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

The 2nd phase Mathematics exam of ITA 2023 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 computer check).

Answers

Question Topic Answer
1 Homogeneous linear system $b = 2a$; $\{(-15t, -16t, 17t) : t \in \mathbb{R}\}$
2 Matrices: when $M$ is invertible $\alpha \notin \{0, 2, 3 - \sqrt{15}, 3 + \sqrt{15}\}$
3 Logarithmic inequality with a variable base $]-2, 0[$
4 Polynomial division and roots $q = x^6 + x^5 - x - 1$, $r = 0$; roots $\operatorname{cis}\frac{\pi}{5}, \operatorname{cis}\frac{3\pi}{5}, \operatorname{cis}\frac{7\pi}{5}, \operatorname{cis}\frac{9\pi}{5}$
5 Trigonometry: sum to product $\sin(\alpha - \beta) = \frac{2AB}{A^2 + B^2}$
6 Radius of the incircle $\frac{6\sqrt{7} - 3\sqrt{3}}{10}$
7 Ellipse: a 60° angle between the focal radii $\left(\pm\frac{4\sqrt{2}}{3}, \pm\frac{1}{3}\right)$
8 Equilateral cylinder with a sphere $400\pi$ cm² and $\frac{4000\pi}{3}$ cm³
9 Law of sines: sides of the triangle $6 + \sqrt{11}$ and $6 - \sqrt{11}$
10 Counting: pairs of diagonals of a decagon 175

Question 1: When the homogeneous system has a non-trivial solution

Statement

Let $a$ and $b$ be positive real numbers. Consider the linear system in the unknowns $x$, $y$ and $z$:

$$\begin{cases} -ax + by + az = 0 \\ b^2 x + a^3 y + 4a^2 z = 0 \\ 4a^2 x + a^3 y + b^2 z = 0 \end{cases}$$

Given that this system has a non-trivial solution, find $b$ in terms of $a$. Find the solution set of the system for $a = \frac{1}{2}$.

Solution

1. Zero determinant. A homogeneous system has a non-trivial solution exactly when the determinant of the coefficient matrix is zero. Subtracting the 3rd row from the 2nd (which does not change the determinant):

$$\det\begin{pmatrix} -a & b & a \\ b^2 & a^3 & 4a^2 \\ 4a^2 & a^3 & b^2 \end{pmatrix} = \det\begin{pmatrix} -a & b & a \\ b^2 - 4a^2 & 0 & 4a^2 - b^2 \\ 4a^2 & a^3 & b^2 \end{pmatrix} = (b^2 - 4a^2)\det\begin{pmatrix} -a & b & a \\ 1 & 0 & -1 \\ 4a^2 & a^3 & b^2 \end{pmatrix}$$

The last determinant, expanded along the 1st row, is

$$-a(0 + a^3) - b(b^2 + 4a^2) + a(a^3 - 0) = -b(b^2 + 4a^2)$$

Therefore

$$\det = -b\,(b^2 - 4a^2)(b^2 + 4a^2) = b\,(2a - b)(2a + b)(4a^2 + b^2)$$

2. The value of $b$. With $a, b > 0$, the factors $b$, $2a + b$ and $4a^2 + b^2$ are positive. That leaves $2a - b = 0$:

$$b = 2a$$

3. The case $a = \frac{1}{2}$. Then $b = 1$ and the system is

$$\begin{cases} -\frac{1}{2}x + y + \frac{1}{2}z = 0 \\ x + \frac{1}{8}y + z = 0 \\ x + \frac{1}{8}y + z = 0 \end{cases}$$

The last two equations are identical. From the 1st, $y = \frac{x - z}{2}$; in the 2nd:

$$x + \frac{x - z}{16} + z = 0 \quad\Rightarrow\quad 17x + 15z = 0 \quad\Rightarrow\quad x = -\frac{15z}{17}$$

and $y = \frac{1}{2}\left(-\frac{15z}{17} - z\right) = -\frac{16z}{17}$. Setting $z = 17t$, the solutions are $(x, y, z) = (-15t, -16t, 17t)$.

Checking with $t = 1$, that is, $(-15, -16, 17)$: $\frac{15}{2} - 16 + \frac{17}{2} = 0$ ✔ and $-15 - 2 + 17 = 0$ ✔

Answer: $b = 2a$. For $a = \frac{1}{2}$, the solution set is $\{(-15t, -16t, 17t) : t \in \mathbb{R}\}$.

