This page details the solutions to the Maths Challenge in the Bridgian Herald,
started in Issue 5 (2025-03-31) and maintained by the Devious One.
Issue 5 – Maths Challenge 1
Level 1
‘Define numerical bounds for the area of the shaded region
– i.e., two numbers, a and b, where
a ≤ A ≤ b.’
There are many ways to do this level; it is implicity assumed that a tighter bound is better, although this is never stated.
The lazy solution is to give a as 0 and b as the area of the circle.
Area of a circle with radius r: πr2
= π(6 cm)2
= 36π cm2.
Therefore, to 2 decimal places, the numerical bound is
0 cm2 ≤ A ≤ 113.10 cm2.
Another, more tightly bounded way is to consider a rectangle outside and a rhombus within the shape, as on the right.
Because the sides of the rhombus are radii of the circles, they also have a length of 6 cm.
The top and bottom sides of the rectangle are also 6 cm long, divided into two 3 cm segments by the circle-intersection.
This forms a right-angled triangle with half the rectangle’s height; using Pythagorus’ theorem, h2 + (3 cm)2 = (6 cm)2
∴ h2 + 9 cm2 = 36 cm2
∴ h2 = 27 cm2
∴ h = √27 cm
= 3√3 cm.
Therefore, the area of the rectangle is
2(3√3 cm) × 6 cm = 36√3 cm2.
The area of the rhombus is exactly half this – i.e., 18√3 cm2.
To 2 decimal places, this makes the inequality 31.18 cm2 ≤ A ≤ 62.35 cm2.
Level 2
‘Compute the exact area, A, of the shaded
region, giving units with your answer.’
Consider one quarter of the area of intersection between the circles, bounded by the line between the circles’ centres and its perpendicular bisector through the intersection-points.
Using the top-left quarter as in the diagram opposite, this is part of a sector of the right circle, with a right-angled triangle removed.
As its lines are all radii of the circles, the triangle between the two centres and the upper intersection point is equilateral, so all its angles must be 60°.
This means that the area of the sector is a sixth of the whole circle’s.
πr2 ÷ 6
= π(6 cm)2 ÷ 6
= 36π cm2 ÷ 6
= 6π cm2
The area of the triangle, meanwhile, is given by half its base × its height.
The base is half the radius of the circle, 3 cm, owing to the original symmetry across the perpendicular bisector.
As the triangle is right-angled (see ‘perpendicular’), its height can be calculated with Pythagorus thus: h2 + (3 cm)2 = (6 cm)2
∴ h2 + 9 cm2 = 36 cm2
∴ h2 = 27 cm2
∴ h = √27 cm
= 3√3 cm.
Therefore, the area of the triangle is ½(3 cm)(3√3 cm) = (9√3 ∕ 2) cm2.
So the area of the shaded quarter must be (6π − 9√3 ∕ 2) cm2.
Multiplying this by 4 for the full area yields (24π − 18√3) cm2.
Level 3
‘Generalise this expression for any two circles of radius r,
with distance d between their centres.’
This is easy in theory, but annoying in practice.
It requires going through the above steps but using r and d rather than 6 cm.
First, divide the area into quarters as before.
Looking at the top-left quarter again (see opposite), the angle of the sector can now be calculated using cosine.
The cosine of an angle, cos(θ), can be found by dividing the hypotenuse by the side adjacent to θ in a right-angled triangle.
This equates to cos(θ) = d ∕ 2 ÷ r, so θ = cos-1(d ∕ 2r).
The portion of the circle taken up by the sector is θ ∕ 2π, measuring the angle in radians.
Therefore, the area of the sector must be:
πr2 × (θ ∕ 2π)
= r2θ ∕ 2
= r2cos-1(d ∕ 2r) ∕ 2
The height of the triangle can again be found using Pythagorus’ theorem as follows: h2 + (d ∕ 2)2 = r2
∴ h2 = r2 − d2 ∕ 4
∴ h = √(r2 − d2 ∕ 4)
The area of the triangle is thus ½(d ∕ 2)√(r2 − d2 ∕ 4)
= d√(r2 − d2 ∕ 4) ∕ 4.
Subtracting the triangle from the sector yields r2cos-1(d ∕ 2r) ∕ 2 − d√(r2 − d2 ∕ 4) ∕ 4.
