ITA 2026 – 1st phase: Mathematics solved

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

The 1st phase of ITA 2026 had 48 multiple-choice questions: 12 in Mathematics, 12 in Physics, 12 in Chemistry and 12 in English. This post covers the mathematics questions, numbered 1 to 12 as in the exam.

Exam conventions. $i^2 = -1$; $\bar{z}$ is the conjugate of $z$; $]a, b[$ is the open interval $a < x < b$.

Answer key

Question 1: z − w real with w at argument π/6

Statement

Let $z, w \in \mathbb{C}$ be such that $z - w = -2\sqrt{3}$ and $w \ne 0$. If $z$ has modulus $2\sqrt{3}$ and $w$ has argument $\pi/6$, then the real part of $z$ is

• A) $-2\sqrt{3}$.
• B) $-\sqrt{3}$.
• C) $0$.
• D) $\sqrt{3}$.
• E) $2\sqrt{3}$.

Solution

1. Writing $w$. With $|w| = \rho > 0$ and argument $\frac{\pi}{6}$: $w = \rho\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$. Then

2. Using $|z| = 2\sqrt{3}$.

Since $\rho \ne 0$, $\rho = 6$.

3. Real part. $\operatorname{Re}(z) = 3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$.

Checking: $z = \sqrt{3} + 3i$ has modulus $\sqrt{3 + 9} = 2\sqrt{3}$ ✔

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

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Question 2: log₄₅ 7 from two approximations

Statement

Using the approximate values

the option that best approximates the decimal representation of $\log_{45} 7$ is

• A) $0.4886$.
• B) $0.5112$.
• C) $0.5193$.
• D) $0.5224$.
• E) $0.5385$.

Solution

1. Factoring. $13.72 = \frac{1372}{100} = \frac{4\cdot 343}{100} = \frac{7^3}{5^2}$ and $6125 = 125\cdot 49 = 5^3\cdot 7^2$.

2. A system. With $u = \log_{45} 7$ and $v = \log_{45} 5$:

3. Solving. Multiplying the 1st by 3 and the 2nd by 2 and adding:

Checking: $v = \frac{3u - 0.6880}{2} = 0.4228$, and indeed $\log_{45} 5 = \frac{\ln 5}{\ln 45} \approx 0.4228$ ✔

Answer: option B ($0.5112$).

To review, see Properties of logarithms.

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Question 3: The matrix aᵢⱼ = 1 + (1 − i)(1 − j)

Statement

Let $A$ be the $100 \times 100$ matrix whose entries are given by

Consider the following statements:

• I. $A$ is a symmetric matrix.
• II. Each row of $A$ forms an arithmetic progression.
• III. $A$ is a singular matrix.

Which is (are) TRUE:

• A) Only I.
• B) Only II.
• C) Only I and II.
• D) Only I and III.
• E) I, II and III.

Solution

I. True. Swapping $i$ and $j$ does not change $1 + (1 - i)(1 - j)$.

II. True. For a fixed row $i$, $a_{ij} = (2 - i) - (1 - i)\,j$ is a first-degree function of $j$: the terms form an arithmetic progression with difference $i - 1$.

III. True. By II, each row is an arithmetic progression; so the 3rd column equals twice the 2nd minus the 1st ($a_{i3} = 2a_{i2} - a_{i1}$). Linearly dependent columns give determinant zero. (In fact every column is a combination of the first two: $A$ has rank 2.)

Answer: option E (I, II and III).

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Question 4: The polygon of the solutions of iz² = 2z̄

Statement

The vertices of a polygon are all the non-zero complex numbers satisfying the equation

The area of this polygon is

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

Solution

1. The modulus. Taking moduli: $|z|^2 = 2|z|$, and since $z \ne 0$, $|z| = 2$.

2. The argument. With $z = 2\operatorname{cis}\theta$: $i\cdot 4\operatorname{cis}2\theta = 4\operatorname{cis}(-\theta)$, that is,

Three solutions: $\theta = -\frac{\pi}{6}, \frac{\pi}{2}, \frac{7\pi}{6}$, that is, $z = \sqrt{3} - i$, $2i$, $-\sqrt{3} - i$.

