Mathematical Olympiad Challenges

Note that \(2730 = 2 \times 3 \times 5 \times 7 \times 13\). By Fermat's little theorem, for each prime \(p \in \{2, 3, 5, 7, 13\}\), we have \(n^p \equiv n \pmod{p}\). Show \(n^{13} \equiv n \pmod{p}\) for each of these primes by iterating the Frobenius endomorphism or by direct computation of orders.

WLOG assume \(a \geq b \geq c \geq 0\). Then the first term is non-negative and the second is non-positive. Group the first and second terms: \(a^t(a-b)(a-c) + b^t(b-a)(b-c) = (a-b)[a^t(a-c) - b^t(b-c)] \geq 0\) since \(a \geq b\) implies both factors in the bracket are favorable. The third term is non-negative since \(c \leq a\) and \(c \leq b\).

Each row has 10 cells with 2 colors, so there are \(\binom{10}{2} = 45\) pairs of columns. Each pair can be colored in 4 ways (the two cells in that pair). But each row contributes at least \(\binom{k}{2}\) monochromatic pairs where \(k\) is the majority color count. By pigeonhole across 10 rows, two rows must share a monochromatic pair of columns in the same color.

Let \(M\) be the midpoint of \(BC\). The reflection \(H'\) of \(H\) over \(M\) satisfies \(H' = 2M - H\). Using vectors with \(O\) as origin and the fact that \(\vec{OH} = \vec{OA} + \vec{OB} + \vec{OC}\), show that \(|OH'| = R\) (the circumradius). Alternatively, show \(H'\) is the point diametrically opposite \(A\) on the circumcircle.

Set \(x = y = 0\) to get \(f(0) = 0\) or \(f(0) = 1\). If \(f(0) = 0\), then \(f \equiv 0\). If \(f(0) = 1\), set \(x = 0\) to get \(f\) is even. Set \(y = x\): \(f(2x) + f(0) = 2f(x)^2\), so \(f(2x) = 2f(x)^2 - 1\). This resembles the identity \(\cos(2\theta) = 2\cos^2(\theta) - 1\). Solutions: \(f(x) = \cos(ax)\) for any \(a\), and \(f(x) = \cosh(ax)\).

Suppose there are finitely many: \(p_1, p_2, \ldots, p_n\). Consider \(N = 4p_1 p_2 \cdots p_n - 1\). Then \(N \equiv 3 \pmod{4}\). Every integer \(\equiv 3 \pmod{4}\) must have at least one prime factor \(\equiv 3 \pmod{4}\) (since \((4a+1)(4b+1) \equiv 1 \pmod{4}\)). But \(N\) is not divisible by any \(p_i\) — contradiction.

Write \(\frac{a_i}{S - a_i} = \frac{1}{(S/a_i) - 1}\). By Cauchy-Schwarz (Titu's lemma / Engel form): \(\sum \frac{a_i}{S - a_i} = \sum \frac{a_i^2}{a_i S - a_i^2}\). Alternatively, note \(f(x) = \frac{x}{S - x}\) is convex and apply Jensen's inequality with the constraint that the average of the \(a_i\) is \(S/n\).

Consider the team with the most wins (say team \(A\) with the most points). For any team \(T\): if \(A\) beat \(T\), done. If \(T\) beat \(A\), then \(T\) beat a team with the maximum number of wins. Among \(A\)'s wins, at least one of them must have beaten \(T\) (otherwise \(T\) beat all teams that \(A\) beat, giving \(T\) more wins — a contradiction).

For the proof, use the inversive or trigonometric approach: apply the law of cosines in triangles \(ABC\) and \(ACD\), using the fact that opposite angles sum to \(180°\). For \(\sin(15°)\): inscribe a suitable cyclic quadrilateral in a unit circle (e.g., vertices at \(0°, 60°, 120°, 180°\) or use a rectangle with one diagonal).

Let \(a^2 + b = m^2\) and \(b^2 + a = n^2\). Then \(m^2 - a^2 = b\) and \(n^2 - b^2 = a\). So \((m - a)(m + a) = b\) and \((n - b)(n + b) = a\). Since \(a, b \geq 1\), we have \(m > a\) and \(n > b\), so \(m \geq a + 1\) and \(n \geq b + 1\). This gives \(b \geq 2a + 1\) and \(a \geq 2b + 1\), which leads to a contradiction for large values. Check small cases systematically.

Set \(x = 0\): \(f(f(y)) = f(0) + y\), so \(f\) is injective (since \(f(f(y_1)) = f(f(y_2))\) implies \(y_1 = y_2\)). Let \(c = f(0)\). Then \(f(f(y)) = y + c\). Set \(y = 0\) in the original: \(f(x + c) = f(x)\), so \(c = 0\) by injectivity... Actually \(f(f(y)) = y\). Then the original gives \(f(x + f(y)) = f(x) + y\). Substituting \(x = 0\): \(f(f(y)) = y\). So \(f\) is an involution. Show \(f(x) = x\) or \(f(x) = -x\)... but check which work.

For the Engel form, it follows directly from Cauchy-Schwarz: \(\left(\sum \frac{a_i^2}{b_i}\right)\left(\sum b_i\right) \geq \left(\sum a_i\right)^2\). For Nesbitt: write \(\frac{a}{b+c} = \frac{a^2}{a(b+c)}\). Apply Titu's: \(\sum \frac{a^2}{a(b+c)} \geq \frac{(a+b+c)^2}{a(b+c) + b(a+c) + c(a+b)} = \frac{(a+b+c)^2}{2(ab+bc+ca)}\). Then show \((a+b+c)^2 \geq 3(ab+bc+ca)\).

Consider the cells containing \(1, n+1, 2n+1, \ldots, (n-1)n+1\) (there are \(n\) such cells). Any path from the cell containing 1 to the cell containing \(n^2\) along adjacent cells must pass through cells with these values in some order. Apply the pigeonhole principle to the \(n+1\) values \(1, n+1, 2n+1, \ldots, n^2\) and the grid paths between them. Use the fact that in any 2-coloring of the grid like a checkerboard, adjacent cells have different colors.

Use the area method: \(\frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[PBD]}{[PCD]} = \frac{[ABP]}{[ACP]}\). Similarly for the other ratios. Multiply them together and observe massive cancellation. For the converse, assume the product equals 1, let \(AP \cap BC = D\), \(BP \cap AC = E\), then show the line through \(C\) and the intersection of \(AE\) and \(BD\) must pass through \(F\).

It suffices to prove \(x^4 + y^4 = z^2\) has no positive integer solutions (which is a stronger statement). Assume a minimal solution exists. Write \((x^2)^2 + (y^2)^2 = z^2\), so this is a Pythagorean triple. WLOG \(\gcd(x,y) = 1\) and \(x\) odd. Then \(x^2 = m^2 - n^2\), \(y^2 = 2mn\), \(z = m^2 + n^2\) for some \(m > n > 0\). From \(y^2 = 2mn\) with \(\gcd(m,n) = 1\), deduce \(m = a^2\), \(n = 2b^2\) (or vice versa). Then \(x^2 + 4b^4 = a^4\), yielding a smaller solution — contradiction by infinite descent.