Finally, multiplying by 4 gives the final area as:
4(r2cos-1(d ∕ 2r) ∕ 2 − d√(r2 − d2 ∕ 4) ∕ 4)
= .
Remember that cos-1(θ) takes inputs in radians.
Level 4
‘Check your answer with comparison to
an integral, over x, of the equation for a
circle: x2 + y2 = r2.’
Why should these be related?
Well, as crudely shown on the right, a quarter of the area can be represented as the area under a curve: particularly, the curve defined by a semicircle.
Rearranging x2 + y2 = r2 gives y = √(r2 − x2).
However, this equation has x = 0 being the centre, whereas in the setup opposite, x = 0 would represent the left end of the diameter.
Replacing x with x − r, y = √(r2 − (x − r)2)
= √(r2 − (x2 − 2rx + r2))
= √(2rx − x2)
Consider the distance between the semicircle’s left edge and the bisector.
As the full radius is r and the part to the right of the bisector is half the distance d between the centres, this must be r − d ∕ 2.
As the area is 4 times the area of the integrated wedge, and calling this length λ,
dA ∕ dλ = 4√(2rλ − λ2)
This can be converted into an integral: A = ∫0λ 4√(2rx − x2) dx
However, the derivative is actually more important, as dA ∕ dλ can be calculated from Level 3’s equation.
Differentiation time! This looks complicated, but a few applications of the chain rule and we’ll be done:
dA ∕ dλ = d ∕ dλ (2r2cos-1(1 − λ ∕ r) − 2(r − λ)√(2rλ − λ2))
= d ∕ dλ 2r2cos-1(1 − λ ∕ r) − d ∕ dλ 2(r − λ)√(2rλ − λ2)
Note that because these are subtracted, the derivative of the total area comes out as the derivative of the wedge’s area minus the derivative of the triangle’s area.
What is d ∕ dx cos-1x?
Start with the identity of an inverse function:
cos(cos-1x) ≡ x
d ∕ dx cos(cos-1x) ≡ dx ∕ dx ≡ 1
As d ∕ dθ cos θ = -sin θ, using the chain rule,
-sin(cos-1x) × d ∕ dx cos-1x ≡ 1
d ∕ dx cos-1x ≡ -1 ∕ sin(cos-1x)
Because sin θ is always positive in the range of cos-1x, this can be rearranged as
-1 ∕ √(sin2(cos-1x))
≡ -1 ∕ √(1 − cos2(cos-1x))
≡ -1 ∕ √(1 − x2)
∴ d ∕ dx cos-1x = -1 ∕ √(1 − x2)
First, the wedge area:
d ∕ dλ 2r2cos-1(1 − λ ∕ r)
= 2r2 × d ∕ dλ cos-1(1 − λ ∕ r)
d ∕ dx cos-1x = -1 ∕ √(1 − x2), so, using the chain rule,
2r2 × d ∕ dλ cos-1(1 − λ ∕ r)
= 2r2 × -1 ∕ √(1 − (1 − λ ∕ r)2) × d ∕ dλ 1 − λ ∕ r
= 2r2 × -1 ∕ √(1 − (1 − 2λ ∕ r + λ2 ∕ r2)) × -1 ∕ r
= 2r2 ∕ r√(2λ ∕ r − λ2 ∕ r2)
= 2r2 ∕ √(2rλ − λ2)
Now for the triangle area:
d ∕ dλ 2(r − λ)√(2rλ − λ2)
= d ∕ dλ (2r − 2λ)√(2rλ − λ2)
The product rule says that d ∕ dx f(x)g(x) ≡ f(x)g′(x) + g(x)f′(x), so the prior expression can be converted to:
= ((2r − 2λ) × d ∕ dλ √(2rλ − λ2)) + (√(2rλ − λ2) × d ∕ dλ 2r − 2λ)
Dealing with these one at a time,
(2r − 2λ) × d ∕ dλ √(2rλ − λ2)
= (2r − 2λ)(1 ∕ 2√(2rλ − λ2))(2r − 2λ)
= (2r − 2λ)2 ∕ 2√(2rλ − λ2)
= 2(r − λ)2 ∕ √(2rλ − λ2)
= (2r2 − 4rλ + 2λ2) ∕ √(2rλ − λ2), and
√(2rλ − λ2) × d ∕ dλ 2r − 2λ
= √(2rλ − λ2) × (-2)
= -2√(2rλ − λ2)
= -2(2rλ − λ2) ∕ √(2rλ − λ2)
= (2λ2 − 4rλ) ∕ √(2rλ − λ2).
