Difference between revisions of "1996 AHSME Problems/Problem 28"
(→Solution 2) |
(→Solution 1) |
||
Line 28: | Line 28: | ||
<cmath>\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,</cmath> so <math>3x + 3y + 4z - 12 = 0.</math> The equation for the distance of a point <math>(a,b,c)</math> to a plane <math>Ax + By + Cz + D = 0</math> is given by: | <cmath>\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,</cmath> so <math>3x + 3y + 4z - 12 = 0.</math> The equation for the distance of a point <math>(a,b,c)</math> to a plane <math>Ax + By + Cz + D = 0</math> is given by: | ||
− | <cmath>\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}.</cmath> | + | <cmath>\frac{|Aa + Bb + Cc + D|}{\sqrt{A^2 + B^2 + C^2}}.</cmath> |
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where <math>a=b=c=0</math>) to the plane is given by: | Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where <math>a=b=c=0</math>) to the plane is given by: |
Revision as of 19:39, 17 December 2020
Contents
Problem
On a rectangular parallelepiped, vertices , , and are adjacent to vertex . The perpendicular distance from to the plane containing , , and is closest to
Solution 1
By placing the cube in a coordinate system such that is at the origin, , , and , we find that the equation of plane is:
so The equation for the distance of a point to a plane is given by:
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where ) to the plane is given by:
Since , this number should be just a little over , and the correct answer is .
Note that the equation above for the distance from a point to a plane is a 3D analogue of the 2D case of the distance formula, where you take the distance from a point to a line. In the 2D case, both and are set equal to .
Solution 2
Let be the desired distance. Recall that the volume of a pyramid is given by , where is the area of the base and is the height. Consider pyramid . Letting be the base, the volume of is given by , but if we let be the base, the volume is given by . Clearly, these two volumes must be equal, so we get the equation . Thus, to find , we just need to find .
By the Pythagorean Theorem, , , .
The altitude to in triangle has length , so . Then or about . The answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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.