Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AHSME Problems/Problem 9"

(Created page with "==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...")
 
(→‎Solution)
 
(3 intermediate revisions by one other user not shown)
Line 30: Line 30:
 
==Solution==
 
==Solution==
  
 +
Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is <math>\sqrt{8^2+10^2} = \sqrt{164},</math> which is between <math>\sqrt{144}</math> and <math>\sqrt{169},</math> so the answer is still <math>\textbf{(C)}.</math> -hailstone
  
 +
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 <math>45</math> degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that <math>S= 4\sqrt{2}+5\sqrt{2}</math>, with <math>4\sqrt{2}</math> being the length of the leg formed by the side of the square with length <math>8</math> and <math>5\sqrt{2}</math> being the length of the leg formed by the side of the square with length <math>10</math>. Adding these up yields <math>9\sqrt{2}</math>.
 +
 +
We have that <math>\sqrt{2}\approx 1.414\approx 1.4</math>. That means that <math>9\sqrt{2}\approx 12.6</math>, which rounds up to <math>\boxed{\textbf{C)} 13}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 07:24, 12 July 2022

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

Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is $\sqrt{8^2+10^2} = \sqrt{164},$ which is between $\sqrt{144}$ and $\sqrt{169},$ so the answer is still $\textbf{(C)}.$ -hailstone

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 $\boxed{\textbf{C)} 13}$.

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_9&oldid=175836"