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Question 2: For which α is the matrix M invertible?

Statement

Consider the following matrices:

$$A = \begin{pmatrix} 1 & -2 \\ -2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 6 \\ 6 & 0 \end{pmatrix} \quad\text{and}\quad C = \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix}.$$

Find the numbers $\alpha \in \mathbb{R}$ such that the matrix $M = \alpha^2 A + \alpha B + C$ is invertible.

Solution

1. Building $M$.

$$M = \begin{pmatrix} \alpha^2 + 3 & -2\alpha^2 + 6\alpha + 3 \\ -2\alpha^2 + 6\alpha + 3 & \alpha^2 + 3 \end{pmatrix}$$

2. Determinant as a difference of squares. With $p = \alpha^2 + 3$ and $q = -2\alpha^2 + 6\alpha + 3$, $\det M = p^2 - q^2 = (p - q)(p + q)$:

$$p - q = 3\alpha^2 - 6\alpha = 3\alpha(\alpha - 2), \qquad p + q = -\alpha^2 + 6\alpha + 6$$

3. Zeros. $\det M = 0$ when $\alpha = 0$, $\alpha = 2$ or $\alpha^2 - 6\alpha - 6 = 0$, that is, $\alpha = 3 \pm \sqrt{15}$.

Checking $\alpha = 2$: $M = \begin{pmatrix} 7 & 7 \\ 7 & 7 \end{pmatrix}$, whose determinant is 0 ✔

Answer: $M$ is invertible for every $\alpha \in \mathbb{R}$ with $\alpha \notin \{0,\ 2,\ 3 - \sqrt{15},\ 3 + \sqrt{15}\}$.

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Question 3: A logarithmic inequality with a variable base

Statement

Find the solution set of the inequality

$$\log_{2-x}\left(-\sqrt[3]{x^2 + 2x - 3}\right) > \log_{2-x}\left(\sqrt[3]{3}\right).$$

Solution

1. Conditions of existence.

  • Base: $2 - x > 0$ and $2 - x \ne 1$, that is, $x < 2$ and $x \ne 1$.
  • Argument: $-\sqrt[3]{x^2 + 2x - 3} > 0 \iff x^2 + 2x - 3 < 0 \iff (x + 3)(x - 1) < 0 \iff -3 < x < 1$.

Together: $-3 < x < 1$. On this interval the base $2 - x$ is greater than 1, so the logarithm is increasing and the inequality is preserved.

2. Solving.

$$-\sqrt[3]{x^2 + 2x - 3} > \sqrt[3]{3} \iff \sqrt[3]{x^2 + 2x - 3} < -\sqrt[3]{3} = \sqrt[3]{-3}$$

The cube root is increasing, so $x^2 + 2x - 3 < -3$, that is, $x(x + 2) < 0$: $-2 < x < 0$.

3. Intersection. $]-2, 0[ \;\cap\; ]-3, 1[ \;=\; ]-2, 0[$.

Checking with $x = -1$: base 3, argument $-\sqrt[3]{-4} = \sqrt[3]{4} > \sqrt[3]{3}$ ✔. With $x = 0.5$: $-\sqrt[3]{-1.75} = \sqrt[3]{1.75} < \sqrt[3]{3}$, and indeed it is left out.

Answer: $S = \;]-2, 0[$.

To review, see Properties of logarithms.

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Question 4: Dividing x¹⁰ − 1 by x⁴ − x³ + x² − x + 1

Statement

Consider the polynomial $p(x) = x^4 - x^3 + x^2 - x + 1$. Find the quotient and the remainder of the division of the polynomial $q(x) = x^{10} - 1$ by $p(x)$, and find all the complex roots of $p(x)$.

Solution

1. Recognising $p$. $p(x)$ is the sum of a geometric progression with ratio $-x$:

$$(x + 1)\,p(x) = x^5 + 1$$

(check: $x^5 - x^4 + x^3 - x^2 + x + x^4 - x^3 + x^2 - x + 1 = x^5 + 1$.)

