ITA 2022 – 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 2022 had 70 multiple-choice questions: 15 in Physics, 15 in Portuguese, 10 in English, 15 in Mathematics and 15 in Chemistry. This post covers the mathematics questions, numbered 41 to 55 as in the exam.

Exam conventions. $\mathbb{N} = \{1, 2, 3, \dots\}$; $i$ is the imaginary unit ($i^2 = -1$); $M_n(\mathbb{R})$ is the set of $n \times n$ real matrices; $\overline{AB}$ is the segment with endpoints $A$ and $B$, and $m(\overline{AB})$ is its length; $A\hat{O}B$ is the angle with vertex $O$ formed by the rays $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Answer key

Question 41: A sum of logarithms in bases 120 and 14400

Statement

If

we can state that

• A) $x = 2$.
• B) $x = 3$.
• C) $x = 4$.
• D) $x = 5$.
• E) $x = 6$.

Solution

1. A single base. Note that $14400 = 120^2$. By the change of base,

So $2\log_{14400} 125 = 3\log_{120} 5$.

2. Combining the terms. Moving the coefficients into the exponents,

3. Spotting a cube. $2^9 \cdot 3^3 \cdot 5^3 = (2^3 \cdot 3 \cdot 5)^3 = 120^3$. Hence

Answer: option B ($x = 3$).

To review, see Properties of logarithms.

Back to the answer key

Question 42: A point equidistant from A and C

Statement

Consider a triangle with vertices $A$, $B$ and $C$, right-angled at $B$. Let $r$ be the line through $A$ and $C$, and let $O$ be a point equidistant from $A$ and $C$ on the same side as $B$ with respect to $r$. Given that $m(\overline{AO}) = 85$, $m(\overline{AB}) = 10$ and $m(\overline{BC}) = 24$, the distance from $O$ to $r$ is

• A) 64.
• B) 66.
• C) 74.
• D) 76.
• E) 84.

Solution

1. The hypotenuse. In the triangle with the right angle at $B$,

2. Where $O$ lies. The points equidistant from $A$ and $C$ form the perpendicular bisector of $\overline{AC}$. It meets $r$ perpendicularly at the midpoint $M$ of $\overline{AC}$. So the distance from $O$ to $r$ is $m(\overline{OM})$, and triangle $AMO$ is right-angled at $M$, with $m(\overline{AM}) = 13$.

3. Pythagoras again.

The phrase “on the same side as $B$” only picks one of two symmetric points; the distance is the same.

Answer: option E (84).

To review, see The Pythagorean theorem.

Back to the answer key

Question 43: When are two systems equivalent?

Statement

Let $m \in \mathbb{R}$. Consider the linear systems

Choose the correct option:

• A) There is no $m \in \mathbb{R}$ such that $S_1$ is equivalent to $S_2$.
• B) There is exactly one $m > 0$ such that $S_1$ is equivalent to $S_2$.
• C) There is exactly one $m < 0$ such that $S_1$ is equivalent to $S_2$.
• D) There are exactly two distinct values of $m$ such that $S_1$ is equivalent to $S_2$.
• E) There are infinitely many distinct values of $m$ such that $S_1$ is equivalent to $S_2$.

Solution

Two systems are equivalent when they have the same solution set. Let us solve each one in terms of $m$.

1. Solving $S_1$. From the 1st equation, $y = 4x - 2$. Substituting into the 3rd:

In the 2nd equation: $-16x + m^2(4x - 2) + 2 = -10$, that is,

• If $m = \pm 2$, this becomes $0 \cdot x = -4$: $S_1$ is inconsistent.
• If $m \ne \pm 2$, the solution is unique: $x = \dfrac{m^2 - 6}{2(m^2 - 4)}$, $y = 4x - 2$, $z = 2$.

2. Solving $S_2$. From the 2nd equation, $y = z - \frac{14}{5}$. Substituting into the 3rd:

and, from the 1st equation, $10x = m^2 + m - 1 - z = 3$, so $x = \frac{3}{10}$. Thus $S_2$ always has a unique solution (its determinant is $-700$, never zero).

