# Thread: Collinear points formed by a measurement on a square's perimeter

1. ## Collinear points formed by a measurement on a square's perimeter

Let a square ABCD with sides of length 1 be given. A point X on BC is at distance d
from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX
meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at
S. If points P, Q, R and S are collinear, determine d.

I have started and found the diagram is symmetrical through AC, X must be (1,d) and Y must be (1-d,0). PR or QS must be at right angles to AC due to the need for a collinear (straight) line through them at the end. I believe the gradients should be the same for AC and QS, is this correct?? I'm stuck on what to do next and also need to clarify so far.
Cheers

2. ## Re: Collinear points formed by a measurement on a square's perimeter

The gradient of SR is 1 (or -1) due to symmetry, so one has to equate the gradient of RP to 1. Let R' be the projection of R to AP. Since RR' = 1, it should be that R'P = AP - AR' = 1. It is left to express AP and AR' through d using similar triangles.

3. ## Re: Collinear points formed by a measurement on a square's perimeter

Originally Posted by pikachu26134
Let a square ABCD with sides of length 1 be given. A point X on BC is at distance d
from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX
meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at
S. If points P, Q, R and S are collinear, determine d.

I have started and found the diagram is symmetrical through AC, X must be (1,d) and Y must be (1-d,0). PR or QS must be at right angles to AC due to the need for a collinear (straight) line through them at the end. I believe the gradients should be the same for AC and QS, is this correct?? I'm stuck on what to do next and also need to clarify so far.
Cheers
1. Use proportions in similar triangles:
The blue triangles will yield:

$\displaystyle \dfrac{1+s}{s+d}=\dfrac{2+s}{1+s}$

and the grey triangles will yield:

$\displaystyle \dfrac ds=\dfrac1{1+s}$

2. Solve this system of equations for s and d.

I've got $\displaystyle d=\frac32 - \frac12 \cdot \sqrt{5}$ and $\displaystyle s= \frac12 \cdot \sqrt{5} - \frac12$