Question:
A sequence $(a_n)$ is defined as follows:
\[a_{i + 1} = \frac{1}{1 - a_i}\]for $i \ge 1.$  If $a_3 = a_1,$ compute $(a_9)^9.$

To solve the problem, we need to understand the behavior of the sequence \((a_n)\) defined by the recurrence relation \(a_{i+1} = \frac{1}{1 - a_i}\) for \(i \ge 1\). We are given that \(a_3 = a_1\), and we need to find \((a_9)^9\).

First, let's compute the first few terms of the sequence to see if there is a pattern.

Starting with \(a_1\), we have:
\[a_2 = \frac{1}{1 - a_1}.\]

Next, we compute \(a_3\):
\[a_3 = \frac{1}{1 - a_2} = \frac{1}{1 - \frac{1}{1 - a_1}} = \frac{1}{\frac{(1 - a_1) - 1}{1 - a_1}} = \frac{1}{\frac{-a_1}{1 - a_1}} = \frac{1 - a_1}{-a_1} = \frac{a_1 - 1}{a_1} = 1 - \frac{1}{a_1}.\]

We are given that \(a_3 = a_1\). Therefore, we have:
\[a_1 = 1 - \frac{1}{a_1}.\]

To solve for \(a_1\), we rearrange the equation:
\[a_1 + \frac{1}{a_1} = 1.\]

Multiplying both sides by \(a_1\) gives:
\[a_1^2 + 1 = a_1,\]
or
\[a_1^2 - a_1 + 1 = 0.\]

This is a quadratic equation. The solutions to this equation are:
\[a_1 = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}.\]

So, \(a_1\) can be either \(\frac{1 + i\sqrt{3}}{2}\) or \(\frac{1 - i\sqrt{3}}{2}\). Let's denote these two possible values as \(\omega\) and \(\omega^2\), where \(\omega = \frac{1 + i\sqrt{3}}{2}\) and \(\omega^2 = \frac{1 - i\sqrt{3}}{2}\). Note that \(\omega\) and \(\omega^2\) are the non-real cube roots of unity, satisfying \(\omega^3 = 1\) and \(\omega^2 + \omega + 1 = 0\).

Now, let's check the periodicity of the sequence. We already know that \(a_3 = a_1\). Let's compute \(a_4\):
\[a_4 = \frac{1}{1 - a_3} = \frac{1}{1 - a_1} = a_2.\]

Similarly, \(a_5 = \frac{1}{1 - a_4} = \frac{1}{1 - a_2} = a_3 = a_1\). Therefore, the sequence is periodic with period 2. That is, \(a_{2k+1} = a_1\) and \(a_{2k} = a_2\) for all integers \(k\).

Since 9 is odd, \(a_9 = a_1\). Therefore, \((a_9)^9 = (a_1)^9\). Since \(a_1\) is a cube root of unity, we have \(a_1^3 = 1\). Therefore, \(a_1^9 = (a_1^3)^3 = 1^3 = 1\).

Thus, the answer is \(\boxed{1}\).


Question: Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$.  If $\angle BAC = 54^\circ$ and $\angle ABC = 52^\circ$, then what is $\angle AHB$?

To find the measure of angle \( \angle AHB \) in triangle \( \triangle ABC \) where the altitudes \( \overline{AD} \) and \( \overline{BE} \) intersect at the orthocenter \( H \), we can use the properties of the orthocenter and the angles in the triangle.

First, let's recall that the orthocenter \( H \) is the point where the altitudes of the triangle intersect. The key property we will use is that the measure of the angle formed by two altitudes is equal to 180 degrees minus the measure of the angle opposite the vertex where the altitudes intersect.

In this problem, we need to find \( \angle AHB \). According to the property mentioned, \( \angle AHB = 180^\circ - \angle ACB \).

To use this, we first need to find the measure of \( \angle ACB \). We know the measures of \( \angle BAC \) and \( \angle ABC \):

\[
\angle BAC = 54^\circ
\]
\[
\angle ABC = 52^\circ
\]

The sum of the angles in a triangle is 180 degrees. Therefore, we can find \( \angle ACB \) as follows:

\[
\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 54^\circ - 52^\circ = 74^\circ
\]

Now, using the property of the orthocenter, we have:

\[
\angle AHB = 180^\circ - \angle ACB = 180^\circ - 74^\circ = 106^\circ
\]

Therefore, the measure of \( \angle AHB \) is \( \boxed{106^\circ} \).