3. The area. It is an equilateral triangle inscribed in the circle of radius 2, with side $2\sqrt{3}$:

Checking with $z = 2i$: $i(2i)^2 = -4i$ and $2\overline{2i} = -4i$ ✔

Answer: option C ($3\sqrt{3}$).

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Question 5: From the hexagram to the inscribed square

Statement

A circle is divided into six equal parts by the points $A$, $B$, $C$, $D$, $E$, $F$, named in this order. The points $A$, $C$, $E$ are the vertices of an equilateral triangle, and the points $B$, $D$, $F$ are the vertices of a second equilateral triangle. The overlap of these two triangles defines a six-pointed star called a hexagram. If the area of this hexagram is $25\sqrt{3}$ cm², then the area of the square inscribed in the circle is

• A) 10 cm².
• B) 25 cm².
• C) 40 cm².
• D) 50 cm².
• E) 75 cm².

Solution

1. The star is a triangle plus three points. The sides of triangle $BDF$ cut each side of $ACE$ into three equal parts. Each point of the star outside $ACE$ is an equilateral triangle with side $\frac{1}{3}$ of the side of $ACE$, with area $\frac{1}{9}$ of that of $ACE$. So

where $T$ is the area of $ACE$.

2. In terms of the radius. The equilateral triangle inscribed in a circle of radius $R$ has side $R\sqrt{3}$ and area $T = \frac{3\sqrt{3}}{4}R^2$. Then

3. The inscribed square has diagonal $2R$, so its area is $\frac{(2R)^2}{2} = 2R^2 = 50$ cm².

Answer: option D (50 cm²).

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Question 6: Sections of the cube and planes

Question annulled by ITA. According to the official answer key, statement III admits two readings with opposite verdicts; that is why the question was annulled.

Statement

Consider the following statements:

• I. A cube has a plane section that is a pentagon.
• II. Let $\alpha$, $\beta$ and $\eta$ be three distinct planes, pairwise intersecting. Let $r$ be the line of intersection of $\alpha$ and $\beta$, $s$ the line of intersection of $\beta$ and $\eta$, and $t$ the line of intersection of $\alpha$ and $\eta$. If $r$, $s$ and $t$ are pairwise distinct parallel lines, then there is a unique line parallel to $r$ that is equidistant from $\alpha$, $\beta$ and $\eta$.
• III. Two intersecting planes are perpendicular if every line of one of them is perpendicular to the other plane.

Which is (are) TRUE:

• A) Only I.
• B) Only I and II.
• C) Only I and III.
• D) Only II and III.
• E) None of the statements.

Comment

I. True. A plane that cuts five of the six faces of the cube produces a pentagon. For example, in the cube $[0, 1]^3$, the plane $x + y + 2z = 2.5$ cuts the faces $x = 1$, $y = 1$, $z = 1$, $x = 0$ and $y = 0$ (and not the face $z = 0$, where it would need $x + y = 2.5$).

II. False. The three planes form an “infinite triangular prism”. In a cross-section perpendicular to $r$, the planes become the three lines containing the sides of a triangle, and a line parallel to $r$ becomes a point. The points equidistant from the three lines are the incentre and the three excentres: there are four such lines, not one.

III. Ambiguous. If “every line of one of them is perpendicular to the other plane” is read literally, the hypothesis never holds (the line of intersection lies in both planes), and the statement would be vacuously true, or false as a criterion, depending on the reading. With III true the answer would be C; with III false, A. This ambiguity is what led to the annulment.

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Question 7: Rotating the vector 31 times

Statement

Consider the matrices

The matrix $A^{31} \times B$ is equal to

• A) $\begin{bmatrix} -\frac{\sqrt{2}}{4}(\sqrt{3} - 1) \\ -\frac{\sqrt{2}}{4}(1 + \sqrt{3}) \end{bmatrix}$.
• B) $\begin{bmatrix} -\frac{\sqrt{2}}{4}(1 - \sqrt{3}) \\ -\frac{\sqrt{2}}{4}(1 + \sqrt{3}) \end{bmatrix}$.
• C) $\begin{bmatrix} \frac{1 - \sqrt{6}}{2\sqrt{2}} \\ \frac{1 + \sqrt{6}}{2\sqrt{2}} \end{bmatrix}$.
• D) $\begin{bmatrix} \frac{\sqrt{6} - 1}{2\sqrt{2}} \\ \frac{1 + \sqrt{6}}{2\sqrt{2}} \end{bmatrix}$.
• E) none of the previous options.

