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The 42nd National Undergraduate Mathematics Contest, Field 1

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The 42nd National Undergraduate Mathematics Contest

November 2, 2024 (10:30 - 13:00)

1.

For the following matrix $A$, find real numbers $a, b, c$ such that $A^5 = aA^2 + bA + cI$. (Here, $I$ is the $3 \times 3$ identity matrix.)

Solution.

We can find the following relation.

∴ $A^5 = A^2 \times (-3A) = (-3)^2 A$, so $a = 0$, $b = 9$, $c = 0$

2.

In 3-dimensional coordinate space, let $S$ be the surface given by $4z^2 = x^2 + y^2 - 1$. Show that the surface $S$ contains exactly two lines passing through the point $P(1, 2, 1)$, and if the acute angle between the two lines is $\theta$, find the value of $\cos \theta$.

Solution.

$S$: The visualization result of $4z^2 = x^2 + y^2 - 1$ is as follows.

A line passing through $P(1, 2, 1)$ can be written as $(at + 1, bt + 2, ct + 1)$, $t \in \mathbb{R}$, so

$4(ct + 1)^2 = (at + 1)^2 + (bt + 2)^2 - 1 \Leftrightarrow 4c^2 = a^2 + b^2$ (Equation 1), $8c = 2a + 4b$ (Equation 2)

To normalize $(a, b, c)$, let $a^2 + b^2 + c^2 = 1$ (Equation 3), and since the direction vector can be taken in both directions, restrict to $c \ge 0$ (Equation 4).

From (Equation 1) and (Equation 2), we obtain $c^2 = 1/5$, and from (Equation 4), $c = \sqrt{1/5}$.

From $a^2 + b^2 = 4/5$, we can set $a = \sqrt{4/5}\cos \varphi$, $b = \sqrt{4/5}\sin \varphi$.

We can obtain $2a + 4b = 2\sqrt{4/5}\cos \varphi + 4\sqrt{4/5}\sin \varphi = \sqrt{(2\sqrt{4/5})^2 + (4\sqrt{4/5})^2}\sin(\varphi+\alpha) = 4\sin(\varphi+\alpha) = 8c = 8/\sqrt{5}$.

Since $4 > 8/\sqrt{5}$, $(a, b, c) = (\sqrt{4/5}\cos \varphi, \sqrt{4/5}\sin \varphi, \sqrt{1/5})$ exists in exactly 2 ways.

Now, to determine the number of solutions exactly, ignore (Equation 3), (Equation 4) introduced separately and focus only on (Equation 1), (Equation 2).

Then we can easily find $(a, b, c) = (0, 2, 1), (8, 6, 5)$. (Footnote: If it is not easy to find, you can determine the solutions using $\varphi$ and $\alpha$ found above.)

Therefore, $\cos \theta$ is as follows.

3.

Find the real number $N$ satisfying the following equation.

Solution.

4.

Find the tens digit and the ones digit of $37^{2024}$, respectively.

Solution.

Euler’s totient function of $100$ is as follows.

Therefore, since $37$ and $100$ are coprime, by Euler’s theorem the following holds.

$37^{\varphi(100)} \equiv 37^{40} \equiv 1 \pmod{100}$

Therefore, the following holds.

5.

Let $f(x)$ be a real function defined on all real numbers, twice differentiable, and $f’‘(x)$ is continuous. Also, for all $x$, $f’(x) > 0$ and $f’‘(x) < 0$. Prove that for any positive $t$, the following holds.

Solution.

I referred to the official solution.

6.

For real numbers $a_1, a_2, \cdots, a_n$ satisfying the two conditions below, find the following maximum value. (Assume $n \ge 3$.)

Non-official Solution.

At first, I attempted the following approach using a quadratic form matrix $A$, but due to the condition $a_1 + a_2 + \cdots + a_n = 0$, it was difficult to proceed further.

However, from the official solution, it is clear that it is related to Linear Algebra.

Since the objective function and constraints are differentiable for all $(a_1, \cdots, a_n)$, the maximum of the objective function is determined at a local maximum.

Let us find a local maximum using the Lagrange multiplier method under multiple equality constraints (Lagrange multiplier).

Now, borrowing an idea from Group Theory, let us understand the characteristics of $f(a, b) = \sum_i a_{i+a}a_{i+b}$, which has cyclic properties under the $\sum$ operation.

○ $f(0, 0) = 1$

○ $f(a, b) = f(b, a)$: Commutative law

○ $f(a+c, b+c) = f(a, b)$

○ $f(a+c, b) = f(a-c, b)$

○ $f(a, b+c) = f(a, b-c)$

○ $\sum_c f(a+c, b) = 0$ (∵ $a_1 + a_2 + \cdots + a_n)^2 = 0)$

○ $\sum_c f(a, b+c) = 0$ (∵ $a_1 + a_2 + \cdots + a_n)^2 = 0)$

If we mark the values $f(a, b)$ on an $n \times n$ matrix at $(a, b)$, it is $1$ along the right-going diagonal direction, and the values appear symmetrically in a plus-shape about that diagonal.

Moreover, since the sum of all function values in any chosen row or column becomes $0$, we may conjecture that $f$ is cyclic.

Therefore, for a cyclic function $g$, we can set $f(a, a \pm c) = g(2\pi c/n)$, with $g(0) = 1$.

