Source: ITA – 1st phase exam and official answer key. Files on the ITA past exams page.

The 1st phase of ITA 2023 had 60 multiple-choice questions, 12 per subject (Physics, Portuguese, English, Mathematics and Chemistry). This post covers the mathematics questions, numbered 37 to 48 as in the exam.

Exam conventions. $\mathbb{N} = \{1, 2, 3, \dots\}$; $i^2 = -1$; $M_n(\mathbb{R})$ is the set of $n \times n$ real matrices; $\overline{AB}$ is the segment and $AB$ the line through $A$ and $B$; $m(\overline{AB})$ is the length of $\overline{AB}$; $A\hat{O}B$ is the angle with vertex $O$; $\det A$ is the determinant of $A$.

Answer key

Question Topic Answer
37 Determinants and an equation with a unique solution C
38 Complex numbers: iterating $f(z) = z^2 + i$ E
39 Convex polygons and arithmetic progression B
40 Polynomials: harmonic mean of the roots C
41 Exponential functions: inequalities B
42 Plane geometry: rectangle inscribed in a triangle D
43 Complex numbers: sum of powers D
44 Hyperbola: triangle with the foci A
45 Probability: the coin game B
46 Solid of revolution: two cones E
47 Trigonometric equation A
48 Polynomials: coefficients of powers that are multiples of 3 C

Question 37: An equation with determinants and a unique solution

Statement

Consider $A \in M_3(\mathbb{R})$ such that there is a unique real number $x$ satisfying the equation $\det(\sqrt[3]{2}\,x^2 A) + \det(xA^3) = \det A^2$. Then $x + \det A$ is

  • A) $-5$.
  • B) $-4$.
  • C) $-3$.
  • D) $-2$.
  • E) $-1$.

Solution

Let $d = \det A$. For a $3 \times 3$ matrix, $\det(kM) = k^3\det M$.

1. Translating.

$$\det(\sqrt[3]{2}\,x^2 A) = \left(\sqrt[3]{2}\,x^2\right)^3 d = 2x^6 d, \qquad \det(xA^3) = x^3 d^3, \qquad \det A^2 = d^2$$

The equation is $2d\,x^6 + d^3 x^3 = d^2$.

2. $d$ cannot be zero. If $d = 0$, the equation becomes $0 = 0$ and every $x$ works, contradicting uniqueness. With $d \ne 0$, divide by $d$:

$$2x^6 + d^2 x^3 - d = 0$$

3. Unique solution. With $t = x^3$ (each real $t$ corresponds to exactly one real $x$), we get $2t^2 + d^2 t - d = 0$. For there to be a single $x$, this quadratic equation must have a single real root: zero discriminant.

$$\Delta = d^4 + 8d = d(d^3 + 8) = 0 \quad\Rightarrow\quad d = -2$$

4. The value of $x$. With $d = -2$: $2t^2 + 4t + 2 = 0 \Rightarrow (t + 1)^2 = 0 \Rightarrow t = -1$, and $x = -1$.

$$x + \det A = -1 + (-2) = -3$$

Answer: option C ($-3$).

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Question 38: Applying f 2023 times

Statement

Let $z \in \mathbb{C}$ and $f(z) = z^2 + i$. For each $n \in \mathbb{N}$, define $f^{(1)}(z) = f(z)$ and $f^{(n)}(z) = f(f^{(n-1)}(z))$. Then $f^{(2023)}(0)$ is

  • A) $1 - i$.
  • B) $i - 1$.
  • C) $-1 - i$.
  • D) $i$.
  • E) $-i$.

Solution

Computing the first iterates starting from $0$:

  • $f^{(1)}(0) = 0^2 + i = i$
  • $f^{(2)}(0) = i^2 + i = -1 + i$
  • $f^{(3)}(0) = (-1 + i)^2 + i = (1 - 2i - 1) + i = -i$
  • $f^{(4)}(0) = (-i)^2 + i = -1 + i$
  • $f^{(5)}(0) = -i$

From $n = 2$ on the values alternate: $-1 + i$ for even $n$ and $-i$ for odd $n$. Since 2023 is odd,

$$f^{(2023)}(0) = -i$$

Answer: option E ($-i$).

