ISI 2026 UGA

This problem is elegantly solved using the principle of complementary counting. Instead of directly counting the pairs of non-consecutive integers, it is computationally simpler to count the total number of all possible pairs and subtract the number of consecutive pairs.

First, the total number of unordered pairs $\{i, j\}$ that can be chosen from the set $S = \{1, 2, \dots, 25\}$ is given by the binomial coefficient: $\binom{25}{2} = \frac{25 \times 24}{2} = 300$

Next, we identify the consecutive pairs. A consecutive pair from the set must be of the form $\{k, k+1\}$. For both $k$ and $k+1$ to be in $S$, $k$ can take any integer value from $1$ up to $24$. This explicitly defines the set of consecutive pairs as: $\{\{1, 2\}, \{2, 3\}, \{3, 4\}, \dots, \{24, 25\}\}$ There are exactly $24$ such pairs.

Finally, we subtract the number of consecutive pairs from the total number of pairs to find the number of non-consecutive pairs: $300 - 24 = 276$

Correct Option: (C)

For the quadratic equation $ax^2 + bx + c = 0$ to possess equal roots, its discriminant must be identically zero. That is: $\Delta = b^2 - 4ac = 0 \implies b^2 = 4ac$ Since $a, b, c$ are the outcomes of a standard fair die, they are integers such that $a, b, c \in \{1, 2, 3, 4, 5, 6\}$.

The condition $b^2 = 4ac = (2\sqrt{ac})^2$ indicates that $b$ must be an even integer. Let us exhaustively analyze the cases for the possible even values of $b$:

• Case 1 ($b = 2$): $2^2 = 4ac \implies 4 = 4ac \implies ac = 1$ The only integers in our sample space that satisfy this product are $a = 1$ and $c = 1$. Thus, we have exactly $1$ outcome: $(1, 2, 1)$.
• Case 2 ($b = 4$): $4^2 = 4ac \implies 16 = 4ac \implies ac = 4$ The integer pairs $(a, c)$ from our sample space whose product is $4$ are $(1, 4)$, $(4, 1)$, and $(2, 2)$. This yields $3$ outcomes: $(1, 4, 4)$, $(4, 4, 1)$, and $(2, 4, 2)$.
• Case 3 ($b = 6$): $6^2 = 4ac \implies 36 = 4ac \implies ac = 9$ The only pair in our sample space whose product is $9$ is $(3, 3)$. This gives $1$ outcome: $(3, 6, 3)$.

Summing these up, there are $1 + 3 + 1 = 5$ favorable outcomes that result in equal roots. The total number of possible outcomes when throwing a six-sided die three times is $6 \times 6 \times 6 = 6^3 = 216$.

The probability is the ratio of favorable outcomes to total outcomes: $P = \frac{5}{6^3}$

Correct Option: (C)

This problem is a classic application of positional numeral systems. Let us express the numbers in the sequence in base-3 (ternary). $\begin{aligned} 1 &= 3^0 \to (1)_3 \\ 3 &= 3^1 \to (10)_3 \\ 4 &= 3^1 + 3^0 \to (11)_3 \\ 9 &= 3^2 \to (100)_3 \\ 10 &= 3^2 + 3^0 \to (101)_3 \\ 12 &= 3^2 + 3^1 \to (110)_3 \\ 13 &= 3^2 + 3^1 + 3^0 \to (111)_3 \end{aligned}$ Notice a striking pattern: the ternary representations of the terms in this sequence contain exclusively the digits $0$ and $1$.

In fact, there is a natural bijection between this sequence and the positive integers. If we interpret the ternary digits as a binary (base-2) number, we generate the natural numbers $1, 2, 3, 4, 5, 6, 7, \dots$ in their exact canonical order. Thus, the $n^{\text{th}}$ term of our sequence corresponds to the number whose ternary representation is identical to the binary representation of $n$.

We are tasked with finding the $66^{\text{th}}$ term. First, we convert $66$ to base-2: $66 = 64 + 2 = 2^6 + 2^1 \implies (1000010)_2$

Now, we treat this string of digits as a base-3 number to find the actual value in our sequence: $(1000010)_3 = 1 \cdot 3^6 + 1 \cdot 3^1 = 729 + 3 = 732$

Correct Option: (D)

Let the required limit be $L = \lim_{n \to \infty} \left(f\left(\frac{1}{n}\right)\right)^n$. Since the base approaches $f(0) = 1$ and the exponent approaches $\infty$, we are dealing with an indeterminate form of type $1^\infty$.

