Difference between revisions of "1997 AIME Problems/Problem 15"
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-Solution by '''thecmd999''' | -Solution by '''thecmd999''' | ||
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+ | ==Solution 4 (Fast, no trig)== | ||
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+ | Clearly one vertex of the equilateral triangle is on a vertex of the rectangle, and the other two are lying on two other sides. Let <math>m</math> be the side length of the triangle, and let the rectangle be partitioned into the equilateral triangle, a right triangle with sides 11, <math>y</math>, <math>m</math>, a right triangle with sides 10, <math>x</math>, <math>m</math>, and a right triangle with sides <math>11-x</math>, <math>10-y</math>, <math>m</math>. Simple area analysis nets | ||
+ | <cmath>110=\frac{\sqrt{3}}{4}m^2+\frac{11}{2}y+\frac{10}{2}x+\frac{(11-x)(10-y)}{2}\implies 110-\frac{\sqrt{3}}{2}m^2=xy</cmath> | ||
+ | By the Pythagorean Theorem, <math>11^2+y^2=m^2</math> and <math>10^2+x^2=m^2</math>, so <math>x^2y^2=(m^2-10^2)(m^2-11^2)</math>. Thus, | ||
+ | <cmath>(110-\frac{\sqrt{3}}{2}m^2)^2=(m^2-10^2)(m^2-11^2)</cmath> | ||
+ | <cmath>110^2-110\sqrt{3}m^2+\frac34m^4=m^4-221m^2+110^2</cmath> | ||
+ | Obviously <math>m^2\neq0</math> so we can divide by <math>m^2</math> after cancellation: | ||
+ | <cmath>-110\sqrt{3}+221=\frac14m^2</cmath> | ||
+ | The area of the triangle is <math>\frac{\sqrt{3}}{4}m^2</math>, so the finish is simple. | ||
+ | <cmath>\frac{\sqrt{3}}{4}m^2=221\sqrt{3}-330\implies p+q+r=221+3+330=\boxed{554}</cmath> | ||
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+ | - clarkculus | ||
== See also == | == See also == |
Latest revision as of 20:39, 20 July 2024
Contents
[hide]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
Solution 4 (Fast, no trig)
Clearly one vertex of the equilateral triangle is on a vertex of the rectangle, and the other two are lying on two other sides. Let be the side length of the triangle, and let the rectangle be partitioned into the equilateral triangle, a right triangle with sides 11, , , a right triangle with sides 10, , , and a right triangle with sides , , . Simple area analysis nets By the Pythagorean Theorem, and , so . Thus, Obviously so we can divide by after cancellation: The area of the triangle is , so the finish is simple.
- clarkculus
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.