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Question 39: Convex polygons with angles in arithmetic progression

Statement

Consider the statements:

  • I. If $P$ is a convex polygon with $n$ equal sides, then $P$ is a regular polygon.
  • II. Let $P$ be a convex polygon with 6 sides. If its interior angles, listed in increasing order, form an arithmetic progression, then the sum of the smallest and the largest interior angle of $P$ is $240^\circ$.
  • III. There is a convex polygon with 100 sides whose interior angles, listed in increasing order, form an arithmetic progression with common difference $r = 1^\circ$.

Which is (are) always true:

  • A) only I.
  • B) only II.
  • C) only III.
  • D) only II and III.
  • E) none.

Solution

I. False. A rhombus that is not a square has 4 equal sides but different angles.

II. True. The sum of the interior angles of a hexagon is $(6 - 2)\cdot 180^\circ = 720^\circ$. In an arithmetic progression with 6 terms, the sum is $\frac{(a_1 + a_6)\cdot 6}{2} = 3(a_1 + a_6)$. Hence $a_1 + a_6 = 240^\circ$.

III. False. The sum of the angles is $98 \cdot 180^\circ = 17640^\circ$. With common difference $1^\circ$:

$$100a_1 + \frac{99 \cdot 100}{2} = 17640 \quad\Rightarrow\quad a_1 = 126.9^\circ$$

and the largest angle would be $a_{100} = 126.9^\circ + 99^\circ = 225.9^\circ > 180^\circ$. A convex polygon has no interior angle greater than $180^\circ$.

Answer: option B (only II).

To review, see Arithmetic progression.

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Question 40: Harmonic mean of the roots

Statement

The harmonic mean of $n$ positive real numbers $a_1, a_2, \dots, a_n$ is

$$H = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}}.$$

Given that the polynomial $p(x) = 30x^3 - 113x^2 + 108x - 30$ has three positive real roots, the harmonic mean of the roots of $p(x)$ is

  • A) $\frac{2}{3}$.
  • B) $\frac{5}{18}$.
  • C) $\frac{5}{6}$.
  • D) $1$.
  • E) $3$.

Solution

Let $r_1, r_2, r_3$ be the roots. By Vieta’s formulas,

$$r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{108}{30}, \qquad r_1 r_2 r_3 = \frac{30}{30} = 1$$

The sum of the reciprocals is

$$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{r_2 r_3 + r_1 r_3 + r_1 r_2}{r_1 r_2 r_3} = \frac{108}{30} = \frac{18}{5}$$

So

$$H = \frac{3}{18/5} = \frac{15}{18} = \frac{5}{6}$$

Answer: option C ($\frac{5}{6}$).

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Question 41: Comparing 9ˣ, 3ˣ and 2ˣ

Statement

Let $f$ and $g$ be real functions defined as follows: $f(x) = 3^{2x}$ and $g(x) = 3^x - 2^x$. Consider the statements:

  • I. $g(x) \ge 0$ for all $x \in \mathbb{R}$.
  • II. $f(x) \ge g(x)$ for all $x \in \mathbb{R}$.
  • III. $f(x) + g(x) \ge 0$ for all $x \in \mathbb{R}$.

Which is (are) always true:

  • A) only I.
  • B) only II.
  • C) only III.
  • D) all of them.
  • E) none.

Solution

Note that $f(x) = 9^x$. For $x > 0$ the larger base gives the larger power ($9^x > 3^x > 2^x$); for $x < 0$ it is the other way round ($2^x > 3^x > 9^x$).

I. False. For $x = -1$: $g(-1) = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} < 0$.

II. True.

  • If $x \ge 0$: $9^x \ge 3^x > 3^x - 2^x$.
  • If $x < 0$: $g(x) = 3^x - 2^x < 0 < 9^x$.