To evaluate this, we take the natural logarithm of both sides. By the continuity of the logarithm function, we can commute the limit and the logarithm: $\ln(L) = \lim_{n \to \infty} n \ln\left(f\left(\frac{1}{n}\right)\right)$ Let us perform a change of variables by substituting $x = \frac{1}{n}$. As $n \to \infty$, we have $x \to 0^+$. The limit transforms to: $\ln(L) = \lim_{x \to 0^+} \frac{\ln(f(x))}{x}$ As $x \to 0$, the numerator approaches $\ln(f(0)) = \ln(1) = 0$, and the denominator approaches $0$. This presents a $0/0$ indeterminate form, allowing us to deploy L’Hôpital’s Rule. Alternatively, we recognize this precisely as the limit definition of the derivative of the composite function $g(x) = \ln(f(x))$ evaluated at $x = 0$: $\lim_{x \to 0} \frac{\ln(f(x)) - \ln(f(0))}{x - 0} = \left. \frac{d}{dx}[\ln(f(x))] \right|_{x=0}$ Applying the chain rule, we differentiate the function: $\frac{d}{dx}[\ln(f(x))] = \frac{f'(x)}{f(x)}$ Evaluating this derivative at $x = 0$ yields: $\frac{f'(0)}{f(0)} = \frac{2}{1} = 2$ Therefore, we conclude that $\ln(L) = 2$, which implies $L = e^2$.

Correct Option: (D)

We seek to maximize the objective function $Z = 3y - 2x$ over the feasible region $A$. Let us first understand the geometry of region $A$ defined by the system of inequalities: $\begin{aligned} y &\le 1 - x \\ y &\le x - 1 \end{aligned}$ By combining these constraints, we find that $y$ is bounded above by the minimum of two linear expressions: $y \le \min(1 - x, x - 1)$.

We can deduce the upper bound of $y$ regardless of $x$. Summing the two inequalities (1) and (2) yields: $2y \le (1 - x) + (x - 1) \implies 2y \le 0 \implies y \le 0$ This indicates that the region $A$ lies entirely in the lower half-plane (or on the $x$-axis). The maximum possible value of $y$ is $0$, which occurs exclusively at the intersection of the two boundary lines: $1 - x = x - 1 \implies 2x = 2 \implies x = 1$ Thus, the vertex of this unbounded, downward-opening region is $(1, 0)$.

To maximize $Z = 3y - 2x$, observe that we can algebraically manipulate $Z$ to incorporate the constraints. Using the second inequality ($y - x \le -1$), we multiply by $2$: $2y - 2x \le -2$ Adding $y$ to both sides yields: $3y - 2x \le y - 2$ Since we have already established that $y \le 0$ globally in region $A$, it follows rigorously that: $Z = 3y - 2x \le y - 2 \le 0 - 2 = -2$ This theoretical maximum of $-2$ is perfectly attainable at the vertex $(x, y) = (1, 0)$, where $Z = 3(0) - 2(1) = -2$.

Correct Option: (A)

This is an archetypal application of the Pigeonhole Principle integrated with elementary number theory.

By the Fundamental Theorem of Arithmetic, a positive integer is a perfect square if and only if every prime factor in its prime factorization possesses an even exponent.

Any element $x \in A$ is uniquely characterized by its exponents $(a, b, c, d)$. Let us take two elements $x_1, x_2 \in A$ with exponents $(a_1, b_1, c_1, d_1)$ and $(a_2, b_2, c_2, d_2)$. Their product is: $x_1 x_2 = 2^{a_1+a_2} 3^{b_1+b_2} 5^{c_1+c_2} 7^{d_1+d_2}$ For $x_1 x_2$ to be a perfect square, the sums $a_1+a_2, b_1+b_2, c_1+c_2, d_1+d_2$ must all be even. This condition is equivalent to stating that $a_1$ and $a_2$ have the same parity, $b_1$ and $b_2$ have the same parity, and so forth.

We can define a parity mapping for any element $x \in A$ to a 4-tuple of its exponent parities: $f(x) = (a \bmod 2, b \bmod 2, c \bmod 2, d \bmod 2) \in (\mathbb{Z}/2\mathbb{Z})^4$ Each component of the parity vector can be either $0$ (even) or $1$ (odd). Consequently, the total number of distinct possible parity vectors is: $2 \times 2 \times 2 \times 2 = 2^4 = 16$ According to the Pigeonhole Principle, if we choose $N$ elements and $N > 16$, then at least two elements must map to the exact same parity vector. The sum of two integers with the same parity is invariably even. Hence, the product of these two distinct elements will inevitably have even exponents across all four prime bases, confirming it is a perfect square.

