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 I Problems/Problem 6"

 
(Fixed the cubes to not just be three black hexagons (which draw(unitcube) inconveniently does))
 
(10 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
The sum of the areas of all triangles whose vertices are also vertices of a <math>1</math> by <math>1</math> by <math>1</math> cube is <math> m + \sqrt{n} + \sqrt{p}, </math> where <math> m, n, </math> and <math> p </math> are [[integer]]s. Find <math> m + n + p. </math>
  
 
== Solution ==
 
== Solution ==
 +
<center>
 +
<asy>
 +
size(120);
 +
import three;
 +
unitsize(1cm);
 +
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
 +
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
 +
draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue);
 +
</asy>
 +
 +
<asy>
 +
size(120);
 +
import three;
 +
unitsize(1cm);
 +
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
 +
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
 +
draw((0,1,1)--(1,0,0)--(1,1,0)--(0,1,1), blue);
 +
</asy>
 +
 +
<asy>
 +
size(120);
 +
import three;
 +
unitsize(1cm);
 +
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
 +
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
 +
draw((1,0,0)--(0,1,0)--(1,1,1)--(1,0,0), blue);
 +
</asy>
 +
</center>
 +
Since there are <math>8</math> [[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 <math>6</math> faces, so there are <math>24</math> triangles of the first type. Each of these is a [[right triangle]] with legs of length <math>1</math>, so each triangle of the first type has area <math>\frac 12</math>.
 +
 +
Each edge of the cube is a side of exactly <math>2</math> of the triangles of the second type, and there are <math>12</math> edges, so there are <math>24</math> triangles of the second type. Each of these is a right triangle with legs of length <math>1</math> and <math>\sqrt 2</math>, so each triangle of the second type has area <math>\frac{\sqrt{2}}{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 + 288 + 48 = \boxed{348}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2003 AIME I Problems]]
+
{{AIME box|year=2003|n=I|num-b=5|num-a=7}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:35, 27 December 2021

Problem

The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers. Find $m + n + p.$

Solution

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue); [/asy]

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((0,1,1)--(1,0,0)--(1,1,0)--(0,1,1), blue); [/asy]

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,0,0)--(0,1,0)--(1,1,1)--(1,0,0), blue); [/asy]

Since there are $8$ vertices of a cube, there are ${8 \choose 3} = 56$ 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 ${4\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$, so each triangle of the first type has area $\frac 12$.

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 of these is a right triangle with legs of length $1$ and $\sqrt 2$, so each triangle of the second type has area $\frac{\sqrt{2}}{2}$.

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 of the these is an equilateral triangle with sides of length $\sqrt 2$, so each triangle of the third type has area $\frac{\sqrt 3}2$.

Thus the total area of all these triangles is $24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}$ and the answer is $12 + 288 + 48 = \boxed{348}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_6&oldid=168584"