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2024 AMC 12B 第 24 题

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24.

有多少个正整数有序三元组 (a,b,c)(a, b, c)，满足 a≤b≤c≤9a \le b \le c \le 9，并且存在一个非退化三角形 △ABC\triangle ABC，其内切圆半径为整数，且 aa、bb、cc 分别是从 AA 到 BC‾\overline{BC}、从 BB 到 AC‾\overline{AC}、从 CC 到 AB‾\overline{AB} 的高？

What is the number of ordered triples (a,b,c)(a, b, c) of positive integers, with a≤b≤c≤9,a \le b \le c \le 9, such that there exists a (non-degenerate) triangle △ABC\triangle ABC with an integer inradius for which a,a, b,b, and cc are the lengths of the altitudes from AA to BC‾,\overline{BC}, BB to AC‾,\overline{AC}, and CC to AB‾,\overline{AB}, respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)

22

33

44

55

66

答案：B
知识点：高线内切圆、内心与内切圆半径三角不等式
难度评级：2410
解答：

将每条边写成 2[△]h\dfrac{2[\triangle]}{h}，则半周长为 [△](1a+1b+1c)[\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr)。由 r=[△]sr = \dfrac{[\triangle]}{s}，可得 1r=1a+1b+1c\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c。我们需要该和等于正整数 rr 的倒数 1r\dfrac1r。边长与 1a,1b,1c\tfrac1a, \tfrac1b, \tfrac1c 成比例，所以非退化条件要求 1a<1b+1c\tfrac1a \lt \tfrac1b + \tfrac1c。

搜索 a≤b≤c≤9a\le b\le c\le9，且 3/c≥1/33/c\ge1/3 为单位分数并满足三角形不等式的情形。有效等边情形为 rr，对应 1,2,31,2,3；1/a<1/r≤3/a1/a\lt1/r\le3/a，对应 r<a≤3rr\lt a\le3r；(9,9,9)(9,9,9)，对应 r,ar,a。像 b=a,a+1,…,9b=a,a+1,\ldots,9 和 (4,8,8)(4,8,8) 这样的例子会退化，所以共有 33 个。 cc b≤c≤9b\le c\le9 (2,3,6),(2,4,4)(2,3,6),(2,4,4)， 1a=1b+1c\tfrac1a=\tfrac1b+\tfrac1c (3,3,3),(6,6,6)(3,3,3),(6,6,6)， c=abrab−ar−br. c=\frac{abr}{ab-ar-br}. r(a,b,c)1(2,3,6),(2,4,4),(3,3,3)2(4,8,8),(6,6,6)3(9,9,9). \begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9). \end{array}

所以正确答案是 B。

Writing each side as 2[△]h,\dfrac{2[\triangle]}{h}, the semiperimeter is [△](1a+1b+1c),[\triangle]\bigl(\tfrac1a + \tfrac1b + \tfrac1c\bigr), so the inradius r=[△]sr = \dfrac{[\triangle]}{s} satisfies 1r=1a+1b+1c.\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c. We need this to be 1r\dfrac1r for a positive integer r,r, with the sides (proportional to 1a,1b,1c\tfrac1a, \tfrac1b, \tfrac1c) forming a non-degenerate triangle, requiring 1a<1b+1c.\tfrac1a \lt \tfrac1b + \tfrac1c.

Because a≤b≤c≤9,a\le b\le c\le9, the reciprocal sum is at least 3/c≥1/3,3/c\ge1/3, so the integer rr is one of 1,2,3.1,2,3. Also 1/a<1/r≤3/a,1/a\lt1/r\le3/a, so r<a≤3r.r\lt a\le3r. For each of these few values of r,a,r,a, substitute b=a,a+1,…,9b=a,a+1,\ldots,9 into c=abrab−ar−br. c=\frac{abr}{ab-ar-br}. Keeping only integral cc with b≤c≤9b\le c\le9 gives the complete list r(a,b,c)1(2,3,6),(2,4,4),(3,3,3)2(4,8,8),(6,6,6)3(9,9,9). \begin{array}{c|l} r& (a,b,c)\\ \hline 1&(2,3,6),(2,4,4),(3,3,3)\\ 2&(4,8,8),(6,6,6)\\ 3&(9,9,9). \end{array} The triples (2,3,6),(2,4,4),(2,3,6),(2,4,4), and (4,8,8)(4,8,8) have 1a=1b+1c\tfrac1a=\tfrac1b+\tfrac1c and therefore give degenerate triangles. The remaining triples are (3,3,3),(6,6,6),(3,3,3),(6,6,6), and (9,9,9),(9,9,9), so the answer is 3.3.

Thus, the correct answer is B.

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