2. The division. Since $x^{10} - 1 = (x^5 - 1)(x^5 + 1) = (x^5 - 1)(x + 1)\,p(x)$,

$$x^{10} - 1 = \underbrace{(x^6 + x^5 - x - 1)}_{\text{quotient}}\cdot p(x) + \underbrace{0}_{\text{remainder}}$$

3. The roots. The roots of $p$ are the roots of $x^5 = -1$ other than $-1$. Since $-1 = \operatorname{cis}\pi$, the fifth roots are $\operatorname{cis}\frac{\pi + 2k\pi}{5}$, $k = 0, \dots, 4$; removing $k = 2$ (which gives $-1$):

$$x \in \left\{\operatorname{cis}\frac{\pi}{5},\ \operatorname{cis}\frac{3\pi}{5},\ \operatorname{cis}\frac{7\pi}{5},\ \operatorname{cis}\frac{9\pi}{5}\right\}$$

In algebraic form, using $\cos 36^\circ = \frac{1 + \sqrt{5}}{4}$ and $\cos 108^\circ = \frac{1 - \sqrt{5}}{4}$:

$$x = \frac{1 + \sqrt{5}}{4} \pm \frac{\sqrt{10 - 2\sqrt{5}}}{4}\,i \qquad\text{and}\qquad x = \frac{1 - \sqrt{5}}{4} \pm \frac{\sqrt{10 + 2\sqrt{5}}}{4}\,i$$

Checking with the sum of the roots: the conjugate pairs add up to $2\cdot\frac{1 + \sqrt{5}}{4} + 2\cdot\frac{1 - \sqrt{5}}{4} = 1$, which is what Vieta requires for $x^4 - x^3 + \dots$ ✔

Answer: quotient $x^6 + x^5 - x - 1$, remainder $0$; roots $\operatorname{cis}\frac{\pi}{5}$, $\operatorname{cis}\frac{3\pi}{5}$, $\operatorname{cis}\frac{7\pi}{5}$ and $\operatorname{cis}\frac{9\pi}{5}$.

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Question 5: sin(α − β) from A and B

Statement

Let $A = \cos(\alpha) + \cos(\beta)$ and $B = \sin(\alpha) - \sin(\beta)$ with $\alpha, \beta \in \mathbb{R}$. Compute $\sin(\alpha - \beta)$ in terms of $A$ and $B$, given that $A$ and $B$ are not both zero.

Solution

1. Sum to product. With $u = \frac{\alpha + \beta}{2}$ and $v = \frac{\alpha - \beta}{2}$:

$$A = 2\cos u\cos v, \qquad B = 2\cos u\sin v$$

Since $A$ and $B$ are not both zero, $\cos u \ne 0$.

2. The formula. Then $A^2 + B^2 = 4\cos^2 u$ and $2AB = 8\cos^2 u\sin v\cos v = 4\cos^2 u\sin 2v$. Dividing:

$$\frac{2AB}{A^2 + B^2} = \sin 2v = \sin(\alpha - \beta)$$

3. Checking with an example. $\alpha = \frac{\pi}{2}$, $\beta = 0$: $A = 1$, $B = 1$, and the formula gives $\frac{2}{2} = 1 = \sin\frac{\pi}{2}$ ✔

Answer: $\sin(\alpha - \beta) = \dfrac{2AB}{A^2 + B^2}$.

To review, see Sum-to-product and product-to-sum formulas.

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Question 6: The incircle of triangle BCD

Statement

Consider a triangle $ABC$ with $m(\overline{AB}) = 4$, $m(\overline{AC}) = 5$ and $B\hat{A}C = 60^\circ$. Let $D$ be a point on side $\overline{AB}$ with $m(\overline{AD}) = 1$. Find the radius of the circle inscribed in triangle $BCD$.

Solution

1. Sides of $BCD$. $m(\overline{DB}) = 4 - 1 = 3$. By the law of cosines with the $60^\circ$ angle at $A$:

$$m(\overline{CD})^2 = 1 + 25 - 2\cdot 1\cdot 5\cdot\tfrac{1}{2} = 21, \qquad m(\overline{BC})^2 = 16 + 25 - 2\cdot 4\cdot 5\cdot\tfrac{1}{2} = 21$$

Triangle $BCD$ is isosceles: sides $3$, $\sqrt{21}$, $\sqrt{21}$.

