I finally did it
by djmathman, Sep 2, 2012, 12:36 AM
I remember in one of my posts back in November I wrote a passage that went something along the lines of
Look who bucked that prediction.
Quote:
Therefore, I will solve my first AIME #15 in 
.
.Look who bucked that prediction.
1997 AIME #15 wrote:
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 
.
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 
.
and 
.
at the origin and 
on the positive imaginary axis. It is clear that the maximum possible area of such a triangle will occur when the triangle is inscribed in the rectangle. Hence, place points 
and 
on 
and 
respectively such that 
is equilateral. For some positive reals 
and 
,
and 
.
rotation of ![\[a+10i=e^{\pi i/3}(11+bi).\]](http://latex.artofproblemsolving.com/0/5/a/05abc96389b5a140fd284cfd1a0bc8521625ffb9.png)
,
.
to the origin is![\[\left|11+(20-11\sqrt{3})i\right|^2=11^2+(20-11\sqrt{3})^2=121+400-440\sqrt{3}+363=884-440\sqrt{3}.\]](http://latex.artofproblemsolving.com/d/1/1/d11ae62554b2d371d2c6e4333be8e3e3744c7c41.png)
.
is precisely what we calculated above, our final answer is 
.This post has been edited 1 time. Last edited by djmathman, Apr 6, 2015, 3:09 AM
Reason: align tags
Reason: align tags
April 2025
November 2024