### Instruction: $\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$. Point $C$ does not coincide with either $A$ or $B$. Let $s=AC+BC$. Then, for all permissible positions of $C$:
$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$.### Response: The problem is asking me to compare the sum of two sides of a triangle inscribed in a semicircle to the square of the radius of the semicircle.
I know that a semicircle has 180 degrees of arc measure, so the angle at $C$ must be a right angle by the inscribed angle theorem.
This means that $\triangle ABC$ is a right triangle, and I can use the Pythagorean theorem to relate the lengths of its sides.
Let $x=AC$ and $y=BC$.
Then, by the Pythagorean theorem, $x^2+y^2=AB^2$.
But $AB$ is also the diameter of the semicircle, so $AB=2r$.
Thus, $x^2+y^2=4r^2$.
I want to compare this to $s^2=(x+y)^2$, which is the square of the sum of the two sides.
I can expand $(x+y)^2$ as $x^2+2xy+y^2$.
Then, using the previous equation, I can substitute $4r^2$ for $x^2+y^2$.
This gives me $(x+y)^2=4r^2+2xy$.
Rearranging, I get $2xy=(x+y)^2-4r^2$.
I notice that the left side of this equation is always positive, since $x$ and $y$ are positive lengths.
This means that the right side of this equation must also be positive, or else there would be no solution for $x$ and $y$.
Therefore, I can conclude that $(x+y)^2>4r^2$, or $s^2>4r^2$.
This eliminates choices D and E, which say that $s^2$ is less than or equal to $4r^2$.
To decide between choices A, B, and C, I need to find the maximum possible value of $s^2$.
I recall that for a fixed perimeter, a right triangle has the largest area when it is isosceles, meaning that its two legs are equal.
So, I will assume that $x=y$, and see what happens to $s^2$.
If $x=y$, then $x^2+y^2=2x^2=4r^2$, or $x^2=2r^2$.
Then, $s^2=(x+y)^2=4x^2=8r^2$.
This is the largest possible value of $s^2$, since any other position of $C$ would make the triangle less isosceles and reduce its area.
Therefore, the answer is choice A, $s^2\le8r^2$.