Knight through the board

Problem:

Starting from the top-left corner of the chessboard, can we move a knight to the bottom-right corner such that every square is visited once?

Solution:

No. Each step, the square the knight is in switches color. There are 63 (i.e. odd) steps to make, hence the last square must be of a different color from that of the starting point. However the top-left corner and the bottom-right corner have the same color. Contradiction.

Comments

p - p^2-2 / p+2

J595 Mathematical Reflection

J595 Mathematical Reflection \(\sqrt[3]{(x-1)^2} - \sqrt[3]{2(x-5)^2} + \sqrt[3]{(x-7)^2} = \sqrt[3]{4x} \) By Titu Adreescu. My Solution (potentially wrong): Let \(A = \sqrt[3]{(x-1)^2}\) , \(B = \sqrt[3]{2(x-5)^2}\), \(C = \sqrt[3]{(x-7)^2} \), \(D = \sqrt[3]{4x}\). Hence we have \(A + C = B + D\). Notice that \(A^3 + C^3 = B^3 + D^3 \). Also notice that \( (A+C)^3 = (B+D)^3\) yields \( AC = BD \). Also by factoring out \(A^3 + C^3 = (A+C)(A^2-AC-C^2)\) and doing the same for \(B^3+D^3\) gives us \(A^2 + C^2 = B^2 + D^2\). Lemma 1 : \((A^3 - C^3)^2 = (B^3 - D^3)^2\). Proof: Note that \((A^3+C^3)^2 - (A^3-C^3)^2 = 4A^3C^3 = 4B^3D^3 = (B^3+D^3)^2 - (B^3-D^3)^2\) and the conclusion follows. \( \square \) Using Lemma 1 we get \( ( (x-1)^2 - (x-7)^2)^2 = (2(x-5)^2 - 4x)^2\) and solving for x gives us \( x = (9 \pm 4\sqrt{2}) , (3 \pm 2\sqrt{2}) \). \(\square\)

232 | 20^n + 16^n - 3^n - 1

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