# Thread: Parallel Lines Of Equal Length Within Square: smallest possible value for angle SPA?

1. ## Parallel Lines Of Equal Length Within Square: smallest possible value for angle SPA?

Hi all,

I'm a first time forum poster, so if I'm doing anything wrong then just let me know and I will fix it.

The question: The square PQRS has points A and B within its area such that PA, RB and AB are of equal length. Lines PA and RB are parallel. What is the lowest possible value for angle SPA?

My thoughts so far: So farm I've come up with two possible routes to a solution. The first involves creating some expressions for certain values within the question involving trig, and then creating and solving a trig inequality. The second involves creating a polynomial equation where y = angle SPA, and then performing optimisation to find the lowest value. The second option seems (to me at least) to be more promising. The only progress that I have made, however, is forming some basic trig expressions for a few lengths. I don't think that I have anything which is actually useful.

I'm looking for someone to point me in the right direction, and perhaps to help me start to solve this question, as I would like to figure most of it out for myself. Any help at all would be much appreciated.

2. Originally Posted by JSITD
… The only progress that I have made … is forming some basic trig expressions for a few lengths. I don't think that I have anything which is actually useful.
Can you share what you've tried, so far?

Did you draw some rough sketches, to envision the different possible arrangements of PA, RB, and AB (not just when angle SPA is minimized). I drew a couple, thinking in terms of segment PA rotating about P and segment RB rotating about R. The angles that these segments make with their respective vertical side of the square are alternate interior angles, so angle SPA must be congruent with angle QRB.

In this view, you've got two circular arcs (same radius) inside PQRS. Point A is on the arc in the upper left of the square, and point B is on the arc in the bottom right. The distance AB must be the same as the radii. So, I was thinking of creating a function for distance AB, in terms of angle SPA, to see which angles lead to AB lengths matching the radii.

This is just a first thought, based on two rough sketches. I didn't think about what any functions might look like (could get messy).

3. What I've attempted with regard to diagrams is essentially the same as what you've been describing. I've also been looking at some trig related to it, by creating right angled triangles between points P, A and E, where E is a point on PS so that triangle PAE is right angled at E. I then created some expressions for the sides of this triangle using SOHCAHTOA and some trig identities, and I've been trying to fit them together to come up with an inequality for the angle, but I haven't got particularly far with it. It's getting quite late in my time zone, so I think I'll look at the problem again tomorrow morning, and see what else I can do.

4. Originally Posted by JSITD
… created some expressions for the sides of [various triangles] …
Good. Other members who take an interest in this exercise might like to compare your expressions to what they might be doing -- if you're willing to share.

5. My thinking was to consider the symmetry of the figure. Then, if your coursework has included loci, look at the locus of A from P and another point found from the symmetry.

6. Here's what I have so far:
In the triangle PAE, where C = hypotenuse, A = adjacent, and B = opposite (relative to angle SPA):

cosθ = A/C
A = Ccos
θ

sin
θ = B/C
B = Csin
θ

(Ccosθ)2 + (Csinθ)2 = C2

I'm struggling to use any of this to actually find an answer, however. The issue is that I can't find a way to incorporate line PQ into any of my equations.

Another idea could be to use vectors, but again I'm not sure how this would work. I'm still pretty stuck on this question.

7. Does the term "locus of a point" mean anything to you?

8. Yes, I do understand the term, and it helped me to visualise the problem, though I was able to solve it another way. Using some advice from another site, I imagined point M, the midpoint of line AB. From there, I could construct an equation in terms of cos(APM), and through optimisation minimise the cosine of the angle, thus maximising the angle itself. As angles APM and SPA add to give 45 degrees, by subtracting the maximum possible value for APM from 45, I reached the solution that the minimum possible value of SPA was 15 degrees.

9. I won't post details of my method. It gave the same answer 15 deg.

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