers to non-math or unrelated topics (e.g., politics, jokes, personal advice)

Based on the above rules, the query is scored on the scale of 100. This score is then used to reject queries with less than score of 80 with an output explanation.

Output Guard Rails

This node uses a LLM to determine weather:

• The response correctly answers the math question
• The explanation is step-by-step and educational
• The answer is safe, non-toxic, and stays in the math domain
• There are no hallucinations, irrelevant info, or broken logic

Similar to the input guard rails, the responses are scored on the scale of 100 and is forwarded to the output corrector node to resolve the mistakes.

Screen Shot

JEE Advanced Math Benchmark

Question 1

If

$$ f(x) = \left|\begin{array}{ccc} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos(x) & \cos(x) & -\sin(x) \\ \sin(x) & \sin(x) & \cos(x) \end{array}\right| $$

then:

[A] $f'(x) = 0$ at exactly three points in $(-\pi, \pi)$
[B] $f'(x) = 0$ at more than three points in $(-\pi, \pi)$
[C] $f(x)$ attains its maximum at $x = 0$
[D] $f(x)$ attains its minimum at $x = 0$

Answer

Let $f(x)=\left|\begin{array}{ccc}\cos (2 x) & \cos (2 x) & \sin (2 x) \ -\cos x & \cos x & -\sin x \ \sin x & \sin x & \cos x\end{array}\right|$.

We can simplify the determinant by performing the column operation

$C_1 \rightarrow C_1 - C_2$: $f(x)=\left|\begin{array}{ccc}0 & \cos (2 x) & \sin (2 x) \ -2\cos x & \cos x & -\sin x \ 0 & \sin x & \cos x\end{array}\right|$

Now, we can expand the determinant along the first column:

$f(x) = 0 \cdot C_{11} + (-1)(-2\cos x) \left|\begin{array}{cc}\cos (2 x) & \sin (2 x) \ \sin x & \cos x\end{array}\right| + 0 \cdot C_{31}$

$f(x) = 2\cos x [\cos(2x)\cos x - \sin(2x)\sin x]$

$f(x) = 2\cos x [\cos(2x+x)]$

$f(x) = 2\cos x \cos(3x)$

Now, we need to find the derivative of $f(x)$:

$f'(x) = 2[-\sin x \cos(3x) - 3\cos x \sin(3x)]$

$f'(x) = -2[\sin x \cos(3x) + 3\cos x \sin(3x)]$

Using the product-to-sum identities, we have:

$\sin x \cos(3x) = \frac{1}{2}[\sin(x+3x) + \sin(x-3x)] = \frac{1}{2}[\sin(4x) - \sin(2x)]$

$\cos x \sin(3x) = \frac{1}{2}[\sin(3x+x) + \sin(3x-x)] = \frac{1}{2}[\sin(4x) + \sin(2x)]$

$f'(x) = -2[\frac{1}{2}(\sin(4x) - \sin(2x)) + \frac{3}{2}(\sin(4x) + \sin(2x))]$

$f'(x) = -[\sin(4x) - \sin(2x) + 3\sin(4x) + 3\sin(2x)]$

$f'(x) = -[4\sin(4x) + 2\sin(2x)]$

$f'(x) = -2[2\sin(4x) + \sin(2x)]$

$f'(x) = -2[2(2\sin(2x)\cos(2x)) + \sin(2x)]$

$f'(x) = -2\sin(2x)[4\cos(2x) + 1]$

We want to find when $f'(x) = 0$, so we need to solve:

$-2\sin(2x)[4\cos(2x) + 1] = 0$

This means either $\sin(2x) = 0$ or $4\cos(2x) + 1 = 0$.

Case 1: $\sin(2x) = 0$

$2x = n\pi$, where $n$ is an integer.

$x = \frac{n\pi}{2}$

In the interval $(-\pi, \pi)$, we have $x = -\frac{\pi}{2}, 0, \frac{\pi}{2}$.

Case 2: $4\cos(2x) + 1 = 0$

$\cos(2x) = -\frac{1}{4}$

$2x = \arccos(-\frac{1}{4}) + 2n\pi$ or $2x = -\arccos(-\frac{1}{4}) + 2n\pi$

$x = \frac{1}{2}\arccos(-\frac{1}{4}) + n\pi$ or $x = -\frac{1}{2}\arccos(-\frac{1}{4}) + n\pi$

Since $\arccos(-\frac{1}{4}) \approx 1.823$, we have $\frac{1}{2}\arccos(-\frac{1}{4}) \approx 0.9115$.

Then $x \approx 0.9115 + n\pi$ or $x \approx -0.9115 + n\pi$.

In the interval $(-\pi, \pi)$, we have $x \approx -0.9115, 0.9115, -0.9115 + \pi \approx 2.23, 0.9115 - \pi \approx -2.23$.

So, $x \approx -2.23, -0.9115, 0.9115, 2.23$.

In total, we have $x = -\frac{\pi}{2}, 0, \frac{\pi}{2}, -2.23, -0.9115, 0.9115, 2.23$.

These are 7 points in $(-\pi, \pi)$.

Therefore, $f'(x) = 0$ at more than three points in $(-\pi, \pi)$.

Final Answer: The final answer is $\boxed{[B]}$

Question 2

Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1. Then what is the inradius of the triangle ABC?

Answer

Let the sides of the triangle be $a, b, c$ such that $a < b < c$.

Since the sides are in arithmetic progression, we can write $a = b-d$ and $c = b+d$ for some $d > 0$.