To guarantee that at least two such elements exist, we require one more element than the total number of parity states: $N = 16 + 1 = 2^4 + 1 = 17$

Correct Option: (B)

Given $z \in \mathbb{C}$ with $|z| = r$, we can express $z$ in polar form as $z = r e^{i\theta} = r(\cos\theta + i\sin\theta)$, where $\theta \in \mathbb{R}$.

We substitute this into the expression $z + r$: $\begin{aligned} z + r &= r(\cos\theta + 1) + i r \sin\theta \\ &= r\left(2\cos^2\frac{\theta}{2}\right) + i r\left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right) \\ &= 2r\cos\frac{\theta}{2} \left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right) \\ &= 2r\cos\frac{\theta}{2} e^{i\theta/2} \end{aligned}$ We require $(z+r)^{100}$ to be a real number. Let us evaluate this power: $(z+r)^{100} = \left(2r\cos\frac{\theta}{2}\right)^{100} e^{i 50\theta}$ For this complex number to be purely real, its imaginary part must be zero. Since $\left(2r\cos\frac{\theta}{2}\right)^{100}$ is always a real scalar, we must have either:

1. The scalar coefficient is zero: $\cos\frac{\theta}{2} = 0 \implies \frac{\theta}{2} = \frac{\pi}{2} + k\pi \implies \theta = \pi + 2k\pi$. In this case, $z = -r$, and $(z+r)^{100} = 0 \in \mathbb{R}$.
2. The exponential component is real: $e^{i 50\theta} \in \mathbb{R} \implies \sin(50\theta) = 0$. Solving for $\theta$: $50\theta = k\pi \implies \theta = \frac{k\pi}{50}, \quad \text{for } k \in \mathbb{Z}$ Notice that the first condition ($\theta = \pi$) is seamlessly incorporated into the second condition when $k = 50$, since $\frac{50\pi}{50} = \pi$. Therefore, the condition $\theta = \frac{k\pi}{50}$ perfectly encapsulates all valid solutions.

Thus, the set $S$ consists of all complex numbers of the form: $z = r e^{i \frac{k\pi}{50}}, \quad k \in \mathbb{Z}$

Correct Option: (B)

The set $D$ is generated by the ternary quadratic form $Q(x, y, z) = a(x^2 - y^2) + bz^2$ defined over the rational numbers $\mathbb{Q}$.

To understand the range of this quadratic form, let us observe its behavior when we simply restrict $z = 0$. The expression then reduces to the binary form: $a(x^2 - y^2) = a(x - y)(x + y)$ We want to determine if this expression can generate any arbitrary rational number $q \in \mathbb{Q}$. Let us set up the equation: $a(x - y)(x + y) = q$ Let us perform a linear change of variables: $w = x - y$ and $v = x + y$. Since $w$ and $v$ are linear combinations of $x, y \in \mathbb{Q}$, they must also be rational. Conversely, given rational $w$ and $v$, we can uniquely invert the system to find: $x = \frac{w + v}{2}, \quad y = \frac{v - w}{2}$ Because the rational numbers are a field (closed under addition and division by $2$), $x$ and $y$ are guaranteed to be rational for any choice of $w, v \in \mathbb{Q}$.

Now the equation simplifies to: $awv = q \implies wv = \frac{q}{a}$ For any given $q \in \mathbb{Q}$ and non-zero $a \in \mathbb{Q}$, $\frac{q}{a}$ is a well-defined rational number. We can simply choose $w = 1 \in \mathbb{Q}$, which immediately forces $v = \frac{q}{a} \in \mathbb{Q}$. This demonstrates that for any $q \in \mathbb{Q}$, there exist valid rational choices for $x$ and $y$ (along with $z=0$) such that the expression exactly evaluates to $q$.

Thus, the set $D$ contains all of $\mathbb{Q}$. Since $D \subseteq \mathbb{Q}$ by closure, we conclude $D = \mathbb{Q}$.

Correct Option: (A)

This problem elegantly models a symmetric 1-dimensional random walk. The sequence $k_0, k_1, \dots, k_{10}$ traces the positions of a particle that starts at the origin ($k_0 = 0$) and must return to the origin after exactly $10$ steps ($k_{10} = 0$).

