geometrytriangles
There is an isosceles triangle with base $a=10$ and sides $b=13$. A square is inscribed inside of this triangle such that two of it's vertices are touching base and two of them are touching sides. What is the length of a side of the square?
The solution is $60/11$, but I don't know how to arrive at it.
As $DF=DE\sqrt 2$ and the angle $\angle EDF=45^{\circ},$ the point $F$ is obtained from $E$ through the rotation composed with the homothety (common center $D$, angle and ratio as above).
Construct in this transformation the image of the side that should contain $E.$ Its intersection (if it exists) with the side that doesn't contain $D$ is $F.$
Consider the following diagram.
Here $D$ is an arbitrary but fixed point on the ellipse, without loss of generality in the first quadrant. $E$ and $G$ are variable points counterclockwise of $D$, and $E'$ and $G'$ are their $90^\circ$ rotations. $DEE'$ and $DGG'$ are then isosceles right-angled triangles. We find the following:
So, since motions are continuous, there must be a position for the second point where the third point lies exactly on the ellipse, making an inscribed triangle. Hence there are infinitely many right-angled isosceles triangles inscribed in the ellipse.
As Misha Lavrov says, the exact position for $E$ can be found by rotating the ellipse $90^\circ$ around $D$. $E$ is another ellipse intersection (there may be up to three of them):
Best Answer
Let the isoceles triangle be denoted by $\Delta ABC$, with the vertex at point $B$. Let the square be denoted by $\square DEFG$, with the vertices $D,E$ lying on the sides of the triangle and the vertices $F,G$ lying on the base. Let $r$ denote the length of the side of the square.
Note that $\Delta ABC$ is similar to $\Delta DBE$, so since the height of $\Delta ABC$ is $12$ and the height of $\Delta DBE$ is $12 - r$, we have $$\frac{r}{10} = \frac{12-r}{12}.$$ Solving for $r$ gives $r = 12\cdot 10/22 = 60/11$.