Since the triangle is obtuse, $a^2 + b^2 < c^2$.

The angles $A, B, C$ are opposite to sides $a, b, c$ respectively. Since $a < b < c$, we have $A < B < C$.

The largest angle is $C$ and the smallest angle is $A$. We are given that $C - A = \frac{\pi}{2}$.

Also, $A + B + C = \pi$. Substituting $C = A + \frac{\pi}{2}$, we get $A + B + A + \frac{\pi}{2} = \pi$, so $2A + B = \frac{\pi}{2}$.

Thus $B = \frac{\pi}{2} - 2A$. Since $A, B, C > 0$, we have $0 < A < \frac{\pi}{4}$.

Using the Law of Sines, we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the circumradius.

Since $R = 1$, we have $a = 2\sin A$, $b = 2\sin B = 2\sin(\frac{\pi}{2} - 2A) = 2\cos(2A)$, and $c = 2\sin C = 2\sin(A + \frac{\pi}{2}) = 2\cos A$.

Since $a, b, c$ are in arithmetic progression, $a + c = 2b$, so $2\sin A + 2\cos A = 2(2\cos(2A))$.

Thus $\sin A + \cos A = 2\cos(2A) = 2(\cos^2 A - \sin^2 A) = 2(\cos A - \sin A)(\cos A + \sin A)$.

Since $\sin A + \cos A \neq 0$, we can divide by $\sin A + \cos A$ to get $1 = 2(\cos A - \sin A)$.

So $\cos A - \sin A = \frac{1}{2}$. Squaring both sides, we get $\cos^2 A - 2\sin A \cos A + \sin^2 A = \frac{1}{4}$.

$1 - 2\sin A \cos A = \frac{1}{4}$, so $2\sin A \cos A = \frac{3}{4}$.

Thus $\sin(2A) = \frac{3}{4}$.

Since $\cos(2A) = \cos^2 A - \sin^2 A = \frac{1}{2}$, we have $\cos(2A) = \frac{1}{2}$.

Then $\cos(2A) = \pm \sqrt{1 - \sin^2(2A)} = \pm \sqrt{1 - \frac{9}{16}} = \pm \sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4}$.

Since $\cos A - \sin A = \frac{1}{2} > 0$, we have $A < \frac{\pi}{4}$. Thus $2A < \frac{\pi}{2}$, so $\cos(2A) > 0$.

Therefore $\cos(2A) = \frac{\sqrt{7}}{4}$. But we found that $\cos(2A) = \frac{1}{2}(\cos A - \sin A) = \frac{\sqrt{7}}{4}$.

Since $\cos(2A) = \frac{\sqrt{7}}{4}$, $b = 2\cos(2A) = \frac{\sqrt{7}}{2}$.

Since $\sin(2A) = \frac{3}{4}$, $2A = \arcsin(\frac{3}{4})$.

We have $A = \frac{1}{2} \arcsin(\frac{3}{4})$.

Also, we have $\cos A - \sin A = \frac{1}{2}$. Multiplying by $\frac{1}{\sqrt{2}}$, we get $\frac{1}{\sqrt{2}}\cos A - \frac{1}{\sqrt{2}}\sin A = \frac{1}{2\sqrt{2}}$.

$\cos(\frac{\pi}{4})\cos A - \sin(\frac{\pi}{4})\sin A = \frac{1}{2\sqrt{2}}$.

$\cos(A + \frac{\pi}{4}) = \frac{1}{2\sqrt{2}}$. So $A + \frac{\pi}{4} = \arccos(\frac{1}{2\sqrt{2}})$.

$A = \arccos(\frac{1}{2\sqrt{2}}) - \frac{\pi}{4}$.

We have $a = 2\sin A$, $b = \frac{\sqrt{7}}{2}$, $c = 2\cos A$.

The semiperimeter $s = \frac{a+b+c}{2} = \frac{2\sin A + \frac{\sqrt{7}}{2} + 2\cos A}{2} = \sin A + \cos A + \frac{\sqrt{7}}{4} = \frac{1}{2} + \frac{\sqrt{7}}{4} + \frac{\sqrt{7}}{4} = \frac{1}{2} + \frac{\sqrt{7}}{2} = \frac{2+\sqrt{7}}{4}$

The area of the triangle is $\frac{abc}{4R} = \frac{2\sin A \cdot \frac{\sqrt{7}}{2} \cdot 2\cos A}{4} = \frac{4\sin A \cos A \cdot \frac{\sqrt{7}}{2}}{4} = \frac{\sqrt{7}}{2} \sin A \cos A = \frac{\sqrt{7}}{4} \sin(2A) = \frac{\sqrt{7}}{4} \cdot \frac{3}{4} = \frac{3\sqrt{7}}{16}$.

The inradius $r = \frac{\text{Area}}{s} = \frac{\frac{3\sqrt{7}}{16}}{\frac{2+\sqrt{7}}{4}} = \frac{3\sqrt{7}}{16} \cdot \frac{4}{2+\sqrt{7}} = \frac{3\sqrt{7}}{4(2+\sqrt{7})} = \frac{3\sqrt{7}(2-\sqrt{7})}{4(4-7)} = \frac{6\sqrt{7} - 21}{-12} = \frac{21-6\sqrt{7}}{12} = \frac{7-2\sqrt{7}}{4}$.

Final Answer: The final answer is $\boxed{\frac{7-2\sqrt{7}}{4}}$