AMC10 2023 A

AMC10 2023 A · Q17

AMC10 2023 A · Q17. It mainly tests Triangles (properties), Pythagorean theorem.

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$. Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$?

设 $ABCD$ 为矩形，$AB = 30$，$BC = 28$。点 $P$ 和 $Q$ 分别位于 $\overline{BC}$ 和 $\overline{CD}$ 上，使得 $\triangle{ABP}$、$\triangle{PCQ}$ 和 $\triangle{QDA}$ 的所有边长均为整数。求 $\triangle{APQ}$ 的周长。

(A) 84 84

(B) 86 86

(D) 90 90

(E) 92 92

Answer

Correct choice: (A)

正确答案：（A）

Solution

We know that all side lengths are integers, so we can test Pythagorean triples for all triangles. First, we focus on $\triangle{ABP}$. The length of $AB$ is $30$, and the possible (small enough) Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the length of the longer leg is a factor of $30$. Testing these, we get that only $(8, 15, 17)$ is a valid solution. Thus, we know that $BP = 16$ and $AP = 34$. Next, we move on to $\triangle{QDA}$. The length of $AD$ is $28$, and the small enough triples are $(3, 4, 5)$ and $(7, 24, 25)$. Testing again, we get that $(3, 4, 5)$ is our triple. We get the value of $DQ = 21$, and $AQ = 35$. We know that $CQ = CD - DQ$ which is $9$, and $CP = BC - BP$ which is $12$. $\triangle{CPQ}$ is therefore a right triangle with side length ratios ${3, 4, 5}$, and the hypotenuse is equal to $15$. $\triangle{APQ}$ has side lengths $34, 35,$ and $15,$ so the perimeter is equal to $34 + 35 + 15 = \boxed{\textbf{(A) } 84}.$

我们知道所有边长均为整数，因此可以测试所有三角形的勾股三元组。首先关注 $\triangle{ABP}$。$AB$ 的长度为 $30$，可能的（足够小）勾股三元组为 $(3, 4, 5)$、$(5, 12, 13)$、$(8, 15, 17)$，其中较长直角边的因子是30的因子。测试这些，只有 $(8, 15, 17)$ 是有效解。因此，$BP = 16$，$AP = 34$。接下来是 $\triangle{QDA}$。$AD$ 的长度为 $28$，足够小的三元组为 $(3, 4, 5)$ 和 $(7, 24, 25)$。再次测试，得到 $(3, 4, 5)$ 是我们的三元组。得到 $DQ = 21$，$AQ = 35$。我们知道 $CQ = CD - DQ = 9$，$CP = BC - BP = 12$。因此 $\triangle{CPQ}$ 是边长比例为 $3, 4, 5$ 的直角三角形，斜边为 $15$。 $\triangle{APQ}$ 的边长为 $34, 35, 15$，周长为 $34 + 35 + 15 = \boxed{\textbf{(A)} 84}$。

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