# 1988 AHSME Problems/Problem 9

## Problem $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); label("Figure 2", (E+F)/2, S); label("10'", (I+J)/2, S); label("8'", (12,12)); label("8'", (L+M)/2, S); label("10'", (42,11)); label("table", (5,12)); label("table", (36,11)); [/asy]$

An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below. The owners desire to move the table to the position shown in Figure $2$. The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart? $\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

## Solution

We begin by thinking about the motion of the table. As it moves, the table will have it's maximum height and width when the rectangle's sides form $45$ degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that $S= 4\sqrt{2}+5\sqrt{2}$, with $4\sqrt{2}$ being the length of the leg formed by the side of the square with length $8$ and $5\sqrt{2}$ being the length of the leg formed by the side of the square with length $10$. Adding these up yields $9\sqrt{2}$.

We have that $\sqrt{2}\approx 1.414\approx 1.4$. That means that $9\sqrt{2}\approx 12.6$, which rounds up to $\boxed{\textbf{C)} 13}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 