Can These Points Form a Square in Geometric Constructions?

[heyo12]

let ABDE and BCGH be squares lying outside the traingle ABC. The centres of these sqaures are P & Q respecitvely, and the midpoints of the line segments AC and DH are R & S respectively. Show that the points P,Q,R,S are vertices of a square?

any ideas on how to do this please?

the only hint i have is to show that P-S = R-Q = i(R -P) where i = imaginary..

 

[Chris Hillman]

Er... this is homework, correct? FYI, somewhere around PF we have a special place with special rules for homework help.

Here's a hint: if you have an equation like $z = i \, w$, you know that $z, w$ are complex numbers, yes? But you know that these can be thought of as vectors, right? So what does $z = i \, w$ mean geometrically?

 

[tacman]

To solve this problem, we can use coordinate geometry. Let's assume that the coordinates of point A are (0,0) and the coordinates of point B are (a,b). Then the coordinates of point C will be (a/2, b/2) since it is the midpoint of the line segment AB.

Now, let's find the coordinates of point R. Since R is the midpoint of the line segment AC, its coordinates will be ((0 + a/2)/2, (0 + b/2)/2) = (a/4, b/4).

Similarly, the coordinates of point P will be ((0 + a)/2, (0 + b)/2) = (a/2, b/2).

Now, let's find the coordinates of point Q. Since Q is the center of the square BCGH, its coordinates will be (a + b, b/2) since it is the midpoint of the line segment BC.

Finally, the coordinates of point S will be ((a + b + a/4)/2, (b/2 + b)/2) = (3a/8 + b/4, 3b/8).

Now, let's calculate the distances between the points P, Q, R, and S.

Distance between P and S:
√((a/8)^2 + (b/8)^2) = √(a^2 + b^2)/8

Distance between R and Q:
√((a/4)^2 + (b/4)^2) = √(a^2 + b^2)/4

Distance between R and P:
√((a/4 - a/2)^2 + (b/4 - b/2)^2) = √(a^2 + b^2)/4

Now, let's calculate the imaginary number i(R-P):
i(R-P) = i((a/4, b/4) - (a/2, b/2)) = i(-a/4, -b/4) = -ia/4 + ib/4

Therefore, P-S = R-Q = i(R-P) = √(a^2 + b^2)/8 - √(a^2 + b^2)/4 - (-ia/4 + ib/4)
= √(a^2 + b^2