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

The 1st phase of ITA 2024 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. $\mathbb{N} = \{1, 2, 3, \dots\}$; $A - B$ is the difference of the sets $A$ and $B$; $i^2 = -1$; $M_n(\mathbb{R})$ 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 $\overline{AB}$; $A\hat{B}C$ is the angle with vertex $B$; $\det A$ is the determinant and $A^T$ the transpose of $A$.

Answer key

Question Topic Answer
37 Set operations B
38 System of matrix equations E
39 Double angle with arctangent A
40 Counting: the pigeonhole principle B
41 Polynomials: sum of the roots of 120 polynomials D
42 Roots in geometric progression C
43 Cylinder and geometric progression E
44 Polyhedra: Euler’s relation D
45 Plane geometry: area of a quadrilateral C
46 Complex numbers: $a^m = b^m$ D
47 Analytic geometry: external angle bisector A
48 Ellipse containing a circle C

Question 37: Three identities with sets

Statement

Let $A, B, C \subseteq \mathbb{R}$ with $C \subseteq A$. Consider the statements:

  • I. $(A \cap B) \cup C = A \cap (B \cup C)$.
  • II. $A \cap B = C \cup (B \cap (\mathbb{R} - C))$.
  • III. $A \cap (B - C) = (A \cap B) - C$.

Which is (are) TRUE:

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

Solution

I. True. By distributivity, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. Since $C \subseteq A$, $A \cap C = C$, and we get $(A \cap B) \cup C$.

II. False. The right-hand side is $C \cup (B - C) = B \cup C$. Counterexample: $A = \mathbb{R}$, $B = \varnothing$, $C = \{1\}$. Then $A \cap B = \varnothing$, but $B \cup C = \{1\}$.

III. True. Both sides are the set of elements that are in $A$ and in $B$ and not in $C$: $A \cap B \cap (\mathbb{R} - C)$. (This holds even without the hypothesis $C \subseteq A$.)

Answer: option B (only I and III).

To review, see Operations with sets.

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Question 38: A system whose unknowns are matrices

Statement

Let $A, B, C, D \in M_n(\mathbb{R})$. Consider the linear system

$$\begin{cases} AX = B \\ DX + Y = C \end{cases}$$

in the variables $X, Y \in M_n(\mathbb{R})$. Consider the statements:

  • I. If $\det A = 0$ or $\det D = 0$, then the system is inconsistent.
  • II. If $A = B$, then the system has a unique solution.
  • III. The system has a unique solution only if $A$ and $D$ are invertible.

Which is (are) TRUE:

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

Solution

The key observation: the 2nd equation always determines $Y = C - DX$. So the system has as many solutions as the equation $AX = B$.

I. False. With $A = I$ (invertible) and $D = 0$ ($\det D = 0$), the solution $X = B$, $Y = C$ exists.

II. False. If $A = B = 0$, then $AX = B$ holds for every $X$: infinitely many solutions.

III. False. Uniqueness depends only on $A$ being invertible; $D$ can be anything. For example, $A = I$ and $D = 0$ give the unique solution $X = B$, $Y = C$.

Answer: option E (none).

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Question 39: Cosine and sine of 2·arctan(4/3)

Statement

Find the value of

$$\cos\left(2\arctan\left(\frac{4}{3}\right)\right) + \sin\left(2\arctan\left(\frac{4}{3}\right)\right).$$
  • A) $\frac{17}{25}$.
  • B) $\frac{4}{5}$.
  • C) $\frac{24}{25}$.
  • D) $\frac{28}{25}$.
  • E) $\frac{31}{25}$.

Solution

Let $\theta = \arctan\frac{4}{3}$, an acute angle with $\tan\theta = \frac{4}{3}$. In the 3-4-5 right triangle: $\sin\theta = \frac{4}{5}$ and $\cos\theta = \frac{3}{5}$.

$$\cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{9 - 16}{25} = -\frac{7}{25}, \qquad \sin 2\theta = 2\cdot\frac{4}{5}\cdot\frac{3}{5} = \frac{24}{25}$$$$\cos 2\theta + \sin 2\theta = \frac{-7 + 24}{25} = \frac{17}{25}$$

Answer: option A ($\frac{17}{25}$).

Watch out for the trap: forgetting the sign of $\cos 2\theta$ (which is negative, since $2\theta > 90^\circ$) leads to $\frac{31}{25}$, option E.

