Difference between revisions of "2003 AIME I Problems/Problem 6"
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Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are <math>8</math> vertices of the cube, so there are <math>8</math> triangles of the third type. Each of the these is an [[equilateral triangle]] with sides of length <math>\sqrt 2</math>, so each triangle of the third type has area <math>\frac{\sqrt 3}2</math>. | Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are <math>8</math> vertices of the cube, so there are <math>8</math> triangles of the third type. Each of the these is an [[equilateral triangle]] with sides of length <math>\sqrt 2</math>, so each triangle of the third type has area <math>\frac{\sqrt 3}2</math>. | ||
| − | Thus the total area of all these triangles is <math>24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}</math> and the answer is <math>12 + | + | Thus the total area of all these triangles is <math>24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}</math> and the answer is <math>12 + 288 + 48 = \boxed{348}</math>. |
== See also == | == See also == | ||
Revision as of 20:38, 25 August 2017
Problem
The sum of the areas of all triangles whose vertices are also vertices of a 
by by cube is 
where 
and 
are integers. Find 
Solution
![[asy] size(120); defaultpen(linewidth(0.5)); import three; draw(unitcube); draw((1,0,0)--(1,0,1)--(1,1,1)--cycle,linewidth(0.9)); [/asy]](http://latex.artofproblemsolving.com/f/7/2/f72d4a32cf0a4820ada9b1d9fbcf2bf73a7ec6fd.png)
![[asy] size(120); defaultpen(linewidth(0.5)); import three; draw(unitcube); draw((1,0,0)--(0,1,0)--(0,1,1)--cycle,linewidth(0.9)); [/asy]](http://latex.artofproblemsolving.com/0/4/a/04a88fde297a2f9ae5f508ef99173192f76ea700.png)
![[asy] size(120); defaultpen(linewidth(0.5)); import three; draw(unitcube); draw((1,0,0)--(0,1,0)--(1,1,1)--cycle,linewidth(0.9)); [/asy]](http://latex.artofproblemsolving.com/4/7/6/47663db7d52fbd190e6789c0a156ebc71d1c43c4.png)
Since there are 
vertices of a cube, there are 
total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.
Each face of the cube contains 
triangles of the first type, and there are 
faces, so there are 
triangles of the first type. Each of these is a right triangle with legs of length , so each triangle of the first type has area 
.
Each edge of the cube is a side of exactly 
of the triangles of the second type, and there are 
edges, so there are triangles of the second type. Each of these is a right triangle with legs of length and 
, so each triangle of the second type has area 
.
Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are vertices of the cube, so there are triangles of the third type. Each of the these is an equilateral triangle with sides of length , so each triangle of the third type has area 
.
Thus the total area of all these triangles is 
and the answer is 
.
See also
| 2003 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
