AIME 1991 · 第 14 题

Solution

Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$.

Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$. Subtracting these equations give $y^2-81y-112\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into the first equation gives $x=105$, so $x+y+z=105+144+135=\boxed{384}$.

Solution 2

Let $\theta$ be the inscribed angle in each of the 5 sides of length 81, so $d \sin \theta = 81$. Since the inscribed angles sum to $\pi$, $d \sin 5\theta = d \sin (\pi - 5\theta) = 31$.

Now consider the Chebyshev polynomials that put $\dfrac{\sin n\theta}{\sin \theta}$ in terms of $\cos \theta$:

$\dfrac{\sin 2\theta}{\sin \theta} = 2 \cos \theta, \dfrac{\sin 3\theta}{\sin \theta} = 4 \cos^2 \theta - 1$ $\dfrac{\sin 4\theta}{\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta, \dfrac{\sin 5\theta}{\sin \theta} = 16 \cos^4 \theta - 12 \cos^2 \theta + 1$

The sum of the diagonals is $d\sin 2\theta + d\sin 3\theta + d\sin 4\theta$, which becomes $(d \sin \theta)(8\cos^3 \theta + 4\cos^2 \theta - 2\cos \theta - 1)$, and we're given $16 \cos^4 \theta - 12 \cos^2 \theta + 1 = \dfrac{31}{81}$

Solve for $\cos \theta$: $16 \cos^4 \theta - 12 \cos^2 \theta + \dfrac{9}{4} = \dfrac{9}{4} - \dfrac{50}{81}$

$\left(4 \cos^2 \theta - \dfrac{3}{2}\right)^2 = \dfrac{729}{324} - \dfrac{200}{324} = \left(\dfrac{23}{18}\right)^2$

$8 \cos^2 \theta - 3 = \dfrac{23}{9}$ or $-\dfrac{23}{9}$, so $8 \cos^2 \theta = \dfrac{50}{9}$ or $\dfrac{4}{9}$

$\cos^2 \theta = \dfrac{25}{36}$ or $\dfrac{1}{18}$, which means $\cos \theta$ must be $\dfrac{5}{6}$ if $5 \theta < \pi$.

Now $(d \sin \theta)\left(8\cos^3 \theta + 4\cos^2 \theta - 2\cos \theta - 1\right) = 81\left(8 \cdot \dfrac{125}{216} + 4 \cdot \dfrac{25}{36} - 2 \cdot \dfrac{5}{6} - 1\right)$

$= 3 \left(8 \cdot \dfrac{125}{8} + 4 \cdot \dfrac{75}{4} - 2 \cdot \dfrac{45}{2} - 27\right) = 3(125 + 75 - 45 - 27) = \boxed{384}$

Solution 3 (Risky)

This solution refers to the diagram given in Solution 1.

Applying Ptolemy’s Theorem to quadrilateral $ADEF$, we obtain

$$81y + 81^2 = z^2.$$ Applying Ptolemy’s Theorem to quadrilateral $ABCF$, we obtain

$$31y + 81^2 = x^2.$$ Since this is an AIME problem and the question asks for the sum of the lengths of the three diagonals, we may assume that all diagonal lengths are integers.

Rewriting the first equation,

$$81y = z^2 - 81^2 = (z+81)(z-81).$$ Because both $y$ and $z$ are integers, $z$ must be a multiple of $9$. Additionally, we observe that $y > z$, so we test the smallest such value of $z$ satisfying this condition, namely $z = 135$. Substituting gives

$$y = \frac{135^2 - 81^2}{81} = 144.$$ Substituting $y = 144$ into the second equation,

$$x^2 = 31(144) + 81^2 = 11025,$$ so $x = 105$.

Finally, summing the three diagonal lengths, $105 + 135 + 144 = \boxed{384}.$

~Voidling

Video Solution by OmegaLearn

https://youtu.be/DVuf-uXjfzY?t=522

~ pi_is_3.14