Area#12429

Area (Editorial)

Problem Author:	Hasan Saleh (MathModel)
Prepared by:	Hasan Saleh (MathModel)
Edited by:	Hasan Saleh (MathModel)

Solution Each square has an area of $m^{2}$ , but because every square after the first is horizontally aligned with its leftmost edge at the center of the previous one, it only contributes a "new" rectangular area of width $\frac{m}{2}$ \frac{m}{2} and height $m$ . Summing the initial square and the $n - 1$ subsequent additions yields a total area of $m^{2} + (n - 1) \frac{m ^{2}}{2} = \frac{m ^{2} ( n + 1 )}{2}$ m^2 + (n-1)\frac{m^2}{2} = \frac{m^2(n+1)}{2}. Discarding the fractional part to satisfy the integer area requirement, the final formula is:

Area = ⌊ \frac{m ^{2} ( n + 1 )}{2} ⌋

\text{Area} = \left\lfloor \frac{m^2(n+1)}{2} \right\rfloor

\frac{m}{2}

m^{2} + (n - 1) \frac{m ^{2}}{2} = \frac{m ^{2} ( n + 1 )}{2}

Area = ⌊ \frac{m ^{2} ( n + 1 )}{2} ⌋