AIME 2001 I · 第 13 题

Solution

Solution 1

Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree arcs, and let $CD$ be the chord of the $3d$-degree arc. Also let $x$ be equal to the chord length of the $3d$-degree arc. Hence, the length of the chords, $AD$ and $BC$, of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem.

Using Ptolemy's theorem,

$$AB \cdot CD + AC \cdot BD = AD \cdot BC$$ $$\iff 22x + 22 \cdot 22 = (x + 20)^2$$ $$\iff 22x + 484 = x^2 + 40x + 400$$ $$\iff x^2 + 18x - 84 = 0.$$ We can then apply the quadratic formula to find the positive root of this equation (since polygons obviously cannot have sides of negative length):

$$x = \frac{-18 + \sqrt{18^2 + 4 \cdot 84}}{2} = \frac{-18 + \sqrt{660}}{2}.$$ This simplifies to $x = \frac{-18 + 2\sqrt{165}}{2} = -9 + \sqrt{165}$. Thus the answer is $9 + 165 = \boxed{174}$.

Solution 2

Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information,

$$2R\sin z=22, \quad\text{and}$$ $$2R(\sin 2z-\sin 3z)=20.$$ Dividing the latter equation by the former gives

$$\frac{2\sin z\cos z-\left(3\cos^2z\sin z-\sin^3 z\right)}{\sin z}=2\cos z-\left(3\cos^2z-\sin^2z\right)=1+2\cos z-4\cos^2z=\frac{10}{11}$$ $$\iff 4\cos^2z-2\cos z-\frac{1}{11}=0. \qquad (*)$$ We want to find

$$\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).$$ From $(*),$ this is equivalent to $44\cos z-20$. Using the quadratic formula, we deduce that this expression equals $-9+\sqrt{165}$, so our answer is $\boxed{174}$.

Solution 3

Let $z=\frac{d}{2}$, $R$ be the circumradius, and $a$ be the length of a $3d$-degree chord. Using the extended sine law, we obtain:

$$22=2R\sin(z),$$ $$20+a=2R\sin(2z), \quad\text{and}$$ $$a=2R\sin(3z).$$ Dividing the second equation by the first, and using the double angle formula, we obtain $\cos(z)=\frac{20+a}{44}$. Now, using the triple angle formula, we can rewrite the third equation as follows:

$$a=2R \sin(3z)=\frac{22}{\sin(z)}\left(3\sin(z)-4\sin^3(z)\right) = 22\left(3-4\sin^2(z)\right) = 22\left(4\cos^2(z)-1\right) = \frac{(20+a)^2}{22}-22,$$ and solving this quadratic equation gives the answer as $\boxed{174}$.