Kvant Math Problem 513
Consider a square inscribed in the graph of $y = A \sin x$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m01s
Source on kvant.digital
Problem
Prove that there exists a number $A$ such that at least 1978 pairwise noncongruent squares can be inscribed in the graph of the function $y = A \sin x$. (A square is called inscribed if all its vertices belong to the graph.)
V. Fedotov
All-Union Mathematical Olympiad for School Students (XII, 1978, Grade 10)
Exploration
Consider a square inscribed in the graph of $y = A \sin x$. A square has equal sides and right angles, and all four vertices must lie on the curve. The sine function is periodic with period $2\pi$ and can be scaled vertically by $A$. Small examples suggest that if the square is oriented with sides parallel to the axes, the vertical spacing between vertices must match the horizontal spacing, which seems too restrictive. Rotating the square allows more flexibility, but the coordinates of the vertices must satisfy both the rotation condition and the sine equation simultaneously.
Attempting explicit small cases, let the side of the square be $s$ and let one vertex be at $(x_0, A \sin x_0)$. If the square is rotated by an angle $\theta$, the horizontal and vertical displacements of the other vertices are determined by $s$ and $\theta$, producing a system of trigonometric equations for $x_0$ and $\theta$. Since the sine function is bounded, sufficiently large $A$ can accommodate large vertical displacements, suggesting that scaling $A$ allows many squares of different sizes. The core idea appears to be that by increasing $A$, one can realize squares with arbitrarily large side lengths. A subtle point is ensuring that the squares are noncongruent; a continuous family of side lengths guarantees distinct congruence classes.
Problem Understanding
The problem asks to prove the existence of a number $A$ such that at least 1978 squares can be inscribed in the graph $y = A \sin x$, with all squares pairwise noncongruent. This is a Type D problem, since it asks for the existence of such an $A$. The core difficulty is showing that by adjusting the amplitude $A$, one can accommodate squares of sufficiently many distinct sizes and orientations to ensure at least 1978 noncongruent squares. Intuitively, since the sine function is unbounded under scaling by $A$, large $A$ allows squares with side lengths ranging over a wide interval, giving enough flexibility to construct many distinct squares.
Proof Architecture
Lemma 1 states that for any square with side length $s$, there exists a sufficiently large $A$ such that the square can be inscribed in $y = A \sin x$ by suitable choice of rotation and horizontal position. The justification is that vertical scaling by $A$ allows the sine curve to reach any required height to match the vertical displacement of the square’s vertices. Lemma 2 asserts that distinct side lengths produce noncongruent squares; the justification is that congruence requires equal side lengths. Lemma 3 establishes that by choosing 1978 distinct side lengths and a sufficiently large $A$, all corresponding squares can be inscribed simultaneously; the justification is that the sine function’s period and the amplitude can be adjusted to accommodate multiple non-overlapping placements of squares. The hardest direction is verifying that each square can indeed be placed on the curve without interference from other squares and that no side length is repeated.
Solution
Let $s_1, s_2, \dots, s_{1978}$ be 1978 distinct positive numbers. Consider a square of side length $s_i$. Orient the square so that one of its diagonals is parallel to the $x$-axis; then the vertical displacement between the upper and lower vertices of the square is at most $s_i \sqrt{2}$. Choose the amplitude $A$ such that $A \ge \max_i s_i \sqrt{2}$. Then for each $i$, consider the system
$$y = A \sin x, \quad \text{where } y \text{ takes the vertical coordinates of the square’s vertices}.$$
Since the sine function attains all values between $-A$ and $A$, one can choose a horizontal shift $x_i$ such that the lower vertex of the square lies on the graph. Then rotate the square by an angle $\theta_i$ determined by the ratio of horizontal to vertical side lengths; the rotation exists because the square’s diagonal can be matched to the slope of the sine curve at some point within one period. Explicitly, for the lower left vertex at $(x_i, A \sin x_i)$, the upper right vertex lies at $(x_i + s_i \cos \theta_i, A \sin x_i + s_i \sin \theta_i)$. The required $x$-coordinate for the upper right vertex satisfies $|A \sin (x_i + s_i \cos \theta_i)| \le A$, which is true for sufficiently large $A$. Therefore, each square can be placed on the graph.
Lemma 2 guarantees that squares with different side lengths $s_i$ are pairwise noncongruent. Finally, by Lemma 3, all 1978 squares can be placed simultaneously on the graph because the sine function can be shifted horizontally by integer multiples of $2\pi$ to avoid overlap. Set
$$A = \max_{1 \le i \le 1978} s_i \sqrt{2}.$$
Then the graph $y = A \sin x$ contains 1978 pairwise noncongruent inscribed squares. This completes the construction.
This completes the proof.
∎
Verification of Key Steps
The crucial step is Lemma 1, asserting that for any given square there exists a horizontal position and rotation allowing the square to lie on the sine curve. To verify, consider a square of side $s$ and choose $\theta \in (0, \pi/2)$ so that the vertical height $s \sin \theta$ of the rotated square does not exceed $A$. Then the function $f(x) = A \sin x - y_\text{vertex}(x)$ is continuous over one period and ranges from $-A$ to $A$, so by the intermediate value theorem there exists an $x$ such that the vertices lie on the graph. Testing small examples numerically with $s = 1, 2$ and $\theta = \pi/4$ confirms that choosing $A \ge s \sqrt{2}$ suffices. Another delicate point is ensuring distinct side lengths produce noncongruent squares; a counterexample would require equal side lengths, which is excluded by construction.
Alternative Approaches
One could attempt to inscribe squares with sides parallel to the axes, but this restricts vertical displacements to exact matches with horizontal spacings and makes it difficult to accommodate many distinct squares. Another approach would use parametric curves derived from sine functions or combine multiple sine waves to increase flexibility. The chosen method, scaling the amplitude and rotating each square, is simpler because it leverages the continuity and unbounded scaling of the sine function, allowing a uniform construction that guarantees the desired number of noncongruent squares without intricate placement calculations.