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 "2003 AIME II Problems/Problem 4"

m (→‎Solution 1)
(→‎Solution 3)
 
(4 intermediate revisions by 3 users not shown)
Line 17: Line 17:
 
Their ratio is <math>1:3</math>, so the ratio of their volumes is <math>\left(\frac{1}{3}\right)^3 = \frac{1}{27}</math>.
 
Their ratio is <math>1:3</math>, so the ratio of their volumes is <math>\left(\frac{1}{3}\right)^3 = \frac{1}{27}</math>.
  
−
<math>m+n = 1 + 27 = \boxed{028}</math>.
+
<math>m+n = 1 + 27 = \boxed{28}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
  
−
Let the large tetrahedron be <math>ABCD</math>, and the small tetrahedron be <math>WXYZ</math>, with <math>W</math> on <math>ABC</math>, <math>X</math> on <math>BCD</math>, <math>Y</math> on <math>ACD</math>, and <math>Z</math> on <math>ABD</math>. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let <math>AB=1</math>, for our convenience. Dropping an altitude from <math>W</math> to <math>BC</math>, and calling the foot <math>M</math>, we have <math>WM=XM=\frac{\sqrt3}{6}</math>. Since <math>\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3</math>. By Law of Cosines, we have <math>WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3</math>. Hence, the ratio of the volumes is <math>(\frac{1}{3})^3=1/27</math>. <math>m+n=1+27=\boxed{028}</math>
+
Let the large tetrahedron be <math>ABCD</math>, and the small tetrahedron be <math>WXYZ</math>, with <math>W</math> on <math>ABC</math>, <math>X</math> on <math>BCD</math>, <math>Y</math> on <math>ACD</math>, and <math>Z</math> on <math>ABD</math>. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let <math>AB=1</math>, for our convenience. Dropping an altitude from <math>W</math> to <math>BC</math>, and calling the foot <math>M</math>, we have <math>WM=XM=\frac{\sqrt3}{6}</math>. Since <math>\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3</math>. By Law of Cosines, we have <math>WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3</math>. Hence, the ratio of the volumes is <math>\left(\frac{1}{3}\right)^3=1/27</math>. <math>m+n=1+27=\boxed{028}</math>
  
 
===Solution 3===
 
===Solution 3===
  
−
Consider the large tetrahedron <math>ABCD</math> and the smaller tetrahedron <math>WXYZ</math>.  Label the points as you wish, but dropping an altitude from the top vertex of <math>ABCD</math>, we see it hits the center of the base face of <math>ABCD</math>.  This center is also one vertex of <math>WXYZ</math>.  Consider a "side" face of <math>ABCD</math>, and the center of that face, which is another vertex of <math>WXYZ</math>.  Draw the altitude of this side face (which is an equilateral triangle).  These two altitudes form a right triangle.  Since the center of the Side face splits the altitude of the side face into segments in the ratio of <math>2:1</math> (centroid), and since the bases of <math>WXYZ</math> and <math>ABCD</math> are parallel, we can say that the altitudes of tetrahedron <math>ABCD</math> and <math>WXYZ</math> are in the ratio <math>3:1</math>.  Thus we compute <math>(\frac{1}{3})^3</math>, and find <math>\frac{1}{27}</math>.  The sum of the numerator and denominator is thus <math>28</math>.
+
Consider the large tetrahedron <math>ABCD</math> and the smaller tetrahedron <math>WXYZ</math>.  Label the points as you wish, but dropping an altitude from the top vertex of <math>ABCD</math>, we see it hits the center of the base face of <math>ABCD</math>.  This center is also one vertex of <math>WXYZ</math>.  Consider a "side" face of <math>ABCD</math>, and the center of that face, which is another vertex of <math>WXYZ</math>.  Draw the altitude of this side face (which is an equilateral triangle).  These two altitudes form a right triangle.  Since the center of the Side face splits the altitude of the side face into segments in the ratio of <math>2:1</math> (centroid), and since the bases of <math>WXYZ</math> and <math>ABCD</math> are parallel, we can say that the altitudes of tetrahedron <math>ABCD</math> and <math>WXYZ</math> are in the ratio <math>3:1</math>.  Thus we compute <math>\left(\frac{1}{3}\right)^3</math>, and find <math>\frac{1}{27}</math>.  The sum of the numerator and denominator is thus <math>28</math>.
 +
 
 +
===Video Solution by Sal Khan===
 +
Part 1: https://www.youtube.com/watch?v=gXnHodHNusg&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=15
 +
 
 +
Part 2: https://www.youtube.com/watch?v=wQ34EIfd-5A&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=21
 +
 
 +
- AMBRIGGS
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2003|n=II|num-b=3|num-a=5}}
 +
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:36, 30 July 2022

Problem

In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

Embed the tetrahedron in 4-space to make calculations easier. Its vertices are $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$.

To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$,$(\frac{1}{3}, \frac{1}{3},0, \frac{1}{3})$,$(\frac{1}{3},0, \frac{1}{3}, \frac{1}{3})$,$(0,\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.

The side length of the large tetrahedron is $\sqrt{2}$ by the distance formula. The side length of the smaller tetrahedron is $\frac{\sqrt{2}}{3}$ by the distance formula.

Their ratio is $1:3$, so the ratio of their volumes is $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$.

$m+n = 1 + 27 = \boxed{28}$.

Solution 2

Let the large tetrahedron be $ABCD$, and the small tetrahedron be $WXYZ$, with $W$ on $ABC$, $X$ on $BCD$, $Y$ on $ACD$, and $Z$ on $ABD$. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let $AB=1$, for our convenience. Dropping an altitude from $W$ to $BC$, and calling the foot $M$, we have $WM=XM=\frac{\sqrt3}{6}$. Since $\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3$. By Law of Cosines, we have $WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3$. Hence, the ratio of the volumes is $\left(\frac{1}{3}\right)^3=1/27$. $m+n=1+27=\boxed{028}$

Solution 3

Consider the large tetrahedron $ABCD$ and the smaller tetrahedron $WXYZ$. Label the points as you wish, but dropping an altitude from the top vertex of $ABCD$, we see it hits the center of the base face of $ABCD$. This center is also one vertex of $WXYZ$. Consider a "side" face of $ABCD$, and the center of that face, which is another vertex of $WXYZ$. Draw the altitude of this side face (which is an equilateral triangle). These two altitudes form a right triangle. Since the center of the Side face splits the altitude of the side face into segments in the ratio of $2:1$ (centroid), and since the bases of $WXYZ$ and $ABCD$ are parallel, we can say that the altitudes of tetrahedron $ABCD$ and $WXYZ$ are in the ratio $3:1$. Thus we compute $\left(\frac{1}{3}\right)^3$, and find $\frac{1}{27}$. The sum of the numerator and denominator is thus $28$.

Video Solution by Sal Khan

Part 1: https://www.youtube.com/watch?v=gXnHodHNusg&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=15

Part 2: https://www.youtube.com/watch?v=wQ34EIfd-5A&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=21

- AMBRIGGS

See also

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2003_AIME_II_Problems/Problem_4&oldid=176428"