ITA 2026 – 2nd phase: Mathematics solved

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

The 2nd phase Mathematics exam of ITA 2026 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 01: Lattice points in the octahedron

Statement

Consider a regular octahedron centred at the origin, with three vertices at $(5, 0, 0)$, $(0, 5, 0)$ and $(0, 0, 5)$. How many points in the interior and on the surface of this solid have all integer coordinates?

Solution

1. The solid. The octahedron with vertices $(\pm 5, 0, 0)$, $(0, \pm 5, 0)$, $(0, 0, \pm 5)$ is the set $|x| + |y| + |z| \le 5$.

2. Slicing by $z$. For each integer $z$ with $|z| \le 5$, we count the integer pairs with $|x| + |y| \le k$, where $k = 5 - |z|$. This “diamond” has $2k^2 + 2k + 1$ lattice points (for example, $k = 1$: the 5 points $(0,0)$, $(\pm1, 0)$, $(0, \pm1)$).

3. Total.

Checking by computer: counting all integer $(x, y, z)$ with $|x| + |y| + |z| \le 5$, we get 231 ✔

Answer: 231 points.

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Question 02: When entry (1, 3) of Aⁿ exceeds 119

Statement

Consider

Find the smallest $n \in \mathbb{N}$ such that the entry in the first row and third column of $A^n$ is greater than 119.

Solution

1. Splitting off the identity. $A = I + N$, with

2. Binomial theorem. $I$ commutes with $N$, so

and entry $(1, 3)$ is $2n + \frac{n(n - 1)}{2} = \frac{n^2 + 3n}{2}$.

3. The inequality. $\frac{n^2 + 3n}{2} > 119 \iff n^2 + 3n > 238$. For $n = 14$: $196 + 42 = 238$, which is not greater. For $n = 15$: $225 + 45 = 270 > 238$.

Checking: entry $(1,3)$ is $119$ in $A^{14}$ and $135$ in $A^{15}$ ✔

Answer: $n = 15$.

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Question 03: The area of parallelogram ABCD

Statement

The diagonals of parallelogram $ABCD$ lie on the lines $r: 2y - x = 2$ and $s: x + 2y = 8$. The lines $r$ and $s$ meet at the point $X$. Given that $C = (8, 5)$ and that the equation of the circle inscribed in triangle $ABX$ is

find the area of $ABCD$.

Solution

1. The centre $X$. Adding the equations of $r$ and $s$: $4y = 10$, $y = \frac{5}{2}$ and $x = 3$. So $X = \left(3, \frac{5}{2}\right)$.

2. The vertex $A$. $C = (8, 5)$ lies on $r$ ($10 - 8 = 2$), so $AC$ is the diagonal on $r$. The diagonals bisect each other: $A = 2X - C = (-2, 0)$.

3. The side $AB$. $B$ is on $s$, and $AB$ is the line through $A$ tangent to the circle with centre $\left(\frac{1}{2}, \frac{5}{2}\right)$ and radius $\frac{\sqrt{5}}{2}$, other than $r$ (which is also tangent). For $y = k(x + 2)$:

So $k = \frac{1}{2}$ (the line $r$) or $k = 2$. Hence $AB: y = 2x + 4$, and with $s$: $x + 2(2x + 4) = 8 \Rightarrow x = 0$, $B = (0, 4)$.

Checking: the incentre of $A(-2, 0)$, $B(0, 4)$, $X\left(3, \frac52\right)$, computed as the mean of the vertices weighted by the opposite sides, is $\left(\frac{1}{2}, \frac{5}{2}\right)$ ✔

4. The area. The diagonals split the parallelogram into four triangles of equal area. With $\overrightarrow{AB} = (2, 4)$ and $\overrightarrow{AX} = \left(5, \frac{5}{2}\right)$:

(The vertices are $A(-2, 0)$, $B(0, 4)$, $C(8, 5)$ and $D = 2X - B = (6, 1)$.)

Answer: the area of $ABCD$ is 30.

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Question 04: How much of the target is red?

Statement

A target is painted on a wall, made of a central disc and a sequence of concentric rings alternating between red and white. The central disc, painted red, has radius $\alpha$ units. Each subsequent circle has a radius one unit larger than the previous one: $\alpha + 1$, then $\alpha + 2$ and so on. Find the total red area in terms of $\alpha$, given that the target has 33 circles.

Solution

1. The regions. The 33 circles have radii $\alpha, \alpha + 1, \dots, \alpha + 32$. The red region is the central disc and the rings between $\alpha + k$ and $\alpha + k + 1$ with odd $k$ ($k = 1, 3, \dots, 31$): 16 rings.

