ITA 2025 – 2nd phase: Mathematics solved

Source: ITA – 2nd phase Mathematics exam. Files on the ITA past exams page.

The 2nd phase Mathematics exam of ITA 2025 has 10 written questions. ITA does not publish official solutions for the written part (the 2nd phase answer key only covers the multiple-choice Portuguese questions); the solutions below are ours, and each answer was checked by a second route (substitution, another formula or a computer check).

Answers

Question 1: The remainder of the division by x² − x + 1

Statement

Find the real values $a$ and $b$ such that the polynomial $p(x) = x^{57} + ax^{14} + bx^7 + 1$, when divided by $x^2 - x + 1$, leaves remainder $2x + 1$.

Solution

1. The roots of the divisor. Let $\omega$ be a root of $x^2 - x + 1$. Since $(x + 1)(x^2 - x + 1) = x^3 + 1$, we have $\omega^3 = -1$ and $\omega^6 = 1$; moreover, $\omega^2 = \omega - 1$.

2. Value of $p(\omega)$. If $p(x) = (x^2 - x + 1)Q(x) + 2x + 1$, then $p(\omega) = 2\omega + 1$. Reducing the exponents modulo 6:

• $\omega^{57} = \omega^{54}\cdot\omega^3 = -1$;
• $\omega^{14} = \omega^{12}\cdot\omega^2 = \omega - 1$;
• $\omega^7 = \omega$.

3. Matching. $(a + b)\omega - a = 2\omega + 1$. Since $\omega$ is not real and $a$, $b$ are real, the coefficients must match: $a + b = 2$ and $-a = 1$. So $a = -1$ and $b = 3$. (The other root, $\bar{\omega}$, gives the same equations.)

Checking by computer: the remainder of $x^{57} - x^{14} + 3x^7 + 1$ divided by $x^2 - x + 1$ is $2x + 1$ ✔

Answer: $a = -1$ and $b = 3$.

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Question 2: For which r is the maximum of y/x equal to 1?

Statement

Let $E$ be an ellipse with focal axis on the axis $O_x$ of the Cartesian coordinate system. The centre of $E$ is the point $(r, 0)$, with $r > 0$, its eccentricity is $\frac{\sqrt{2}}{2}$, and its semi-major axis measures $\sqrt{2}$. Considering the points $(x, y) \in E$, find the value of $r$ for which $\dfrac{y}{x}$ has maximum value 1.

Solution

1. The ellipse. $a = \sqrt{2}$ and $e = \frac{c}{a} = \frac{\sqrt{2}}{2}$ give $c = 1$ and $b^2 = a^2 - c^2 = 1$:

2. Interpretation. $\frac{y}{x}$ is the slope of the line joining the origin to the point $(x, y)$. The maximum occurs on the line $y = kx$ tangent to the ellipse (and it only exists if the origin is outside it, that is, $r > \sqrt{2}$).

3. Tangency. Substituting $y = kx$:

Zero discriminant: $r^2 = 4\left(\frac{1}{2} + k^2\right)\left(\frac{r^2}{2} - 1\right) = (1 + 2k^2)(r^2 - 2)$, that is,

4. Maximum equal to 1. $k = 1 \Rightarrow r^2 - 2 = 1 \Rightarrow r = \sqrt{3}$ (and indeed $\sqrt{3} > \sqrt{2}$).

Checking numerically: going around the ellipse with $r = \sqrt{3}$, the largest value of $y/x$ found is $0.99999\ldots$ ✔

Answer: $r = \sqrt{3}$.

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Question 3: sin(α + β) from two equations

Statement

Let $\alpha, \beta \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ be such that

Compute the value of $\sin(\alpha + \beta)$.

Solution

1. Isolating $\beta$. Subtracting the 1st equation from the 2nd:

2. Finding $\beta$. Squaring, $1 - 2\sin\beta\cos\beta = \frac{1}{4}$, so $\sin\beta\cos\beta = \frac{3}{8} > 0$. On the given interval, $\cos\beta \le 0$; for the product to be positive, $\sin\beta < 0$ as well. With $s = \sin\beta$ and $\cos\beta = s + \frac{1}{2}$:

Since $s < 0$: $\sin\beta = \frac{-1 - \sqrt{7}}{4}$ and $\cos\beta = \frac{1 - \sqrt{7}}{4}$ (both negative ✔).

3. Finding $\alpha$. $\sin\alpha = \sin\beta + \frac{1}{4} = -\frac{\sqrt{7}}{4}$. On the given interval, $\cos\alpha \le 0$: $\cos\alpha = -\sqrt{1 - \frac{7}{16}} = -\frac{3}{4}$.

