Difference between revisions of "1997 AIME Problems/Problem 15"
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Thus, the area of the triangle is <math>\frac{s^2\sqrt{3}}{4} =(a^2+121)\frac{\sqrt{3}}{4}</math>, which yields the same answer as above. | Thus, the area of the triangle is <math>\frac{s^2\sqrt{3}}{4} =(a^2+121)\frac{\sqrt{3}}{4}</math>, which yields the same answer as above. | ||
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| + | ===Solution 3=== | ||
| + | |||
| + | Here we present a synthetic solution. Since <math>\angle{BAD}=90</math> and <math>\angle{EAF}=60</math>, it follows that <math>\angle{DAF}+\angle{BAE}=90-60=30</math>. Rotate triangle <math>ADF</math> <math>60</math> degrees clockwise. Note that the image of <math>AF</math> is <math>AE</math>. Let the image of <math>D</math> be <math>D'</math>. Since angles are preserved under rotation, <math>\angle{DAF}=\angle{D'AE}</math>. It follows that <math>\angle{D'AE}+\angle{BAE}=\angle{D'AB}=30</math>. Since <math>\angle{ADF}=\angle{ABE}=90</math>, it follows that quadrilateral <math>ABED'</math> is cyclic with circumdiameter <math>AE=s</math> and thus circumradius <math>\frac{s}{2}</math>. Let <math>O</math> be its circumcenter. By Inscribed Angles, <math>\angle{BOD'}=2\angle{BAD}=60</math>. By the definition of circle, <math>OB=OD'</math>. It follows that triangle <math>OBD'</math> is equilateral. Therefore, <math>BD'=r=\frac{s}{2}</math>. Applying the Law of Cosines to triangle <math>ABD'</math>, <math>\frac{s}{2}=\sqrt{</math> | ||
== See also == | == See also == | ||
Revision as of 01:16, 15 February 2013
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
Solution 1
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 2
This is a trigonometric re-statement of the above. Let 
; by alternate interior angles, 
. Let 
and the side of the equilateral triangle be , so 
by the Pythagorean Theorem. Now 
. This reduces to 
.
Thus, the area of the triangle is 
, which yields the same answer as above.
Solution 3
Here we present a synthetic solution. Since 
and 
, it follows that 
. Rotate triangle 
degrees clockwise. Note that the image of 
is 
. Let the image of 
be 
. Since angles are preserved under rotation, 
. It follows that 
. Since 
, it follows that quadrilateral 
is cyclic with circumdiameter 
and thus circumradius 
. Let 
be its circumcenter. By Inscribed Angles, 
. By the definition of circle, 
. It follows that triangle 
is equilateral. Therefore, 
. Applying the Law of Cosines to triangle 
, $\frac{s}{2}=\sqrt{$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
