Difference between revisions of "2001 AIME II Problems/Problem 12"
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On the first construction, <math>P_1</math>, four new tetrahedra will be constructed with side lengths <math>\frac 12</math> of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume <math>\left(\frac 12\right)^3 = \frac 18</math>. The total volume added here is then <math>\Delta P_1 = 4 \cdot \frac 18 = \frac 12</math>. | On the first construction, <math>P_1</math>, four new tetrahedra will be constructed with side lengths <math>\frac 12</math> of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume <math>\left(\frac 12\right)^3 = \frac 18</math>. The total volume added here is then <math>\Delta P_1 = 4 \cdot \frac 18 = \frac 12</math>. | ||
| − | We now note that for each midpoint triangle we construct in step <math>P_{i}</math>, there are now <math>6</math> places to construct new midpoint triangles for step <math>P_{i+1}</math>. The outward tetrahedron for the midpoint triangle provides <math>3</math> of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other <math>3</math>. However, the volume of the tetrahedra being constructed decrease by a factor of <math>\frac 18</math>. Thus we have the recursion <math>\Delta P_{i+1} = \frac{6}{8} \Delta P_i</math>, and so <math>\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1</math>. | + | We now note that for each midpoint triangle we construct in step <math>P_{i}</math>, there are now <math>6</math> places to construct new midpoint triangles for step <math>P_{i+1}</math>. The outward tetrahedron for the midpoint triangle provides <math>3</math> of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other <math>3</math>. This is because if you read this question carefully, it asks to add new tetrahedra to each face of <math>P_{i}</math> which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of <math>\frac 18</math>. Thus we have the recursion <math>\Delta P_{i+1} = \frac{6}{8} \Delta P_i</math>, and so <math>\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1</math>. |
| − | The volume of <math>P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}</math>, and <math>m+n=\boxed{101}</math>. Note that the summation was in fact a [[geometric series]]. | + | The volume of <math>P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}</math>, and <math>m+n=\boxed{101}</math>. Note that the summation was in fact a [[geometric series]]. |
== See also == | == See also == | ||
Revision as of 15:12, 20 September 2017
Problem
Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra 
is defined recursively as follows: 
is a regular tetrahedron whose volume is 1. To obtain 
, replace the midpoint triangle of every face of by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of 
is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
On the first construction, 
, four new tetrahedra will be constructed with side lengths 
of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume 
. The total volume added here is then 
.
We now note that for each midpoint triangle we construct in step , there are now 
places to construct new midpoint triangles for step 
. The outward tetrahedron for the midpoint triangle provides 
of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other . This is because if you read this question carefully, it asks to add new tetrahedra to each face of which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of 
. Thus we have the recursion 
, and so 
.
The volume of 
, and 
. Note that the summation was in fact a geometric series.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