3. Matching the solutions. Since $S_2$ always has exactly one solution, $S_1$ must too (so $m \ne \pm 2$), and the solutions must coincide.

• $z$: $m^2 + m - 4 = 2 \Rightarrow m^2 + m - 6 = 0 \Rightarrow m = 2$ or $m = -3$. We discard $m = 2$.
• $x$ with $m = -3$: $\dfrac{9 - 6}{2(9 - 4)} = \dfrac{3}{10}$ ✔
• $y$: in $S_1$, $4 \cdot \frac{3}{10} - 2 = -\frac{4}{5}$; in $S_2$, $2 - \frac{14}{5} = -\frac{4}{5}$ ✔

Only $m = -3$ works: both systems have the solution $\left(\frac{3}{10}, -\frac{4}{5}, 2\right)$.

Answer: option C (exactly one $m < 0$, namely $m = -3$).

Watch out for the trap: the equation for $z$ gives two values, $2$ and $-3$, which leads to option D. But with $m = 2$ the system $S_1$ has no solution, while $S_2$ does.

Back to the answer key

Question 44: Real sum, difference, product and quotient

Statement

Let $z_1, z_2 \in \mathbb{C}$ with $z_2 \ne 0$. Consider the statements:

• I. If $z_1 + z_2 \in \mathbb{R}$ and $z_1 - z_2 \in \mathbb{R}$ then $z_1 \in \mathbb{R}$ and $z_2 \in \mathbb{R}$.
• II. If $z_1 \cdot z_2 \in \mathbb{R}$ and $z_1/z_2 \in \mathbb{R}$ then $z_1 \in \mathbb{R}$ and $z_2 \in \mathbb{R}$.
• III. If $z_1 + z_2 \in \mathbb{R}$ and $z_1 \cdot z_2 \in \mathbb{R}$ then $z_1 \in \mathbb{R}$ and $z_2 \in \mathbb{R}$.

Which is (are) always true:

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

Solution

I. True. If $s = z_1 + z_2$ and $d = z_1 - z_2$ are real, then

II. False. Counterexample: $z_1 = z_2 = i$. Then $z_1 z_2 = -1$ and $z_1/z_2 = 1$ are real, but $i \notin \mathbb{R}$.

III. False. Counterexample: $z_1 = i$, $z_2 = -i$. Then $z_1 + z_2 = 0$ and $z_1 z_2 = 1$ are real. (In general, $z_1$ and $z_2$ are the roots of $t^2 - st + p = 0$ with real $s$, $p$, and these roots may be complex conjugates.)

Answer: option A (only I).

Back to the answer key

Question 45: The quadrilateral of the roots

Statement

Consider the polynomial $p(z) = z^4 - 6z^3 + 14z^2 - 6z + 13$ and note that $p(i) = 0$. Consider, in the complex plane, the quadrilateral whose vertices are the roots of $p(z)$. The area of this quadrilateral is

• A) 4.
• B) 6.
• C) 8.
• D) 9.
• E) 10.

Solution

1. A second root for free. The coefficients are real, so the conjugate $-i$ is also a root. Hence $(z - i)(z + i) = z^2 + 1$ divides $p(z)$:

(check by multiplying: $z^4 - 6z^3 + 13z^2 + z^2 - 6z + 13$.)

2. The other roots. $z^2 - 6z + 13 = 0 \Rightarrow z = \dfrac{6 \pm \sqrt{36 - 52}}{2} = 3 \pm 2i$.

3. The figure. The vertices are $(0, 1)$, $(0, -1)$, $(3, 2)$ and $(3, -2)$. It is a trapezoid with vertical parallel sides of lengths $2$ (at $x = 0$) and $4$ (at $x = 3$) and height $3$:

Answer: option D (9).