Because the area is zero when λ = 0, this means that A = ∫0λ 4√(2rx − x2) dx.
This is the same integral from earlier, so the answer has been verified using an alternate method.
Issue 6 – Maths Challenge 2
‘A person throws a 500-gram ball from a height of 0.25 metres above a flat ground, at a speed of 10 metres per second (m s-1) forwards and 2 m s-1 upwards.
The ball flies under 10 m s-2 gravitational acceleration, with no other forces affecting it.’
Level 1
‘Define an equation for the vertical speed of the ball (v) in terms of the time, t seconds, from when the ball is thrown.
10 m s-2 of acceleration changes the velocity by 10 metres per second, every second.’
We know from the question that the ball is accelerating downwards at a constant rate of 10 m s-2.
This means that the equation must be of the form v = -10t + c, where c is a constant.
At the start of the throw (t = 0), the velocity is 2 m s-1 upwards, so v = 2.
∴ 2 = -10(0) + c
∴ 2 = c
∴ c = 2
∴ v = -10t + 2.
Level 2
‘Plot a velocity–time (v–t) graph for the vertical movement of the ball.
Hence or otherwise, find the time when the ball is at its highest, and when it hits the ground.
(Remember that it starts at 0.25m up.)’
The highest point of the ball is where its vertical velocity goes from positive to negative.
As seen on the graph, this is at the time t = 0.2; i.e., 0.2 seconds after the ball is thrown.
The ball hits the ground when its height becomes zero.
Since the displacement can be given by the area under the curve, this means that it is the point where the triangle below the x-axis equals the height reached at the peak.
So first, what is the height at the peak?
The area under the top-left triangle is ½ × 0.2 s × 2 m s-1
= 0.2 m.
Add this to the initial height to get 0.2 + 0.25 = 0.45 m.
To compute the time when the ball hits the ground, we need to find the width (b) of the triangle with an area of 0.45.
Because the gradient of the curve is -10, h = 10b.
Therefore, the area of the triangle is ½ × b × 10b = 5b2.
For an area of 0.45,
5b2 = 0.45
∴ b2 = 0.09
∴ b = √0.09 = 0.3
Adding this to the t-value of the highest point yields t = 0.2 + 0.3 = 0.5
∴ the ball hits the ground 0.5 seconds after it is thrown.
Level 3
‘By computing the area under the curve, derive an expression for the coordinates (x, y) of the ball in terms of t.
For reference, its initial position would be represented as (0, 0.25).’
The x-coordinate of the ball is the easiest to calculate.
As the ball is travelling at 10 m s-1, the value of x changes by 10 times any change in t.
In other words, the gradient of a displacement–time graph for the horizontal direction would be 10.
Using a variation on the equation y = mx + c, x = 10t + c, for some constant c.
When t = 0, x = 0.
∴ 0 = 10(0) + c
∴ c = 0
So x = 10t.
The y-coordinate is slightly more difficult.
Start by getting the velocity in terms of t.
The gradient is -10 and the y-intercept is 2, so v = -10t + 2.
Consider the area under the curve between two values of t: this can be composed of a rectangle and a triangle as shown opposite.
The area of the rectangle is (t2 − t1)v2, and the area of the triangle is ½(t2 − t1)(v2 − v1).
Setting t1 to 0 and v1 to 2, the area under the curve = ½t2(v2 + 2).
Because y begins at 0.25, y = ½t(-10t + 2 + 2) + 0.25
= -5t2 + 2t + 0.25.
∴ (x, y) = (10t, -5t2 + 2t + 0.25).
Level 4
‘The action of an object’s path is defined as the integral of its kinetic energy minus its potential energy, over time.
Using the equations Ek = ½mv2 and Ep = mgh, find the action of the ball’s throw up to when it hits the ground. Give your
answer to 3 significant figures, with appropriate units.’
