Geometry

[smittty]

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


OA:

SPOILER: E

[DWarrior]

Coordinates for the first side must be (0,0) to (m,n) where m and n are integers, then the other vertices will also have integer coordinates. You want m and n such that the line will have a length of 10, so m^2+n^2=10^2.

Make a table for m:
m | m^2 | n^2
0 | 0 | 100
1 | 1 | 99
2 | 4 | 96
3 | 9 | 91
4 | 16 | 84
5 | 25 | 75
6 | 36 | 64
7 | 49 | 51
8 | 64 | 36
9 | 81 | 19
10 | 100 | 0

When m is 0 or 100, the sides will be parallel/perpendicular to the axis and there will be 4 such rotations of the square.

When m is 6, n is 8 and that square will have 4 rotations.

When m is 8, n is 6 and that square will also have 4 rotations. Don't forget, vertex (8,6) looks similar to (6,8) but the squares are not the same, so each has 3 unique rotations.

12, E

[lurkman]

Dwarrior,

Nice solution!

[krovvidy]

Nice approach Dwarrior ... really useful ...

[Lhomme]

dude, you rock. Thanks a lot

[smittty]

Thanks... I now have a new appreciation for sqaures and coordinate geometry.

[ndcruz]

I am not sure how if you have 6, or 8 as a length you can say that the side is 10. Except if you are looking at a special right triangle with corresponding ratios 6:8:10.

Doesnt a sqaure have to have 4 equal length sides?

Can someone explain with a picture please?

Thanks

[ramgmat]

yes ndcruz , i had the same doubt as you had.....still not clear

[bayarea83]

area of the given square is 100
==> l^2 = 100 where l is the length on one side
given that one vertice is (0,0) lets assume the other vertice is (m,n)

solving for 1st quadrant the m & n are positive we can represent length of l
==> (0-m)^2 + (0-n)^2 = 100=10^2
==> 6^2 + 8^2 = 10^2
so one vertice is (6,8) similarly other can be (8,6)
also one more is obviously (10,0)

so 3 vertice is 1st quadrant
==> for all 4 quardrant we can have 4*3 = 12 squares.

[ndcruz]

I finally figured it out, THANKS guys.