Problems

2020 Paper 2 Q4

D: 1500.0 B: 1500.0

triangle inequality proof inequality functions geometric construction counterexample

Given that $a$, $b$ and $c$ are the lengths of the sides of a triangle, explain why $c < a + b$, $a < b + c$ and $b < a + c$.
Use a diagram to show that the converse of the result in part (i) also holds: if $a$, $b$ and $c$ are positive numbers such that $c < a + b$, $a < b + c$ and $b < c + a$ then it is possible to construct a triangle with sides of length $a$, $b$ and $c$.
When $a$, $b$ and $c$ are the lengths of the sides of a triangle, determine in each case whether the following sets of three lengths can
- always
- sometimes but not always
- never
form the sides of a triangle. Prove your claims. (A) $a+1$, $b+1$, $c+1$. (B) $\dfrac{a}{b}$, $\dfrac{b}{c}$, $\dfrac{c}{a}$. (C) $|a-b|$, $|b-c|$, $|c-a|$. (D) $a^2 + bc$, $b^2 + ca$, $c^2 + ab$.
Let $\mathrm{f}$ be a function defined on the positive real numbers and such that, whenever $x > y > 0$, \[\mathrm{f}(x) > \mathrm{f}(y) > 0 \quad \text{but} \quad \frac{\mathrm{f}(x)}{x} < \frac{\mathrm{f}(y)}{y}.\] Show that, whenever $a$, $b$ and $c$ are the lengths of the sides of a triangle, then $\mathrm{f}(a)$, $\mathrm{f}(b)$ and $\mathrm{f}(c)$ can also be the lengths of the sides of a triangle.

Solution:

Not that unless a side is the largest side, it is clearly shorter than the sum of the other two sides (since it's greater than or equal to one on its own). Note also that the distance from one vertex to the other (say $c$) is shorter than going via the other vertex $a+b$, therefore $c < a+b$.
Draw a line of the length of the largest number, say $c$, then since $c < a+b$ we must have circles radius $a$ and $b$ at the endpoints cross, and at their intersection we have a vertex of a $c$-$a$-$b$ triangle.
(A) always. Suppose $c$ is the longest side, then $c < a+b \Rightarrow c+1 < a + 1 + b+1$ so $(a+1,b+1,c+1)$ are still sides of a triangle. (B) sometimes, but not always. $(1,1,1) \to (1,1,1)$ is still a triangle, but $(10, 10, 1) \to (1, 10, \frac{1}{10})$ is not a triangle since $10 > 1 + \frac{1}{10}$ (C) never, suppose $a \leq b \leq c$ then the sides are $b-a, c-b, c-a$ but $c-a = (c-b)+(b-a)$ so the triangle inequality cannot be satisfied. (D) always - without loss of generality let $c$ be the longest side, and since every term is homogeneous degree $2$ we can divide through by $c^2$ to see we have the sides $a^2+b, b^2+a, 1+ab$ and note that $1 + ab < a+b +ab < a+b+a^2+b^2$, also $a^2+b < 1 + b < 1 + (a+b)b = 1 + b^2 + ab < (1+ab)+(b^2+a)$.
Suppose $f$ is increasing and $\f(x)/x$ is decreasing, and suppose $a,b,c$ are side-lengths of a triangle. Wlog $c$ is the longest side, then note $f(c) > f(b), f(a)$, so it suffices to prove that $f(c) < f(a)+f(b)$ \begin{align*} \frac{f(c)}{c} < \frac{f(a)}{a}: && f(a) &> \frac{a}{c} f(c) \\ \frac{f(c)}{c} < \frac{f(b)}{b}: && f(b) &> \frac{b}{c} f(c) \\ \Rightarrow && f(a)+f(b) &> f(c) \underbrace{\left ( \frac{a+b}{c} \right)}_{>1} \\ &&&> f(c) \end{align*} as required

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2019 Paper 2 Q3

D: 1500.0 B: 1500.0

polynomials triangle inequality proof by induction roots of polynomials inequality geometric series bound on roots rational root theorem

