Difference between revisions of "2006 AIME I Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | + | Square <math> ABCD </math> has sides of length 1. Points <math> E </math> and <math> F </math> are on <math> \overline{BC} </math> and <math> \overline{CD}, </math> respectively, so that <math> \triangle AEF </math> is equilateral. A square with vertex <math> B </math> has sides that are parallel to those of <math> ABCD </math> and a vertex on <math> \overline{AE}. </math> The length of a side of this smaller square is <math>\frac{a-\sqrt{b}}{c}, </math> where <math> a, b, </math> and <math> c </math> are positive integers and <math> b</math> is not divisible by the square of any prime. Find <math> a+b+c. </math> | |
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== Solution == | == Solution == | ||
Revision as of 13:10, 25 September 2007
Problem
Square 
has sides of length 1. Points 
and 
are on 
and 
respectively, so that 
is equilateral. A square with vertex 
has sides that are parallel to those of and a vertex on 
The length of a side of this smaller square is 
where 
and 
are positive integers and 
is not divisible by the square of any prime. Find 
Solution
Numbers of the form 
can be written as 
. There are 
such numbers. Each digit will appear in each place value 
times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is 
.
Alternatively, for every number, , there will be exactly one other number, such that when they are added together, the sum is 
, or, more precisely, 1. As an example, 
.
Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is 
.
See also
| 2006 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 | ||
