Let $AC$ be a diameter of a circle $\omega$ of radius $1$, and let $D$ be the point on $AC$ such that $CD = \frac{1}{5}$. Let $B$ be the point on $\omega$ such that $DB$ is perpendicular to $AC$, and let $E$ be the midpoint of $DB$. Compute length $AE$.
Since the radius of the circle is $1$, that means point $B$ lies on the circle of $O$ with diameter $OD=1$. Thus, since $\triangle ODB$ is a $3-4-5$ right triangle, $OB=\frac{4}{5}$. Furthermore, $DE=\frac{1}{2}$ and $BD=\frac{3}{5}$, so in $\triangle DEB$, $DO > 1$. Now, since $\triangle ADO$ and $\triangle BAD$ share angles $\angle BDA = \angle ODA$, $\angle ADO = \angle BAD$, so $\triangle ADO \sim \triangle BAD$. Hence $\frac{DO}{AD}=\frac{AD}{3}$. Since $DO > 1$, $AD<\frac{1}{3}$. Thus, we can conclude that $\triangle ABC$ is obtuse at $\angle A$. Since the altitude to the hypotenuse of a right triangle is the geometric mean of the two products each outside segment makes with the hypotenuse, and we know that the length of BD is $\frac{3}{5}$, which is smaller than the other segment length $2\cdot \frac{4}{5}=\frac{8}{5}$, we find that $DE=\frac{1}{2}$. Now, we may solve for $AE$: $\triangle ADO \sim \triangle BAD$, so $\frac{AO}{AD}=\frac{AD}{3}$. Solving gives $AD^{2}=\frac{1}{15}$. Thus, $OD^{2}=\frac{2}{15}$. Clearly, $DE=1-\frac{2}{15}=\frac{3}{5}$, so $AE=\frac{1}{3} \cdot \left(\frac{3}{5}\right)=\boxed{\frac{1}{5}}$.
DB^2 + (4/5)^2 = 1, so DB = 3/5
so DE = 3/10
AE^2 = (1 + 4/5)^2 + (3/10)^2 = 333/100
AE = √333/10 ≈ 1.825
To compute the length $AE$, we can form a right triangle $ADC$. Here's how:
1. Draw a circle $\omega$ with radius $1$ and a diameter $AC$. Let's label the center of the circle as $O$.
2. Since $CD = \frac{1}{5}$ and $AC$ is a diameter, $AD = AC - CD = 2 - \frac{1}{5} = \frac{9}{5}$.
3. Draw a segment $DE \perp AC$ with $D$ on $AC$ and $E$ on $DB$. Since $DE$ is perpendicular to $AC$, $DE$ is the height of the right triangle $ADC$.
4. Since $E$ is the midpoint of $DB$, $DE = \frac{1}{2} DB$. Therefore, $DB = 2 DE$.
5. Since $DB$ is perpendicular to $AC$, $DB$ is the diameter of the circle $\omega$. Therefore, $DB = 2$.
6. Substituting $DB = 2$ into $DB = 2 DE$, we get $2 = 2 DE$. Dividing both sides by $2$, we have $DE = 1$.
7. Since $DE$ is the height of right triangle $ADC$, and $AD = \frac{9}{5}$, we can now use the Pythagorean theorem to find $AC$.
Using the Pythagorean theorem:
$AC^2 = AD^2 + DE^2$
$AC^2 = \left(\frac{9}{5}\right)^2 + 1^2$
$AC^2 = \frac{81}{25} + 1$
$AC^2 = \frac{81 + 25}{25}$
$AC^2 = \frac{106}{25}$
Taking the square root of both sides, we get:
$AC = \frac{\sqrt{106}}{5}$
8. Now that we know the length of $AC$, we can find the length of $AE$. Since $E$ is the midpoint of $DB$, we have $AE = AD - DE = \frac{9}{5} - 1 = \frac{4}{5}$.
Therefore, the length $AE$ is $\frac{4}{5}$.