Difference between revisions of "2003 AIME I Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | Since there are 8 [[vertex | vertices]] of a [[cube]], there are <math>{8 \choose 3} = 56</math> total [[triangle]]s 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 [[edge]]s 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. | + | Since there are 8 [[vertex | vertices]] of a [[cube (geometry) | cube]], there are <math>{8 \choose 3} = 56</math> total [[triangle]]s 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 [[edge]]s 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 <math>{4\choose 3} = 4</math> triangles of the first type, and there are 6 faces, so there are 24 triangles of the first type. | Each face of the cube contains <math>{4\choose 3} = 4</math> triangles of the first type, and there are 6 faces, so there are 24 triangles of the first type. | ||
Revision as of 12:27, 30 October 2006
Problem
The sum of the areas of all triangles whose vertices are also vertices of a 1 by 1 by 1 cube is 
where 
and 
are integers. Find 
Solution
Since there are 8 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 6 faces, so there are 24 triangles of the first type.
Each edge of the cube is a side of exactly 2 of the triangles of the second type, and there are 12 edges, so there are 24 triangles of the second type.
Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are 8 vertices of the cube, so there are 8 triangles of the third type.
Each triangle of the first type is a right triangle with legs of length 1, so each triangle of the first type has area 
.
Each triangle of the second type is a right triangle with legs of length 1 and 
, so each triangle of the second type has area 
.
Each triangle of the third type is an equilateral triangle withs 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 
.
