1999 AMC 12 第 19 题

先试着解答 1999 AMC 12 第 19 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1999 AMC 12 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

考虑所有满足下列条件的三角形 ABCABCAB=ACAB = AC,点 DD 在线段 AC\overline{AC} 上,且 BDAC\overline{BD} \perp \overline{AC}ADADCDCD 均为整数,且 BD2=57BD^2 = 57。在所有这样的三角形中,ACAC 的最小可能值是

Consider all triangles ABCABC satisfying the following conditions: AB=AC,AB = AC, DD is a point on AC\overline{AC} for which BDAC,\overline{BD} \perp \overline{AC}, ADAD and CDCD are integers, and BD2=57.BD^2 = 57. Among all such triangles, the smallest possible value of ACAC is

99

1010

1111

1212

1313

答案:C
知识点:等腰三角形勾股定理丢番图方程
难度评级:1810
解答:

AD=nAD = nCD=mCD = m。因为 ADB\triangle ADBDD 处为直角,AB2=n2+57AB^2 = n^2 + 57。又 AB=AC=m+nAB = AC = m + n,所以 化简为 m(m+2n)=57m(m + 2n) = 57(m+n)2=n2+57, (m + n)^2 = n^2 + 57,

正整数解为 m=1,n=28m = 1, n = 28,此时 AC=29AC = 29;以及 m=3,n=8m = 3, n = 8,此时 AC=11AC = 11。因此 ACAC 的最小可能值为 1111

所以正确答案是 C

Let AD=nAD = n and CD=m.CD = m. Since ADB\triangle ADB is right-angled at D,D, AB2=n2+57.AB^2 = n^2 + 57. Also AB=AC=m+n,AB = AC = m + n, so (m+n)2=n2+57, (m + n)^2 = n^2 + 57, which simplifies to m(m+2n)=57.m(m + 2n) = 57.

The positive integer solutions are m=1,n=28m = 1, n = 28 (giving AC=29AC = 29) and m=3,n=8m = 3, n = 8 (giving AC=11AC = 11). The smallest possible value of ACAC is 11.11.

Thus, the correct answer is C.

← 第 18 题#18
完整试卷

其他年份的第 19 题