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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
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<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>
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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,
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<cmath>(110-\frac{\sqrt{3}}{2}m^2)^2=(m^2-10^2)(m^2-11^2)</cmath>
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<cmath>110^2-110\sqrt{3}m^2+\frac34m^4=m^4-221m^2+110^2</cmath>
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Obviously <math>m^2\neq0</math> so we can divide by <math>m^2</math> after cancellation:
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<cmath>-110\sqrt{3}+221=\frac14m^2</cmath>
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The area of the triangle is <math>\frac{\sqrt{3}}{4}m^2</math>, so the finish is simple.
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<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

Problem

The sides of rectangle $ABCD$ have lengths $10$ and $11$. An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$. The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$, where $p$, $q$, and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$.

Solution 1 (Coordinate Bash)

Consider points on the complex plane $A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)$. 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 $A$, and the other two points $E$ and $F$ on $BC$ and $CD$, respectively. Let $E (11,a)$ and $F (b, 10)$. Since it's equilateral, then $E\cdot\text{cis}60^{\circ} = F$, so $(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i$, and expanding we get $\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i$.

1997 AIME-15a.PNG

We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})$. Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}$. Using the area formula $\frac{s^2\sqrt{3}}{4}$, the area of the equilateral triangle is $\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330$. Thus $p + q + r = 221 + 3 + 330 = \boxed{554}$.

Solution 2

This is a trigonometric re-statement of the above. Let $x = \angle EAB$; by alternate interior angles, $\angle DFA=60+x$. Let $a = EB$ and the side of the equilateral triangle be $s$, so $s= \sqrt{a^2+121}$ by the Pythagorean Theorem. Now $\frac{10}{s} = \sin(60+x)= \sin {60} \cos x+ \cos {60} \sin x = \left(\frac{\sqrt{3}}2\right)\left(\frac{11}s\right)+\left(\frac 12\right)\left( \frac as \right)$. This reduces to $a=20-11\sqrt{3}$.

Thus, the area of the triangle is $\frac{s^2\sqrt{3}}{4} =(a^2+121)\frac{\sqrt{3}}{4}$, which yields the same answer as above.

Solution 3

Since $\angle{BAD}=90$ and $\angle{EAF}=60$, it follows that $\angle{DAF}+\angle{BAE}=90-60=30$. Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$. Let the image of $D$ be $D'$. Since angles are preserved under rotation, $\angle{DAF}=\angle{D'AE}$. It follows that $\angle{D'AE}+\angle{BAE}=\angle{D'AB}=30$. Since $\angle{ADF}=\angle{ABE}=90$, it follows that quadrilateral $ABED'$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\frac{s}{2}$. Let $O$ be its circumcenter. By Inscribed Angles, $\angle{BOD'}=2\angle{BAD'}=60$. By the definition of circle, $OB=OD'$. It follows that triangle $OBD'$ is equilateral. Therefore, $BD'=r=\frac{s}{2}$. Applying the Law of Cosines to triangle $ABD'$, $\frac{s}{2}=\sqrt{10^2+11^2-(2)(10)(11)(\cos{30})}$. Squaring and multiplying by $\sqrt{3}$ yields $\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}}$

-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 $m$ be the side length of the triangle, and let the rectangle be partitioned into the equilateral triangle, a right triangle with sides 11, $y$, $m$, a right triangle with sides 10, $x$, $m$, and a right triangle with sides $11-x$, $10-y$, $m$. Simple area analysis nets \[110=\frac{\sqrt{3}}{4}m^2+\frac{11}{2}y+\frac{10}{2}x+\frac{(11-x)(10-y)}{2}\implies 110-\frac{\sqrt{3}}{2}m^2=xy\] By the Pythagorean Theorem, $11^2+y^2=m^2$ and $10^2+x^2=m^2$, so $x^2y^2=(m^2-10^2)(m^2-11^2)$. Thus, \[(110-\frac{\sqrt{3}}{2}m^2)^2=(m^2-10^2)(m^2-11^2)\] \[110^2-110\sqrt{3}m^2+\frac34m^4=m^4-221m^2+110^2\] Obviously $m^2\neq0$ so we can divide by $m^2$ after cancellation: \[-110\sqrt{3}+221=\frac14m^2\] The area of the triangle is $\frac{\sqrt{3}}{4}m^2$, so the finish is simple. \[\frac{\sqrt{3}}{4}m^2=221\sqrt{3}-330\implies p+q+r=221+3+330=\boxed{554}\]

- clarkculus

See also

1997 AIME (Problems • Answer Key • Resources)
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