4. The sine of the sum.

Checking numerically: $\alpha \approx 3.864$ and $\beta \approx 4.288$ (both in $[\frac{\pi}{2}, \frac{3\pi}{2}] \approx [1.571,\ 4.712]$) satisfy both equations, and $\sin(\alpha + \beta) \approx 0.9557 = \frac{5 + \sqrt{7}}{8}$ ✔

Answer: $\sin(\alpha + \beta) = \dfrac{5 + \sqrt{7}}{8}$.

To review, see Addition and subtraction formulas.

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Question 4: The excircle opposite A

Statement

Let $ABC$ be a triangle with sides $m(\overline{AB}) = 6$, $m(\overline{AC}) = 10$ and $m(\overline{BC}) = 14$. Compute the radius of the circle outside triangle $ABC$ that is tangent simultaneously to the segment $\overline{BC}$ and to the lines $AB$ and $AC$.

Solution

This is the excircle opposite $A$. Its radius is $r_a = \dfrac{S}{p - a}$, where $S$ is the area, $p$ the semiperimeter and $a = m(\overline{BC})$.

1. The angle at $A$. By the law of cosines,

2. Area. $S = \frac{1}{2}\cdot 6\cdot 10\cdot\sin120^\circ = 15\sqrt{3}$.

3. The radius. $p = \frac{6 + 10 + 14}{2} = 15$ and $p - a = 1$:

Checking with another formula: the centre of the excircle lies on the bisector of $\hat{A}$, and $r_a = p\cdot\tan\frac{\hat{A}}{2} = 15\tan60^\circ = 15\sqrt{3}$ ✔ (Heron also confirms the area: $\sqrt{15\cdot 1\cdot 9\cdot 5} = \sqrt{675} = 15\sqrt{3}$.)

Answer: $15\sqrt{3}$.

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Question 5: The first digit of 3¹⁰⁰

Statement

Using the approximations $\log_{10} 2 = 0.3010$, $\log_{10} 3 = 0.4771$ and $\log_{10} 7 = 0.8450$, find the first digit (from left to right) of the number $3^{100}$.

Solution

1. Characteristic and mantissa.

So $3^{100} = 10^{47}\cdot 10^{0.71}$, and the first digit is the integer part of $10^{0.71}$.

2. Comparing with known logarithms.

Since $0.6990 < 0.71 < 0.7781$, we have $5 < 10^{0.71} < 6$.

Checking: $3^{100} = 515377520732011331036461129765621272702107522001$, which indeed starts with 5 (and has 48 digits, as the characteristic 47 says) ✔

Answer: the first digit is 5.

To review, see Introduction to the table of logarithms.

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Question 6: When is the fourth head most likely?

Statement

A fair coin is tossed $n$ times. Find the values of $n$ that maximise the probability that the fourth head appears exactly on the $n$-th toss.

Solution

1. The probability. The fourth head appears on toss $n$ when the $n$-th toss is heads and exactly 3 heads appear in the previous $n - 1$ tosses:

2. Comparing neighbours.

This ratio is $> 1$ for $n < 6$, $= 1$ for $n = 6$ and $< 1$ for $n > 6$. So $P$ grows up to $n = 6$, ties at $P(6) = P(7)$ and then decreases.

3. The values.

and, for example, $P(5) = \frac{4}{32} = \frac{8}{64}$ and $P(8) = \frac{35}{256} \approx \frac{8.75}{64}$, which are smaller.

Answer: $n = 6$ and $n = 7$ (maximum probability $\frac{5}{32}$).

To review, see Introduction to probability.

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Question 7: A double root and a right triangle

Statement

Consider the polynomial $p(x) = x^3 + ax^2 + b$. Find the real values $a$ and $b$, given that:

• I. $p(x)$ has a real double root;
• II. The points $(x_1, 0)$, $(x_2, 0)$ and $(0, b)$ are the vertices of a right triangle, where $x_1$ and $x_2$ are distinct roots of $p(x)$.

Solution

1. The double root. A double root $r$ makes both $p$ and $p'(x) = 3x^2 + 2ax = x(3x + 2a)$ vanish.

• If $r = 0$, then $b = 0$, and the point $(0, b) = (0, 0)$ lies on the $x$-axis together with $(x_1, 0)$ and $(x_2, 0)$: there is no triangle. Discarded.
• So $r = -\frac{2a}{3}$ (with $a \ne 0$) and $p(r) = 0$ gives $-\frac{8a^3}{27} + \frac{4a^3}{9} + b = 0$, that is, $b = -\frac{4a^3}{27}$.

2. The distinct roots. The sum of the roots is $-a$: $2r + s = -a$, so the third root is $s = \frac{a}{3}$. The distinct roots are $x_1 = -\frac{2a}{3}$ and $x_2 = \frac{a}{3}$, of opposite signs.