Back to the answer key

Question 46: Three statements about matrices

Statement

Let $n \ge 2$ and $A, B \in M_n(\mathbb{R})$. Consider the following statements:

• I. If $AB \ne BA$ then either $A$ or $B$ is not invertible.
• II. If $AB = 0$ then $BA = 0$.
• III. If $A^T = -A^2$ and $A$ is invertible then $\det(A) = -1$.

Which is (are) true:

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

Solution

I. False. Two invertible matrices may fail to commute. For example,

and both have determinant 1.

II. False. With

III. False. Taking determinants in $A^T = -A^2$ (recall that $\det(A^T) = \det A$ and $\det(-M) = (-1)^n \det M$):

(we divided by $\det A \ne 0$). For even $n$, $\det A = 1$. And such a matrix exists: the rotation by $-60^\circ$,

satisfies $A^T = R(60^\circ)$ and $-A^2 = -R(-120^\circ) = R(60^\circ)$, with $\det A = 1$.

Answer: option E (none of the statements).

Back to the answer key

Question 47: Tangents in arithmetic progression

Statement

Let $x, r \in \mathbb{R}$ and suppose that

About $\tan(x - r)$, $\tan(x)$ and $\tan(x + r)$, in this order, we can state that:

• A) They never form an arithmetic progression.
• B) They can form an arithmetic progression only if $r = 0$.
• C) They can form an arithmetic progression only if $r = 0$ or $r = \sqrt{3}/3$.
• D) They can form an arithmetic progression for infinitely many distinct values of $r$.
• E) They form an arithmetic progression for every $x$ and $r$ as in the statement.

Solution

Three numbers are in arithmetic progression (AP) when the middle one is the mean of the other two: $\tan(x - r) + \tan(x + r) = 2\tan x$.

1. Adding the tangents. With $\tan a + \tan b = \dfrac{\sin(a + b)}{\cos a \cos b}$:

Since $\sin 2x = 2\sin x\cos x$, this is equivalent to

2. The two cases. Using $\cos(x - r)\cos(x + r) = \cos^2 x - \sin^2 r$, the bracket equals $\sin^2 r$. So the condition is

• If $\sin x = 0$, that is, $x = 0$ (the only possible value in the interval), any $r$ with $0 \le r < \frac{\pi}{2}$ works: the numbers $\tan(-r)$, $0$, $\tan r$ are in AP.
• If $x \ne 0$, we need $r = 0$ (a constant AP).

So infinitely many values of $r$ produce an AP (all with $x = 0$). But not always: with $x = \frac{\pi}{6}$ and $r = \frac{\pi}{6}$, for example, we get $0, \frac{\sqrt{3}}{3}, \sqrt{3}$, which is not an AP.

Answer: option D.

To review, see Addition and subtraction formulas.

Back to the answer key

Question 48: Where is the centre of the hyperbola?

Statement

Let $b \in \mathbb{R}$ be such that the equation

defines a hyperbola. About the centre $C$ of this hyperbola we can state:

• A) $C \in \{(x, y) \in \mathbb{R}^2 \mid x^2/9 + y^2/12 < 1\}$.
• B) $C \in \{(x, y) \in \mathbb{R}^2 \mid x^2/4 + y^2/2 > 1\}$.
• C) $C \in \{(x, y) \in \mathbb{R}^2 \mid x^2/9 - y^2/2 < 1\}$.
• D) $C \in \{(x, y) \in \mathbb{R}^2 \mid 3x^2 - 2y^2 > 1\}$.
• E) None of the previous options.

Solution

1. Completing the squares. $x^2 - 6bx = (x - 3b)^2 - 9b^2$ and $y^2 - 2by = (y - b)^2 - b^2$. Substituting:

The constant terms add up to $-9b^2 + b^2 - b^4 + b^4 + 8b^2 - 1 = -1$. We get

2. When it is a hyperbola. The coefficient of $(y - b)^2$ must be positive: $1 - b^2 > 0$, that is, $-1 < b < 1$. The centre is $C = (3b, b)$.