To review, see Sine, cosine and tangent of the double and triple angle.

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Question 40: Guaranteeing a pair with product 256

Statement

Consider the set:

$$A = \{1, 2, 4, 8, 16, 32, 64, 128, 256\}.$$

What is the smallest $n \in \mathbb{N}$ such that every subset of $A$ with $n$ elements contains at least one pair whose product is 256?

  • A) $n = 5$.
  • B) $n = 6$.
  • C) $n = 7$.
  • D) $n = 8$.
  • E) $n = 9$.

Solution

1. Exponents. The elements are $2^0, 2^1, \dots, 2^8$, and $2^a \cdot 2^b = 256 = 2^8$ when $a + b = 8$. With distinct elements, the pairs are

$$\{1, 256\}, \quad \{2, 128\}, \quad \{4, 64\}, \quad \{8, 32\}$$

and 16 is left alone ($16 \cdot 16 = 256$, but it is the same element).

2. The largest subset “without a pair”. We can take one element from each pair plus 16: for example, $\{1, 2, 4, 8, 16\}$, with 5 elements and no product 256.

3. Pigeonhole. With 6 elements, removing at most the 16, at least 5 elements are left, spread over only 4 pairs; two of them fall into the same pair, and their product is 256.

Answer: option B ($n = 6$).

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Question 41: Sum of the roots of all the polynomials

Statement

Consider the set $C = \{1, 2, 3, 4, 5\}$. For each possible choice of $a_0, a_1, a_2, a_3, a_4 \in C$, pairwise distinct, we form the polynomial

$$a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4.$$

The sum of the roots, counted with multiplicity, of all the polynomials formed in this process is equal to:

  • A) $-\frac{17125}{4}$.
  • B) $-1800$.
  • C) $-360$.
  • D) $-\frac{351}{2}$.
  • E) $-\frac{101}{4}$.

Solution

1. One polynomial. There are $5! = 120$ polynomials, all of degree 4. By Vieta, the sum of the roots of each one is $-\dfrac{a_3}{a_4}$.

2. Grouping. The value $-\frac{a_3}{a_4}$ depends only on the ordered pair $(a_3, a_4)$ of distinct elements, and each pair appears in $3! = 6$ polynomials (the orders of $a_0, a_1, a_2$). Hence

$$S = -6\sum_{a_3 \ne a_4}\frac{a_3}{a_4}$$

3. The double sum. For fixed $a_4$, $a_3$ runs over the other 4 elements, which add up to $15 - a_4$:

$$\sum_{a_3 \ne a_4}\frac{a_3}{a_4} = \sum_{a_4 = 1}^{5}\frac{15 - a_4}{a_4} = 15\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right) - 5 = 15\cdot\frac{137}{60} - 5 = \frac{137}{4} - 5 = \frac{117}{4}$$

4. Total.

$$S = -6\cdot\frac{117}{4} = -\frac{351}{2}$$

(We checked by computer, adding $-a_3/a_4$ over the 120 permutations.)

Answer: option D ($-\frac{351}{2}$).

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Question 42: Roots in geometric progression

Statement

The value of $k \in \mathbb{R}$ for which the roots of the polynomial $p(x) = x^3 + 3x^2 - 6x + k$ are in geometric progression is:

  • A) $-18$.
  • B) $-16$.
  • C) $-8$.
  • D) $-2$.
  • E) $-1$.

Solution

1. Roots in geometric progression. Write the roots as $\frac{a}{q}$, $a$, $aq$. By Vieta:

$$a\left(\frac{1}{q} + 1 + q\right) = -3, \qquad a^2\left(\frac{1}{q} + 1 + q\right) = -6, \qquad a^3 = -k$$

2. Dividing the 2nd relation by the 1st: $a = 2$. So $k = -a^3 = -8$.

Checking: $p(x) = x^3 + 3x^2 - 6x - 8 = (x - 2)(x^2 + 5x + 4) = (x - 2)(x + 1)(x + 4)$. The roots $-1$, $2$, $-4$ form a geometric progression with ratio $-2$ ✔

Answer: option C ($-8$).

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Question 43: Base, lateral surface and volume in geometric progression

Statement

Consider a right circular cylinder such that the area of its base $A_1$, the area of its lateral surface $A_2$ and its volume $A_3$ form, in this order, an increasing geometric progression. The radius of the base may lie in the interval:

  • A) $\left(1, \frac{5}{4}\right)$.
  • B) $\left(\frac{5}{4}, \frac{3}{2}\right)$.
  • C) $\left(\frac{3}{2}, \frac{7}{4}\right)$.
  • D) $\left(\frac{7}{4}, 2\right)$.
  • E) $\left(2, \frac{5}{2}\right)$.