According to representation theory, any real 1-dimensional representation of a finite cyclic group (an abelian group) must ultimately be expressible in the form of $\cos$ or $\sin$.

That is, every element of a cyclic group is expressed as a power of a fixed element, and when considering complex functions as well, $e^{i\theta}$ becomes that element.

Therefore, considering $g(0) = 1$, we have $f(a, a \pm c) = \cos(2\pi c/n)$.

Even without using representation theory, we can reach a similar conclusion.

Even if an arbitrary function $f$ is given, it can be approximated by values similar to $f(a, a \pm c) = \cos(2\pi c/n)$.

But to satisfy $f(a, a) = 1$, $f(a, a+1) \simeq \cos(2\pi/n)$, $\cdots$, $f(a, a+(n-1)) \simeq \cos(2(n-1)\pi/n)$, we must determine $(a_1, \cdots, a_n)$.

Since there are $n$ equations and $n$ variables, the solution is uniquely determined, and the previously found $f(a, a \pm c) = \cos(2\pi c/n)$ is exactly that solution.

Therefore, $(\text{quasi-inequality}) \le f(a, a+1) = \cos(2\pi/n)$, and the equality condition is as follows.

For reference, when $a_1 = 1/\sqrt{2}$, $a_2 = -1/\sqrt{2}$, $a_3 = \cdots = a_n = 0$, the given conditions are satisfied, but it is not the maximum because it does not satisfy the Lagrange multiplier conditions.

Official Solution.

Let $V = \mathbb{C}^n$, and define the linear map $\phi$: $V \to V$ as follows.

$\phi(a_1, a_2, \cdots, a_n) = (a_2, a_3, \cdots, a_n, a_1)$

For the $n$-th root of unity $\zeta = e^{2\pi i/n} = \cos(2\pi / n) + i\sin(2\pi/n)$ and $1 \le k \le n$,

let $v_k = (\zeta^k, \zeta^{2k}, \cdots, \zeta^{nk})$.

(For example, $v_n = (1, 1, \cdots, 1)$.) Then the following holds.

○ $\phi(v_k) = \zeta^k v_k$. That is, $v_k$ is an eigenvector of $\phi$ (with eigenvalue $\zeta^k)$. In particular, the vectors ${v_1, v_2, \cdots, v_n}$ form a basis of the complex vector space $V = \mathbb{C}^n$.

○ $v_k \cdot v_k^* = (\zeta^k, \zeta^{2k}, \cdots, \zeta^{nk}) \cdot (\zeta^{-k}, \zeta^{-2k}, \cdots, \zeta^{-nk}) = n$ (where $v^*$ is the complex conjugate of $v$)

○ $j := k_1 - k_2 \ne 0 \Rightarrow v_{k_1} \cdot v_{k_2}^* = \zeta^j + \zeta^{2j} + \cdots + \zeta^{nj} = 0$ (where $v^*$ is the complex conjugate of $v$)

Consider an arbitrary vector $v$ satisfying the conditions of the problem as follows.

⑴ $v = (a_1, a_2, \cdots, a_n) \in \mathbb{R}^n$

⑵ $a_1 + a_2 + \cdots + a_n = 0$

⑶ $a_1^2 + a_2^2 + \cdots + a_n^2 = 1$

Suppose the coordinate expression of $v$ with respect to the basis ${v_1, v_2, \cdots, v_n}$ is as follows.

By condition ⑵,

By condition ⑶,

From this we obtain the following.

Now,

Since this value is real,

Therefore, the maximum of $a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n + a_na_1$ is $\cos(2\pi/n)$, and this value is attained when

$c_2 = c_3 = \cdots = c_{n-2} = 0$,

so $v$ is of the form $v = c_1 v_1 + c_{n-1} v_{n-1}$. In this case,

$v = c_1(\zeta, \zeta^2, \cdots, \zeta^n) + c_{n-1}(\zeta^, \zeta^{2}, \cdots, \zeta^{*n}) \in \mathbb{R}^n$

so we can see that $c_{n-1} = c_1^*$. Also, since $n(\left|c_1\right|^2 + \left|c_{n-1}\right|^2) = 1$,

we can see it is of the form

In particular, when $c = 1/\sqrt{2n}$, choosing

yields the maximum value $\cos(2\pi/n)$.

7.

For a bounded closed block region $D$ on the 2-dimensional coordinate plane, suppose the boundary $\beta$ of $D$ satisfies the following.

⑴ $\beta$ is a smooth simple closed curve.

⑵ For any point $O$ on $\beta$, the circle of radius $1$ centered at $O$ meets $\beta$ at exactly two points.

⑶ The tangent line drawn at any point $O$ on $\beta$ does not meet $\beta$ at any other point except $O$.

Let $M$ be the midpoint of a rod $PQ$ of length $1$. If we rotate the rod one full turn so that the endpoints $P, Q$ always lie on $\beta$ (i.e., move the two points $P, Q$ continuously once around along the curve $\beta)$, suppose the locus of $M$ becomes a simple closed curve $\gamma$. Show that the area of the region between the two curves $\beta$ and $\gamma$ is $\pi/4$.

Solution.

Given the triangle as above, by Pappus’s median theorem, $a^2 + b^2 = 2(c^2 + d^2)$ holds.

Therefore, the following holds.

Input: 2024.12.06 19:02