ITA 2022 – 2nd phase: Mathematics solved

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

The 2nd phase Mathematics exam of ITA 2022 has 10 written questions. ITA does not publish official solutions for the 2nd phase; the solutions below are ours, and each answer was checked by a second route (substitution, another formula or a numerical check).

The notation conventions are the same as in the 1st phase: $i^2 = -1$, $m(\overline{AB})$ is the length of segment $\overline{AB}$ and $A\hat{O}B$ is the angle with vertex $O$.

Answers

Question 1: The sides of the rectangle with perimeter 8

Statement

Let $x \in \mathbb{R}$. Consider a rectangle $R$ with sides measuring $a = 9x^2 - 5x^4$ and $b = 8x - 8x^3$. Given that the perimeter of $R$ is 8, find $a$ and $b$.

Solution

1. The equation. Perimeter 8 means $a + b = 4$:

2. Factoring. Trying rational roots, $x = 1$ and $x = -1$ make the polynomial vanish ($5 + 8 - 9 - 8 + 4 = 0$ and $5 - 8 - 9 + 8 + 4 = 0$). Dividing by $x^2 - 1$:

The roots are $x \in \left\{1, -1, -2, \frac{2}{5}\right\}$.

3. Sides must be positive.

For $x = \frac{2}{5}$: $a = 9\cdot\frac{4}{25} - 5\cdot\frac{16}{625} = \frac{36}{25} - \frac{16}{125} = \frac{164}{125}$ and $b = \frac{16}{5} - 8\cdot\frac{8}{125} = \frac{400 - 64}{125} = \frac{336}{125}$.

Checking: $a + b = \frac{500}{125} = 4$ ✔

Answer: $a = \dfrac{164}{125}$ and $b = \dfrac{336}{125}$.

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Question 2: Two zero imaginary parts

Statement

Let $z \in \mathbb{C}$ and denote by $\operatorname{Im}(z)$ the imaginary part of $z$. Find all possible $z \in \mathbb{C}$ with $\operatorname{Im}(z) \ne 0$ such that simultaneously $\operatorname{Im}(z^3) = 0$ and $\operatorname{Im}\left((1 + z)^3\right) = 0$.

Solution

1. Imaginary part of a cube. For $w = u + vi$,

so $\operatorname{Im}(w^3) = v(3u^2 - v^2)$.

2. The two conditions. With $z = x + yi$, $y \ne 0$, and $1 + z = (1 + x) + yi$:

3. Solving. $x^2 = (1 + x)^2 \Rightarrow 0 = 1 + 2x \Rightarrow x = -\frac{1}{2}$, and $y^2 = \frac{3}{4}$, $y = \pm\frac{\sqrt{3}}{2}$.

Checking geometrically: $z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \operatorname{cis} 120^\circ$, so $z^3 = \operatorname{cis} 360^\circ = 1$; and $1 + z = \frac{1}{2} + \frac{\sqrt{3}}{2}i = \operatorname{cis} 60^\circ$, so $(1 + z)^3 = \operatorname{cis} 180^\circ = -1$. Both real ✔

Answer: $z = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ or $z = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$.

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Question 3: Sum of the entries of a recursive matrix

Statement

Let $A$ be the matrix with 5 rows and 10 columns whose entries $a_{n,m}$ are given by

Find the sum of all the entries of $A$.

Solution

1. Closed formula. In row $n$, we start at 1 and add $n$ at each column: it is an arithmetic progression with common difference $n$. Hence

2. Summing row $n$. With $m$ from 1 to 10:

3. Summing the rows.

Checking with row 1: $1, 2, \dots, 10$ adds up to 55, and the formula gives $10 + 45 = 55$ ✔

Answer: 725.

To review, see Arithmetic progression.

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Question 4: Winning tic-tac-toe on the third move

Statement

In tic-tac-toe, two players compete on an ordered board with 3 rows and 3 columns. The players take turns marking an unoccupied cell until one of them occupies a whole row, column or diagonal and is declared the winner. How many different board configurations correspond to a win by the first player on their third move?

Solution

At the first player’s (X’s) third move, 5 marks have been made: 3 by X and 2 by the second player (O).

1. X’s cells. X wins exactly on the 3rd move if their 3 cells form a row, column or diagonal. There are $3 + 3 + 2 = 8$ choices.

2. O’s cells. O’s 2 cells are any two of the remaining 6: $\binom{6}{2} = 15$. The game cannot have ended earlier: X cannot win with fewer than 3 cells, and O, with only 2 cells, cannot either.

3. Total.

A configuration is the final picture of the board, not the order of the moves; that is why we do not multiply by the possible orders. (We checked by computer, simulating every game: there are exactly 120 distinct final boards with X winning on the 5th mark.)

Answer: 120 configurations.

