Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>. | Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>. | ||
− | Its surface area is <cmath>2lw+2lh+2wh=384,</cmath> | + | Its surface area is <cmath>2lw+2lh+2wh=384,</cmath> |
and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath> | and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath> | ||
Latest revision as of 19:34, 17 August 2020
Contents
Problem
A rectangular box is inscribed in a sphere of radius . The surface area of is 384, and the sum of the lengths of its 12 edges is 112. What is ?
Solution
Box P has dimensions , , and . Its surface area is and the sum of all its edges is
The diameter of the sphere is the space diagonal of the prism, which is Notice that so the diameter is The radius is half of the diameter, so
Solution 2
As in the previous solution, we have that and , and the diameter of the sphere is the space diagonal of the prism, .
Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that . (This essentially means that we have an infinitesimally thin box.) We now have that and , and we are solving for . Because this means that so the space diagonal is . Since the diameter of the sphere is , the radius is . ~emerald_block
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.