Difference between revisions of "1997 AIME Problems/Problem 15"
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==Solution 3== | ==Solution 3== | ||
− | 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{10^2+11^2-(2)(10)(11)(\cos{30})}</math>. Squaring and multiplying by <math>\sqrt{3}</math> yields <math>\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}}</math> | + | 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{10^2+11^2-(2)(10)(11)(\cos{30})}</math>. Squaring and multiplying by <math>\sqrt{3}</math> yields <math>\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}}</math> |
-Solution by '''thecmd999''' | -Solution by '''thecmd999''' |
Revision as of 21:49, 17 July 2023
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 1 (Coordinate Bash)
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
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 , . Squaring and multiplying by yields
-Solution by thecmd999
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.