Question

Raiders 队和 Wildcats 队的一场高中篮球赛在第一节结束时打平。Raiders 队四节的得分构成递增等比数列，Wildcats 队四节的得分构成递增等差数列。第四节结束时，Raiders 队以一分获胜。两队得分都不超过 $100$ 分。上半场两队总共得了多少分？

A high school basketball game between the Raiders and the Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?

$30$

$31$

$32$

$33$

$34$

答案：E

知识点：等比数列等差数列极限情形界定

难度评级：2180

小提示：

设 Raiders 得分为 $a, ar, ar^2, ar^3$ ，Wildcats 得分为 $a, a+d, a+2d, a+3d$ ，第一节同为 $a$

Let the Raiders score $a, ar, ar^2, ar^3$ and the Wildcats $a, a+d, a+2d, a+3d,$ tied at $a$

大提示：

总分相差 $1$ ，且两队的总分都低于 $100$ ，这迫使各项取值较小

The totals differ by $1$ and both stay under $100,$ which forces small values

解答：

设 Raiders 四节的得分依次为 $a, ar, ar^2, ar^3$ （递增等比数列， $r\gt1$ ），Wildcats 的得分依次为 $a, a+d, a+2d, a+3d$ （递增等差数列），两队第一节都得 $a$ 。

将 $r=\frac{m}{n}$ 写成最简分数，其中 $m\gt n$ 。因为 $ar^3$ 是整数，所以 $a$ 是 $n^3$ 的倍数；令 $a=A n^3$ 。Raiders 的总分为 $R=A(n^3+mn^2+m^2n+m^3)\text{。}$ 由于 $m^3\lt R\le100$ ，可知 $m\le4$ 。数对 $(4,3)$ 已使括号内的和达到 $175$ ，所以互质数对 $(m,n)$ 只可能是 $(2,1),(3,1),(3,2),(4,1)$ 。

相应的 $\frac{R}{A}$ 分别为 $15,40,65,85$ 。对于 $(4,1)$ 和 $(3,2)$ ，上界迫使 $A=1$ ，但由 $R-1=4a+6d$ 得到的 $d$ 不是整数。对于 $(3,1)$ ，会得到 $36A-1=6d$ ，这在模 $6$ 意义下不可能。

当 $(m,n)=(2,1)$ 时， $R=15A$ 且 $a=A$ ，所以 $11A-1=6d$ 。因而 $A\equiv5\pmod6$ ，再由 $R\le100$ 得 $A=5$ 。此时 $d=9$ ，Raiders 的得分是 $5,10,20,40$ ，Wildcats 的得分是 $5,14,23,32$ 。Raiders 以 $75$ 比 $74$ 获胜。

上半场总分为 $(5+10)+(5+14)=34$ 。

因此，正确答案是 E。

Let the Raiders score $a, ar, ar^2, ar^3$ (increasing geometric, $r\gt1$ ) and the Wildcats $a, a+d, a+2d, a+3d$ (increasing arithmetic), tied in the first quarter at $a.$

Write $r=\frac{m}{n}$ in lowest terms, with $m\gt n.$ Since $ar^3$ is an integer, $a$ is divisible by $n^3;$ put $a=A n^3.$ The Raiders’ total is $R=A(n^3+mn^2+m^2n+m^3).$ Since $m^3\lt R\le100,$ we have $m\le4.$ The pair $(4,3)$ already makes the parenthesized sum $175,$ so the only possible coprime pairs $(m,n)$ are $(2,1),(3,1),(3,2),(4,1).$

The corresponding base values of $\frac{R}{A}$ are $15,40,65,85.$ For $(4,1)$ and $(3,2),$ the bound forces $A=1,$ but $R-1=4a+6d$ gives a nonintegral $d.$ For $(3,1),$ it would give $36A-1=6d,$ which is impossible modulo $6.$

For $(m,n)=(2,1),$ we have $R=15A$ and $a=A,$ so $11A-1=6d.$ Thus $A\equiv5\pmod6,$ and $R\le100$ forces $A=5.$ Then $d=9,$ giving Raiders scores $5,10,20,40$ and Wildcats scores $5,14,23,32.$ The Raiders win $75$ to $74.$

The first-half total is $(5+10)+(5+14)=34.$

Thus, the correct answer is E.

2010 AMC 12B 第 19 题

答案：E

难度评级：2180

小提示：

大提示：

解答：

其他年份的第 19 题