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 $AE$.
3 answers
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
so DE = 3/10
AE^2 = (1 + 4/5)^2 + (3/10)^2 = 333/100
AE = √333/10 ≈ 1.825
There is a mistake in the calculation for $AE$. The equation $DB^2 + (4/5)^2 = 1$ should be $DB^2 + (3/5)^2 = 1$, which gives $DB = 4/5$. Therefore, $DE = 4/10 = 2/5$. The correct equation for $AE$ is $AE^2 = (1 + 4/5)^2 + (2/5)^2 = 53/25$. Taking the square root gives $AE = \sqrt{53}/5$.