2. Area. The altitude to the base $\overline{DB}$ is $\sqrt{21 - \frac{9}{4}} = \frac{5\sqrt{3}}{2}$, so

$$S = \frac{1}{2}\cdot 3\cdot\frac{5\sqrt{3}}{2} = \frac{15\sqrt{3}}{4}$$

Checking: $S_{ABC} - S_{ADC} = \frac{1}{2}\cdot 4\cdot 5\sin60^\circ - \frac{1}{2}\cdot 1\cdot 5\sin60^\circ = \frac{15}{2}\cdot\frac{\sqrt{3}}{2}$ ✔

3. The radius. $S = p\cdot r$, with semiperimeter $p = \frac{3 + 2\sqrt{21}}{2}$:

$$r = \frac{15\sqrt{3}/4}{(3 + 2\sqrt{21})/2} = \frac{15\sqrt{3}}{2(3 + 2\sqrt{21})}\cdot\frac{2\sqrt{21} - 3}{2\sqrt{21} - 3} = \frac{15\sqrt{3}(2\sqrt{21} - 3)}{150} = \frac{6\sqrt{7} - 3\sqrt{3}}{10}$$

(we used $\sqrt{3}\cdot\sqrt{21} = 3\sqrt{7}$). Numerically, $r \approx 1.068$.

Answer: $r = \dfrac{6\sqrt{7} - 3\sqrt{3}}{10}$.

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Question 7: Points of the ellipse that see the foci at 60°

Statement

Find the points $P$ on the ellipse $E$ given by the equation $\dfrac{x^2}{4} + y^2 = 1$ such that the line segments joining $P$ to the foci of $E$ form an angle of $60^\circ$.

Solution

1. Elements of the ellipse. $a = 2$, $b = 1$, $c = \sqrt{4 - 1} = \sqrt{3}$. Foci $F_{1,2} = (\pm\sqrt{3}, 0)$, with $m(\overline{F_1F_2}) = 2\sqrt{3}$.

2. The focal radii. Let $m = m(\overline{PF_1})$ and $n = m(\overline{PF_2})$. Then $m + n = 2a = 4$ and, by the law of cosines with the $60^\circ$ angle at $P$,

$$12 = m^2 + n^2 - mn = (m + n)^2 - 3mn = 16 - 3mn \quad\Rightarrow\quad mn = \frac{4}{3}$$

3. Area in two ways.

$$S = \frac{1}{2}mn\sin60^\circ = \frac{1}{2}\cdot\frac{4}{3}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{3}, \qquad S = \frac{1}{2}\cdot 2\sqrt{3}\cdot|y| = \sqrt{3}\,|y|$$

Hence $|y| = \frac{1}{3}$ and, from the equation of the ellipse, $x^2 = 4\left(1 - \frac{1}{9}\right) = \frac{32}{9}$, $|x| = \frac{4\sqrt{2}}{3}$.

Checking with the focal radii: $m$ and $n$ are the roots of $t^2 - 4t + \frac{4}{3} = 0$, that is, $2 \pm \frac{2\sqrt{6}}{3}$; and for the ellipse, $m = a + \frac{c}{a}x = 2 + \frac{\sqrt{3}}{2}x$. With $x = \frac{4\sqrt{2}}{3}$, $m = 2 + \frac{2\sqrt{6}}{3}$ ✔

Answer: the four points $\left(\pm\dfrac{4\sqrt{2}}{3},\ \pm\dfrac{1}{3}\right)$ (all sign combinations).

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Question 8: The sphere that makes the water rise to the brim

Statement

An equilateral cylinder stands on one of its bases and is partially filled with water. When a sphere is placed inside it, touching the bottom, the water level reaches the height of the cylinder. If the radius of the sphere equals the radius of the base of the cylinder and the volume of water is $2000\frac{\pi}{3}$ cm³, find the lateral surface area of the cylinder and the volume of the sphere.

Solution

1. The configuration. Equilateral cylinder: height $h = 2R$. The sphere of radius $R$ has diameter $2R = h$; touching the bottom, it also touches the top, and lies entirely inside the cylinder.

