Suppose I have a triangular land-plot, but some part of it (the yellow part) is unusable. I want to build a square house on the usable (white) part. The house may be rotated (but must be square). What is the maximum area I can build?
NOTES:
- Both triangles in the picture are right-angled and isosceles.
- When $0<y<1$, the white part is a quadrilateral whose angles are: 90, 135, 90, 45. Its area is $\frac{1+y(2-y)}{4}$.
EDIT (following a hint by Ross Millikan): We can get a lower bound by considering two possible locations: an axis-aligned square in the lower-left corner (pink), and a 45-degree-rotated square in the top corner (green):
- In the lower-left corner, the vertical side of the square can be at most $y$, and the diagonal can be at most $\sqrt{\frac{1}{2}}$, therefore the side can be at most $min(y,\frac{1}{2})$.
- In the top corner, the vertical diagonal can be at most $\frac{1}{2}(1+y)$, and the horizontal diagonal can be at most $1-y$, therefore the side can be at most $\sqrt{\frac{1}{2}} min (\frac{1}{2}(1+y),1-y)$. This expression is maximized when $y=\frac{1}{3}$, and the side length is $\sqrt{\frac{1}{2}} \frac{2}{3} \approx 0.47 $. When $y>\frac{1}{3}$, a square at the top corner must become smaller because of the horizontal diagonal, but, we don't have to push the square towards the new top corner - we can leave it at its previous position $(\frac{1}{3},\frac{2}{3})$, and have the same side-length 0.47 .
The following graph (drawn by http://rechneronline.de/function-graphs/) shows the side lengths of both squares:
And the following graph shows the areas of both squares, relative to the area of the white part:
NOW, MY QUESTION IS: Is there a way to build a larger house (a larger bounded square)?
My intuition is that there must be a way. If in some cases the largest square is at 0 degrees, and in other cases the largest square is rotated at 45 degrees, then between these cases, there must be a way to build a square rotated between 0 to 45 degrees, with a larger area.
If $y=\sqrt{2}-1 \approx 0.414$ the two maximal squares (flush with the left side and with the diagonal) have the same sidelengths. This was obtained by equating the two formulas in the range $(1/3,1/2)$ wherein they are respectively $y$ and $(1/\sqrt{2})(1-y).$ Note that this $y$ lies in $(1/3,1/2)$ as it should. Each of these squares can be realized with one corner at $(0,y),$ and a sketch reveals there is room to rotate either one of them a bit into the white region. After that one could expand the square.
I took the square aligned to the vertical, and supposed we kept its corner at $(0,y)$ and took another corner at $(z,0)$. Then the remaining two corners are at $(y+z,z)$ (the rightmost corner, tipped up from the $x$ axis) and at $(y,y+z)$ (the topmost corner). In order for these two corners to remain in the white region we need both $y+2z \le 1$ and $2y+z \le 1$. It seems clear the restriction from the upper corner will be the more restrictive, at least for small $z$. So I set $2y+z=1$ and got $$z=3-2\sqrt{2},\ y=\sqrt{2}-1,$$ giving the sidelength of $\sqrt{y^2+z^2}=\sqrt{20-14\sqrt{2}}$ or approximately $0.4483,$ which beats the $0.414$ from the aligned squares by a small amount.
By the way I did check the fourth corner is in the white area, since $y+2z=5-3\sqrt{2} \approx 0.757.$
To check that nothing has entered the yellow region, note that the angle of "rotation", that is, the angle $OYZ$ where $O=(0,0),Y=(0,y),Z=(z,0)$ is $\arctan(z/y)=\pi/8.$ This means the rotated square is situated so that its right angle cuts the 135 degree angle at $Y$ in half, with $45/2$ degrees on either side. Thus the angle made by the top side of the rotated square with the vertical is $22/2+90<135.$ So nothing has ended up in the yellow region.
ADDED: The OP has asked more generally about the "tipped square" case when $1/3<y<1/2.$ With the same square as above, having one corner at $(0,y)$ and another at $(z,0)$, we have again the other two corners at $(y+z,z)$ and $(y,y+z)$ where the latter is uppermost. For this upper point to lie on or below $x+y=1$ requires $2y+z\le 1$, so we just define $z=1-2y$ and check things. Note first that $z \ge 0$ since $y \in (1/3,1/2).$ We also need to check that the lower rotated corner $(y+z,z)$ lies below $x+y=1$, and with $z$ taken as $1-2y$ this condition is exactly $y>1/3$ so we're OK on that score.
Finally we need to check that the top side of the square has not rotated into the forbidden yellow area; this will be so iff $z/y \le 1$ since the rotation angle is $\arctan(z/y)$ and the angle from the vertical to the topmost vertex of the rotated square is $\pi/2+\arctan(y/x).$ Now $z/y<1$ becomes $y>1/3$ once $z=1-2y$ has been put in, so we now have checked all so as to show the stretched/rotated square lies entirely in the white region. Its sidelength is $$\sqrt{y^2+z^2}=\sqrt{5y^2-4y+1}.$$
It is interesting to look at the three graphs on the interval $(1/3,1/2)$ where all three bounds apply. That is, we graph each of $y,\ (1/\sqrt{2})(1-y),\ \sqrt{5y^2-4y+1}$ over $[1/3,1/2].$ We find that the third curve (which is the upper branch of a hyperbola) lies above each of the others in the interior of $[1/3,1/2]$ and agrees with the maximum of the first two at the two endpoints, and furthermore has the same tangents at those endpoints as may be checked via derivatives.
Conclusion: For $1/3<y<1/2$ we can always do better by a rotation/stretch from the "aligned squares" mentioned in thee question, and the above describes how much better at such $y$ values. We should emphasize that this is not a proof that some other orientation/choice of where the square is to be placed might beat this rotated one. A more general analysis would be necessary for that, and seems to me would be difficult even to set up conveniently.