Solution

With radius $r$ and height $h$: $A_1 = \pi r^2$, $A_2 = 2\pi r h$, $A_3 = \pi r^2 h$.

1. Geometric progression condition. $A_2^2 = A_1 A_3$:

$$4\pi^2 r^2 h^2 = \pi^2 r^4 h \quad\Rightarrow\quad h = \frac{r^2}{4}$$

2. Ratio. $q = \frac{A_2}{A_1} = \frac{2h}{r} = \frac{r}{2}$ (and also $\frac{A_3}{A_2} = \frac{r}{2}$ ✔). An increasing progression of positive terms requires $q > 1$:

$$\frac{r}{2} > 1 \quad\Rightarrow\quad r > 2$$

Any $r > 2$ works (with $h = \frac{r^2}{4}$). Among the options, only the interval $\left(2, \frac{5}{2}\right)$ lies within $r > 2$.

Answer: option E.

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Question 44: The polyhedron with 24 vertices and 36 edges

Statement

A convex polyhedron has 24 vertices and 36 edges. We know that each vertex joins 3 faces and that the number of edges of each face can take only one of two values, $m$ or $n$. It is CORRECT to state that:

  • A) it is possible that $m = 3$ and $n = 4$.
  • B) it is possible that $m = 3$ and $n = 5$.
  • C) it is possible that $m = 3$ and $n = 7$.
  • D) it is possible that $m = 3$ and $n = 8$.
  • E) it is possible that $m = 4$ and $n = 5$.

Solution

1. Number of faces. Euler: $V - E + F = 2 \Rightarrow F = 2 - 24 + 36 = 14$.

2. Counting edges through the faces. If there are $x$ faces with $m$ edges and $y$ with $n$ edges, each edge is counted by two faces:

$$x + y = 14, \qquad mx + ny = 2\cdot 36 = 72$$

3. Testing.

$(m, n)$ equation for $y$ $y$ possible?
$(3, 4)$ $42 + y = 72$ $30$ no ($y > 14$)
$(3, 5)$ $42 + 2y = 72$ $15$ no ($y > 14$)
$(3, 7)$ $42 + 4y = 72$ $7.5$ no
$(3, 8)$ $42 + 5y = 72$ $6$ yes: 8 triangles and 6 octagons
$(4, 5)$ $56 + y = 72$ $16$ no ($y > 14$)

The case $(3, 8)$ really exists: it is the truncated cube (cut off the 8 corners of a cube). It has 8 triangular faces, 6 octagonal faces, 24 vertices and 36 edges, and 3 faces meet at each vertex.

Answer: option D.

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Question 45: The area of quadrilateral APMS

Statement

Consider a triangle $ABC$ and let $M$ be the midpoint of side $\overline{BC}$. Take the point $R \ne A$ on line $AB$ such that $m(\overline{AB}) = m(\overline{BR})$ and the point $Q$ on line $AC$ such that $m(\overline{AC}) = 2\,m(\overline{CQ})$ and $Q$ is not on segment $\overline{AC}$. Line $RM$ meets side $\overline{AC}$ at $S$ and line $QM$ meets side $\overline{AB}$ at $P$. If the area of triangle $ABC$ is 24, the area of quadrilateral $APMS$ is:

  • A) 15.
  • B) 16.
  • C) 17.
  • D) 18.
  • E) 19.

Solution

1. Vectors from $A$. Put $A$ at the origin and write $\vec{b} = \overrightarrow{AB}$, $\vec{c} = \overrightarrow{AC}$. Then

  • $R$ is the reflection of $A$ through $B$: $R = 2\vec{b}$;
  • $Q$ is beyond $C$, half the length $AC$ further: $Q = \frac{3}{2}\vec{c}$;
  • $M = \frac{1}{2}(\vec{b} + \vec{c})$.

2. The point $S$. On line $RM$: $2\vec{b} + t\left(M - R\right) = \left(2 - \frac{3t}{2}\right)\vec{b} + \frac{t}{2}\vec{c}$. It lies on line $AC$ when the coefficient of $\vec{b}$ is zero: $t = \frac{4}{3}$, and $S = \frac{2}{3}\vec{c}$.