2. Area of a ring.

3. Adding up. With $k = 2m - 1$, $m = 1, \dots, 16$: $2\alpha + 2k + 1 = 2\alpha + 4m - 1$. So

Checking with the total: the white area is the sum of the rings with even $k$ ($k = 0, 2, \dots, 30$), $\pi(32\alpha + 496)$; red + white $= \pi(\alpha^2 + 64\alpha + 1024) = \pi(\alpha + 32)^2$, the area of the largest circle ✔

Answer: $\pi\left(\alpha^2 + 32\alpha + 528\right)$.

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Question 05: Four spheres inside a sphere

Statement

Four identical spheres, pairwise tangent, are simultaneously inscribed in a single larger sphere, so that the centres of the smaller spheres form a regular tetrahedron. Compute the ratio between the volume of the larger sphere and the sum of the volumes of the four smaller spheres.

Solution

1. The tetrahedron of the centres. Pairwise tangent spheres of radius $r$ have centres at distance $2r$: a regular tetrahedron with edge $2r$. The circumradius of a regular tetrahedron with edge $\ell$ is $\frac{\ell\sqrt{6}}{4}$; here, $\frac{2r\sqrt{6}}{4} = \frac{r\sqrt{6}}{2}$.

2. The larger sphere. It touches the small ones on their outer side, along the line from the centre of the configuration to their centres:

3. The ratio.

Since $(\sqrt{6} + 2)^3 = 6\sqrt{6} + 3\cdot 6\cdot 2 + 3\sqrt{6}\cdot 4 + 8 = 44 + 18\sqrt{6}$,

Answer: $\dfrac{22 + 9\sqrt{6}}{16}$.

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Question 06: Triangular matrices and column sums

Statement

Given the real numbers $\alpha_1, \alpha_2, \dots, \alpha_{n-1}$, define the matrices $A = (a_{ij})$ and $B = (b_{ij})$, both of order $n \times n$, by

Suppose that the sum of the entries of column $j$ of the matrix $AB$ equals $n - j$.

• a) Find the determinant of $AB$.
• b) Find the trace of $AB$ in terms of $n$.
• c) Assuming $n = 5$, find all possible values of $\alpha_1, \alpha_2, \alpha_3, \alpha_4 \in \mathbb{R}$.

Comment

Note: read literally, the statement is inconsistent: no choice of the $\alpha$’s makes the columns of $AB$ have sums $n - j$. We show why and then solve the version that makes the problem coherent (column sums of $A$). The solution published by the Poliedro school reaches the same inconsistency.

1. The columns of $A$. Column $j$ of $A$ is $(0, \dots, 0, 1, \alpha_j, \alpha_j^2, \dots, \alpha_j^{\,n-j-1})$, with the 1 in row $j + 1$. Its sum is

2. The columns of $AB$. Column $j$ of $B$ has 1 in rows $1, \dots, j$. So column $j$ of $AB$ is the sum of columns $1, \dots, j$ of $A$, and its sum is

3. The contradiction. Since $S_n = 0$, we have $c_n = c_{n-1}$. But the hypothesis requires $c_{n-1} = 1$ and $c_n = 0$. Impossible. (And even ignoring the last column, $c_1 = n - 1$ and $c_2 = n - 2$ would force $S_2 = -1$; for $n = 5$, $1 + \alpha_2 + \alpha_2^2 = -1$ has no real solution.)

a) Holds on any reading. $A$ is lower triangular with zero diagonal, so $\det A = 0$ and

Coherent version: column sums of $A$ equal to $n - j$. Suppose $S_j = n - j$ for all $j$ (for $j = n$ this says $0 = 0$, and for $j = n - 1$, $S_{n-1} = 1$, which always holds).

b) Entry $(i, i)$ of $AB$ is $\sum_{k \le i} a_{ik}$, the sum of row $i$ of $A$ (the terms with $k \ge i$ are zero). So the trace of $AB$ is the sum of all the entries of $A$:

c) For $n = 5$:

• $S_1 = 1 + \alpha_1 + \alpha_1^2 + \alpha_1^3 = 4 \Rightarrow (\alpha_1 - 1)(\alpha_1^2 + 2\alpha_1 + 3) = 0 \Rightarrow \alpha_1 = 1$ (the quadratic factor has no real root);
• $S_2 = 1 + \alpha_2 + \alpha_2^2 = 3 \Rightarrow (\alpha_2 - 1)(\alpha_2 + 2) = 0 \Rightarrow \alpha_2 = 1$ or $\alpha_2 = -2$;
• $S_3 = 1 + \alpha_3 = 2 \Rightarrow \alpha_3 = 1$;
• $S_4 = 1$ always: $\alpha_4$ only appears as $\alpha_4^0$, so it can be any real number.

