OMTC Prepare – the invariants

Syllabus and Format

You can download the format and syllabus of the OMTC here (last updated: 28th February, 2025).

The Oxford Mathematics Team Challenge is a competition between teams composed of four students, which consists of four rounds. One round is individual, and the other three are team rounds: teamwork is key!

The syllabus consists of most of the pure AS Level Mathematics, with the omission of all calculus, and with the addition of various topics from the GCSE and A-Level Mathematics curricula, among others.

Problem of the Day

Each weekday until the OMTC, we’ll share a problem of the day so that you can start the fun early! Solutions are available when you click the dropdown button.

Monday, 3rd March

A bug walks all day and sleeps all night. On the first day, it starts at point

O

, faces east, and walks a distance of

5

units due east. Each night the bug rotates

60

counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point

P

. Then

OP^2=\frac mn

, where

m

and

n

are relatively prime positive integers. Find

m+n

. (from AIME, 2020)

The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\tfrac58+\tfrac5{64}-\cdots=\tfrac{40}9$ units east. Thus, it goes $\tfrac{40/9}2=\tfrac{20}9$ units northeast and $\frac{10}9$ units northwest. Because of the $30-60-90$ right triangles in its path, the bug travels a total of $\frac{40}9+\frac{10}9-\frac59=\frac{45}9=5$ units due east and a total of $\tfrac{10\sqrt3}9+\tfrac{5\sqrt3}9=\tfrac{15\sqrt3}9=\tfrac{5\sqrt3}3$ units due north. Now, $OP^2=5^2+\tfrac{5^2}3=\tfrac{100}3$ by the Pythagorean Theorem, and the answer is $103$ .

Tuesday, 4th March

The amount

2.5

is split into two nonnegative real numbers uniformly at random, for instance, into

2.143

and

.357

, or into

\sqrt3

and

2.5-\sqrt3

. Then each number is rounded to its nearest integer, for instance,

2

and

0

in the first case above,

2

and

1

in the second. What is the probability that the two integers sum to

3

? (from AHSME, 1987)

Let $x$ be the smaller of the two numbers and $y$ the greater. Then either:

$x=0$ and $y=2.5$ , hence we round to $0$ and $3$ and get $3$
$0<x<0.5$ so that $2<y<2.5$ , hence we round to $0$ and $2$ and get $2$
$0.5\le x\le 1$ and $1.5\le y\le 2$ , hence we round to $1$ and $2$ and get $3$
$1<x\le1.25$ and $1.25\le y<1.5$ , hence we round to $1$ and $1$ and get $2$

Putting the potential values for $x$ on a line, we see that this line has length $1.25$ . The parts where the sum is $3$ have total length $0+0.5=0.5$ , so that we have a $\tfrac{0.5}{1.25}=\tfrac25$ chance of getting $3$ .

Wednesday, 5th March

A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola

x^2-y^2=2000^2

? (from AIME II, 2000)

(x-y)(x+y)=2000^2=2^8\cdot 5^6

Note that $(x-y)$ and $(x+y)$ are either both even or both odd. Since they multiply to an even number, they must both be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$ . We have $2^6\cdot 5^6$ left. Since there are $7\cdot 7=49$ factors of $2^6\cdot5^6$ , and since both $x$ and $y$ can be negative, this gives us $49\cdot2=98$ lattice points.

Thursday, 6th March

Let

a

and

b

be randomly selected three-digit integers and suppose

a > b

. We say that

a

is clearly bigger
than

b

if each digit of

a

is larger than the corresponding digit of

b

. If the probability that

a

is clearly
bigger than

b

\frac{m}{n}

, where

m

and

n

are relatively prime integers, compute

m + n

. (from Online Math Open Fall 2014)

Compute

\frac{\binom{9}{2} \cdot \binom{10}{2}^2}{\binom{900}{2}} = \frac{36 \cdot 45^2}{450 \cdot 899} = \frac{162}{899}.

The numerator represents the number of ways to select a pair of hundreds digits, tens digits, and unit
digits in a pair of clearly bigger numbers. The denominator represents the total number of pairs $a > b$
of three-digit numbers. Hence the answer is $162 + 899 = \boxed{1061}$ .

Friday, 7th March

A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure

x

. Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is

\frac{14}{25}

. Find the difference between the largest and smallest possible values of

x

. (from AIME I, 2017)

The probability that the chord doesn’t intersect the triangle is $\frac{11}{25}$ . The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\frac{2x}{360}=\frac x{180}$ (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), and the probability that a point is chosen on the arc between the two base angles is $\frac{180-2x}{180}$ . Therefore, we can write $2\left(\frac x{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}$ . This simplifies to $x^2-120x+3024=0$ (note that the simplification is quite tedious) which factors as $(x-84)(x-36)=0$ . So $x=84,36$ . The difference between these is $\boxed{48}$ .

Tip 1: It might help to simplify the algebra if you make the substitution $y = \frac{x}{180}$ , solve for $y$ and then multiply by $180$ to get $x$ .

Tip 2: We actually do not need to spend time factoring $x^2-120x+3024$ . Since the problem asks for $|x_1-x_2|$ , where $x_1$ and $x_2$ are the roots of the quadratic, we can note that $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$ . Vieta’s gives us $x_1+x_2=120$ and $x_1 x_2 = 3024$ . Plugging this into the above equation and simplifying gives us $(x_1-x_2)^2=2304$ , or $|x_1-x_2|=48$ .

Our answer is then $\boxed{48}$ .

Frequently Asked Questions

What kinds of maths can I expect to come up? We can’t tell you! A lot of the maths will be familiar and just use familiar tools in creative ways. Some of the maths will be new, but will be accessible to you. Make sure you’re comfortable with using the mathematicals tools in the syllabus.