A fat man who was reborn from the dead? He transforms into a perfect male god in a week.

……

Question 1 (45 points)

Let n be a positive integer, and a₁, a₂, ..., aₙ be real numbers satisfying ∑_{i=1}^n a_i = 0 and ∑_{i=1}^n a_i² = 1. Prove that for any real number x, ∑_{i=1}^n (a_i - x)² ≥ n/(n-1).

After reading the question, three solutions immediately popped into Jiang Chen's mind.

Solution 1: Use Cauchy's inequality and the AM-GM inequality directly, and you can solve it in three steps.

Solution 2: Transform it into a quadratic function extremum problem and use the discriminant.

Solution 3: Use the Lagrange multiplier method (although it is beyond the scope of the curriculum, it is concise).

He chose the first option and began writing.

"Based on the conditions ∑a_i=0, ∑a_i²=1, then ∑(a_i-x)²=∑a_i² - 2x∑a_i + nx² = 1 + nx²."

"It needs to be proven that 1 + nx² ≥ n/(n-1), that is, nx² ≥ 1/(n-1)."

"By Cauchy's inequality: (∑a_i²)(∑1²) ≥ (∑a_i)², i.e., n≥0, this always holds. But we need to find another inequality..."

"Consider ∑(a_i - ā)² = ∑a_i² - nā² = 1, where ā=0, so ∑a_i²=1 is given."

"Actually, we can directly use ∑(a_i-x)² = ∑a_i² - 2x∑a_i + nx² = 1 + nx², which takes the minimum value of 1 when x=0. But do we need to prove 1 ≥ n/(n-1)? No, when n＞1, 1＜n/(n-1), so we need to adjust our thinking..."

Jiang Chen stopped writing and looked at the question again.

Oh, I read it wrong.

The goal is not to prove ∑(a_i-x)² ≥ n/(n-1), but to prove that ∑(a_i-x)² ≥ n/(n-1) holds for any x.

That makes it even simpler.

Let f(x) = ∑(a_i-x)² = nx² - 2(∑a_i)x + ∑a_i² = nx² + 1 (because ∑a_i=0).

"This is a quadratic function of x, opening upwards, with a minimum value of 1 (when x=0)."

"We need to prove that f(x) ≥ n/(n-1) holds for any x, i.e., prove the minimum value 1 ≥ n/(n-1)? etc., 1 ≥ n/(n-1) if and only if n≤2..."

Jiang Chen frowned.

Is there something wrong with this question?

He read the question carefully again.

Then he understood.

"I see, I misunderstood. The conditions are ∑a_i=0 and ∑a_i²=1, but a_i is a real number, which can be positive or negative."

"What we need to prove is ∑(a_i-x)² ≥ n/(n-1), which means nx² + 1 ≥ n/(n-1), or nx² ≥ 1/(n-1)."

"This is not always true, because x can be 0. So... does the problem imply a range of values ​​for x? No, the problem says 'for any real number x,' so the inequality is not true."

Jiang Chen fell into deep thought.

Three seconds later, he realized what was happening.

"Damn, I fell for the question setter's trick."

"The correct understanding of this question is: we need to prove that there exists a constant C that is independent of {a_i} such that ∑(a_i-x)² ≥ C holds for any x and any {a_i} that satisfies the condition, and then find the maximum value of C."

"And the maximum value of C is n/(n-1)."

"Therefore, the proof consists of two steps: first, proving that ∑(a_i-x)² ≥ n/(n-1) holds for all {a_i} and some specific x that satisfy the condition; second, proving that this lower bound is tight, that is, there exists a set of {a_i} and x such that the equality holds."

Having figured this out, Jiang Chen smiled.

"The person who set the question is quite something; they even play word games."

He picked up his pen and rewrote:

"Proof: By Cauchy's inequality, (∑_{i=1}^n a_i)² ≤ n∑_{i=1}^n a_i², that is, 0 ≤ n·1, which always holds, but this inequality cannot directly lead to the desired conclusion."

「考虑固定{a_i}，令f(x)=∑(a_i-x)²=nx²+1，最小值为1（当x=0时）。但1可能小于n/(n-1)（当n＞2时），故需考虑调整{a_i}使下界最大化。」

"In fact, given the conditions ∑a_i=0 and ∑a_i²=1, we can obtain ∑_{i≠j} a_i a_j = -1/2 (expanding (∑a_i)²=0)."