3. The right angle. Two vertices are on the $x$-axis and the third, $(0, b)$, on the $y$-axis. The right angle cannot be at $(x_1, 0)$ or at $(x_2, 0)$ (the side to $(0, b)$ would have to be vertical, requiring $x_1 = 0$ or $x_2 = 0$). So it is at $(0, b)$, and by the metric relation (altitude squared = product of the projections):

4. Solving. With $b = -\frac{4a^3}{27}$: $\frac{16a^6}{729} = \frac{2a^2}{9}$, so $a^4 = \frac{81}{8}$ and

And $b = -\frac{4a^3}{27} = -\frac{4a}{27}\cdot\frac{9\sqrt{2}}{4} = -\frac{\sqrt{2}}{3}a = \mp\frac{\sqrt{2}}{2}\sqrt[4]{2} = \mp\frac{1}{\sqrt[4]{2}}$.

Checking with $a = \frac{3}{2}\sqrt[4]{2} \approx 1.784$ and $b \approx -0.841$: the roots of $x^3 + ax^2 + b$ are $\approx -1.189$ (double) and $\approx 0.595$, and $b^2 \approx 0.7071 = 1.189\cdot 0.595$ ✔

Answer: $a = \dfrac{3\sqrt[4]{2}}{2}$ and $b = -\dfrac{1}{\sqrt[4]{2}}$, or $a = -\dfrac{3\sqrt[4]{2}}{2}$ and $b = \dfrac{1}{\sqrt[4]{2}}$.

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Question 8: The sum of the determinants of max{i, j}

Statement

Let $A_k = (a_{ij})$ be a square matrix of order $k$, where $a_{ij} = \max\{i, j\}$ for all $i, j$ in $\{1, 2, \dots, k\}$. Find $\displaystyle\sum_{k=1}^{2025}\det(A_k)$.

Solution

1. The matrix. For example,

2. Row reduction. Replace each row $L_i$ ($i < k$) by $L_i - L_{i+1}$ (using the original rows; the determinant does not change). Entry $(i, j)$ becomes $\max\{i, j\} - \max\{i + 1, j\}$:

• for $j \le i$: $i - (i + 1) = -1$;
• for $j \ge i + 1$: $j - j = 0$ (for $j = i+1$, $(i+1) - (i+1) = 0$).

Row $i$ becomes $(-1, \dots, -1, 0, \dots, 0)$, with $i$ entries equal to $-1$, and the last row is still $(k, k, \dots, k)$. The matrix is lower triangular, with diagonal $-1, -1, \dots, -1, k$:

(Check: $\det A_1 = 1$, $\det A_2 = 2 - 4 = -2$.)

3. Adding up.

Checking by computer: the determinants of $A_1$ to $A_6$ are $1, -2, 3, -4, 5, -6$ ✔

Answer: 1013.

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Question 9: Integer matrices whose inverse is their transpose

Statement

Find the number of invertible $5 \times 5$ matrices with integer entries that satisfy $A^{-1} = A^T$.

Solution

1. What the condition says. $A^{-1} = A^T$ is equivalent to $A^T A = I$: entry $(i, j)$ of $A^T A$ is the dot product of column $i$ with column $j$. So the columns have length 1 and are pairwise perpendicular.

2. Integer columns of length 1. The sum of the squares of the entries of each column is 1, and the entries are integers: exactly one entry is $\pm 1$ and the others are 0.

3. Perpendicularity. Two columns with their $\pm 1$ in the same row would have dot product $\pm 1 \ne 0$. So the $\pm 1$’s lie in different rows: $A$ is a signed permutation matrix.

4. Counting. There are $5! = 120$ ways to choose the row of the $\pm 1$ in each column and $2^5 = 32$ choices of signs. Every such matrix satisfies $A^T A = I$.

Answer: 3840 matrices.

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Question 10: The shadow of the cube along the diagonal

Statement

Compute the area of the orthogonal projection of a cube with edge 2 onto a plane perpendicular to one of the diagonals of the cube.

Solution

1. The shape of the shadow. Looking at the cube along the diagonal, we see three faces, and the six vertices not on the diagonal form the outline: the projection is a regular hexagon.

2. From the visible faces. The three visible faces cover the projection without overlapping. Each face has area 4 and its normal makes with the diagonal an angle $\theta$ with $\cos\theta = \frac{1}{\sqrt{3}}$ (the diagonal has direction $(1,1,1)$ and the normal, for example, $(0,0,1)$). The projected area of a plane figure is its original area times $\cos\theta$:

3. Checking with the hexagon. Each edge of the cube makes with the diagonal an angle with cosine $\frac{1}{\sqrt{3}}$, so it projects with length $2\sqrt{1 - \frac{1}{3}} = \frac{2\sqrt{6}}{3}$, which is the side of the hexagon:

Answer: $4\sqrt{3}$.

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