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The 43rd National University Student Mathematics Competition, Section 2

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The 43rd National University Student Mathematics Competition

November 1, 2025 (10:30 - 13:00)

1.

A differentiable function of two variables $f:\mathbb{R}^2 \to \mathbb{R}$ satisfying all of the following conditions is given.

For real numbers $a,b$ satisfying $a^2+b^2=1$, find the maximum and minimum possible values of the following limit.

Solution.

This problem is solved by using directional derivatives.

$\nabla f(0,0)\cdot \mathbf{v} = D_{\mathbf{v}} f(0,0)=\lim_{t\to 0}\frac{f(t\mathbf{v})-f(0,0)}{t}$

∴ $2=(\frac{\partial f}{\partial x}\vert_{x=0},\frac{\partial f}{\partial y}\vert_{y=0})\cdot(1,1),\ -1=(\frac{\partial f}{\partial x}\vert_{x=0},\frac{\partial f}{\partial y}\vert_{y=0})\cdot(1,2)$

∴ $\frac{\partial f}{\partial x}\vert_{x=0}=5,\ \frac{\partial f}{\partial y}\vert_{y=0}=-3$

$\lim_{t\to 0}\frac{f(at,bt)}{t}=\nabla f(0,0)\cdot(a,b)=5a-3b=5\sin\theta-3\cos\theta=\sqrt{5^2+3^2}\sin(\theta+\alpha)$

∴ The maximum value of (the target expression) is $\sqrt{34}$, and the minimum value is $-\sqrt{34}$.

2.

Find the solution $f$ of the following differential equation.

Solution.

Let $g(x):=f’(x)+f(x)$, then $g(1)=0$.

$g’(x)+g(x)=2x-2e^{-x}\ \Leftrightarrow\ e^x g’(x)+e^x g(x)=2xe^x-2$

∴ $e^x g(x)=e^x f’(x)+e^x f(x)=-2x-1+2$ (∵ $g(1)=0$)

∴ $e^x f(x)=-2\ln\left|x\right|+2x-2$ (∵ $f(1)=0$)

∴ $f(x)=2e^{-x}(-\ln\left|x\right|+x-1)$

3.

Five points $A,B,C,D,E$ lying on a plane in a 3-dimensional vector space satisfy all of the following conditions.

$\left|\mathbf{AB}\right|=\left|\mathbf{BC}\right|=\left|\mathbf{CD}\right|=\left|\mathbf{DE}\right|=\left|\mathbf{EA}\right|=1$

$\mathbf{AB}\times\mathbf{BC}=\mathbf{BC}\times\mathbf{CD}=\mathbf{CD}\times\mathbf{DE}=\mathbf{DE}\times\mathbf{EA}=\mathbf{EA}\times\mathbf{AB}\ne\mathbf{0}$

$\mathbf{AB}\cdot\mathbf{BC}=\mathbf{BC}\cdot\mathbf{CD}=\mathbf{CD}\cdot\mathbf{DE}=\mathbf{DE}\cdot\mathbf{EA}=\mathbf{EA}\cdot\mathbf{AB}$

⑴ Show that $\mathbf{CE}$ is a scalar multiple of $\mathbf{AB}$.

⑵ Find all possible values of the scalar $k$ such that $\mathbf{CE}=k\mathbf{AB}$, and for each case describe what kind of figure $ABCDE$ is.

Solution.

⑴

⑵

Since $\mathbf{AB}\times\mathbf{BC}=\mathbf{BC}\times\mathbf{CD}$ and $\mathbf{AB}\cdot\mathbf{BC}=\mathbf{BC}\cdot\mathbf{CD}$, we have $\angle ABC=\angle BCD$.

This is because $\sin\angle ABC=\sin\angle BCD$ and $\cos\angle ABC=\cos\angle ABC$.

Therefore, we can see that $AB\to BC\to CD\to DE\to EA$ rotates by a constant angle.

Pay attention to the following figure.

Let that constant angle be $\theta_2$. Then $\theta_1-5(\pi-\theta_2)\equiv\theta_1\pmod{2\pi}\ \Leftrightarrow\ 5\theta_2\equiv\pi\pmod{2\pi}\ \Leftrightarrow\ \theta_2=\pi/5,\ 3\pi/5$.

If $\theta_2=3\pi/5$, it becomes a regular pentagon. Applying the law of cosines in $\triangle ABE$ gives $1^2+1^2-2\cos(3\pi/5)=\left|k\right|^2$, and applying the law of cosines in $\triangle BCE$ gives $1^2+\left|k\right|^2-2k\cos(2\pi/5)=\left|k\right|^2$. Since $\cos(3\pi/5)+\cos(2\pi/5)=0$, solving simultaneously yields $\left|k\right|^3-2\left|k\right|-1=(\left|k\right|+1)(\left|k\right|^2-\left|k\right|-1)=0$. Hence $\left|k\right|=(1+\sqrt{5})/2$. Considering the direction of vectors, we can obtain $k=(-1-\sqrt{5})/2$.

