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Difference between revisions of "1996 AHSME Problems/Problem 28"
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==Problem==
 
==Problem==
  
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On a <math> 4\times 4\times 3 </math> rectangular [[parallelepiped]], vertices <math>A</math>, <math>B</math>, and <math>C</math> are adjacent to vertex <math>D</math>. The perpendicular distance from <math>D</math> to the plane containing
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On a <math> 4\times 4\times 3 </math> rectangular parallelepiped, vertices <math>A</math>, <math>B</math>, and <math>C</math> are adjacent to vertex <math>D</math>. The perpendicular distance from <math>D</math> to the plane containing
 
<math>A</math>, <math>B</math>, and <math>C</math> is closest to  
 
<math>A</math>, <math>B</math>, and <math>C</math> is closest to  
  
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<math> \text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9 </math>
 
<math> \text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9 </math>
  
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==Solution==
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==Solution 1==
  
 
By placing the cube in a coordinate system such that <math>D</math> is at the origin, <math>A(0,0,3)</math>, <math>B(4,0,0)</math>, and <math>C(0,4,0)</math>, we find that the equation of plane <math>ABC</math> is:
 
By placing the cube in a coordinate system such that <math>D</math> is at the origin, <math>A(0,0,3)</math>, <math>B(4,0,0)</math>, and <math>C(0,4,0)</math>, we find that the equation of plane <math>ABC</math> is:
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<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:
  
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<cmath>\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}.</cmath>
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<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:
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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 <math>c</math> and <math>C</math> are set equal to <math>0</math>.
 
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 <math>c</math> and <math>C</math> are set equal to <math>0</math>.
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==Solution 2==
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Let <math>x</math> be the desired distance. Recall that the volume of a pyramid is given by <math>\frac{1}{3}\cdot h \cdot B</math>, where <math>B</math> is the area of the base and <math>h</math> is the height. Consider pyramid <math>ABCD</math>. Letting <math>ABC</math> be the base, the volume of <math>ABCD</math> is given by <math>\frac{1}{3} \cdot x \cdot [ABC]</math>, but if we let <math>BCD</math> be the base, the volume is given by <math>\frac{1}{3} \cdot [BCD]\cdot [AD] = \frac{1}{3} \cdot [\frac{1}{2} \cdot 4 \cdot 4] \cdot 3 = 8</math>. Clearly, these two volumes must be equal, so we get the equation <math>\frac{1}{3}\cdot x \cdot[ABC]=8</math>. Thus, to find <math>x</math>, we just need to find <math>[ABC]</math>.
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By the Pythagorean Theorem, <math>AB=\sqrt{AD^2+DB^2}=5</math>, <math>AC=\sqrt{AD^2+DC^2}=5</math>, <math>BC=\sqrt{BD^2+DC^2}=4\sqrt{2}</math>.
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The altitude to <math>BC</math> in triangle <math>ABC</math> has length <math>\sqrt{AC^2-\frac{BC}{2}^2}=\sqrt{17}</math>, so <math>[ABC]=\frac{1}{2}\cdot 4\sqrt{2} \cdot \sqrt{17} = 2\sqrt{34}</math>. Then <math>x=\frac{24}{[ABC]}=\frac{24}{2\sqrt{34}}=\frac{6\sqrt{34}}{17}</math> or about <math>2.1</math>. The answer is <math>\boxed{C}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 09:21, 16 September 2022

Problem

On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is closest to

[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]

$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$

Solution 1

By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$, $B(4,0,0)$, and $C(0,4,0)$, we find that the equation of plane $ABC$ is:

\[\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by:

\[\frac{|Aa + Bb + Cc + D|}{\sqrt{A^2 + B^2 + C^2}}.\]

Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$) to the plane is given by:

\[\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.\]

Since $\sqrt{34} < 6$, this number should be just a little over $2$, and the correct answer is $\boxed{\text{(C)}}$.

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 $c$ and $C$ are set equal to $0$.

Solution 2

Let $x$ be the desired distance. Recall that the volume of a pyramid is given by $\frac{1}{3}\cdot h \cdot B$, where $B$ is the area of the base and $h$ is the height. Consider pyramid $ABCD$. Letting $ABC$ be the base, the volume of $ABCD$ is given by $\frac{1}{3} \cdot x \cdot [ABC]$, but if we let $BCD$ be the base, the volume is given by $\frac{1}{3} \cdot [BCD]\cdot [AD] = \frac{1}{3} \cdot [\frac{1}{2} \cdot 4 \cdot 4] \cdot 3 = 8$. Clearly, these two volumes must be equal, so we get the equation $\frac{1}{3}\cdot x \cdot[ABC]=8$. Thus, to find $x$, we just need to find $[ABC]$.

By the Pythagorean Theorem, $AB=\sqrt{AD^2+DB^2}=5$, $AC=\sqrt{AD^2+DC^2}=5$, $BC=\sqrt{BD^2+DC^2}=4\sqrt{2}$.

The altitude to $BC$ in triangle $ABC$ has length $\sqrt{AC^2-\frac{BC}{2}^2}=\sqrt{17}$, so $[ABC]=\frac{1}{2}\cdot 4\sqrt{2} \cdot \sqrt{17} = 2\sqrt{34}$. Then $x=\frac{24}{[ABC]}=\frac{24}{2\sqrt{34}}=\frac{6\sqrt{34}}{17}$ or about $2.1$. The answer is $\boxed{C}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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