III. False. For $x = -1$: $\frac{1}{9} + \frac{1}{3} - \frac{1}{2} = \frac{2 + 6 - 9}{18} = -\frac{1}{18} < 0$.

Answer: option B (only II).

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Question 42: The shortest segment EF

Statement

Let $ABC$ be a right triangle such that $B\hat{A}C = 30^\circ$. Consider a point $D$ on the hypotenuse $\overline{AC}$ and lines $r$ and $s$ through $D$, parallel to the sides $\overline{AB}$ and $\overline{BC}$, respectively. If $E = r \cap \overline{BC}$, $F = s \cap \overline{AB}$ and $m(\overline{BC}) = 1$, the smallest possible value of $m(\overline{EF})$ is

  • A) $\frac{\sqrt{2}}{5}$.
  • B) $\frac{\sqrt{2}}{2}$.
  • C) $\frac{\sqrt{3}}{3}$.
  • D) $\frac{\sqrt{3}}{2}$.
  • E) $\sqrt{3}$.

Solution

1. The triangle. The hypotenuse is $\overline{AC}$, so the right angle is at $B$. With $\hat{A} = 30^\circ$ and $m(\overline{BC}) = 1$ (the leg opposite $30^\circ$): $m(\overline{AC}) = 2$ and $m(\overline{AB}) = \sqrt{3}$.

2. A rectangle. $DE \parallel AB$ and $DF \parallel BC$, and $AB \perp BC$. So $BEDF$ is a rectangle, and its diagonals are equal:

$$m(\overline{EF}) = m(\overline{BD})$$

3. Minimising. $D$ runs along the hypotenuse; the shortest distance from $B$ to a point of $\overline{AC}$ is the altitude to the hypotenuse:

$$m(\overline{BD})_{\min} = \frac{m(\overline{AB})\cdot m(\overline{BC})}{m(\overline{AC})} = \frac{\sqrt{3}\cdot 1}{2} = \frac{\sqrt{3}}{2}$$

The foot of this altitude lies between $A$ and $C$ (the angles at $A$ and $C$ are acute), so it is a valid point.

Answer: option D ($\frac{\sqrt{3}}{2}$).

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Question 43: Adding up |z²ⁿ⁺¹ + z̄²ⁿ⁺¹|

Statement

Given $z = 5 - 5i \in \mathbb{C}$, define $f(n) = \left|z^{(2n+1)} + \bar{z}^{(2n+1)}\right|$ for each $n \in \mathbb{N}$. The sum of $f(n)$ for $n$ from 1 to 20 is

  • A) $250(50^{21} - 1)/49$.
  • B) $500\sqrt{2}(50^{20} - 1)/49$.
  • C) $1000(50^{21} - 1)/49$.
  • D) $500(50^{20} - 1)/49$.
  • E) none of the previous options.

Solution

1. Polar form. $|z| = \sqrt{25 + 25} = 5\sqrt{2}$ and the argument is $-\frac{\pi}{4}$: $z = 5\sqrt{2}\,\operatorname{cis}\left(-\frac{\pi}{4}\right)$.

2. Adding the conjugate. For any $w$, $w + \bar{w} = 2\operatorname{Re}(w)$. With $k = 2n + 1$,

$$z^k + \bar{z}^k = 2\,(5\sqrt{2})^k\cos\left(\frac{k\pi}{4}\right)$$

For odd $k$, $\left|\cos\frac{k\pi}{4}\right| = \frac{\sqrt{2}}{2}$. So

$$f(n) = 2\,(5\sqrt{2})^{2n+1}\cdot\frac{\sqrt{2}}{2} = (5\sqrt{2})^{2n}\cdot 5\sqrt{2}\cdot\sqrt{2} = 10\cdot 50^n$$

Checking with $n = 1$: $z^3 = 125(1 - i)^3 = 125(-2 - 2i)$, so $z^3 + \bar{z}^3 = -500$ and $f(1) = 500 = 10 \cdot 50$ ✔

3. Sum of the geometric progression. $f(n)$ is a geometric progression with first term 500 and ratio 50:

$$\sum_{n=1}^{20} 10\cdot 50^n = 500\cdot\frac{50^{20} - 1}{50 - 1} = \frac{500\,(50^{20} - 1)}{49}$$

Answer: option D.

