geometrytriangles
Find the maximum area of a rectangle placed in a right angle triangle $\triangle ABC$.
My Work:
Now, how to continue?
Hint:
The hyperbola whose asymptotes are $DA$ and $DC$ and passes through $F$ is tangent to $AC$.
The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us another Diophantine constraint. Draw the altitude $BH$ from $B$ onto $AC$:
Since $\triangle BCD$ is isosceles, $H$ is the midpoint of $CD$, so $CH = 1$; by looking at the right triangle $\triangle BHC$, we get $\cos \angle C = \frac{CH}{BC} = \frac1a$. The Law of Cosines tells us that $$c^2 = a^2 + b^2 - 2b.$$
Substituting the first equation into the second yields $$c^2 = (b^2 + c^2 - bc) + b^2 - 2b \implies bc = 2b^2 - 2b \implies c=2(b-1),$$ and substituting this value of $c$ back into the first equation tells us that $$a^2 = b^2 + (2b-2)^2 - b(2b-2) \implies a^2 - 3(b-1)^2 = 1,$$ which is a Pell equation. The simplest solution to $x^2-3y^2=1$ is $x=2,y=1$, and all other solutions are obtained by taking the coefficients of $1$ and $\sqrt3$ in $(2 + \sqrt3)^k$ for some $k$.
If we use $x=2$ and $y=1$, then this gives us $a=b=c=2$, which is certainly a solution to the problem: in this case, $\triangle ABC$ is equilateral and $D=A$. This triangle has area $\sqrt 3$.
If you don't like this solution, because it's a somewhat degenerate case, then $(2+\sqrt3)^2 = 7+4\sqrt3$ yields $x=7$ and $y=4$ as the next solution. In this case, $a=7$, $b=5$, and $c=8$, which is the triangle I used for the diagram. By Heron's formula, $\triangle ABC$ has area $10 \sqrt 3$.
One of these, depending on whether you find the equilateral solution acceptable or not, is the solution with the smallest area, but we can find infinitely many triangles with integer sides that satisfy the conditions, by taking higher powers of $2+\sqrt 3$ to solve Pell's equation.
Best Answer
The area of the Rectangle is half the area of the triangle, and this is even universal and not dependent on the triangle be right angled.
Pick the base of the triangle that is longest that has length $b$ and associated height $h$. Now you can place on there a triangle of width $b/2$ and height $h/2$.
There must be a position where it fits since at half height the triangle has still half its width by the interception theorem.
So the area of the rectangle is given by $\frac{b}{2} \frac{h}{2} = b h / 4$ and as such has half the area of the triangle.
For illustration look at http://jwilson.coe.uga.edu/emt725/Class/Pearman/rectangle.triangle/rect.tri.html