1997 AIME Problems/Problem 15


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

See also

1997 AIME (ProblemsAnswer KeyResources)
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