AMC12 2015 A

AMC12 2015 A · Q19

AMC12 2015 A · Q19. It mainly tests Factoring, Arithmetic misc.

For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, 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$p$，存在一个四边形$ABCD$，其边长均为正整数，周长为$p$，在$B$和$C$处为直角，且$AB=2$、$CD=AD$。问满足$p<2015$的$p$共有多少个不同取值？

(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）：在所有这样的四边形中，$CD \ge AB$。令$E$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$x = 2z$，则$z^2 = 1 + y$。该四边形的周长为$x + 2y + 6 = 2z^2 + 2z + 4$。取递增的正整数$z$即可得到所需四边形，且周长递增。当$z = 31$时周长为1988；当$z = 32$时周长为2116。因此共有31个这样的四边形。

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