For any two real numbers $x_1$ and $x_2$, show that $$|x_1 + x_2| \leq |x_1| + |x_2|.$$ Show further that, for any real numbers $x_1, x_2, \ldots, x_n$, $$|x_1 + x_2 + \cdots + x_n| \leq |x_1| + |x_2| + \cdots + |x_n|.$$

The polynomial f is defined by $$f(x) = 1 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + x^n$$ where the coefficients are real and satisfy $|a_i| \leq A$ for $i = 1, 2, \ldots, n-1$, where $A \geq 1$.
1. If $|x| < 1$, show that $$|f(x) - 1| \leq \frac{A|x|}{1 - |x|}.$$
2. Let $\omega$ be a real root of f, so that $f(\omega) = 0$. In the case $|\omega| < 1$, show that $$\frac{1}{1 + A} \leq |\omega| \leq 1 + A. \quad (*)$$
3. Show further that the inequalities $(*)$ also hold if $|\omega| \geq 1$.
Find the integer root or roots of the quintic equation $$135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0.$$

Solution: Claim: $|x_1 + x_2| \leq |x_1| + |x_2|$ Proof: Case 1: $x_1, x_2 \geq 0$. The inequality is equivalent to $|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|$ so it's an equality. Case 2: $x_1, x_2 \leq 0$. The inequality is equivalent to $|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2$, so it's also an equality in this case. Case 3: (wlog) $|x_1| \geq |x_2| > 0$ and $x_1x_2 < 0$ then $|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|$ We can prove this by induction, we've already proven the base case and: $|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|$

$\,$ \begin{align*} && |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\ &&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\ &&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\ &&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\ &&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\ &&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\ &&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\ &&&\leq \frac{A|x|}{1-|x|} \end{align*}
If $f(\omega) = 0$ then \begin{align*} && 1 & \leq \frac{A|\omega|}{1-|\omega|} \\ \Leftrightarrow && 1-|\omega| &\leq A |\omega| \\ \Leftrightarrow && 1 &\leq (1+A) |\omega| \\ \Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\ \end{align*} We also know $\omega \leq 1 < 1 + A$
If $\omega$ is a root of $f(x)$ then $1/\omega$ is a root of $1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n$ and so $1/\omega$ satisfies that inequality, ie \begin{align*} && \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\ \Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A} \end{align*}
First notice that it's equivalent to: $0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1$ therefore all integer roots must be between $-2,-1$ and $1$ and $2$. $1$ doesn't work. $-1$ works. Clearly $2$ cannot work by parity argument, therefore the only integer root is $-1$.

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2019 Paper 2 Q11

D: 1500.0 B: 1500.0

probability combinatorics triangle inequality counting summation uniform distribution discrete random variables

The three integers $n_1$, $n_2$ and $n_3$ satisfy $0 < n_1 < n_2 < n_3$ and $n_1 + n_2 > n_3$. Find the number of ways of choosing the pair of numbers $n_1$ and $n_2$ in the cases $n_3 = 9$ and $n_3 = 10$. Given that $n_3 = 2n + 1$, where $n$ is a positive integer, write down an expression (which you need not prove is correct) for the number of ways of choosing the pair of numbers $n_1$ and $n_2$. Simplify your expression. Write down and simplify the corresponding expression when $n_3 = 2n$, where $n$ is a positive integer.
You have $N$ rods, of lengths $1, 2, 3, \ldots, N$ (one rod of each length). You take the rod of length $N$, and choose two more rods at random from the remainder, each choice of two being equally likely. Show that, in the case $N = 2n + 1$ where $n$ is a positive integer, the probability that these three rods can form a triangle (of non-zero area) is $$\frac{n - 1}{2n - 1}.$$ Find the corresponding probability in the case $N = 2n$, where $n$ is a positive integer.
You have $2M + 1$ rods, of lengths $1, 2, 3, \ldots, 2M + 1$ (one rod of each length), where $M$ is a positive integer. You choose three at random, each choice of three being equally likely. Show that the probability that the rods can form a triangle (of non-zero area) is $$\frac{(4M + 1)(M - 1)}{2(2M + 1)(2M - 1)}.$$ Note: $\sum_{k=1}^{K} k^2 = \frac{1}{6}K(K + 1)(2K + 1)$.

