Difference between revisions of "1997 AIME Problems/Problem 15"
Kevinr9828 (talk | contribs) (→See also) |
(→Solution) |
||
| Line 8: | Line 8: | ||
We can then set the real and imaginary parts equal, and solve for <math>(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})</math>. Hence a side <math>s</math> of the equilateral triangle can be found by <math>s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}</math>. Using the area formula <math>\frac{s^2\sqrt{3}}{4}</math>, the area of the equilateral triangle is <math>\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330</math>. Thus <math>p + q + r = 221 + 3 + 330 = \boxed{554}</math>. | We can then set the real and imaginary parts equal, and solve for <math>(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})</math>. Hence a side <math>s</math> of the equilateral triangle can be found by <math>s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}</math>. Using the area formula <math>\frac{s^2\sqrt{3}}{4}</math>, the area of the equilateral triangle is <math>\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330</math>. Thus <math>p + q + r = 221 + 3 + 330 = \boxed{554}</math>. | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | alternatively | ||
| + | |||
| + | Let angle EAB =x . then angle DFA=60+x. Let the side of the equilateral triangle be s so s= (a^2+121)^ .5 | ||
| + | |||
| + | therefore sinx=a/s and sin(60+x)=10/s... | ||
| + | therfore 10/s =sin(60+x)=sin60cosx+cos60sinx =(3^.5/2)(11/s)+(1/2)(a/s) | ||
| + | multiplying both sides by 2s we get 20=11(3^.5)+a so a=20-11(3^.5) | ||
| + | |||
| + | sustituting this into area=(a^2+121)(3^.5/4) we get the desired answer | ||
== See also == | == See also == | ||
Revision as of 11:08, 23 July 2008
Contents
Problem
The sides of rectangle 
have lengths 
and 
. An equilateral triangle is drawn so that no point of the triangle lies outside . The maximum possible area of such a triangle can be written in the form 
, where 
, 
, and 
are positive integers, and is not divisible by the square of any prime number. Find 
.
Solution
Consider points on the complex plane 
. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at 
, and the other two points 
and 
on 
and 
, respectively. Let 
and 
. Since it's equilateral, then 
, so 
, and expanding we get 
.
We can then set the real and imaginary parts equal, and solve for 
. Hence a side 
of the equilateral triangle can be found by 
. Using the area formula 
, the area of the equilateral triangle is 
. Thus 
.
Solution
alternatively
Let angle EAB =x . then angle DFA=60+x. Let the side of the equilateral triangle be s so s= (a^2+121)^ .5
therefore sinx=a/s and sin(60+x)=10/s... therfore 10/s =sin(60+x)=sin60cosx+cos60sinx =(3^.5/2)(11/s)+(1/2)(a/s) multiplying both sides by 2s we get 20=11(3^.5)+a so a=20-11(3^.5)
sustituting this into area=(a^2+121)(3^.5/4) we get the desired answer
See also
| 1997 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
alternatively
Let angle EAB =x . then angle DFA=60+x. Let the side of the equilateral triangle be s so s= (a^2+121)^ .5
therefore sinx=a/s and sin(60+x)=10/s... therfore 10/s =sin(60+x)=sin60cosx+cos60sinx =(3^.5/2)(11/s)+(1/2)(a/s) multiplying both sides by 2s we get 20=11(3^.5)+a so a=20-11(3^.5)
sustituting this into area=(a^2+121)(3^.5/4) we get the desired answer