"Then ∑(a_i-x)² = 1 + nx², to maximize its lower bound, is it equivalent to finding the minimum value of 1 + nx²? No, it should be finding min_x max_{a_i} ∑(a_i-x)² among all {a_i} that satisfy the condition? No, that's not it either..."

Jiang Chen shook his head.

"Damn, this problem is more complicated than I thought."

However, it's just "troublesome," not "difficult."

He changed his approach.

"Let's just use the Lagrange multiplier method. It's beyond our scope, but who cares?"

"Consider the optimization problem: Given ∑a_i=0, ∑a_i²=1, find the lower bound of min_x max_{a_i} ∑(a_i-x)²."

"Given a fixed value x, find the maximum value of max_{a_i} ∑(a_i-x)² under the constraints..."

"By taking the partial derivative of the Lagrange function L = ∑(a_i-x)² + λ∑a_i + μ(∑a_i²-1), we get 2(a_i-x) + λ + 2μa_i = 0, and solving for a_i, we get a_i = (2x-λ)/(2+2μ)..."

Substituting the constraints λ and μ, we finally obtain the worst-case scenario ∑(a_i-x)² = n/(n-1)……

In three minutes, he wrote a whole page of densely packed text.

After finishing writing, Jiang Chen breathed a sigh of relief.

"Done."

He glanced at the time: 9:43.

……

On the podium, proctor Liu Yue kept her eyes fixed on Jiang Chen.

When she saw Jiang Chen finish the first question in three minutes, her eyes almost popped out of their sockets.

"This...how is this possible?"

She couldn't help but step down from the podium, pretending to inspect the area, and walk behind Jiang Chen.

Then she saw Jiang Chen's answer.

From initial misunderstandings to reanalysis, and finally to a complete solution using the Lagrange multiplier method...

The steps are rigorous and the logic is clear.

And... all correct.

Liu Yue felt that her worldview had been shaken.

This is the first question in the second round!

A 45-point essay question!

In previous years, even top students would take twenty or thirty minutes to complete this type of question.

This Jiang Chen... three minutes?

Did they even use university-level mathematics methods?

She took a deep breath, forced herself to calm down, and continued watching Jiang Chen do the second question.

……

Question 2 (45 points)

Let S be a finite set of points in a plane, where no three points are collinear. A "kite" is called a quadrilateral formed by four points A, B, C, D ∈ S, satisfying AB = AD and CB = CD (i.e., two pairs of adjacent sides are equal). Prove that if any four points in S can form a kite, then all points in S are concyclic.

Jiang Chen was stunned for a moment after reading the question.

"The kite is a quadrilateral...and concentric circles..."

Several geometric theorems flashed through his mind instantly.

"Isn't that obvious?"

He picked up his pen and wrote:

"Proof: Take any two points A and B in S. By the condition, for any other two points C and D ∈ S{A,B}, quadrilateral ABCD is a kite."

"In particular, let C be any point in S that is different from A and B. Then there exists a D (which may coincide with C? No, D must be different from A, B, and C) such that AB = AD and CB = CD."

"But the condition states that 'any four points can form a kite,' which means that for any four points, two of them are 'shoulder points' (the common endpoints of adjacent sides), and the other two are 'wing points.'"

"Consider any three points A, B, and C. Given the condition that there exists a point D such that AB = AD and CB = CD, meaning D lies at the intersection of the perpendicular bisectors of AB and BC, is D the circumcenter of triangle ABC? No, the circumcenter is the intersection of the three perpendicular bisectors, but here only two are used..."

"Wait, this question requires careful analysis of its structure."

Jiang Chen stopped writing and thought for a few seconds.

"The fact that any four points can form a kite means that for any four points, two of them are vertices of an isosceles triangle, and the other two are vertices of another isosceles triangle, and these two isosceles triangles share a base? No, a kite is a quadrilateral with two pairs of equally adjacent sides."

"Given four points A, B, C, and D, the kite's structure has two possibilities: either A and C are 'shoulder points' (AB=AD, CB=CD), or B and D are 'shoulder points' (BA=BC, DA=DC)."

