### Instruction: Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$, where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$.### Response: I want to find the largest possible value of $d$, which is the length of a segment on the diameter $\overline{MN}$.
I notice that $d$ depends on the position of $C$ on the semicircle, and it seems that $d$ is maximized when $C$ is as far away from $A$ and $B$ as possible.
That means I should consider the case when $C$ is the antipode of $A$, or the point where the semicircle meets $\overline{MN}$ on the opposite side of $A$.
In that case, $\overline{AC}$ and $\overline{BC}$ are perpendicular to $\overline{MN}$, so they form right triangles with $\overline{MN}$.
I can label the lengths of the sides of these triangles using the Pythagorean theorem and the given information.
Let $x$ be the length of $\overline{AN}$, then $\overline{AM}=\frac{1}{2}-x$ since $A$ is the midpoint of the semicircle.
By the Pythagorean theorem, $\overline{AN}^2+\overline{NC}^2=1^2$, so $\overline{NC}=\sqrt{1-x^2}$.
Similarly, $\overline{MB}^2+\overline{BC}^2=1^2$, and I know that $\overline{MB}=\frac{3}{5}$, so $\overline{BC}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$.
Now I can find $d$ by subtracting the lengths of the segments on $\overline{MN}$ that are not part of it.
I have $d=1-\overline{AM}-\overline{NC}=1-\left(\frac{1}{2}-x\right)-\sqrt{1-x^2}$.
I want to maximize this expression for $d$, so I need to minimize the terms that are subtracted from 1.
The term $\frac{1}{2}-x$ is minimized when $x$ is maximized, and $x$ is maximized when $\overline{AN}$ is the radius of the circle, or $\frac{1}{2}$.
So the optimal value of $x$ is $\frac{1}{2}$, and this makes $\frac{1}{2}-x=0$.
The term $\sqrt{1-x^2}$ is also minimized when $x$ is maximized, since the square root function is decreasing for positive values.
So the optimal value of $\sqrt{1-x^2}$ is $\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$.
Plugging these values into the expression for $d$, I get $d=1-0-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}$.
This is the largest possible value of $d$, and it is in the form $r-s\sqrt{t}$, where $r=2$, $s=1$, and $t=3$.
Therefore, the answer is $r+s+t=2+1+3=6$.