Difference between revisions of "2013 AIME I Problems/Problem 12"
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== Problem 12 == | == Problem 12 == | ||
| − | + | Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^o</math> and <math>\angle Q = 60^o</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | |
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== Solution == | == Solution == | ||
Revision as of 12:14, 17 March 2013
Problem 12
Let 
be a triangle with 
and 
. A regular hexagon 
with side length 1 is drawn inside 
so that side 
lies on 
, side 
lies on 
, and one of the remaining vertices lies on 
. There are positive integers 
and 
such that the area of can be expressed in the form 
, where 
and are relatively prime, and c is not divisible by the square of any prime. Find 
.
Solution
(solution)
See also
| 2013 AIME I (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 | ||