At each step $i$, the particle moves by a displacement $\Delta_i = k_{i+1} - k_i$. The condition $|k_i - k_{i+1}| = 1$ strictly dictates that $\Delta_i \in \{+1, -1\}$. The final position of the particle is the cumulative sum of these $10$ independent displacements: $k_{10} = k_0 + \sum_{i=0}^9 \Delta_i = \sum_{i=0}^9 \Delta_i = 0$ Let $P$ denote the total number of positive steps ($+1$) and $N$ denote the total number of negative steps ($-1$). We have a simple system of linear equations: $\begin{aligned} P + N &= 10 \quad \text{(total number of steps)} \\ P - N &= 0 \quad \text{(net displacement)} \end{aligned}$ Solving this system yields $2P = 10 \implies P = 5$ and consequently $N = 5$.

The problem therefore reduces to finding the number of distinct ways to choose exactly $5$ positive steps out of the $10$ total steps. This is a fundamental combination: $\text{Number of tuples} = \binom{10}{5}$

Correct Option: (B)

To analyze the roots of this equation, we define an auxiliary continuous function $f(x)$ on the closed interval $[0, \pi/2]$: $f(x) = \sin(\cos x) - \log(1 + \sin x)$ Let us first evaluate the function at the boundary points:

• At $x = 0$: $f(0) = \sin(\cos 0) - \log(1 + \sin 0) = \sin(1) - \log(1) = \sin(1) > 0$.
• At $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = \sin\left(\cos \frac{\pi}{2}\right) - \log\left(1 + \sin \frac{\pi}{2}\right) = \sin(0) - \log(2) = -\log(2) < 0$.

Since $f(x)$ is continuous and undergoes a sign change over $(0, \pi/2)$, the Intermediate Value Theorem guarantees the existence of at least one root in this interval.

To determine the exact number of roots, we investigate the monotonicity of $f(x)$ via its first derivative: $f'(x) = -\sin x \cdot \cos(\cos x) - \frac{\cos x}{1 + \sin x}$ For $x \in (0, \pi/2)$, we can tightly bound the trigonometric components:

• $\sin x > 0$ and $\cos x > 0$.
• Since $0 < \cos x < 1$, we are taking the cosine of an angle strictly between $0$ and $1$ radian. In the first quadrant, the cosine function is strictly positive, hence $\cos(\cos x) > 0$.
• The denominator $1 + \sin x > 1 > 0$.

Consequently, both terms in $f'(x)$ are strictly negative: $f'(x) = - (\text{positive}) - (\text{positive}) < 0$ Because $f'(x) < 0$ strictly for all $x \in (0, \pi/2)$, the function $f(x)$ is strictly monotonically decreasing. A strictly decreasing continuous function can intersect the x-axis at most once. Coupling this with the Intermediate Value Theorem, we rigorously conclude there is exactly one solution.

Correct Option: (D)

Let us define the function $f(x) = x \sin\left(\frac{1}{x}\right) - \log x$ for $x > 1$.

We probe the behavior of $f(x)$ at the boundaries of its domain:

• As $x \to 1^+$: $x \sin\left(\frac{1}{x}\right) \to \sin(1) \approx 0.841$, and $\log x \to 0$. Thus, $f(1^+) > 0$.
• At $x = e$: $f(e) = e \sin\left(\frac{1}{e}\right) - \log e$. By the well-known inequality $\sin(t) < t$ for $t > 0$, substituting $t = 1/e$ yields $\sin(1/e) < 1/e$, so $e \sin(1/e) < 1$. Since $\log e = 1$, we get $f(e) < 1 - 1 = 0$.

By the Intermediate Value Theorem, $f(x)$ transitions from positive to negative, confirming at least one root in $(1, e)$.

To prove uniqueness, we examine the derivative $f'(x)$: $f'(x) = \sin\left(\frac{1}{x}\right) + x \cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) - \frac{1}{x} = \sin\left(\frac{1}{x}\right) - \frac{1}{x} \cos\left(\frac{1}{x}\right) - \frac{1}{x}$ To simplify the analysis, we introduce the substitution $t = \frac{1}{x}$. For $x > 1$, we have $t \in (0, 1)$. We define an auxiliary function $g(t)$: $g(t) = \sin t - t \cos t - t$ We study the derivative of $g(t)$ to determine its sign: $g'(t) = \cos t - (\cos t - t \sin t) - 1 = t \sin t - 1$ For $t \in (0, 1)$, $t < 1$ and $\sin t < \sin 1 < 1$. Thus, the product $t \sin t < 1$, which rigorously implies $g'(t) < 0$. Since $g(t)$ is strictly decreasing on $(0, 1)$ and its limit as $t \to 0$ is $g(0) = 0$, it follows that $g(t) < 0$ for all $t \in (0, 1)$.

