British Mathematical Olympiad

Conducting Body: United Kingdom Mathematics Trust (UKMT)
Who It Is For: High-scoring students from the Senior Mathematical Challenge
Exam Papers: Two rounds: BMO1 and BMO2
Question Type: A small number of long, proof-based problems
Duration: 3.5 hours for each round
Marking Focus: Complete mathematical proofs, not just final answers
Held: Annually
Qualifies For: Invitation to BMO2; selection for the IMO training squad

The British Mathematical Olympiad (BMO) is the United Kingdom's national mathematics competition for schools. It is run by the UK Mathematics Trust and entry is by invitation only, extended to the top-scoring students from the Senior Mathematical Challenge. The BMO is a key part of the selection process for the UK team that competes in the International Mathematical Olympiad (IMO).

The competition consists of two rounds, BMO1 and BMO2. Each is a single 3.5-hour paper containing a small number of complex problems. The exam is not a test of curriculum knowledge or calculation speed. Instead, it assesses deep mathematical thinking, ingenuity, and the ability to construct logical arguments.

The primary difficulty of the BMO lies in its emphasis on proof. A correct numerical answer on its own is worth very little. To score well, a student must write a complete and rigorous mathematical proof for their solution. This requires a shift from finding answers to explaining why those answers are correct, a skill that demands practice, patience, and a different approach to problem-solving than is typical in school examinations.

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JEE Advanced 2025

Let $\mathbb{R}$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in \mathbb{R} \times \mathbb{R} : x > 0,\, y > \frac{1}{x},\, 5x - 4y - 1 > 0,\, 4x + 4y - 17 < 0\}$ is

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$x > 0,\quad y > \dfrac1x$ $5x - 4y - 1 > 0,\quad 4x + 4y - 17 < 0$ $\text{To find: area of the region}$ $L_1: y = \dfrac{5x-1}{4},\quad L_2: y = \dfrac{17-4x}{4}$ $H: y = \dfrac1x$ $\dfrac1x < y < \min\!\left(\dfrac{5x-1}{4},\ \dfrac{17-4x}{4}\right)$ $L_1 \le L_2 \iff x \le 2$ $H \cap L_1:\ x = 1,\ y = 1$ $L_1 \cap L_2:\ x = 2,\ y = \dfrac94$ $H \cap L_2:\ x = 4,\ y = \dfrac14$ $A = \int_1^2\!\left(\dfrac{5x-1}{4} - \dfrac1x\right)dx + \int_2^4\!\left(\dfrac{17-4x}{4} - \dfrac1x\right)dx$ $\int_1^2\!\left(\dfrac{5x-1}{4} - \dfrac1x\right)dx = \dfrac{13}{8} - \ln 2$ $\int_2^4\!\left(\dfrac{17-4x}{4} - \dfrac1x\right)dx = \dfrac52 - \ln 2$ $A = \dfrac{13}{8} + \dfrac52 - 2\ln 2 = \dfrac{33}{8} - \ln 4$ $\boxed{\dfrac{33}{8} - \log_e 4} \quad (2)$

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