3. Testing the options with $x = 3b$, $y = b$ and $0 \le b^2 < 1$:

• A) $b^2 + \frac{b^2}{12} = \frac{13}{12}b^2 < 1$ fails for $b^2 \ge \frac{12}{13}$ (for example, $b = 0.99$).
• B) $\frac{9b^2}{4} + \frac{b^2}{2} = \frac{11}{4}b^2 > 1$ fails for $b = 0$.
• C) $\frac{9b^2}{9} - \frac{b^2}{2} = \frac{b^2}{2} < \frac{1}{2} < 1$ ✔ always holds.
• D) $27b^2 - 2b^2 = 25b^2 > 1$ fails for $b = 0$.

Answer: option C.

Back to the answer key

Question 49: The hexagonal pyramid inside the cube

Statement

Let $P$ be a regular pyramid whose apex $V$ is one of the vertices of a cube with edge $l$ and whose base is the hexagon formed by the midpoints of the six edges of the cube that contain neither $V$ nor the vertex opposite to $V$. The radius of the sphere circumscribed about $P$ is

• A) $l\sqrt{2}/12$.
• B) $l\sqrt{3}/12$.
• C) $5l\sqrt{2}/12$.
• D) $5l\sqrt{3}/12$.
• E) $l\sqrt{3}/6$.

Solution

1. Coordinates. Place the cube in $[0, l]^3$ with $V = (0, 0, 0)$; the opposite vertex is $(l, l, l)$. The six edges that touch neither of them have midpoints such as $\left(l, \frac{l}{2}, 0\right)$, $\left(0, l, \frac{l}{2}\right)$ etc. All of them satisfy $x + y + z = \frac{3l}{2}$: the hexagon is the cross-section of the cube by the plane perpendicular to the diagonal through the centre $O = \left(\frac{l}{2}, \frac{l}{2}, \frac{l}{2}\right)$.

2. Measurements of the pyramid.

• Side of the hexagon: the distance between midpoints of neighbouring edges, $\frac{l\sqrt{2}}{2}$. In a regular hexagon the circumradius equals the side: $\rho = \frac{l\sqrt{2}}{2}$.
• Height: $h = m(\overline{VO}) = \frac{l\sqrt{3}}{2}$ (half the diagonal of the cube).

3. The sphere. The centre of the circumscribed sphere lies on the axis $VO$, at a distance $R$ from $V$. It is also at distance $R$ from a vertex of the hexagon:

Note that $R \approx 0.72\,l < h \approx 0.87\,l$: the centre lies between $V$ and the base, as expected.

Answer: option D ($5l\sqrt{3}/12$).

Back to the answer key

Question 50: Lines, planes and equidistant points

Statement

Consider the following statements:

• I. If $\alpha$ and $\beta$ are distinct parallel planes and $r$ is a line such that $r \cap \alpha \ne \varnothing$ then $r \cap \beta \ne \varnothing$.
• II. If $r$ is a line and $P$ and $Q$ are distinct points, then there are infinitely many planes equidistant from $P$ and $Q$ that contain $r$.
• III. Given four points in space, there is a unique point equidistant from them.

Which is (are) true:

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

Solution

I. False. If $r$ is contained in $\alpha$, then $r \cap \alpha \ne \varnothing$, but $r$ is parallel to $\beta$ and does not meet it.

II. False. A plane is equidistant from $P$ and $Q$ when it is parallel to the line $PQ$ or passes through the midpoint $M$ of $\overline{PQ}$. Among the planes containing $r$, in general at most one passes through $M$ and at most one is parallel to $PQ$. For example, take $r$ to be the $Oz$ axis, $P = (1, 0, 0)$ and $Q = (0, 2, 0)$: the only plane through the $Oz$ axis that passes through $M = \left(\frac12, 1, 0\right)$ is $y = 2x$, and the only one parallel to $PQ$ is $y = -2x$. Just two, not infinitely many.

