Consider any triangle ABC such that the midpoint P of side BC is joined to the midpoint Q of side AC by the line segment PQ. Suppose R and S are the projections of Q and P respectively on AB, extended if necessary. What relationship must hold between the sides of the triangle if the figure PQRS is a square?

problem found in https://www.mathematrucker.com/problems_1975-1979.pdf
originally published in the Pi Mu Epsilon Journal
diagram is my own, any mistakes in it are my fault

I tried doing this:
0.5(A+C-B-C) x (A-B) = 0

>not calling the rectangle SPQR

By proportionality, we have AB=2RS, as well as PS=QR=CH/2, H being the projection of C on AB.
To have SPQR be a square, we need AB=CH.
BH²=CB²-AB²; AH²=AC²-AB²
AB=AH+BH=sqrt(CB²-AB²)+sqrt(AC²-AB²)
I feel like there should be a way to get rid of the square roots, but I can't find it.

P4990
>I feel like there should be a way to get rid of the square roots, but I can't find it.

You can isolate one square root at a time and square the equation, but
AB = sqrt(CB²-AB²) + sqrt(AC²-AB²)
AB - sqrt(CB²-AB²) = sqrt(AC²-AB²)
AB² - 2ABsqrt(CB²-AB²) + (CB²-AB²) = AC²-AB²
-2ABsqrt(CB²-AB²) + CB² = AC²-AB²
-2ABsqrt(CB²-AB²) = AC²-AB²-CB²
4(AB)²(CB²-AB²) = (AC²-AB²-CB²)²
it just ends up needlessly complicated.

Alternately, you can use Heron's formula, but it still ends up more complicated than the square root version.

In the case of the square (of side L to make it easy) existing, there is of course the 1:2 similarity between triangles ABC and QPC (since QP and RS and therefore AB are parallel and QC lies on AC and PC lies on BC), so |AB|=2|QP|=2L. And as |QR|=L, |CH|=2L with H as described by P4990. So ABC must always have an area of 4L^2.

Expressing |AC| as a function of |BC| given the base to height ratio is left as an exercise for the reader. I failed to make a geometric construction.

The bottom of the triangle must be the shortest of the three sides. That's as far as I've deduced.

a) The triangle at the top must have identical proportions to the overall triangle. Since the corners of the square are at the midpoints of the two leg-sides, the top triangle's legs are exactly half those lengths, and thus ditto for the bottoms. The bottom of the top triangle is half that of the overall triangle, and since this is one side of the square, all of the others must be as well.

b) The two side triangles are right-angled, meaning that the hypotenuse must be the longest side. This side is the exact same as one of the legs of the small triangle. Meanwhile, both of these two side triangles have one side of the square as one of their smaller sides, which means that the sides of the square must be shorter than the hypotenuse.

c) Since the bottom of the overall triangle is double that of one square side, and the other two sides are double the two hypotenuses, and those are both longer than a square side, the bottom of the overall triangle must thusly also be smaller than the other two sides.

One can also view this from the purview a prospective square in an equilateral triangle. Start by drawing the top side of the square, then the two legs of the square. The top must be the same length as half the sides of the overall triangle, yet the legs must be shorter than this per the right-angled triangle principles. An equilateral triangle requires a short, wide rectangle to fit neatly in while touching the midpoints of the two leg sides - it only fits then that a tall, thin sort of isosceles or scalene triangle could narrow this rectangle into a true square.

The only other thoughts I have related to this are the occult and historiographic implications of a square labelled SPQR, posited in the middle of a triangle. Like summoning Caesar's ghost for a seance.