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2015 Paper 2 Q3

D: 1600.0 B: 1483.4

combinatorics proof by induction triangles triangle inequality counting summation closed form formula

Three rods have lengths $a$, $b$ and $c$, where $a< b< c$. The three rods can be made into a triangle (possibly of zero area) if $a+b\ge c$. Let $T_{n}$ be the number of triangles that can be made with three rods chosen from $n$ rods of lengths $1$, $2$, $3$, $\ldots$ , $n$ (where $n\ge3$). Show that $T_8-T_7 = 2+4+6$ and evaluate $T_8 -T_6$. Write down expressions for $T_{2m}-T_{2m-1}$ and $T_{2m} - T_{2m-2}$. Prove by induction that $T_{2m}=\frac 16 m (m-1)(4m+1)\,$, and find the corresponding result for an odd number of rods.

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2013 Paper 3 Q6

D: 1700.0 B: 1500.0

complex numbers triangle inequality modulus conjugate Argand diagram inequality algebraic manipulation

Let $z$ and $w$ be complex numbers. Use a diagram to show that $\vert z-w \vert \le \vert z\vert + \vert w \vert\,.$ For any complex numbers $z$ and $w$, $E$ is defined by \[ E = zw^* + z^*w +2 \vert zw \vert\,. \]

Show that $\vert z-w\vert^2 = \left( \vert z \vert + \vert w\vert\right)^2 -E\,$, and deduce that $E$ is real and non-negative.
Show that $\vert 1-zw^*\vert^2 = \left ( 1 +\vert zw \vert \right)^2 -E\,$.

Hence show that, if both $\vert z \vert >1$ and $\vert w \vert >1$, then \[ \frac {\vert z-w\vert} {\vert 1-zw^*\vert } \le \frac{\vert z \vert +\vert w\vert }{1+\vert z w \vert}\,. \] Does this inequality also hold if both $\vert z \vert <1$ and $\vert w \vert <1$?

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2002 Paper 3 Q8

D: 1700.0 B: 1469.7

complex numbers unit modulus argument Ptolemy's theorem unit circle geometric proof modulus triangle inequality

Four complex numbers $u_1$, $u_2$, $u_3$ and $u_4$ have unit modulus, and arguments $\theta_1$, $\theta_2$, $\theta_3$ and $\theta_4$, respectively, with $-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi$. Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where $n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,$. Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer $n$. Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

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1998 Paper 2 Q5

D: 1600.0 B: 1470.9

complex numbers modulus triangle inequality proof by induction inequality geometric proof roots of polynomials

Define the modulus of a complex number $z$ and give the geometric interpretation of $\vert\,z_1-z_2\,\vert$ for two complex numbers $z_1$ and $z_2$. On the basis of this interpretation establish the inequality $$\vert\,z_1+z_2\,\vert\le \vert\,z_1\,\vert+\vert\,z_2\,\vert.$$ Use this result to prove, by induction, the corresponding inequality for $\vert\,z_1+\cdots+z_n\,\vert$. The complex numbers $a_1,\,a_2,\,\ldots,\,a_n$ satisfy $|a_i|\le 3$ ($i=1, 2, \ldots , n$). Prove that the equation $$a_1z+a_2z^2\cdots +a_nz^n=1$$ has no solution $z$ with $\vert\,z\,\vert\le 1/4$.