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Question 5: Arccosine and arcsine adding up to π/2

Statement

Consider $\arccos: [-1, 1] \to [0, \pi]$ and $\arcsin: [-1, 1] \to [-\pi/2, \pi/2]$. Find all values of $\arccos(x)$ given that $x$ satisfies

Solution

1. The key identity. For $u \in [-1, 1]$, $\arcsin u + \arccos u = \frac{\pi}{2}$. So $\arcsin\left(x^2 - \frac14\right) = \frac{\pi}{2} - \arccos\left(x^2 - \frac14\right)$ and the equation becomes

2. Arccosine is injective. So $x^4 = x^2 - \frac{1}{4}$, that is,

Both arguments equal $\frac{1}{4}$, inside $[-1, 1]$ ✔

3. The requested values. $\arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ and $\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$.

Answer: $\arccos(x) = \dfrac{\pi}{4}$ or $\arccos(x) = \dfrac{3\pi}{4}$.

To review, see Arcsine, arccosine and arctangent.

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Question 6: The point Q at distance √2 from line s

Statement

Let $A = (0, 1)$. Consider the line $r$ with equation $y = 1 - x/4$ and let $s$ be a line through the origin $O$ that meets $r$ in the 1st quadrant at a point $P$. Find the point $Q$ in the 2nd quadrant that lies on $r$ and is at distance $\sqrt{2}$ from $s$, given that $A\hat{P}O = \theta$ and $\tan(\theta) = \frac{5}{3}$.

Solution

1. Where $A$ is. $A = (0, 1)$ lies on $r$ (for $x = 0$, $y = 1$). So the side $PA$ of angle $\theta$ lies on $r$ and the side $PO$ lies on $s$.

2. Finding $s$. Let $s: y = kx$. At $P$, the vectors pointing to $A$ and to $O$ are parallel to $u = (-1, \frac{1}{4})$ (decreasing $x$ along $r$) and $v = (-1, -k)$. The tangent of the angle between them is

Setting it equal to $\frac{5}{3}$: $3k + \frac{3}{4} = 5 - \frac{5k}{4} \Rightarrow \frac{17k}{4} = \frac{17}{4} \Rightarrow k = 1$. So $s: y = x$ and $P = \left(\frac{4}{5}, \frac{4}{5}\right)$, in the 1st quadrant ✔

Checking with angles: $s$ makes $45^\circ$ with the $x$-axis and $r$ has slope $-\frac14$; the angle at $P$ is $45^\circ + \arctan\frac14$ and

3. The point $Q$. A point of $r$ is $\left(x, 1 - \frac{x}{4}\right)$. Its distance to $s: x - y = 0$ is

So $x = \frac{12}{5}$ or $x = -\frac{4}{5}$. In the 2nd quadrant $x < 0$, so $x = -\frac{4}{5}$ and $y = 1 + \frac{1}{5} = \frac{6}{5} > 0$ ✔

Answer: $Q = \left(-\dfrac{4}{5}, \dfrac{6}{5}\right)$.

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Question 7: Volume of the frustum of a hexagonal pyramid

Statement

Consider $T$ a frustum of a regular pyramid with height $h = 4 + 2\sqrt{3}$ and parallel hexagonal bases. Given that the side of the larger hexagonal base measures $8\sqrt{3}/3$ and that the dihedral angle between the lateral faces and the base of the frustum measures $75^\circ$, find the volume of $T$.

Solution

1. Apothems of the bases. The apothem of a regular hexagon with side $\ell$ is $\frac{\ell\sqrt{3}}{2}$. In the larger base:

2. The dihedral angle. A cut through the axis, perpendicular to a pair of sides, shows a right trapezoid with height $h$ and horizontal “step” $m_1 - m_2$; the angle between the lateral face and the base is an angle of this trapezoid, so

Since $\tan 75^\circ = \tan(45^\circ + 30^\circ) = 2 + \sqrt{3}$ and $h = 2(2 + \sqrt{3})$:

The smaller base has apothem 2, half that of the larger one: its side is $\frac{4\sqrt{3}}{3}$.

3. Areas of the bases. A regular hexagon with side $\ell$ has area $\frac{3\sqrt{3}}{2}\ell^2$:

4. Volume of the frustum.

Checking as a difference of pyramids: since the smaller base has half the side, the whole pyramid has height $2h$ and the frustum is $1 - \frac{1}{8} = \frac{7}{8}$ of it: $\frac{7}{8}\cdot\frac{1}{3}\cdot 32\sqrt{3}\cdot 2h = \frac{56\sqrt{3}\,h}{3}$, the same value ✔

Answer: $V = 112 + \dfrac{224\sqrt{3}}{3} = \dfrac{336 + 224\sqrt{3}}{3}$.

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Question 8: Radius of the cyclic quadrilateral

Statement

Let $Q$ be a quadrilateral with vertices $A$, $B$, $C$ and $D$ whose sides satisfy $m(\overline{AB}) = 5 = m(\overline{CD})$, $m(\overline{BC}) = 3$ and $m(\overline{DA}) = 8$. Given that $Q$ is inscribed in a circle of radius $r$, find $r$.

