求助一道数学难题

127HLLJ

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

答案是12，拜托哪位牛人能帮我讲解一下这道题的算法吗？非常感谢！

jeffery2541

题中的那个area应该是10吧~否则我算不出来哦。。。如果是10的话：
这样的正方形每4个是一组，每组中的4个刚好构成一个大正方形，这个大正方形的对角线交点是原点；
你应该发现，在平面直角坐标系中，只要某定正方形的两个顶点在整数点上，那么另外两个顶点也在整数点上，你画个整数点的格子就能看出来的；
这样，除了原点外的一个顶点的坐标的绝对值可以是（6，8）（8,6），也可以是（0，10）（10,0）；
找到了两个定点可以在的点，简单画个图，连连看，是3个大“田”字正方形，一共12个。

jeffery2541

画图，语言有点抽象。

127HLLJ

非常感谢你这么快帮我解答，这道题area是100没有错，就是说明length是10。外国的网站是如下解释的，可是看了你的解答和国外网站的解答，还是不太理解，麻烦您帮我详细点解答一下好吗？

This question becomes much easier if you visualize/draw it.

Let the origin be O and one of the vertices be A. Now, we are told that length of OA must be 10 (area to be 100). So if the coordinates of A is (x, y) then we would have x^2+y^2=100 (distance from the origin to the point A(x, y) can be found by the formula d^2=x^2+y^2)

Now, x^2+y^2=100 has several integer solutions for x and y, so several positions of vertex A, note that when vertex A has integer coordinates other vertices also have integer coordinates. For example imagine the case when square rests on X-axis to the right of Y-axis, then the vertices are: A(10,0), (10,10), (0,10) and (0,0).

Also you can notice that 100=6^2+8^2 and 100=0^2+10^2, so x can tale 7 values: -10, -8, -6, 0, 6, 8, 10. For x=-10 and x=10, y can take only 1 value 0, but for other values of x, y can take two values positive or negative. For example: when x=6 then y=8 or y=-8. This gives us 1+1+5*2=12 coordinates of point A:

x=10 and y=0, imagine this one to be the square which rests on X-axis and to get the other options rotate OA anticlockwise to get all possible cases;
x=8 and y=6;
x=6 and y=8;
x=0 and y=10;
x=-6 and y=8;
x=-8 and y=6;
x=-10 and y=0;
x=-8 and y=-6;
x=-6 and y=-8;
x=0 and y=-10;
x=6 and y=-8;
x=8 and y=-6.

jeffery2541

忽略我那句无知的话吧，第一眼没仔细看，100是对的。

这个文字表达有点乱，如果可以当面讲肯定没问题。
在提示你一下吧~你先找到距离原点距离为10的整数点，再看看这些点怎样构成如题要求的正方形，相信你一定可以做出来的！加油~

非常感谢你这么快帮我解答，这道题area是100没有错，就是说明length是10。外国的网站是如下解释的，可是看了你的解答和国外网站的解答，还是不太理解，麻烦您帮我详细点解答一下好吗？

This question becomes much easier if you visualize/draw it.

Let the origin be O and one of the vertices be A. Now, we are told that length of OA must be 10 (area to be 100). So if the coordinates of A is (x, y) then we would have x^2+y^2=100 (distance from the origin to the point A(x, y) can be found by the formula d^2=x^2+y^2)

Now, x^2+y^2=100 has several integer solutions for x and y, so several positions of vertex A, note that when vertex A has integer coordinates other vertices also have integer coordinates. For example imagine the case when square rests on X-axis to the right of Y-axis, then the vertices are: A(10,0), (10,10), (0,10) and (0,0).

Also you can notice that 100=6^2+8^2 and 100=0^2+10^2, so x can tale 7 values: -10, -8, -6, 0, 6, 8, 10. For x=-10 and x=10, y can take only 1 value 0, but for other values of x, y can take two values positive or negative. For example: when x=6 then y=8 or y=-8. This gives us 1+1+5*2=12 coordinates of point A:

x=10 and y=0, imagine this one to be the square which rests on X-axis and to get the other options rotate OA anticlockwise to get all possible cases;
x=8 and y=6;
x=6 and y=8;
x=0 and y=10;
x=-6 and y=8;
x=-8 and y=6;
x=-10 and y=0;
x=-8 and y=-6;
x=-6 and y=-8;
x=0 and y=-10;
x=6 and y=-8;
x=8 and y=-6.

-- by 会员 127HLLJ (2012/10/3 19:31:23)