Because $f'(x) = g(1/x) < 0$ for all $x > 1$, the function $f(x)$ is strictly monotonically decreasing. A strictly decreasing function can have at most one real root. Thus, there is exactly one solution.

Correct Option: (D)

In combinatorial terminology, an index $i$ such that $\pi_i > \pi_{i+1}$ in a permutation $\pi$ is called a descent. The problem asks us to find the number of permutations of length $n = 10$ with exactly $k = 1$ descent. The number of permutations of length $n$ with exactly $k$ descents is formalized by the Eulerian numbers $\left\langle \begin{matrix} n \\ k+1 \end{matrix} \right\rangle$.

A permutation with exactly one descent can be viewed as the concatenation of two strictly increasing sequences. To construct such a permutation, we can select a subset of the $n$ integers to form the first increasing block, while the remaining integers automatically form the second increasing block. For a set of $n$ elements, there are $2^n$ possible subsets.

However, we must be careful to avoid overcounting configurations that do not produce exactly one descent:

1. If we choose the empty set or the complete set of $n$ elements, the resulting permutation is entirely strictly increasing (the identity permutation), which has $0$ descents. We must subtract these $2$ cases.
2. We must also exclude subsets that fail to create a descent when concatenated. A descent fails to occur if and only if the maximum element of the first block is smaller than the minimum element of the second block. This happens precisely when the first block consists of the first $j$ smallest integers: $\{1, 2, \dots, j\}$. There are $n - 1$ such proper non-empty prefixes.

Subtracting these invalid configurations, the number of permutations with exactly one descent is compactly given by: $2^n - 2 - (n - 1) = 2^n - n - 1$ Substitute $n = 10$ into our derived formula: $2^{10} - 10 - 1 = 1024 - 11 = 1013$

Correct Option: (C)

This functional equation is a variation of Cauchy’s functional equation. We can solve it systematically by reducing it to a differential equation.

First, let us evaluate the function at the origin. Substituting $x = 0$ and $y = 0$ into the given equation: $f(0+0) = f(0) + f(0) + (0)(0) \implies f(0) = 2f(0) \implies f(0) = 0$ Next, we use the first principles definition of the derivative to find $f'(x)$: $f'(x) = \lim_{y \to 0} \frac{f(x+y) - f(x)}{y}$ From the given functional equation, we can isolate the difference $f(x+y) - f(x)$: $f(x+y) - f(x) = f(y) + xy$ Substituting this back into the limit yields: $f'(x) = \lim_{y \to 0} \frac{f(y) + xy}{y} = \lim_{y \to 0} \left( \frac{f(y)}{y} + x \right) = x + \lim_{y \to 0} \frac{f(y)}{y}$ Notice that the remaining limit is exactly the definition of the derivative at $x = 0$, since $f(0) = 0$: $\lim_{y \to 0} \frac{f(y)}{y} = \lim_{y \to 0} \frac{f(y) - f(0)}{y - 0} = f'(0)$ We are given that $f'(0) = 1$. Therefore, we obtain a simple first-order ordinary differential equation: $f'(x) = x + 1$ Integrating both sides with respect to $x$ gives the general solution for $f(x)$: $f(x) = \int (x + 1) \,dx = \frac{x^2}{2} + x + C$ Using our initial condition $f(0) = 0$, we find $C = 0$. Thus, the uniquely determined function is: $f(x) = \frac{x^2}{2} + x$ Finally, we evaluate this function at $x = 1$: $f(1) = \frac{1^2}{2} + 1 = \frac{3}{2}$

Correct Option: (A)

This problem tests the convergence of iterated function compositions under an asymptotically varying Lipschitz-like bound.

