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

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==Solution==
 
==Solution==
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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</math>
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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
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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>.
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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

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