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Math· olympiad

Omni-MATH

Peking University · 2024

4,428 olympiad-level competition problems across 33+ subdomains and difficulty tiers.

Mathematics
GitHub stars
Task type
open-ended
Modality
text
Access
open
Size
4,428 items
License
Apache-2.0
Metrics
accuracy
domain
Mathematics -> Algebra -> Other
difficulty
8
problem
Let $ n(\ge2) $ be a positive integer. Find the minimum $ m $, so that there exists $x_{ij}(1\le i ,j\le n)$ satisfying: (1)For every $1\le i ,j\le n, x_{ij}=max\{x_{i1},x_{i2},...,x_{ij}\} $ or $ x_{ij}=max\{x_{1j},x_{2j},...,x_{ij}\}.$ (2)For every $1\le i \le n$, there are at most $m$ indices $k$ with $x_{ik}=max\{x_{i1},x_{i2},...,x_{ik}\}.$ (3)For every $1\le j \le n$, there are at most $m$ indices $k$ with $x_{kj}=max\{x_{1j},x_{2j},...,x_{kj}\}.$
solution
Let \( n (\geq 2) \) be a positive integer. We aim to find the minimum \( m \) such that there exists \( x_{ij} \) (for \( 1 \leq i, j \leq n \)) satisfying the following conditions: 1. For every \( 1 \leq i, j \leq n \), \( x_{ij} = \max \{ x_{i1}, x_{i2}, \ldots, x_{ij} \} \) or \( x_{ij} = \max \{ x_{1j}, x_{2j}, \ldots, x_{ij} \} \). 2. For every \( 1 \leq i \leq n \), there are at most \( m \) indices \( k \) such that \( x_{ik} = \max \{ x_{i1}, x_{i2}, \ldots, x_{ik} \} \). 3. For every \( 1 \leq j \leq n \), there are at most \( m \) indices \( k \) such that \( x_{kj} = \max \{ x_{1j …
answer
1 + \left\lceil \frac{n}{2} \right\rceil
source
china_team_selection_test
domain
Mathematics -> Geometry -> Plane Geometry -> Triangulations
difficulty
7
problem
In an acute scalene triangle $ABC$, points $D,E,F$ lie on sides $BC, CA, AB$, respectively, such that $AD \perp BC, BE \perp CA, CF \perp AB$. Altitudes $AD, BE, CF$ meet at orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \perp EF$ and $HQ \perp EF$. Lines $DP$ and $QH$ intersect at point $R$. Compute $HQ/HR$.
solution
In an acute scalene triangle \(ABC\), points \(D, E, F\) lie on sides \(BC, CA, AB\), respectively, such that \(AD \perp BC\), \(BE \perp CA\), \(CF \perp AB\). Altitudes \(AD, BE, CF\) meet at orthocenter \(H\). Points \(P\) and \(Q\) lie on segment \(EF\) such that \(AP \perp EF\) and \(HQ \perp EF\). Lines \(DP\) and \(QH\) intersect at point \(R\). We aim to compute \(\frac{HQ}{HR}\). Note that \(H\) and \(A\) are the incenter and \(D\)-excenter of \(\triangle DEF\), respectively. Thus, \(HQ\) is an inradius of \(\triangle DEF\). Let \(R'\) be the reflection of \(Q\) over \(H\). The homo …
answer
1
source
usa_team_selection_test

Real rows from the Hugging Face datasets server · long values truncated