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2024 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.

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？（回忆：三角形的内切圆半径，是能内接于该三角形的最大圆的半径。）

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
小提示：

因为每条边等于 2[△]高\dfrac{2[\triangle]}{\text{高}}，所以内切圆半径满足 1r=1a+1b+1c\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c。

Since each side equals 2[△]altitude,\dfrac{2[\triangle]}{\text{altitude}}, the inradius satisfies 1r=1a+1b+1c\dfrac1r = \dfrac1a + \dfrac1b + \dfrac1c

大提示：

边长与 1a,1b,1c\tfrac1a, \tfrac1b, \tfrac1c 成比例，所以非退化要求 1a<1b+1c\tfrac1a \lt \tfrac1b + \tfrac1c；寻找 1a+1b+1c\tfrac1a + \tfrac1b + \tfrac1c 为单位分数的三元组。

The sides are proportional to 1a,1b,1c,\tfrac1a, \tfrac1b, \tfrac1c, so non-degeneracy needs 1a<1b+1c;\tfrac1a \lt \tfrac1b + \tfrac1c; seek triples with 1a+1b+1c\tfrac1a + \tfrac1b + \tfrac1c a unit fraction

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。

因为 a≤b≤c≤9a\le b\le c\le9，倒数之和至少为 3c≥13\frac{3}{c}\ge\frac{1}{3}，所以整数 rr 只能是 1,2,31,2,3 之一。又 1a<1r≤3a\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{a}，所以 r<a≤3rr\lt a\le3r。对这少数几组 r,ar,a，把 b=a,a+1,…,9b=a,a+1,\ldots,9 代入 c=abrab−ar−br。 c=\frac{abr}{ab-ar-br}\text{。} 只保留满足 b≤c≤9b\le c\le9 的整数 cc，就得到完整的列表 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}\text{。} 其中三元组 (2,3,6),(2,4,4)(2,3,6),(2,4,4) 和 (4,8,8)(4,8,8) 满足 1a=1b+1c\tfrac1a=\tfrac1b+\tfrac1c，因而给出退化三角形。剩下的三元组是 (3,3,3),(6,6,6)(3,3,3),(6,6,6) 和 (9,9,9)(9,9,9)，所以答案是 33。

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 3c≥13,\frac{3}{c}\ge\frac{1}{3}, so the integer rr is one of 1,2,3.1,2,3. Also 1a<1r≤3a,\frac{1}{a}\lt\frac{1}{r}\le\frac{3}{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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