Difference between revisions of "1997 AIME Problems/Problem 15"
(→See also) |
m (tex) |
||
| Line 2: | Line 2: | ||
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>. | ||
| + | __TOC__ | ||
== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
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>. | ||
| Line 9: | Line 11: | ||
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== | + | ===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>. | ||
| − | + | 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. | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
== See also == | == See also == | ||
Revision as of 11:46, 25 July 2008
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.
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 | ||