Solution

1. Rotation. $A = \begin{bmatrix} \cos 30^\circ & -\sin30^\circ \\ \sin30^\circ & \cos 30^\circ \end{bmatrix}$ rotates vectors by $30^\circ$. So $A^{31}$ rotates by $31\cdot 30^\circ = 930^\circ$, which is the same as $930^\circ - 720^\circ = 210^\circ$.

2. The vector. $B = (\cos 45^\circ, \sin45^\circ)$. Rotating by $210^\circ$, it reaches $(\cos 255^\circ, \sin255^\circ)$.

3. Values. $255^\circ = 180^\circ + 75^\circ$, so

Answer: option A.

To review, see Addition and subtraction formulas.

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Question 8: A polynomial symmetric about the point (2, 0)

Statement

Let $p(x) = x^3 + bx^2 + cx + d$ be a polynomial with real coefficients. If all the roots of $p(x)$ are real and, for all $x \in \mathbb{R}$,

then the smallest possible value of $p(0)$ is

• A) $-10$.
• B) $-8$.
• C) $-6$.
• D) $-4$.
• E) $-2$.

Solution

1. The symmetry. With $q(t) = p(2 + t)$, the condition says $q(t) = -q(-t)$: $q$ is odd. A monic odd polynomial of degree 3 is $q(t) = t^3 + kt$. Hence

2. Real roots. $x = 2$ is a root, and the others come from $(x - 2)^2 = -k$: they are real exactly when $k \le 0$.

3. The value of $p(0)$.

With $k \le 0$, $-2k \ge 0$, and $p(0) \ge -8$, with equality for $k = 0$, that is, $p(x) = (x - 2)^3$ (triple root 2).

Answer: option B ($-8$).

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Question 9: Triangle ABV

Statement

Consider the line $r: 3x + 4y = -15$ and the parabola $P: y = x^2 + x + 6$ with vertex $V$. Let $t$ be the line tangent to $P$ that has negative slope and makes an angle of $45^\circ$ with $r$. If $A$ is the point of tangency of $t$ with $P$ and $B$ is the intersection point of $r$ and $t$, the area of triangle $ABV$ is

• A) $\frac{13}{500}$.
• B) $\frac{25}{4}$.
• C) $\frac{105}{8}$.
• D) $\frac{147}{8}$.
• E) $\frac{126}{5}$.

Solution

1. The vertex. $x_V = -\frac{1}{2}$, $y_V = \frac{1}{4} - \frac{1}{2} + 6 = \frac{23}{4}$.

2. The slope of $t$. Line $r$ has slope $-\frac{3}{4}$. An angle of $45^\circ$ with $r$ requires

Since $m < 0$, $m = -7$.

3. The point of tangency. The derivative of the parabola is $2x + 1 = -7 \Rightarrow x = -4$, and $y = 16 - 4 + 6 = 18$: $A = (-4, 18)$. The tangent is $y = -7x - 10$.

4. The point $B$. $3x + 4(-7x - 10) = -15 \Rightarrow -25x = 25 \Rightarrow x = -1$, $y = -3$: $B = (-1, -3)$.

5. Area. With $\overrightarrow{BA} = (-3, 21)$ and $\overrightarrow{BV} = \left(\frac{1}{2}, \frac{35}{4}\right)$:

Checking the tangency: $x^2 + x + 6 = -7x - 10 \Rightarrow x^2 + 8x + 16 = (x + 4)^2 = 0$, a double root ✔

Answer: option D ($\frac{147}{8}$).

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Question 10: A system with tangents and secants

Statement

Let $x, y \in \left]0, \pi/2\right[$ satisfy the system of equations

The product of all values of $x$ and $y$ that solve this system is

• A) $\frac{\pi^2}{24}$.
• B) $\frac{\pi^4}{324}$.
• C) $\frac{\pi^2}{50}$.
• D) $\frac{\pi^2}{16}$.
• E) $\frac{\pi^2}{18}$.