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Question 44: Triangle formed by a point of the hyperbola and the foci

Statement

Consider the hyperbola $H$ with equation $x^2 - \dfrac{y^2}{4} = 1$. Let $T$ be a triangle with vertices $P$, $F_1$, $F_2$, where $F_1$ and $F_2$ are the foci of $H$ and $P$ is a point on $H$. Given that the perimeter of $T$ is $5\sqrt{5}$, the product of the side lengths of $T$ is

  • A) $\frac{41\sqrt{5}}{2}$.
  • B) $\frac{41}{4}$.
  • C) $\frac{41\sqrt{5}}{4}$.
  • D) $\frac{41}{8}$.
  • E) $\frac{41\sqrt{5}}{8}$.

Solution

1. Elements of the hyperbola. $a = 1$, $b = 2$, $c = \sqrt{a^2 + b^2} = \sqrt{5}$. So $m(\overline{F_1F_2}) = 2\sqrt{5}$.

2. The other two sides. Let $u = m(\overline{PF_1})$ and $v = m(\overline{PF_2})$.

  • Definition of the hyperbola: $|u - v| = 2a = 2$.
  • Perimeter: $u + v = 5\sqrt{5} - 2\sqrt{5} = 3\sqrt{5}$.

3. The product $uv$. Without finding $u$ and $v$:

$$uv = \frac{(u + v)^2 - (u - v)^2}{4} = \frac{45 - 4}{4} = \frac{41}{4}$$

4. Product of the three sides.

$$\frac{41}{4}\cdot 2\sqrt{5} = \frac{41\sqrt{5}}{2}$$

Answer: option A ($\frac{41\sqrt{5}}{2}$).

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Question 45: How long does the coin game last?

Statement

A set of coins is tossed repeatedly. In each toss, all coins that land tails, and only those, are removed. The remaining coins stay for the next toss. The game ends when all coins have been removed. The probability that the game lasts more than three rounds, if it starts with four coins, is

  • A) $1341/4096$.
  • B) $1695/4096$.
  • C) $2049/4096$.
  • D) $2401/4096$.
  • E) $2755/4096$.

Solution

1. One coin at a time. Each coin stays in the game as long as it lands heads. It is still there after 3 rounds if it landed heads 3 times in a row: probability $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$. So it is removed within 3 rounds with probability $\frac{7}{8}$.

2. The complement. The game ends within 3 rounds when all 4 coins are removed within 3 rounds. The coins are independent:

$$P(\text{ends within 3 rounds}) = \left(\frac{7}{8}\right)^4 = \frac{2401}{4096}$$

3. Answer.

$$P(\text{lasts more than 3 rounds}) = 1 - \frac{2401}{4096} = \frac{1695}{4096}$$

Answer: option B ($1695/4096$).

Watch out for the trap: $2401/4096$ (option D) is the probability of the opposite event.

To review, see Introduction to probability.

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Question 46: Spinning the triangle around the hypotenuse

Statement

The hypotenuse of a right triangle measures 10 cm. The volume of the solid generated by rotating this triangle around an axis containing the hypotenuse is $30\pi$ cm³. The perimeter of this triangle, in cm, is

  • A) $10 + 4\sqrt{7}$.
  • B) $10 + 5\sqrt{7}$.
  • C) $10 + 2\sqrt{10}$.
  • D) $10 + 3\sqrt{10}$.
  • E) $10 + 4\sqrt{10}$.

