AMC10 2015 A

AMC10 2015 A · Q24

AMC10 2015 A · Q24. It mainly tests Quadratic equations, Pythagorean theorem.

For some positive integers $p$, quadrilateral $ABCD$ with positive integer side lengths has perimeter $p$, right angles at $B$ and $C$, $AB = 2$, and $CD = AD$. How many different values of $p < 2015$ are possible?

对于某些正整数 $p$，四边形 $ABCD$ 的边长均为正整数，周长为 $p$，在 $B$ 和 $C$ 处为直角，且 $AB = 2$、$CD = AD$。问满足 $p < 2015$ 的 $p$ 有多少种不同的取值？

(A) 30 30

(B) 31 31

(D) 62 62

(E) 63 63

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): In every such quadrilateral, $CD\ge AB$. Let $E$ be the foot of the perpendicular from $A$ to $CD$; then $CE=2$ and $AE=BC$. Let $x=AE$ and $y=DE$; then $AD=2+y$. By the Pythagorean Theorem, $x^2+y^2=(2+y)^2$, or $x^2=4+4y$. Therefore $x$ is even, say $x=2z$, and $z^2=1+y$. The perimeter of the quadrilateral is $x+2y+6=2z^2+2z+4$. Increasing positive integer values of $z$ give the required quadrilaterals, with increasing perimeter. For $z=31$ the perimeter is $1988$, and for $z=32$ the perimeter is $2116$. Therefore there are $31$ such quadrilaterals.

答案（B）：在每个这样的四边形中，$CD\ge AB$。设$E$为从$A$到$CD$的垂足；则$CE=2$且$AE=BC$。令$x=AE$、$y=DE$；则$AD=2+y$。由勾股定理，$x^2+y^2=(2+y)^2$，即$x^2=4+4y$。因此$x$为偶数，设$x=2z$，并且$z^2=1+y$。该四边形的周长为$x+2y+6=2z^2+2z+4$。取$z$为递增的正整数即可得到所需四边形，且其周长递增。当$z=31$时周长为$1988$，当$z=32$时周长为$2116$。因此这样的四边形共有$31$个。

Topics