Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1997 AHSME Problems/Problem 23"
(Deleted an incorrect method and fixed spelling and punctuation)
Line 30: Line 30:
 
Since there are three squares, the figure may appear to be based upon a unit cube.
 
Since there are three squares, the figure may appear to be based upon a unit cube.
  
−
Since there is an equilateral triangle of side length <math>\sqrt{2}</math>, this triangle has the length of a facial (non-spacial) diagonal of a unit cube.
+
Since there is an equilateral triangle of side length <math>\sqrt{2}</math>, this triangle has the length of a facial (non-spatial) diagonal of a unit cube.
  
−
Since there are three isoceles right triangles that are half the area of a face, they might represent faces of the cube that have been cut in half along the diagonal.
+
Since there are three isosceles right triangles that are half the area of a face, they might represent faces of the cube that have been cut in half along the diagonal.
  
−
Putting these three clues together, we can mentally fold the solid back together to form a unit cube that has had one vertex removed.  Along with this vertex, the triangular half of each of the three adjoining faces has been removed.  When the unit cube is sliced like this, it will cut three of the faces into isoceles triangles like <math>A</math>, <math>E</math>, and <math>F</math>.  Additionally, it will create an equilateral triangle that has as its sides three facial diagonals, just like <math>G</math>.
+
Putting these three clues together, we can mentally fold the solid back together to form a unit cube that has had one vertex removed.  Along with this vertex, the triangular half of each of the three adjoining faces has been removed.  When the unit cube is sliced like this, it will cut three of the faces into isosceles triangles like <math>A</math>, <math>E</math>, and <math>F</math>.  Additionally, it will create an equilateral triangle that has as its sides three facial diagonals, just like <math>G</math>.
  
−
So the volume is a unit cube withh a tetrahedron removed.  The tetrahedron has three right triangular faces like <math>A</math>, <math>E</math>, and <math>F</math>, and one equilateral triangular face like <math>G</math>.  The volume of this removed tetrahedron can be calculated in two ways:
+
So the volume is a unit cube with a tetrahedron removed.  The tetrahedron has three right triangular faces like <math>A</math>, <math>E</math>, and <math>F</math>, and one equilateral triangular face like <math>G</math>.  The volume of this removed tetrahedron can be calculated as follows:
  
−
1)  Place one of the three right triangles on the ground as the base.  It has an area of <math>\frac{1}{2}</math>.  The height is the height of one of the other two right triangles - the equilateral triangle is slanted and is not a height.  Thus, the volume of the tetrahedron is <math>\frac{1}{3}\cdot \frac{1}{2}\cdot 1 = \frac{1}{6}</math>, and when this volume is removed, the remaining volume is <math>\frac{5}{6}</math>, which is option <math>\boxed{D}</math>.
+
Place one of the three right triangles on the ground as the base.  It has an area of <math>\frac{1}{2}</math>.  The height is the height of one of the other two right triangles - the equilateral triangle is slanted and is not a height.  Thus, the volume of the tetrahedron is <math>\frac{1}{3}\cdot \frac{1}{2}\cdot 1 = \frac{1}{6}</math>, and when this volume is removed, the remaining volume is <math>\frac{5}{6}</math>, which is option <math>\boxed{D}</math>.
−
 
 
−
2)  Place the equilateral triangle on the ground as the base.  Its side is <math>\sqrt{2}</math>, so its area is <math>s^2\cdot\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}</math>.  The highest point of this tetrahedron is the vertex of the original cube that was removed, which lies directly over the center of the equilateral triangle.  Thus, the distance from the center of the triangle to the vertex of the cube is the height of the tetrahedron.  Since the center of the equilateral triangle is also the center of the cube, the distance from the center of the cube to the vertex will be the distance from <math>(\frac{2}{3}, \frac{2}{3}, \frac{2}{3})</math> to <math>(1,1,1)</math>.  This distance is <math>\frac{1}{3}\cdot \sqrt{3}</math>.  The volume of the tetrahedron is <math>\frac{1}{3}Bh = \frac{1}{3}\cdot \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{3} = \frac{1}{6}</math>, which is the removed volume from the unit cube.  Again, the volume remaining is <math>\boxed{\frac{5}{6}}</math>.
 
−
 
 
−
The center of the cube is ''not'' the center of the equiangular triangle. Thanks overall, though.
 
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=22|num-a=24}}
 
{{AHSME box|year=1997|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:06, 15 February 2018

Problem

[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); draw((-1,1)--(2,1)); draw((-1,0)--(1,0)); draw((-1,1)--(-1,0)); draw((0,-1)--(0,3)); draw((1,2)--(1,0)); draw((-1,1)--(1,1)); draw((0,2)--(1,2)); draw((0,3)--(1,2)); draw((0,-1)--(2,1)); draw((0,-1)--((0,-1) + sqrt(2)*dir(-15))); draw(((0,-1) + sqrt(2)*dir(-15))--(1,0)); label("$\textbf{A}$",foot((0,2),(0,3),(1,2)),SW); label("$\textbf{B}$",midpoint((0,1)--(1,2))); label("$\textbf{C}$",midpoint((-1,0)--(0,1))); label("$\textbf{D}$",midpoint((0,0)--(1,1))); label("$\textbf{E}$",midpoint((1,0)--(2,1)),NW); label("$\textbf{F}$",midpoint((0,-1)--(1,0)),NW); label("$\textbf{G}$",midpoint((0,-1)--(1,0)),2SE);[/asy]

In the figure, polygons $A$, $E$, and $F$ are isosceles right triangles; $B$, $C$, and $D$ are squares with sides of length $1$; and $G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is

$\textbf{(A)}\ 1/2\qquad\textbf{(B)}\ 2/3\qquad\textbf{(C)}\ 3/4\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 4/3$

Solution

Since there are three squares, the figure may appear to be based upon a unit cube.

Since there is an equilateral triangle of side length $\sqrt{2}$, this triangle has the length of a facial (non-spatial) diagonal of a unit cube.

Since there are three isosceles right triangles that are half the area of a face, they might represent faces of the cube that have been cut in half along the diagonal.

Putting these three clues together, we can mentally fold the solid back together to form a unit cube that has had one vertex removed. Along with this vertex, the triangular half of each of the three adjoining faces has been removed. When the unit cube is sliced like this, it will cut three of the faces into isosceles triangles like $A$, $E$, and $F$. Additionally, it will create an equilateral triangle that has as its sides three facial diagonals, just like $G$.

So the volume is a unit cube with a tetrahedron removed. The tetrahedron has three right triangular faces like $A$, $E$, and $F$, and one equilateral triangular face like $G$. The volume of this removed tetrahedron can be calculated as follows:

Place one of the three right triangles on the ground as the base. It has an area of $\frac{1}{2}$. The height is the height of one of the other two right triangles - the equilateral triangle is slanted and is not a height. Thus, the volume of the tetrahedron is $\frac{1}{3}\cdot \frac{1}{2}\cdot 1 = \frac{1}{6}$, and when this volume is removed, the remaining volume is $\frac{5}{6}$, which is option $\boxed{D}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_23&oldid=91283"