2. Volume of water. Water and sphere together fill the cylinder:

$$V_{\text{water}} = \pi R^2\cdot 2R - \frac{4}{3}\pi R^3 = \frac{2}{3}\pi R^3 = \frac{2000\pi}{3} \quad\Rightarrow\quad R^3 = 1000 \quad\Rightarrow\quad R = 10 \text{ cm}$$

3. What is asked.

$$A_{\text{lateral}} = 2\pi R\cdot h = 2\pi\cdot 10\cdot 20 = 400\pi \text{ cm}^2, \qquad V_{\text{sphere}} = \frac{4}{3}\pi\cdot 10^3 = \frac{4000\pi}{3} \text{ cm}^3$$

Checking: the sphere takes up $\frac{2}{3}$ of the cylinder (Archimedes’ result), and the water the remaining $\frac{1}{3}$: $\frac{1}{3}\cdot 2000\pi = \frac{2000\pi}{3}$ ✔

Answer: lateral area $400\pi$ cm² and volume of the sphere $\dfrac{4000\pi}{3}$ cm³.

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Question 9: Perimeter 20 and sum of sines 2

Statement

A triangle has perimeter 20 and its interior angles $\alpha$, $\beta$ and $\gamma$ satisfy $\sin(\alpha) + \sin(\beta) + \sin(\gamma) = 2$. Given that one of the sides of this triangle measures 8, find the lengths of the other two sides.

Solution

1. The circumradius. By the law of sines, each side is $2R$ times the sine of the opposite angle. Adding the three sides:

$$20 = 2R\,(\sin\alpha + \sin\beta + \sin\gamma) = 4R \quad\Rightarrow\quad R = 5$$

2. The angle opposite the side 8. Let $\alpha$ be that angle: $\sin\alpha = \frac{8}{10} = \frac{4}{5}$, so $\cos\alpha = \pm\frac{3}{5}$.

3. The other sides. Let them be $b$ and $c$, with $b + c = 12$. Law of cosines:

$$64 = b^2 + c^2 - 2bc\cos\alpha = (b + c)^2 - 2bc(1 + \cos\alpha) = 144 - 2bc(1 + \cos\alpha)$$
  • If $\cos\alpha = -\frac{3}{5}$: $\frac{4}{5}bc = 80$, $bc = 100$; but with $b + c = 12$ the product is at most $36$. Impossible.
  • If $\cos\alpha = \frac{3}{5}$: $\frac{16}{5}bc = 80$, $bc = 25$. So $b$ and $c$ are the roots of $t^2 - 12t + 25 = 0$: $t = 6 \pm \sqrt{11}$.

Checking: for the triangle with sides $8$, $6 + \sqrt{11}$ and $6 - \sqrt{11}$ we computed (Heron) the circumradius $R = \frac{abc}{4S} = 5$, and the sum of sines $\frac{20}{2\cdot 5} = 2$ ✔. The triangle inequality holds: $(6 - \sqrt{11}) + 8 \approx 10.7 > 6 + \sqrt{11} \approx 9.3$.

Answer: the other sides measure $6 + \sqrt{11}$ and $6 - \sqrt{11}$.

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

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Question 10: Two diagonals of the decagon that do not meet

Statement

In a convex decagon, in how many ways can we choose two diagonals that do not intersect?

Solution

1. Total number of pairs. The decagon has $\frac{10\cdot(10 - 3)}{2} = 35$ diagonals, and $\binom{35}{2} = 595$ pairs of diagonals.

2. Pairs crossing in the interior. Any four vertices form a convex quadrilateral whose two diagonals cross; and each pair crossing in the interior comes from exactly four vertices. That makes $\binom{10}{4} = 210$ pairs.

3. Pairs with a common vertex. These also intersect (at the vertex). From each vertex come 7 diagonals, forming $\binom{7}{2} = 21$ pairs: $10\cdot 21 = 210$ pairs.

4. Those that do not intersect.

$$595 - 210 - 210 = 175$$

We checked by computer, testing all 595 pairs in a regular decagon: 175 disjoint pairs, 210 crossing in the interior and 210 sharing a vertex.

Note on the reading: if “do not intersect” were read as “do not cross inside the decagon” (allowing diagonals with a common vertex), the answer would be $595 - 210 = 385$. On the literal reading, two diagonals with a common endpoint intersect at that point, and the answer is 175.

Answer: 175 ways.

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