The first six questions were multiple-choice. I cannot recall the specific details, but they were actually quite simple. Interested readers can find the exam questions in the PDF attachment of this article (sourced from the "Kong Nian Yuan Xi" mathematics website).
After checking the reference answers in the PDF, BoJone found that all my multiple-choice answers were correct. As for the last three major problems, I only completed the final two. My solutions are somewhat different from those in the PDF, so I am writing them here for discussion.
1. Prove: An inscribed pentagon with equal interior angles is a regular pentagon.
I solved this problem in the final fifteen minutes. Initially, I thought of many complex theorems and methods, but later I realized it could be proven very simply.
The figure below shows a pentagon satisfying the given conditions.
Since each triangle has one vertex at the center of the circle and two vertices on the circle, they are all isosceles triangles. Let the central angles subtended by the sides be a, b, c, d, e.
From the equality of the interior angles, we have: a+b = b+c = c+d = d+e = e+a Furthermore: \begin{aligned} a+b=b+c \Rightarrow a=c; \quad c+d=d+e \Rightarrow c=e \Rightarrow a=c=e \\ b+c=c+d \Rightarrow b=d; \quad d+e=e+a \Rightarrow d=a \Rightarrow b=d=a \end{aligned} Therefore: a=b=c=d=e From this, it follows that \angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOA. Thus, the pentagon is a regular pentagon. Q.E.D.
Note: It seems this can be generalized to: "An inscribed polygon with an odd number of sides and equal interior angles is a regular odd-sided polygon."
2. Prove: Any positive integer power of \sqrt{2}+1 can be expressed in the form \sqrt{s}+\sqrt{s-1}, where s is a positive integer.
This was the first problem I considered (just as the exam papers were being handed out, before the start of the exam). At first, I thought of using mathematical induction, but I found it wasn’t quite working, so I looked for another method. The logic of this solution is similar to the one in the PDF, but the expression is different.
Let (\sqrt{2}+1)^n = a+b\sqrt{2} = \sqrt{a^2} + \sqrt{2b^2}, where a and b are positive integers. It is not difficult to prove that:
When n is odd, (\sqrt{2}-1)^n = -a+b\sqrt{2};
When n is even, (\sqrt{2}-1)^n = a-b\sqrt{2}.
Now we prove:
When n is odd, s=2b^2, which implies 2b^2 - a^2 = 1;
When n is even, s=a^2, which implies a^2 - 2b^2 = 1.
For case (1), we can solve for a and b: \begin{aligned} a &= \frac{(\sqrt{2}+1)^n - (\sqrt{2}-1)^n}{2} \\ \sqrt{2}b &= \frac{(\sqrt{2}+1)^n + (\sqrt{2}-1)^n}{2} \end{aligned} Substituting these into 2b^2 - a^2: \begin{aligned} 2b^2 - a^2 &= \frac{[(\sqrt{2}+1)^n + (\sqrt{2}-1)^n]^2 - [(\sqrt{2}+1)^n - (\sqrt{2}-1)^n]^2}{4} \\ &= \frac{4(\sqrt{2}+1)^n(\sqrt{2}-1)^n}{4} = 1 \end{aligned}
For case (2), we can solve for a and b: \begin{aligned} a &= \frac{(\sqrt{2}+1)^n + (\sqrt{2}-1)^n}{2} \\ \sqrt{2}b &= \frac{(\sqrt{2}+1)^n - (\sqrt{2}-1)^n}{2} \end{aligned} Similarly, substituting these yields a^2 - 2b^2 = 1. Q.E.D.
Attachments:
2012 NATO Independent Recruitment Mathematics - Kong Nian Yuan Xi.pdf
2012 NATO Independent Recruitment Mathematics Reference Answers - Kong Nian Yuan Xi.pdf
When reprinting, please include the original address: https://kexue.fm/archives/1551
For more detailed reprinting matters, please refer to: Scientific Space FAQ