AIME 1985 · 第 9 题

Solution 1

All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle.

This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$. The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$, so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$.

Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle $\alpha$, so by the Law of Cosines,

$$2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}$$ and the answer is $17 + 32 = \boxed{049}$.

Video Solution by Pi Academy (Fast and Easy)

https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D

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Solution 2 (Law of Cosines)

It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$, and using the Law of Cosines, we get:

$$2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}$$ Which, rearranges to:

$$21 = 24\cos\frac{\alpha}{2}$$ And, that gets us:

$$\cos\frac{\alpha}{2} = 7/8$$ Using $\cos 2\theta = 2\cos^2 \theta - 1$, we get that:

$$\cos\alpha = 17/32$$ Which gives an answer of $\boxed{049}$

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Solution 3 (trig)

Using the first diagram above,

$$\sin \frac{\alpha}{2} = \frac{1}{r}$$ $$\sin \frac{\beta}{2} = \frac{1.5}{r}$$ $$\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}$$ by the Pythagorean trig identities,

$$\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}$$ $$\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}$$ so by the composite sine identity

$$\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}$$ multiply both sides by $2r$, then subtract $\sqrt{4-\frac{9}{r^2}}$ from both sides squaring both sides, we get

$$16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}$$ $$\Longrightarrow 16+4=9+8\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}$$ $$\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}$$ plugging this back in,

$$\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}$$ so

$$\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}$$ and the answer is $17+32=\boxed{049}$