Answer (on the coherent reading): a) $\det(AB) = 0$; b) $\operatorname{tr}(AB) = \frac{n(n-1)}{2}$; c) $\alpha_1 = 1$, $\alpha_2 \in \{1, -2\}$, $\alpha_3 = 1$ and $\alpha_4 \in \mathbb{R}$ arbitrary. On the literal reading, no $\alpha$’s satisfy the hypothesis; only item a) survives.

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Question 07: The roots from the remainder

Statement

Let $p(x)$ be a degree-four polynomial with real coefficients. The sum of the roots of $p(x)$ is $\frac{3}{2}$, and their product is $-\frac{5}{2}$. Find all the roots of $p(x)$ in $\mathbb{C}$, given that the remainder of the division of $p(x)$ by $x^3 - x + 4$ is $9x^2 - 4x + 7$.

Solution

1. The form of $p$. The quotient of the division by a degree-3 polynomial has degree 1:

2. Vieta. Sum of the roots: $-\frac{b}{a} = \frac{3}{2}$, so $b = -\frac{3a}{2}$. Product: $\frac{4b + 7}{a} = -\frac{5}{2}$, so $-6a + 7 = -\frac{5a}{2}$, $a = 2$ and $b = -3$.

3. Roots. Trying rational roots: $p(-1) = 2 + 3 + 7 - 7 - 5 = 0$ and $p\left(\frac{1}{2}\right) = \frac{1}{8} - \frac{3}{8} + \frac{7}{4} + \frac{7}{2} - 5 = 0$. Dividing:

and $x^2 - 2x + 5 = 0$ gives $x = 1 \pm 2i$.

Checking: sum $-1 + \frac{1}{2} + 2 = \frac{3}{2}$ ✔; product $(-1)\cdot\frac{1}{2}\cdot 5 = -\frac{5}{2}$ ✔; and the remainder of $p$ divided by $x^3 - x + 4$ is $9x^2 - 4x + 7$ ✔ (checked by computer).

Answer: $-1$, $\dfrac{1}{2}$, $1 + 2i$ and $1 - 2i$.

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Question 08: Three increasing natural numbers adding up to 100

Statement

How many strictly increasing sequences of three natural numbers can be built so that the sum of the three numbers is 100?

Solution

We want $a < b < c$ in $\mathbb{N} = \{1, 2, 3, \dots\}$ with $a + b + c = 100$.

1. Removing the strict inequality. Write $a = x$, $b = x + y$, $c = x + y + z$ with $x, y, z \ge 1$. Then

and each solution corresponds to exactly one sequence.

2. Counting. For each $x$ and $y$ with $3x + 2y \le 99$, there is a unique $z \ge 1$. For fixed $x$, $y$ goes from 1 to $\left\lfloor\frac{99 - 3x}{2}\right\rfloor$:

($x$ goes up to 32, since with $x = 33$ there is no room left for $y \ge 1$.) Grouping the terms with odd $x$ ($48, 45, \dots, 3$: 16 terms, an arithmetic progression with difference $-3$) and with even $x$ ($46, 43, \dots, 1$: 16 terms, difference $-3$):

Checking by computer: counting the triples $a < b < c$ with sum 100 directly, we get 784 ✔. (It also matches the classical formula: the number of partitions of $m$ into 3 parts is the integer nearest to $\frac{m^2}{12}$; here $m = 100 - 3 = 97$ and $\frac{97^2}{12} \approx 784.1$.)

Answer: 784 sequences.

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Question 09: Tangents to the hyperbola from (0, −8)

Statement

Let $H$ be the hyperbola with foci $A = (-6, 0)$, $B = (6, 0)$ and eccentricity 3.

• a) Find the equations of all lines through $C = (0, -8)$ that are tangent to $H$.
• b) For each line of the previous item, find the points of tangency.

Solution

1. The hyperbola. $c = 6$ and $e = \frac{c}{a} = 3$ give $a = 2$ and $b^2 = c^2 - a^2 = 32$:

2. Lines through $C$. The vertical line $x = 0$ does not meet $H$. The others are $y = mx - 8$. Substituting:

If $m^2 = 8$ (lines parallel to the asymptotes), there is a single intersection point, but the line is secant, not tangent. With $m^2 \ne 8$, tangency means zero discriminant:

3. Points of tangency. The double root is $x = \frac{-16m}{2(8 - m^2)} = \frac{-16m}{-32} = \frac{m}{2}$, and $y = m\cdot\frac{m}{2} - 8 = 12 - 8 = 4$.

