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2014 AMC 12B 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

设 ABCDEABCDE 是一个内接于圆的五边形，满足 AB=CD=3AB = CD = 3，BC=DE=10BC = DE = 10，且 AE=14AE = 14。ABCDEABCDE 所有对角线长度之和等于 mn\dfrac{m}{n}，其中 mm 和 nn 是互质正整数。m+nm + n 是多少？

Let ABCDEABCDE be a pentagon inscribed in a circle such that AB=CD=3,AB = CD = 3, BC=DE=10,BC = DE = 10, and AE=14.AE = 14. The sum of the lengths of all diagonals of ABCDEABCDE is equal to mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m + n?

129129

247247

353353

391391

421421

答案：D
知识点：托勒密定理弦多项式
难度评级：2650
小提示：

等弦所对的弧相等，所以 AC=BD=CEAC = BD = CE；设这个公共长度为 xx。

Equal chords subtend equal arcs, so AC=BD=CE;AC = BD = CE; call this xx

大提示：

对 ABCDABCD，BCDEBCDE，和 ABDEABDE 使用托勒密定理，得到关于 x,y=AD,z=BEx, y = AD, z = BE 的方程。

Apply Ptolemy’s theorem to ABCD,ABCD, BCDE,BCDE, and ABDEABDE to get equations in x,y=AD,z=BEx, y = AD, z = BE

解答：

因为弧 AB,CDAB, CD 相等，弧 BC,DEBC, DE 相等，所以弦 AC,BD,CEAC, BD, CE 都相等；设 x=AC=BD=CEx = AC = BD = CE， y=ADy = AD， 且 z=BEz = BE。

对 ABCDABCD、BCDEBCDE 和 ABDEABDE 使用托勒密定理，得到 10y+9=x2,100+3z=x2,30+14x=yz。 \begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz\text{。} \end{gathered} 由前两个方程解出 yy 和 zz，再代入第三个方程，得到 x3−109x−420=0=(x−12)(x+5)(x+7)。 \begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7)\text{。} \end{gathered}

所以 x=12x = 12， y=13510=272y = \tfrac{135}{10} = \tfrac{27}{2}， 且 z=443z = \tfrac{44}{3}。 五条对角线为 AC,BD,CE,AD,BEAC, BD, CE, AD, BE， 它们的和为 3x+y+z=36+272+443=3856。 \begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}\text{。} \end{gathered}

因此 m+n=385+6=391m + n = 385 + 6 = 391， 正确答案是 D。

Because arcs AB,CDAB, CD are equal and arcs BC,DEBC, DE are equal, the chords AC,BD,CEAC, BD, CE are all equal; let x=AC=BD=CE,x = AC = BD = CE, y=AD,y = AD, and z=BE.z = BE.

Ptolemy’s theorem on ABCD,ABCD, BCDE,BCDE, and ABDEABDE gives 10y+9=x2,100+3z=x2,30+14x=yz. \begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz. \end{gathered} Solving the first two for yy and zz and substituting into the third yields x3−109x−420=0=(x−12)(x+5)(x+7). \begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7). \end{gathered}

So x=12,x = 12, y=13510=272,y = \tfrac{135}{10} = \tfrac{27}{2}, and z=443.z = \tfrac{44}{3}. The five diagonals are AC,BD,CE,AD,BE,AC, BD, CE, AD, BE, summing to 3x+y+z=36+272+443=3856. \begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}. \end{gathered}

Thus m+n=385+6=391,m + n = 385 + 6 = 391, and the correct answer is D.

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