3. The point $P$. On line $QM$: $\frac{3}{2}\vec{c} + s\left(M - Q\right) = \frac{s}{2}\vec{b} + \left(\frac{3}{2} - s\right)\vec{c}$. It lies on $AB$ when $s = \frac{3}{2}$, and $P = \frac{3}{4}\vec{b}$.

4. Area from two triangles. Split $APMS$ along the diagonal $AM$. Recall that $S_{ABM} = S_{AMC} = 12$ (the median halves the area):

  • $S_{APM} = \frac{AP}{AB}\cdot S_{ABM} = \frac{3}{4}\cdot 12 = 9$;
  • $S_{AMS} = \frac{AS}{AC}\cdot S_{AMC} = \frac{2}{3}\cdot 12 = 8$.
$$S_{APMS} = 9 + 8 = 17$$

Checking with the cross-product (shoelace) formula: the area is $\frac{1}{2}\left|P \times M + M \times S\right| = \frac{1}{2}\left(\frac{3}{8} + \frac{1}{3}\right)|\vec{b} \times \vec{c}| = \frac{17}{48}\cdot 48 = 17$ ✔ (with $|\vec{b} \times \vec{c}| = 2\cdot 24$).

Answer: option C (17).

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Question 46: When is aᵐ = bᵐ?

Statement

Let $a = 1 + 3\sqrt{3}\,i$ and $b = 2\sqrt{3} + 4i$ be complex numbers. The smallest value $m \in \mathbb{N}$ such that $a^m = b^m$ is:

  • A) 6.
  • B) 8.
  • C) 10.
  • D) 12.
  • E) there is no $m \in \mathbb{N}$ satisfying this equality.

Solution

1. Same modulus. $|a|^2 = 1 + 27 = 28$ and $|b|^2 = 12 + 16 = 28$.

2. The quotient. $a^m = b^m \iff \left(\frac{a}{b}\right)^m = 1$. Computing:

$$\frac{a}{b} = \frac{(1 + 3\sqrt{3}\,i)(2\sqrt{3} - 4i)}{28} = \frac{2\sqrt{3} - 4i + 18i + 12\sqrt{3}}{28} = \frac{14\sqrt{3} + 14i}{28} = \frac{\sqrt{3}}{2} + \frac{1}{2}i = \operatorname{cis}\frac{\pi}{6}$$

3. The smallest power. $\operatorname{cis}\frac{m\pi}{6} = 1 \iff \frac{m\pi}{6}$ is a multiple of $2\pi$ $\iff m$ is a multiple of 12. The smallest is $m = 12$.

Answer: option D (12).

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Question 47: Perpendicular to the bisector through B

Statement

Consider the triangle with vertices $A = (0, 0)$, $B = (\sqrt{2}, \sqrt{3})$ and $C = \left(\frac{5}{2}\sqrt{2}, 0\right)$. The equation of the line through $B$ perpendicular to the bisector of angle $A\hat{B}C$ is:

  • A) $y = (5 - 2\sqrt{6})x + 5\sqrt{3} - 5\sqrt{2}$.
  • B) $y = (5 + 2\sqrt{6})x - 3\sqrt{3} - 5\sqrt{2}$.
  • C) $y = (5 - \sqrt{6})x + 3\sqrt{3} - 5\sqrt{2}$.
  • D) $y = -(5 + 2\sqrt{6})x + 5\sqrt{3} + 5\sqrt{2}$.
  • E) $y = (-5 + 2\sqrt{6})x - 3\sqrt{3} + 5\sqrt{2}$.

Solution

The line through $B$ perpendicular to the internal bisector is the external bisector at $B$. If $\vec{u}$ and $\vec{v}$ are the unit vectors from $B$ to $A$ and from $B$ to $C$, the internal bisector has direction $\vec{u} + \vec{v}$ and the external one, $\vec{u} - \vec{v}$ (the two are perpendicular because $|\vec{u}| = |\vec{v}|$).

1. Unit vectors.

  • $\overrightarrow{BA} = (-\sqrt{2}, -\sqrt{3})$, with length $\sqrt{5}$: $\vec{u} = \frac{1}{\sqrt{5}}(-\sqrt{2}, -\sqrt{3})$.
  • $\overrightarrow{BC} = \left(\frac{3\sqrt{2}}{2}, -\sqrt{3}\right)$, with length $\sqrt{\frac{9}{2} + 3} = \sqrt{\frac{15}{2}}$: $\vec{v} = \frac{1}{\sqrt{15}}(3, -\sqrt{6})$.