"Due to the arbitrariness, for any three points A, B, and C, consider a fourth point D (take another point in S). Then the four points A, B, C, and D form a kite. If A is a shoulder point, then AB = AD and CB = CD; if C is a shoulder point, then BA = BC and DA = DC; the same logic applies if B or D is a shoulder point."

"This will lead to a series of equal relationships..."

Jiang Chen drew several diagrams on the draft paper.

Ten seconds later, his eyes lit up.

"Got it!"

"Lemma: If any four points form a kite, then for any three points A, B, C, at least one of the following holds true: AB=AC, BA=BC, or CA=CB."

"Proof: Take the fourth point D. If A is a shoulder point, then AB=AD and CB=CD, but this has no direct constraint on the relationship between B and C. We need to find another approach..."

"A more direct method: Consider any three points A, B, and C, and take another point D from S. The four points form a kite based on the given conditions. If A and C are shoulder points, then AB=AD and CB=CD. This implies that AB=AD and CB=CD, but the relationship between B and D is unknown."

"Actually, according to the definition of a kite, in quadrilateral ABCD, are A and C the intersection of diagonals? No, a kite usually refers to a quadrilateral with a pair of opposite angles equal and adjacent sides equal..."

Jiang Chen frowned.

This question... is a bit tricky.

But he quickly found a breakthrough.

"From another perspective: the kite quadrilateral is essentially two isosceles triangles sharing a base."

"Let there be four points A, B, C, and D. If A and C are shoulder points, then △ABD and △CBD are both isosceles triangles (AB=AD, CB=CD) and share the same base BD."

"So BD is the perpendicular bisector of AB and AD, and also the perpendicular bisector of CB and CD? No, the perpendicular bisector is a straight line..."

"Actually, since AB=AD, point A lies on the perpendicular bisector of BD; since CB=CD, point C lies on the perpendicular bisector of BD. So, A and C are both on the perpendicular bisector of BD, meaning AC⊥BD and the midpoint of BD lies on AC? No, the perpendicular bisector is the perpendicular bisector of the line..."

Jiang Chen felt like he was being tricked.

"Damn it, geometry problems are such a hassle."

He decided to use brute force with analytic geometry.

"Establish a coordinate system, define the coordinates of the points, express the conditions using the distance formula, and then prove that these points are concyclic..."

"But this would involve too much computation, and the condition of 'any four points' is difficult to express analytically."

Jiang Chen thought for a moment and then changed his approach.

"Proof by contradiction: Assume that S is not concyclic, then there are four points that are not concyclic, which leads to a contradiction."

"Consider four non-circular points A, B, C, and D, which form a kite. Let A and C be the shoulder points. Then AB = AD and CB = CD, meaning A lies on the perpendicular bisector of BD, and C also lies on the perpendicular bisector of BD. So, are A and C symmetrical about the perpendicular bisector of BD? No, they are simply both on the same straight line..."

"Wait, if A and C are both on the perpendicular bisector of BD, then AC is the perpendicular bisector of BD? Then B and D are symmetrical about AC, so AB=AD and CB=CD naturally hold true..."

"Therefore, as long as A and C are on the perpendicular bisector of BD, quadrilateral ABCD is a kite."

"Then the problem becomes: if for any four points there exist two points on the perpendicular bisector of the line connecting the other two points, then all points are concyclic."

Jiang Chen's eyes grew brighter and brighter.

"This is the breakthrough!"

He picked up his pen and wrote quickly:

"Proof: Assume that S is not concyclic. Then there exist four points A, B, C, and D that are not concyclic. Without loss of generality, assume that A, B, and C are not concyclic (because if four points are not concyclic, then three of them must be not concyclic, plus one more)."

"Take the fourth point, D. The kite is formed by conditions A, B, C, and D. Discuss the different cases..."

Three minutes, problem two solved.

Jiang Chen glanced at the time: 9:46.

Two questions, six minutes.

Liu Yue stood behind Jiang Chen, already numb.

For the second question, as a member of the question-setting team, she knew the difficulty of the question... It was a classic problem in combinatorial geometry, adapted from an IMO preliminary question.

The standard answer was two full pages long, using complex case studies and proof by contradiction.

And Jiang Chen... finished it in three minutes?

Moreover, his solution is simpler and more essential than the standard answer.

Liu Yue felt that she had wasted decades of her math studies.

……