Solution

1. Opposite angles are supplementary. In a cyclic quadrilateral, $\hat{B} + \hat{D} = 180^\circ$, so $\cos\hat{B} = -\cos\hat{D}$.

2. The diagonal $AC$ from two triangles. Law of cosines in $ABC$ and in $ACD$:

Equating: $110\cos\hat{D} = 55$, so $\cos\hat{D} = \frac{1}{2}$, $\hat{D} = 60^\circ$ and $AC^2 = 89 - 40 = 49$, $AC = 7$.

Checking with Ptolemy: the quadrilateral is an isosceles trapezoid ($AB = CD$), with equal diagonals, and $AC \cdot BD = AB\cdot CD + BC\cdot DA = 25 + 24 = 49$ gives $AC = 7$ ✔

3. The radius. The circle of $Q$ is the same as that of triangle $ACD$. By the law of sines,

Answer: $r = \dfrac{7\sqrt{3}}{3}$.

To review, see Laws of sines and cosines, projections and areas.

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Question 9: Three circles and a horizontal tangent

Statement

Let $P_1 = (0, 6)$, $P_2 = (1, 5)$ and $P_3 = (2, 6)$, and let $C_1$, $C_2$ and $C_3$ be circles centred at $P_1$, $P_2$ and $P_3$, respectively. Given that there is a horizontal line tangent to $C_1$, $C_2$ and $C_3$, find $C_1 \cap C_2 \cap C_3$ when it is not empty.

Solution

1. The radii. Let $y = t$ be the common tangent line. The radius of each circle is the distance from the centre to the line:

2. Points common to $C_1$ and $C_3$. Both have the same radius and centres symmetric about $x = 1$; their common points lie on that line: $(1, y)$ with $1 + (y - 6)^2 = (6 - t)^2$, that is,

3. Requiring it to be on $C_2$. The point $(1, y)$ is on $C_2$ when $|y - 5| = |5 - t|$. Squaring, with $y - 5 = 1 \pm \sqrt{(6 - t)^2 - 1}$:

Squaring again: $4(t^2 - 12t + 35) = 4t^2 - 44t + 121$, that is, $-4t = -19$, $t = \frac{19}{4}$.

4. The point. With $t = \frac{19}{4}$: $6 - t = \frac{5}{4}$, $\sqrt{\frac{25}{16} - 1} = \frac{3}{4}$, and $2t - 11 = -\frac{3}{2}$ requires the minus sign: $y = 6 - \frac{3}{4} = \frac{21}{4}$.

Checking: the distance from $\left(1, \frac{21}{4}\right)$ to $P_1$ is $\sqrt{1 + \frac{9}{16}} = \frac{5}{4} = r_1$; to $P_3$, also $\frac{5}{4} = r_3$; to $P_2$, $\frac{21}{4} - 5 = \frac{1}{4} = |5 - \frac{19}{4}| = r_2$ ✔. The line $y = \frac{19}{4}$ lies below the three centres and is tangent to all three circles.

For any other horizontal tangent line, the intersection of the three is empty.

Answer: $C_1 \cap C_2 \cap C_3 = \left\{\left(1, \dfrac{21}{4}\right)\right\}$ (with the common tangent $y = \frac{19}{4}$).

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Question 10: Octahedron, cube, octahedron…

Statement

Consider a regular octahedron with edge length $l_1$. Inscribe in this octahedron a cube whose vertices are at the centroids of the faces of the octahedron. Inside this cube inscribe a new regular octahedron with edge length $l_2$ whose vertices are at the centres of the faces of the cube. Continue this process, obtaining a sequence $l_i$ for $i \in \mathbb{N}$. Then find the ratio $l_{10}/l_1$.

Solution

1. Octahedron → cube. Place the octahedron with vertices $(\pm a, 0, 0)$, $(0, \pm a, 0)$, $(0, 0, \pm a)$; its edge is $l_1 = a\sqrt{2}$. The centroid of the face with vertices $(a,0,0)$, $(0,a,0)$, $(0,0,a)$ is $\left(\frac{a}{3}, \frac{a}{3}, \frac{a}{3}\right)$; the eight centroids are $\left(\pm\frac{a}{3}, \pm\frac{a}{3}, \pm\frac{a}{3}\right)$, the vertices of a cube with edge

2. Cube → octahedron. The centres of the faces of the cube are at distance $\frac{c}{2}$ from the centre; the new octahedron has “$a$” equal to $\frac{c}{2}$, so its edge is

3. The sequence. Each step divides the edge by 3: $l_i = \frac{l_1}{3^{i-1}}$ is a geometric progression with ratio $\frac{1}{3}$. Therefore

Answer: $\dfrac{l_{10}}{l_1} = \dfrac{1}{3^9} = \dfrac{1}{19683}$.

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