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Difference between revisions of "2002 AMC 12P Problems/Problem 9"

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== Solution ==
 
== Solution ==
 
We can use the formula for the diagonal of a rectangular prism, or <math>d=\sqrt{a^2+b^2+c^2}</math> The problem gives us <math>a=1, b=8,</math> and <math>d=9.</math> Solving gives us <math>9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16  \implies c=\boxed{\textbf{(D) } 4}.</math>
 
We can use the formula for the diagonal of a rectangular prism, or <math>d=\sqrt{a^2+b^2+c^2}</math> The problem gives us <math>a=1, b=8,</math> and <math>d=9.</math> Solving gives us <math>9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16  \implies c=\boxed{\textbf{(D) } 4}.</math>
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~Minor edits by Astro2010~
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2002|ab=P|num-b=8|num-a=10}}
 
{{AMC12 box|year=2002|ab=P|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

Latest revision as of 20:45, 15 October 2025

The following problem is from both the 2002 AMC 12P #9 and 2002 AMC 10P #16, so both problems redirect to this page.

Problem

Two walls and the ceiling of a room meet at right angles at point $P.$ A fly is in the air one meter from one wall, eight meters from the other wall, and nine meters from point $P$. How many meters is the fly from the ceiling?

$\text{(A) }\sqrt{13} \qquad \text{(B) }\sqrt{14} \qquad \text{(C) }\sqrt{15} \qquad \text{(D) }4 \qquad \text{(E) }\sqrt{17}$

Solution

We can use the formula for the diagonal of a rectangular prism, or $d=\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $d=9.$ Solving gives us $9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16 \implies c=\boxed{\textbf{(D) } 4}.$


~Minor edits by Astro2010~

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
All AMC 10 Problems and Solutions
2002 AMC 12P (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
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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