Let $x \in (-\alpha, \alpha)$ for some $\alpha > 0$ to be determined. We define the sequence of iterates $y_0 = x$, and $y_k = f_k(y_{k-1})$ for $k \ge 1$. We must ensure that $y_n \to 0$ for any sequence of functions satisfying the bound. Assuming $y_k$ remains in $(-1/2, 1/2)$, the given bound can be applied recursively: $\begin{aligned} |y_1| &\le c^{t^1} |x| \\ |y_2| &\le c^{t^2} |y_1| \le c^{t^2} c^{t^1} |x| = c^{t + t^2} |x| \\ &\vdots \\ |y_n| &\le c^{t^n} |y_{n-1}| \le c^{\sum_{k=1}^n t^k} |x| \end{aligned}$ Let $S_n = \sum_{k=1}^n t^k$. The bound on the iterate is $|y_n| \le c^{S_n} |x|$. For $\Omega$ to contain a neighborhood of the origin, the maximal possible magnitude of $y_n$ (which occurs if $f_k(x)$ equals its upper bound $c^{t^k} x$) must converge to 0. Thus, we require $\lim_{n \to \infty} c^{S_n} = 0$.

Let us analyze the behavior of $S_n$:

• Case 1 ($t < 1$): The series is a convergent geometric series. $S_n \to \frac{t}{1-t}$, which is a finite constant $S$. Therefore, $c^{S_n} \to c^S > 0$. If we choose $f_n(x) = c^{t^n} x$, then $y_n \to c^S x$, which does not approach $0$ for any $x \neq 0$. Thus, $t < 1$ fails to guarantee convergence to $0$. Options (C) and (D) are incorrect.
• Case 2 ($t > 1$): The sum $S_n = t\frac{t^n-1}{t-1}$ diverges to $+\infty$ as $n \to \infty$. We want $c^{S_n} \to 0$. Since the exponent $S_n \to +\infty$, the base $c$ must be strictly less than $1$. Thus, we must have $0 < c < 1$.

To rigorously confirm $0 < c < 1$ and $t > 1$ is sufficient, we must verify that $y_n$ never escapes the interval $(-1/2, 1/2)$, as the bound $|f_n(x)| \le c^{t^n} |x|$ only applies there. Since $0 < c < 1$ and $S_n$ is a strictly positive, monotonically increasing sequence, the multiplier sequence $c^{S_n}$ is strictly monotonically decreasing. Its maximum value is the first term: $c^{S_1} = c^t$. Because $0 < c^t < 1$, we can choose our neighborhood radius $\alpha$ such that $\alpha < \frac{1}{2}$. Then for any $x \in (-\alpha, \alpha)$, we have: $|y_n| \le c^{S_n} |x| \le c^t |x| < |x| < \alpha < 1/2$ This unequivocally proves that the sequence $\{y_n\}$ never escapes $(-1/2, 1/2)$, the bounds remain universally valid, and $y_n \to 0$.

Correct Option: (A)

To analyze the asymptotic behavior of the area $a_n$ as $n \to \infty$, we must examine the moduli of the complex vertices. A complex number $z$ raised to the $n$-th power has a modulus $|z|^n$. If $|z| < 1$, then $|z|^n \to 0$ as $n \to \infty$.

Let us compute the squared moduli of the given complex numbers: $|z_1|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\pi}\right)^2 = \frac{1}{4} + \frac{1}{\pi^2}$ Since $\pi > 3$, we know $\pi^2 > 9$, which implies $\frac{1}{\pi^2} < \frac{1}{9} \approx 0.111$. Thus, $|z_1|^2 < 0.25 + 0.112 = 0.362 < 1$. So, $|z_1| < 1$.

Now for $z_2$: $|z_2|^2 = \left(\frac{\pi}{6}\right)^2 + \left(\frac{1}{3}\right)^2 = \frac{\pi^2}{36} + \frac{1}{9}$ Since $\pi < 3.2$, we know $\pi^2 < 10.24$, so $\frac{\pi^2}{36} < \frac{10.24}{36} \approx 0.284$. Thus, $|z_2|^2 < 0.284 + 0.112 = 0.396 < 1$. So, $|z_2| < 1$.

The numbers $z_4$ and $z_3$ are the complex conjugates of $z_1$ and $z_2$ respectively, meaning they have identical moduli: $|z_4| = |z_1| < 1$ and $|z_3| = |z_2| < 1$.

Since the modulus of every single vertex is strictly less than $1$, the distance of every vertex from the origin converges to $0$ as $n \to \infty$: $\lim_{n \to \infty} |z_k^n| = \lim_{n \to \infty} |z_k|^n = 0 \quad \text{for } k=1, 2, 3, 4$ As $n \to \infty$, all four vertices of the quadrilateral collapse into the origin $(0,0)$. Consequently, the quadrilateral degenerates into a single point, and its area $a_n$ must limit to $0$.