If $\theta_2=\pi/5$, it becomes a star shape. By group theory, we obtain $k=(-1+\sqrt{5})/2$, which is irrationally conjugate to $k=(-1-\sqrt{5})/2$.

4.

For a positive integer $n$ and positive numbers $a_{ij}$ $(1\le i,j\le n)$, define $A_n,B_n,C_n$ as follows.

Prove the following inequality.

Solution.

For understanding, first note that $n^2C_n=(1^2+\cdots+1^2)C_n\ge A_n$ can be quickly proved by the Cauchy–Schwarz inequality.

$B_n^2\le n^2C_n^2$ can be proved as follows.

Also, $A_n\le nB_n$ can be proved as follows.

For reference, the rearrangement inequality is an inequality that holds when variables have symmetry. For example, in cases like $\alpha^2+\beta^2+\gamma^2\ge \alpha\beta+\beta\gamma+\gamma\alpha$, it is typical that pairing larger values together tends to make the total value larger.

5.

For a positive integer $n$ and a real number $\alpha=\pi/(n+1)$, define an $n\times n$ matrix $A$ as follows.

Find $\left|\det A\right|$.

Solution.

It is similar to 40th Algebra Competition, Section 2, Problem 3, but it is a different problem because there do not exist $\alpha_i,\beta_j$ such that $\alpha_i-\beta_j=\pi/2-i\times j\times \pi/(n+1)$.

Of note, since linear algebra deals with linear expressions, there is no room for a quadratic term like $i\times j$, so it is impossible to decompose the matrix $A$ into a product of two matrices.

Meanwhile, check the following formula about sums of cosine sequences (cf. 36th Algebra Competition, Section 1, Problem 7).

Therefore $A^{\mathsf{T}}A$ is as follows.

Therefore the determinant of $A$ is as follows.

This kind of solution is inspired by the process of computing the determinant of a Gram matrix.

For reference, we expressed the $(x,y)$-entry of $A$ as $\operatorname{Im}(\exp(ixy\alpha))$ and tried to express it like a Vandermonde matrix, but we could not compute the determinant because of the $\operatorname{Im}$.

6.

Let $A=(a_{jk})$ be a real matrix of size $(n+1)\times(n+1)$, and let $B=(b_{jk})$ be an $n\times n$ matrix whose entries are polynomials in $n$ variables, defined as follows.

Prove the following.

Solution.

When $n=3$, $B$ and $\det B$ can be written as follows.

If we use the concept of substitution appropriately, we can show that $\int\cdots\int \det B,dx_1\cdots dx_n=\det\big((\int b_{jk},dx_j)_{1\le j,k\le n}\big)$, and together with properties of determinants, we can prove (the target expression).

In the middle, we restricted to the case $n=3$ to aid understanding, but it is easy to see that it holds for all $n$ without loss of generality.

7.

For positive integers $n$, $k\ge 2$, define a function $F:(0,1)^n\to \mathbb{R}$ as follows.

Find the range of positive $p$ for which the following integral converges.

Solution.

$F$ has a scaling property. That is, $F(\lambda\mathbf{x})=\lambda^{1-k}F(\mathbf{x})$.

In the region where $\mathbf{x}$ is far from the origin, $F(\mathbf{x})$ does not diverge and is well-bounded by some finite constant.

Therefore, the integral of $F(\mathbf{x})^p$ over that region always converges.

On the other hand, letting $\mathbf{x}=r\theta$, we have $F(\mathbf{x})=F(r\theta)=r^{1-k}F(\theta)$, and since $F(\theta)$ is continuous, we get $C_1 r^{1-k}\le F(\mathbf{x})\le C_2 r^{1-k}$.

Therefore, in the region where $\mathbf{x}$ is close to the origin ($\left|\mathbf{x}\right|<\varepsilon\ll 1$), we have $F(\mathbf{x})^p\sim r^{p(1-k)}$.

Hence (the target expression) $=\int_{(0,1)^n} F(\mathbf{x})^p,d\mathbf{x}\sim \int_{(0,2\pi)}\int_{(0,\varepsilon)} r^{n-1}\cdot r^{p(1-k)},dr,d\theta<\infty$ $\Leftrightarrow\ (n-1)+p(1-k)+1>0\ \Leftrightarrow\ p<\frac{n}{k-1}$.

Here, we used the fact that since $F(\mathbf{x})$ diverges near $\mathbf{x}=0$, we have $\int_{(0,1)^n} F(\mathbf{x})^p,d\mathbf{x}\sim \int_{(0,\varepsilon)^n} F(\mathbf{x})^p,d\mathbf{x}$.

Input: 2025.12.05 20:25