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1993 Paper 1 Q5

D: 1500.0 B: 1516.0

complex numbers modulus triangle inequality proof inequality geometric series roots of polynomials Argand diagram

If $z=x+\mathrm{i}y$ where $x$ and $y$ are real, define $\left|z\right|$ in terms of $x$ and $y$. Show, using your definition, that if $z_{1},z_{2}\in\mathbb{C}$ then $\left|z_{1}z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|.$ Explain, by means of a diagram, or otherwise, why $\left|z_{1}+z_{2}\right|\leqslant\left|z_{1}\right|+\left|z_{2}\right|.$ Suppose that $a_{j}\in\mathbb{C}$ and $\left|a_{j}\right|\leqslant1$ for $j=1,2,\ldots,n.$ Show that, if $\left|z\right|\leqslant\frac{1}{2},$ then \[ \left|a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z\right|<1, \] and deduce that any root $w$ of the equation \[ a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+1=0 \] must satisfy $\left|x\right|>\frac{1}{2}.$

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1991 Paper 1 Q14

D: 1516.0 B: 1457.1

probability combinatorics triangle inequality counting summation arithmetic series enumeration

A set of $2N+1$ rods consists of one of each length $1,2,\ldots,2N,2N+1$, where $N$ is an integer greater than 1. Three different rods are selected from the set. Suppose their lengths are $a,b$ and $c$, where $a > b > c$. Given that $a$ is even and fixed, show, by considering the possible values of $b$, that the number of selections in which a triangle can then be formed from the three rods is \[ 1+3+5+\cdots+(a-3), \] where we allow only non-degenerate triangles (i.e. triangles with non-zero area). Similarly obtain the number of selections in which a triangle may be formed when $a$ takes some fixed odd value. Write down a formula for the number of ways of forming a non-degenerate triangle and verify it for $N=3$. Hence show that, if three rods are drawn at random without replacement, then the probability that they can form a non-degenerate triangle is \[ \frac{(N-1)(4N+1)}{2(4N^{2}-1)}. \]

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1990 Paper 3 Q16

D: 1700.0 B: 1484.0

probability conditional probability geometric probability triangle inequality obtuse triangle uniform distribution bivariate distribution integration

A rod of unit length is cut into pieces of length $X$ and $1-X$; the latter is then cut in half. The random variable $X$ is uniformly distributed over $[0,1].$ For some values of $X$ a triangle can be formed from the three pieces of the rod. Show that the conditional probability that, if a triangle can be formed, it will be obtuse-angled is $3-2\sqrt{2.}$
The bivariate distribution of the random variables $X$ and $Y$ is uniform over the triangle with vertices $(1,0),(1,1)$ and $(0,1).$ A pair of values $x,y$ is chosen at random from this distribution and a (perhaps degenerate) triangle $ABC$ is constructed with $BC=x$ and $CA=y$ and $AB=2-x-y.$ Show that the construction is always possible and that $\angle ABC$ is obtuse if and only if \[ y>\frac{x^{2}-2x+2}{2-x}. \] Deduce that the probability that $\angle ABC$ is obtuse is $3-4\ln2.$

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1988 Paper 1 Q7

D: 1484.0 B: 1500.0

differentiation from first principles quadratic inequality optimisation polynomial triangle inequality Chebyshev polynomial

The function $\mathrm{f}$ is defined by \[ \mathrm{f}(x)=ax^{2}+bx+c. \] Show that \[ \mathrm{f}'(x)=\mathrm{f}(1)\left(x+\tfrac{1}{2}\right)+\mathrm{f}(-1)\left(x-\tfrac{1}{2}\right)-2\mathrm{f}(0)x. \] If $a,b$ and $c$ are real and such that $\left|\mathrm{f}(x)\right|\leqslant1$ for $\left|x\right|\leqslant1$, show that $\left|\mathrm{f}'(x)\right|\leqslant4$ for $\left|x\right|\leqslant1$. Find particular values of $a,b$ and $c$ such that, for the corresponding function $\mathrm{f}$ of the above form $\left|\mathrm{f}(x)\right|\leqslant1$ for all $x$ with $\left|x\right|\leqslant1$ and $\mathrm{f}'(x)=4$ for some $x$ satisfying $\left|x\right|\leqslant1$.

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