Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

Let $n \neq 0$. For every sequence of integers \[ A = a_0,a_1,a_2,\dots, a_n \] satisfying $0 \le a_i \le i$, for $i=0,\dots,n$, define another sequence \[ t(A)= t(a_0), t(a_1), t(a_2), \dots, t(a_n) \] by setting $t(a_i)$ to be the number of terms in the sequence $A$ that precede the term $a_i$ and are different from $a_i$. Show that, starting from any sequence $A$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $B$ such that $t(B) = B$.

Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.

Click for solution MithsApprentice wrote: Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E,$ respectively. Lines $AB$ and $DE$ intersect at $F,$ while lines $BD$ and $CF$ intersect at $M.$ Prove that $MF = MC$ if and only if $MB\cdot MD = MC^{2}$ Nice problem! And here is a nice proof. Proof. Take $G\in BD: \,FG\parallel CD$ We have: $MF=MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram}\\ \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA=\angle GCD\Longleftrightarrow\angle FDA+\angle CGF=180^\circ\\ \Longleftrightarrow \angle ABE+\angle CGF=180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic}\\ \Longleftrightarrow\angle CBM=\angle CBG=\angle CFG=\angle DCF=\angle DCM\\ \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD=MC^{2}$

Let $ a$, $ b$, $ c$ be positive real numbers. Prove that \[ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8. \]

At the vertices of a regular hexagon are written six nonnegative integers whose sum is $2003^{2003}$. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.

These problems are copyright $\copyright$ Mathematical Association of America.