First, we need to write Ek and Ep in terms of t.
By Pythagoras’s theorem, v2 = vx2 + vy2
= 102 + (-10t + 2)2
= 100 + 100t2 − 40t + 4
= 100t2 − 40t + 104.
Let the action of the path be S. Because the ball hits the ground at t = 0.5, S = ∫00.5 (50t2 − 20t + 24.75) dt
= [(50 ∕ 3)t3 − 10t2 + 24.75t]00.5
= (11.958…) − 0
≈ 12.0 J s.
Issue 7 – Maths Challenge 3
Level 1
‘State the size of the angle marked θ.’
ΔACD is equilateral, so θ = 60°.
Level 2
‘Using the cosine rule, a2 = b2 + c2 − 2bc cos A, work out the length of the diagonal AC.’
‘ABCD is inscribed within a circle.
Calculate the radius of the circle and show that all the points are that distance from the centre, O.’
Because A, C, and D are all at a distance r (the radius) from O, ΔAOC ≡ ΔCOD ≡ ΔDOA.
∴ ∠AOC = ∠AOD = ∠DOA.
These are angles around a point, so ∠AOC + ∠AOD + ∠DOA = 360°.
∴ ∠AOC = ∠AOD = ∠DOA = φ = 120°.
By the cosine rule,
AD2 = r2 + r2 − 2r2 cos φ
∴ 6.72 = 2r2 − 2r2(-½)
∴ 44.89 = 3r2
∴ r = √(44.89 ∕ 3) ≈ 3.87
(Exact value: 67√3 ∕ 30.)
Now we need to show that OB = r as well.
The sine rule allows ∠ACB to be worked out as follows:
sin ∠ACB ∕ AB = sin ∠ABC ∕ AC
∴ sin ∠ACB = sin 120°(AB ∕ AC)
= (√3 ∕ 2)(4.5 ∕ 6.7)
∴ ∠ACB = sin-1(45√3 ∕ 134) ≈ 35.6°
Using the cosine rule again,
OB2 = BC2 + r2 − 2rBC cos ∠BCO
= 10.24 + (14.963…) − 2(3.87)(3.2)(0.414…)
= 14.693…
∴ OB = √14.693 ≈ 3.87
∴ OB ≈ r
(A correct proof of this would use exact values, but this is more difficult to calculate than numerical values.)
Level 4
‘The triangle ΔABC was created by transforming a shape under the matrix above.
Use the principle of linearity to deduce the exact area of this shape.’
Using the expression bc sin A for the area of a triangle,
area(ΔABC) = (4.5)(3.2) sin 120°
= 14.4(√3 ∕ 2)
= 7.2√3
= 36√3 ∕ 5.
One result of the principle of linearity is that a matrix will always multiply shapes’ areas by the same amount (the determinant).
The determinant of the matrix above can be calculated by considering the area of a triangle made by the points (0, 0), (0, 1), and (1, 0), which has area ½.
This is transformed into the triangle shown opposite, whose area can be determined by subtracting many rectangles and triangles from a larger rectangle.
Therefore, the determinant of the matrix is 2 ÷ ½ = 4.
So (area of ΔABC) = (area of original shape) * 4
∴ A = (36√3 ∕ 5) ÷ 4 = 9√3 ∕ 5.
Issue 8 – Maths Challenge 4
‘A bag contains 64 counters, which are each either black, red, or white.
One is randomly drawn and placed on every square of an 8×8 chess-board.’
Level 1
‘The probability of the first counter removed being black is 0.375.
How many black counters are in the bag?’
Let the number of black counters be n. Then the probability of drawing a black counter is n ∕ 64.
∴ n ∕ 64 = 0.375
∴ n = 0.375 × 64 = 24
Level 2
‘There turn out to be 10 red counters in the bag.
Draw a probability tree to compute the chance that 2 counters of the same colour are placed on the first two squares.’
There are 10 red counters and 24 black counters at the start; therefore, there must be 30 white counters.
Filling these in and accounting for the counters being removed from the bag, the probability tree opposite can be formed.