III. False. Four coplanar points that do not lie on a common circle have no equidistant point (for example, the vertices of a parallelogram that is not a rectangle). And if they do lie on a circle there are infinitely many (every point on the axis of the circle).

Answer: option A (none of the statements).

Back to the answer key

Question 51: When k + 1 gains a binary digit

Statement

We say that the binary representation of a number $N \in \mathbb{N}$ of the form

is $(abcdefg)_2$, where $a, b, c, d, e, f, g \in \{0, 1\}$ and the digits 0 are omitted up to the first digit 1 from left to right. Let $k$ be an integer with $1 \le k \le 100$. What is the probability that $k$ and $k + 1$ have binary representations with a different number of digits?

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

Solution

1. When the number of digits changes. A number with $n$ binary digits lies between $2^{n-1}$ and $2^n - 1$. Adding 1 increases the number of digits only when $k = 2^n - 1$ (for example, $111_2 + 1 = 1000_2$), that is, when $k + 1$ is a power of 2.

2. Counting. With $2 \le k + 1 \le 101$: $k + 1 \in \{2, 4, 8, 16, 32, 64\}$, six cases ($k = 1, 3, 7, 15, 31, 63$).

Answer: option C (6%).

Back to the answer key

Question 52: Two lines through the vertices of a cube

Statement

Let $A$ be the set of all lines that pass through two distinct vertices of a cube $C$. Choosing two distinct lines of $A$ at random, the probability that these lines intersect at a vertex of $C$ is:

• A) $4/9$.
• B) $1/2$.
• C) $2/3$.
• D) $1/14$.
• E) $3/7$.

Solution

1. How many lines. No three vertices of the cube are collinear, so each pair of vertices gives a different line: $\binom{8}{2} = 28$ lines.

2. Possible outcomes. Pairs of distinct lines: $\binom{28}{2} = 378$.

3. Favourable outcomes. Through each vertex pass 7 lines (one to each of the other vertices), forming $\binom{7}{2} = 21$ pairs that meet at that vertex. Two distinct lines have at most one common point, so no pair is counted twice: $8 \cdot 21 = 168$.

Answer: option A ($4/9$).

To review, see Introduction to probability.

Back to the answer key

Question 53: An inequality of cosines in a triangle

Statement

Let $\alpha$, $\beta$ and $\theta$ be the interior angles of a triangle. If $\cos(\beta + \theta) \le \cos(\alpha + 2\beta)$, we can state that:

• A) The triangle is not isosceles.
• B) The triangle is not right-angled.
• C) The triangle is not acute.
• D) The triangle is not obtuse.
• E) None of the previous items can be guaranteed.

Solution

1. Rewriting with $\alpha + \beta + \theta = \pi$.

• $\beta + \theta = \pi - \alpha$, so $\cos(\beta + \theta) = -\cos\alpha$.
• $\alpha + 2\beta = \pi - (\theta - \beta)$, so $\cos(\alpha + 2\beta) = -\cos(\theta - \beta)$.

The hypothesis becomes

2. Comparing angles. The angles $\alpha$ and $|\theta - \beta|$ lie in $[0, \pi)$, where cosine is decreasing. Hence $\alpha \le |\theta - \beta|$.

3. Conclusion. Say $\theta \ge \beta$ (the other case is symmetric). Then $\alpha + \beta \le \theta$, and since $\alpha + \beta = \pi - \theta$,

The triangle has a right or obtuse angle, that is, it is not acute. The other options can fail: the angles $\theta = 90^\circ$, $\alpha = \beta = 45^\circ$ satisfy the hypothesis (with equality) and give an isosceles right triangle; and $\theta = 120^\circ$, $\alpha = \beta = 30^\circ$ gives an obtuse one.

Answer: option C.

Back to the answer key

Question 54: An equation in powers of cosine

Statement

The number of distinct real solutions of the equation

in the interval $[0, 2\pi[$ is

• A) 2.
• B) 3.
• C) 4.
• D) 5.
• E) 6.