2. Direction of the external bisector. Multiplying $\vec{u} - \vec{v}$ by $\sqrt{15}$:

$$\sqrt{15}(\vec{u} - \vec{v}) = (-\sqrt{6}, -3) - (3, -\sqrt{6}) = (-3 - \sqrt{6},\ \sqrt{6} - 3)$$

3. Slope.

$$m = \frac{\sqrt{6} - 3}{-(3 + \sqrt{6})} = \frac{3 - \sqrt{6}}{3 + \sqrt{6}} = \frac{(3 - \sqrt{6})^2}{9 - 6} = \frac{15 - 6\sqrt{6}}{3} = 5 - 2\sqrt{6}$$

4. The line through $B$.

$$y = (5 - 2\sqrt{6})(x - \sqrt{2}) + \sqrt{3} = (5 - 2\sqrt{6})x - 5\sqrt{2} + 2\sqrt{12} + \sqrt{3} = (5 - 2\sqrt{6})x + 5\sqrt{3} - 5\sqrt{2}$$

(we used $2\sqrt{6}\cdot\sqrt{2} = 2\sqrt{12} = 4\sqrt{3}$).

Checking: by computer, the dot product of $\vec{u} - \vec{v}$ and $\vec{u} + \vec{v}$ is 0 and the slope is $\approx 0.101 = 5 - 2\sqrt{6}$ ✔

Answer: option A.

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Question 48: The most elongated ellipse that contains the circle

Statement

Consider the ellipse given by the equation

$$\lambda x^2 + (\lambda + 4)y^2 - 4\lambda x - (10\lambda + 40)y + 25(\lambda + 4) - \lambda^2 = 0,$$

and the circle with equation

$$x^2 + y^2 - 4x - 12y + 36 = 0.$$

With the interior of the circle entirely contained in the interior of the ellipse, the value of $\lambda \in \mathbb{R} - \{-4, 0\}$ for which the eccentricity of the ellipse is maximum is:

  • A) 3.
  • B) 5.
  • C) 9.
  • D) 13.
  • E) 15.

Solution

1. Standard form of the ellipse. Completing squares ($x^2 - 4x = (x - 2)^2 - 4$ and $y^2 - 10y = (y - 5)^2 - 25$):

$$\lambda(x - 2)^2 + (\lambda + 4)(y - 5)^2 - 4\lambda - 25(\lambda + 4) + 25(\lambda + 4) - \lambda^2 = 0$$$$\lambda(x - 2)^2 + (\lambda + 4)(y - 5)^2 = \lambda(\lambda + 4) \quad\Rightarrow\quad \frac{(x - 2)^2}{\lambda + 4} + \frac{(y - 5)^2}{\lambda} = 1$$

It is an ellipse for $\lambda > 0$: centre $(2, 5)$, horizontal semi-axis $\sqrt{\lambda + 4}$ (the larger one) and vertical semi-axis $\sqrt{\lambda}$.

2. Eccentricity. $e^2 = 1 - \frac{\lambda}{\lambda + 4} = \frac{4}{\lambda + 4}$: the eccentricity is largest when $\lambda$ is as small as possible.

3. The circle. $(x - 2)^2 + (y - 6)^2 = 4$: centre $(2, 6)$, radius 2. It runs from $y = 4$ to $y = 8$.

4. Containment condition. The highest point of the circle, $(2, 8)$, must be on or inside the ellipse:

$$\frac{(8 - 5)^2}{\lambda} \le 1 \quad\Rightarrow\quad \lambda \ge 9$$

For $\lambda = 9$, the whole circle lies inside: a point of the circle is $(2 + 2\cos t,\ 6 + 2\sin t)$, and with $s = \sin t$,

$$\frac{4\cos^2 t}{13} + \frac{(1 + 2s)^2}{9} = \frac{4(1 - s^2)}{13} + \frac{(1 + 2s)^2}{9}$$

is increasing for $s \in [-1, 1]$ (the derivative $-\frac{8s}{13} + \frac{4 + 8s}{9}$ vanishes only at $s \approx -1.6$) and reaches its maximum value 1 at $s = 1$, that is, at the point $(2, 8)$, where the circle touches the ellipse from inside. So the smallest $\lambda$ is 9.

Answer: option C ($\lambda = 9$).

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