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Here's a cool fact about equilateral triangles.
Draw an equilateral triangle. Pick a point inside the triangle. Now calculate (or measure) the perpendicular distance from the point to each side of the triangle. (The perpendicular distance is the distance along a line which is perpendicular to the side; it's the shortest distance from the point to the side.) Add up the three distances. Then pick another point, and do the same. You should get the same sum!
The sum of the perpendicular distances from any point inside a given equilateral triangle to the three sides is a constant.
I'm not a student -- at least not in the formal sense -- so no worries there! I just posted it as an interesting puzzle in its own right, and subsequently thought of the generalisation to a rectangle.
The rectangle problem can be solved without calculus. (Anyway, did you really mean to imply that advanced calculus is boring?!)
In your diagram above, if the square has a vertex at (0,0) and side a, by similar triangles, (4-a)/a = 4/3, so a = 12/7.
If the square has one side on the hypotenuse, the small triangle at the top is similar to the whole 3-4-5 triangle. By area considerations, the height of the 3-4-5 triangle is 12/5. So, by similar triangles, a/5 = (12/5 - a)/(12/5), from which a = 60/37; a tad smaller than 12/7 = 60/35.
If the square is to be inscribed, i.e., each vertex touches the triangle, these are the only two possibilities.
For the inscribed rectangle, suppose the rectangle has vertices (0,0), (0,r), and (s,0), and the triangle has vertices (0,a) and (b,0). Then by similar triangles, as above, a/b = (a - r)/s, so s = b - (b/a)r. Then the area of the rectangle, A = rs = br - (b/a)r² = (b/a)(ar - r²). Completing the square, A = (b/a)[(a²/4) - (r - a/2)²]. So the maximum area of ab/4 occurs when r = a/2, as you said.
The case where the rectangle has one side on the hypotenuse may be handled similarly. Is the maximum area in that case <. =, or > ab/4? I won't spoil the fun by giving that away!
If you have trouble remembering the values for sin and cos of 0°, 30°, 45°, 60°, and 90°, the following device may help. This looks better when written out as a horizontal table, but I'll write it vertically here...
sin 0° = sqrt(0)/2
sin 30° = sqrt(1)/2
sin 45° = sqrt(2)/2
sin 60° = sqrt(3)/2
sin 90° = sqrt(4)/2
cos 0° = sqrt(4)/2
cos 30° = sqrt(3)/2
cos 45° = sqrt(2)/2
cos 60° = sqrt(1)/2
cos 90° = sqrt(0)/2
As an extension, find the largest (in terms of area) such rectangle. Generalise to an a-b-c right triangle.
What is the largest square that can be inscribed in a 3-4-5 triangle?
In a non-right angled triangle, with internal angles a, b, c, tan a + tan b + tan c = tan a * tan b * tan c.
If I have seen farther than others, it is because I was standing on the shoulders of giants.-Albert Einstein
Wasn't this Newton? In a letter to Robert Hooke, I believe. Hooke was reportedly a short man, so the remark may have been partly a joke at Hooke's expense!
There's also the crank scientist's version of this quotation: "If others have seen further than me, it is because giants have stood on my shoulders."