2005 AIME II 第 14 题

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14.

在三角形 ABCABC 中,AB=13AB = 13BC=15BC = 15CA=14CA = 14。点 DDBC\overline{BC} 上,且 CD=6CD = 6。点 EEBC\overline{BC} 上,使得 BAECAD\angle BAE \cong \angle CAD。已知 BE=pqBE = \frac{p}{q},其中 ppqq 是互质正整数,求 qq

In triangle ABC,ABC, AB=13,AB = 13, BC=15,BC = 15, and CA=14.CA = 14. Point DD is on BC\overline{BC} with CD=6.CD = 6. Point EE is on BC\overline{BC} such that BAECAD.\angle BAE \cong \angle CAD. Given that BE=pq,BE = \frac{p}{q}, where pp and qq are relatively prime positive integers, find q.q.

答案:463
知识点:面积比三角学比与比例
难度评级:3060
解答:

线段 ADAD 将对边分成的比满足 BDDC=[ABD][ACD]=12ABADsinBAD12ACADsinCAD=ABsinBADACsinCAD, \begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD}, \end{aligned} 同理,BEEC=ABsinBAEACsinCAE\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}

因为 BAE=CAD\angle BAE = \angle CAD, 也有 BAD=CAE\angle BAD = \angle CAE(二者都是这个公共角加上 EAD\angle EAD),所以两个比值相乘时所有正弦都约去: BDDCBEEC=AB2AC2\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}。 代入 BD=9BD = 9DC=6DC = 6AB=13AB = 13, 和 AC=14AC = 14, 得 BEEC=13214269=169294.\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.

因而 BE=15169169+294=2535463BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}。 由于 463463 是质数, 且不整除 2535=351322535 = 3 \cdot 5 \cdot 13^2, 该分数已最简,所以 q=463q = 463

A cevian ADAD splits the opposite side in the ratio BDDC=[ABD][ACD]=12ABADsinBAD12ACADsinCAD=ABsinBADACsinCAD, \begin{aligned} \frac{BD}{DC} &= \frac{[ABD]}{[ACD]} \\ &= \small \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} \\ &= \frac{AB \sin\angle BAD}{AC \sin\angle CAD}, \end{aligned} and similarly BEEC=ABsinBAEACsinCAE.\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}.

Since BAE=CAD,\angle BAE = \angle CAD, we also have BAD=CAE\angle BAD = \angle CAE (each is that common angle plus EAD\angle EAD), so multiplying the two ratios cancels all the sines: BDDCBEEC=AB2AC2.\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}. With BD=9,BD = 9, DC=6,DC = 6, AB=13,AB = 13, and AC=14,AC = 14, this gives BEEC=13214269=169294.\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.

Hence BE=15169169+294=2535463.BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}. Since 463463 is prime and does not divide 2535=35132,2535 = 3 \cdot 5 \cdot 13^2, the fraction is in lowest terms, and q=463.q = 463.

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