Solution

1. A single variable. Let $c = \cos^2 x \in [0, 1]$. Since $\cos 2x = 2\cos^2 x - 1 = 2c - 1$:

2. Factoring. $c = -1$ is a root ($-1 + 4 - 1 - 2 = 0$), so

3. Only one root works. With $\sqrt{17} \approx 4.12$: $c = \frac{-3 + \sqrt{17}}{2} \approx 0.56$ lies in $[0, 1]$; the others ($-1$ and $\approx -3.56$) are negative.

4. Counting the $x$. $\cos^2 x = 0.56$ gives $\cos x = \pm 0.75$ (approximately), and each value of $\cos x$ strictly between $-1$ and $1$ corresponds to two angles in $[0, 2\pi[$. Total: $2 \cdot 2 = 4$ solutions.

Answer: option C (4).

To review, see Intervals and classical trigonometric equations.

Back to the answer key

Question 55: An inscribed triangle with an acute angle at B

Statement

Let $T$ be a triangle with vertices $A$, $B$ and $C$ with $m(\overline{AB}) = 2\sqrt{5}$ and $m(\overline{BC}) = 6$. Given that $A\hat{B}C$ is acute and $T$ is inscribed in a circle of radius $R = 5$, we can state that:

• A) $m(\overline{AC}) = \sqrt{5}/5$.
• B) $m(\overline{AC}) = 2\sqrt{5}/5$.
• C) $m(\overline{AC}) = 4\sqrt{5}/5$.
• D) $m(\overline{AC}) = 8\sqrt{5}/5$.
• E) $m(\overline{AC}) = 14\sqrt{5}/5$.

Solution

Write $\hat{A}$, $\hat{B}$, $\hat{C}$ for the interior angles.

1. Law of sines. $\dfrac{a}{\sin\hat{A}} = 2R = 10$, with $a = m(\overline{BC})$ etc.:

2. Which cosine? Since $m(\overline{AB}) < m(\overline{BC})$, we have $\hat{C} < \hat{A}$, so $\hat{C}$ is acute: $\cos\hat{C} = \frac{2\sqrt{5}}{5}$. But $\hat{A}$ may be acute ($\cos\hat{A} = \frac{4}{5}$) or obtuse ($\cos\hat{A} = -\frac{4}{5}$).

3. Using that $\hat{B}$ is acute. Since $\hat{B} = \pi - (\hat{A} + \hat{C})$, $\cos\hat{B} = \sin\hat{A}\sin\hat{C} - \cos\hat{A}\cos\hat{C}$.

• If $\cos\hat{A} = \frac{4}{5}$: $\cos\hat{B} = \frac{3\sqrt{5}}{25} - \frac{8\sqrt{5}}{25} = -\frac{\sqrt{5}}{5} < 0$. $\hat{B}$ would be obtuse: discarded.
• If $\cos\hat{A} = -\frac{4}{5}$: $\cos\hat{B} = \frac{3\sqrt{5}}{25} + \frac{8\sqrt{5}}{25} = \frac{11\sqrt{5}}{25} > 0$ ✔

4. The side $AC$. $\sin\hat{B} = \sin(\hat{A} + \hat{C}) = \frac{3}{5}\cdot\frac{2\sqrt{5}}{5} - \frac{4}{5}\cdot\frac{\sqrt{5}}{5} = \frac{2\sqrt{5}}{25}$, and

Checking with the law of cosines: $b^2 = 20 + 36 - 2\cdot 2\sqrt{5}\cdot 6 \cdot \frac{11\sqrt{5}}{25} = 56 - \frac{1320}{25} = \frac{16}{5}$, and $\sqrt{16/5} = \frac{4\sqrt{5}}{5}$ ✔

Answer: option C ($4\sqrt{5}/5$).

Watch out for the trap: choosing $\hat{A}$ acute gives $m(\overline{AC}) = 4\sqrt{5}$, which is not even among the options, but in that case the angle $\hat{B}$ would be obtuse.

To review, see Solving triangles and the ambiguous case.

Back to the answer key