Solution

1. The 2nd equation. With $u = \frac{\sin y}{\cos x} > 0$: $u + \frac{1}{u} = 2 \iff (u - 1)^2 = 0 \iff u = 1$. So $\sin y = \cos x$ and, on the given interval, $y = \frac{\pi}{2} - x$.

2. Simplifying the 1st. Using $\tan\frac{y}{2} = \frac{1 - \cos y}{\sin y}$:

and the equation becomes $\frac{1 - \cos y}{\cos y} = \frac{\cos 2x}{\cos y}$, that is, $1 - \cos y = \cos 2x$.

3. Putting them together. With $\cos y = \sin x$ and $\cos 2x = 1 - 2\sin^2 x$:

On the interval, $\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}$ and $y = \frac{\pi}{3}$.

4. The product. $x\cdot y = \frac{\pi}{6}\cdot\frac{\pi}{3} = \frac{\pi^2}{18}$.

Checking: $\tan 60^\circ\tan 30^\circ = 1$ and $\cos 60^\circ\sec 60^\circ = 1$ ✔; $\frac{\sin60^\circ}{\cos 30^\circ} = 1$, and $1 + 1 = 2$ ✔

Answer: option E ($\frac{\pi^2}{18}$).

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Question 11: Bijections with order constraints

Statement

Let $A = \{1, 2, 3, 4, 5, 6, 7, 8\}$. The number of bijections $F: A \to A$ that satisfy $F(1) < F(5) < F(3)$ and $F(7) < F(2)$ is

• A) 3360.
• B) 4200.
• C) 5040.
• D) 5880.
• E) 6720.

Solution

1. Total. There are $8! = 40320$ bijections from $A$ to $A$.

2. By symmetry. The three values $F(1)$, $F(5)$, $F(3)$ are distinct; of the $3! = 6$ possible orders among them, only one is $F(1) < F(5) < F(3)$. Swapping values among these three positions, the bijections split into groups of 6, and in each group exactly one satisfies the condition. The same goes for $F(7) < F(2)$, with factor 2. Since the positions $\{1, 5, 3\}$ and $\{7, 2\}$ are disjoint, the two factors combine:

3. By direct counting. Choose the 3 values for positions 1, 5, 3 ($\binom{8}{3} = 56$, a single order), the 2 values for positions 7, 2 among the remaining 5 ($\binom{5}{2} = 10$, a single order) and distribute the remaining 3 values ($3! = 6$): $56\cdot 10\cdot 6 = 3360$ ✔

Answer: option A (3360).

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Question 12: Tangent circles with radii in geometric progression

Statement

Consider a positive $m \in \mathbb{R}$. The sequence $r_k$ is an increasing geometric progression of positive terms with ratio $q$ and first term $r_1 = q$. The circles

are pairwise externally tangent, in this order. The expression of $q$ in terms of $m$ is

• A) $\frac{\sqrt{m^2 + 1} + 1}{m}$.
• B) $\left(\frac{\sqrt{m^2 + 1} + 1}{m}\right)^2$.
• C) $\frac{\sqrt{m^2 + 1} - 1}{m}$.
• D) $\left(\frac{\sqrt{m^2 + 1} - 1}{m}\right)^2$.
• E) $\frac{m^2 + 1}{m}$.

Solution

1. The centres. $C_k$ has centre $(r_k, mr_k)$, on the line $y = mx$, and radius $r_k$. The distance between the centres of $C_k$ and $C_{k+1}$ is

2. External tangency. Distance between centres = sum of the radii:

Dividing by $r_k$ (and using $r_{k+1} = q\,r_k$): $(q - 1)\sqrt{1 + m^2} = q + 1$. The condition does not depend on $k$, so it holds for all consecutive pairs.

3. Isolating $q$. With $s = \sqrt{1 + m^2}$:

Checking with $m = \frac{4}{3}$: $s = \frac{5}{3}$, $q = \frac{8/3}{2/3} = 4$, and the formula gives $\left(\frac{8/3}{4/3}\right)^2 = 4$ ✔

Answer: option B.

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