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)== | |
− | |||
− | |||
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> | |
+ | |||
+ | -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 <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> | ||
+ | |||
+ | - clarkculus | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:39, 20 July 2024
Contents
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 |
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