Difference between revisions of "1997 AIME Problems/Problem 15"
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The sides of [[rectangle]] <math>ABCD</math> have lengths <math>10</math> and <math>11</math>. An [[equilateral triangle]] is drawn so that no point of the triangle lies outside <math>ABCD</math>. The maximum possible [[area]] of such a triangle can be written in the form <math>p\sqrt{q}-r</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive integers, and <math>q</math> is not divisible by the square of any prime number. Find <math>p+q+r</math>. | The sides of [[rectangle]] <math>ABCD</math> have lengths <math>10</math> and <math>11</math>. An [[equilateral triangle]] is drawn so that no point of the triangle lies outside <math>ABCD</math>. The maximum possible [[area]] of such a triangle can be written in the form <math>p\sqrt{q}-r</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive integers, and <math>q</math> is not divisible by the square of any prime number. Find <math>p+q+r</math>. | ||
− | + | == Solution 1 (Coordinate Bash)== | |
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Consider points on the [[complex plane]] <math>A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)</math>. 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 <math>A</math>, and the other two points <math>E</math> and <math>F</math> on <math>BC</math> and <math>CD</math>, respectively. Let <math>E (11,a)</math> and <math>F (b, 10)</math>. Since it's equilateral, then <math>E\cdot\text{cis}60^{\circ} = F</math>, so <math>(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i</math>, and expanding we get <math>\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>. | Consider points on the [[complex plane]] <math>A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)</math>. 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 <math>A</math>, and the other two points <math>E</math> and <math>F</math> on <math>BC</math> and <math>CD</math>, respectively. Let <math>E (11,a)</math> and <math>F (b, 10)</math>. Since it's equilateral, then <math>E\cdot\text{cis}60^{\circ} = F</math>, so <math>(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i</math>, and expanding we get <math>\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>. | ||
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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 2== | |
This is a trigonometric re-statement of the above. Let <math>x = \angle EAB</math>; by alternate interior angles, <math>\angle DFA=60+x</math>. Let <math>a = EB</math> and the side of the equilateral triangle be <math>s</math>, so <math>s= \sqrt{a^2+121}</math> by the [[Pythagorean Theorem]]. Now <math>\frac{10}{s} = \sin(60+x)= \sin {60} \cos x+ \cos {60} \sin x = \left(\frac{\sqrt{3}}2\right)\left(\frac{11}s\right)+\left(\frac 12\right)\left( \frac as \right)</math>. This reduces to <math>a=20-11\sqrt{3}</math>. | This is a trigonometric re-statement of the above. Let <math>x = \angle EAB</math>; by alternate interior angles, <math>\angle DFA=60+x</math>. Let <math>a = EB</math> and the side of the equilateral triangle be <math>s</math>, so <math>s= \sqrt{a^2+121}</math> by the [[Pythagorean Theorem]]. Now <math>\frac{10}{s} = \sin(60+x)= \sin {60} \cos x+ \cos {60} \sin x = \left(\frac{\sqrt{3}}2\right)\left(\frac{11}s\right)+\left(\frac 12\right)\left( \frac as \right)</math>. This reduces to <math>a=20-11\sqrt{3}</math>. | ||
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. | ||
− | + | ==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> |
Latest revision as of 20:23, 17 April 2021
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 | ||
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