Solution

1. The solid. Rotating around the hypotenuse, the triangle generates two cones joined at their bases. The common radius is the altitude $h$ to the hypotenuse, and the heights of the cones add up to 10:

$$V = \frac{1}{3}\pi h^2 d_1 + \frac{1}{3}\pi h^2 d_2 = \frac{1}{3}\pi h^2\cdot 10 = 30\pi \quad\Rightarrow\quad h = 3$$

2. The legs. Let $b$ and $c$ be the legs. From the area, $bc = 10 \cdot h = 30$; by Pythagoras, $b^2 + c^2 = 100$. So

$$(b + c)^2 = 100 + 60 = 160 \quad\Rightarrow\quad b + c = 4\sqrt{10}$$

3. Perimeter. $10 + 4\sqrt{10}$ cm.

Checking that the triangle exists: $b$ and $c$ are the roots of $t^2 - 4\sqrt{10}\,t + 30 = 0$, with discriminant $160 - 120 = 40 > 0$ ✔

Answer: option E ($10 + 4\sqrt{10}$).

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Question 47: An equation that needs two maxima at once

Statement

Consider the real function

$$f(x) = \cos(x)\cdot\left[\cos\left(\frac{x}{3}\right) + 2\sin(x)\right] - \sin(x)\cdot\sin\left(\frac{x}{3}\right) - 2,$$

defined on the interval $I = ]-4\pi, 4\pi[$. About the equation $f(x) = 0$, we can state that it

  • A) has no solutions in $I$.
  • B) has a single solution in $I$.
  • C) has exactly two solutions in $I$.
  • D) has exactly three solutions in $I$.
  • E) has exactly four solutions in $I$.

Solution

1. Simplifying. Distributing and grouping:

$$f(x) = \underbrace{\cos x\cos\tfrac{x}{3} - \sin x\sin\tfrac{x}{3}}_{\cos\left(x + \frac{x}{3}\right)} + \underbrace{2\sin x\cos x}_{\sin 2x} - 2 = \cos\frac{4x}{3} + \sin 2x - 2$$

2. Only at the maximum. Since cosine and sine are at most 1, $f(x) = 0$ requires both

$$\cos\frac{4x}{3} = 1 \quad\text{and}\quad \sin 2x = 1$$

3. Incompatible.

  • $\cos\frac{4x}{3} = 1 \Rightarrow \frac{4x}{3} = 2k\pi \Rightarrow x = \frac{3k\pi}{2}$, $k \in \mathbb{Z}$.
  • $\sin 2x = 1 \Rightarrow 2x = \frac{\pi}{2} + 2m\pi \Rightarrow x = \frac{\pi}{4} + m\pi$, $m \in \mathbb{Z}$.

Equating: $\frac{3k}{2} = \frac{1}{4} + m \Rightarrow 6k = 1 + 4m$. The left side is even and the right side is odd: impossible. There is no solution anywhere, let alone in $I$.

Answer: option A (no solutions).

To review, see Addition and subtraction formulas.

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Question 48: Filtering the powers that are multiples of 3

Statement

In the expansion of $\left[1 + x^2 - x^3 + x^4\right]^{10}$, the sum of all the coefficients of the powers that are multiples of 3 is

  • A) 114.
  • B) 228.
  • C) 342.
  • D) 456.
  • E) 570.

Solution

Let $P(x) = (1 + x^2 - x^3 + x^4)^{10} = \sum c_k x^k$. We want $S = c_0 + c_3 + c_6 + \dots$

1. The cube-roots-of-unity filter. Let $\omega = \operatorname{cis}\frac{2\pi}{3}$, with $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Then $1 + \omega^k + \omega^{2k}$ equals 3 if $k$ is a multiple of 3 and 0 otherwise. Hence

$$S = \frac{P(1) + P(\omega) + P(\omega^2)}{3}$$

2. Computing.

  • $P(1) = (1 + 1 - 1 + 1)^{10} = 2^{10} = 1024$.
  • $P(\omega) = (1 + \omega^2 - 1 + \omega)^{10} = (\omega + \omega^2)^{10} = (-1)^{10} = 1$, since $\omega^3 = 1$ and $\omega^4 = \omega$.
  • $P(\omega^2) = 1$, in the same way.
$$S = \frac{1024 + 1 + 1}{3} = 342$$

(Here $x^0$ counts as a power that is a multiple of 3; without it the sum would be 341, which is not among the options.)

Answer: option C (342).

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