• $m = 2\sqrt{6}$: point $(\sqrt{6}, 4)$.
• $m = -2\sqrt{6}$: point $(-\sqrt{6}, 4)$.

Checking: $\frac{6}{4} - \frac{16}{32} = 1$ ✔, and the tangent to $\frac{x^2}{4} - \frac{y^2}{32} = 1$ at $(\sqrt{6}, 4)$ is $\frac{\sqrt{6}\,x}{4} - \frac{4y}{32} = 1$, that is, $y = 2\sqrt{6}\,x - 8$, which passes through $(0, -8)$ ✔

Answer: a) $y = 2\sqrt{6}\,x - 8$ and $y = -2\sqrt{6}\,x - 8$. b) $(\sqrt{6}, 4)$ and $(-\sqrt{6}, 4)$, respectively.

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Question 10: Two circles and the lens between them

Statement

The points $A$ and $B$ define a segment with $m(\overline{AB}) = 6(1 + \sqrt{3})$ cm. The points $C$ and $D$ lie on segment $\overline{AB}$. The circle centred at $C$ passing through $A$ and the circle centred at $D$ passing through $B$ meet at the points $P$ and $Q$, which satisfy

• a) Find the radius of each circle.
• b) Find the area of the region common to the interiors of the two circles.

Solution

1. Coordinates. $A = (0, 0)$, $B = (L, 0)$ with $L = 6(1 + \sqrt{3})$. The centres are on the $x$-axis, so $P$ and $Q$ are symmetric: $P = (u, h)$, $Q = (u, -h)$, and $m(\overline{PQ}) = 2h$.

2. Using the conditions.

• $AP = 2h$: $u^2 + h^2 = 4h^2 \Rightarrow u = \sqrt{3}\,h$ (with $u > 0$; if $u < 0$, the centre $C$ would fall outside the segment).
• $BP = 2h\sqrt{2 + \sqrt{3}}$: $(L - u)^2 + h^2 = 4h^2(2 + \sqrt{3})$, so $(L - u)^2 = h^2(7 + 4\sqrt{3}) = h^2(2 + \sqrt{3})^2$ and $L - u = h(2 + \sqrt{3})$.

Adding: $L = h(2 + 2\sqrt{3}) = 2h(1 + \sqrt{3})$, so $h = 3$, $u = 3\sqrt{3}$ and $P = (3\sqrt{3}, 3)$.

3. a) The radii.

• Centre $C = (R_1, 0)$, through $A$ and $P$: $(3\sqrt{3} - R_1)^2 + 9 = R_1^2 \Rightarrow 36 = 6\sqrt{3}\,R_1 \Rightarrow R_1 = 2\sqrt{3}$ cm.
• Centre $D = (L - R_2, 0)$, through $B$ and $P$: with $w = L - u = 3(2 + \sqrt{3})$, $(w - R_2)^2 + 9 = R_2^2 \Rightarrow R_2 = \frac{w^2 + 9}{2w} = \frac{36(2 + \sqrt{3})}{6(2 + \sqrt{3})} = 6$ cm.

(We used $w^2 = 9(7 + 4\sqrt{3})$, so $w^2 + 9 = 36(2 + \sqrt{3})$.) The centres $C = (2\sqrt{3}, 0) \approx (3.46,\ 0)$ and $D = (6\sqrt{3}, 0) \approx (10.39,\ 0)$ are on the segment ✔

4. b) The lens. The common chord $\overline{PQ}$ has length 6 and lies on the line $x = 3\sqrt{3}$. The common region is the union of two circular segments:

• Circle centred at $C$: the chord is at distance $3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$ from the centre; the half central angle has cosine $\frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$, so the central angle is $120^\circ$. Segment: $\frac{R_1^2}{2}\left(\frac{2\pi}{3} - \sin120^\circ\right) = 6\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 4\pi - 3\sqrt{3}$.
• Circle centred at $D$: the chord is at distance $6\sqrt{3} - 3\sqrt{3} = 3\sqrt{3}$ from the centre; the half angle has cosine $\frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}$, central angle $60^\circ$. Segment: $\frac{36}{2}\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) = 6\pi - 9\sqrt{3}$.

Checking numerically: $AP = 6 = PQ$ and $BP \approx 11.591 = 6\sqrt{2 + \sqrt{3}}$ ✔; a numerical integration of the common region gives $\approx 10.632$ cm² ✔

Answer: a) $2\sqrt{3}$ cm (centre $C$) and $6$ cm (centre $D$). b) $10\pi - 12\sqrt{3}$ cm².

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