Now just add up all the options to get the total probability:
2 blacks: 24 ∕ 64 × 23 ∕ 63 ≈ 0.137
2 reds: 10 ∕ 64 × 9 ∕ 63 ≈ 0.022
2 whites: 30 ∕ 64 × 29 ∕ 63 ≈ 0.216
0.137 + 0.022 + 0.216 = 0.375 [exactly!]
Level 3
‘Calculate the total number of ways in which the counters could be arranged on the chess-board, accounting for the duplicate answers given by indistinguishable counters.
Give your answer in standard form.’
The total number of ways n objects (i.e., counters) can be arranged is given by 1 × 2 × 3 × … × (n − 1) × n = n!
Therefore, 64 distinguishable counters can be arranged in 64! ≈ 1.27 × 1089 unique ways.
However, the 24 black counters could be permuted (shuffled) in any way and there would be no difference.
This means we have to divide by 24!, along with 10! and 30! for the red and white counters.
Plugging this into a calculator,
64! ∕ (24! × 10! × 30!) ≈ 2.12 × 1026.
Level 4
‘Dôsel wagers that the first 32 counters drawn will contain 5 red counters.
By working out the number of permutations (including duplicates) in which this is true, determine the probability that Dôsel wins the bet.’
Firstly, according to Level 3, the number of ways that the 32 counters could be arranged is 32!
= 1 × 2 × 3 × … × 31 × 32
≈ 2.63 × 1035.
The number of permutations for the other 32 counters also need to be considered, producing (32!)2 ≈ 6.92 × 1070.
However, the 10 red counters and the 54 non-red counters could be swapped between the sides, which has not been accounted for.
For the red counters, the number of arrangements is currently (5!)2, but should be (10!).
For the non-red counters, the number of arrangements is currently (27!)2, but should be (54!).
This results in a total number of permutations equal to:
(32!)2 × 10! × 54! ∕ (5!)2(27!)2
≈ 3.40 × 1088
As each permutation is equally likely, the probability of there being 5 red counters in the first 32 can be calculated as:
3.40 × 1088 ∕ 64! = 0.2677…
Therefore, Dôsel has a 26.8% chance of winning the bet.
Issue 9 – Maths Challenge 5
‘A researcher counted the number of plants of Reynoutria japonica, an invasive species, that were present in a greenhouse on the first day of each month between April and September:
1, 2, 7, 24, 53, 98’
Level 1
‘Compute the mean rate of spread, in plants per month, between June and August.’
There are two months between the 1st of June and the 1st of August, and the total change in the number of plants was 53 − 7 = 46.
The mean rate of spread is, therefore 46 ÷ 2 = 23 plants per month.
Level 2
‘Plot a labelled graph of the number of R. japonica plants between April and September, drawing a line of best fit to estimate October’s number of plants.’
See opposite. Shamelessly made in Microsoft Paint.
Level 3
‘Suggest a possible factor which would make your extrapolation invalid, explaining the effect this would have on the true number of plants in October.’
The plants may use up the space/nutrient supply, reducing the true number of plants in October.
Level 4
‘Compute a polynomial (degree 3) line of best fit by minimising the total variance from the data in the sequence,
Σ(ax3 + bx2 + cx + d – y)2,
where x starts at 0 in April. Do not use trial and error.
Bonus challenge: compute the exact value of all the coefficients a, b, c, d,
and calculate an estimate for rate of spread of R. japonica in the middle of May.’
This part is evil. Just warning you now.
Substituting in the values of x and y in the data, σ2 = (d – 1)2
+ (a + b + c + d – 2)2
+ (8a + 4b + 2c + d – 7)2
+ (27a + 9b + 3c + d – 24)2
+ (64a + 16b + 4c + d – 53)2
+ (125a + 25b + 5c + d – 98)2
Don’t expand the brackets. Just don’t.
Now we need to work out the rate of change of σ2 with respect to a, b, c, and d.
This can be worked out using the following formula:
d ∕ dx (ax + b)2 = 2a(ax + b)
∴ ∂(σ2) ∕ ∂a = 2(a + b + c + d – 2)
+ 16(8a + 4b + 2c + d – 7)
+ 54(27a + 9b + 3c + d – 24)
+ 128(64a + 16b + 4c + d – 53)
+ 250(125a + 25b + 5c + d – 98)
= 41030a + 8850b + 1958c + 450d – 32696
Computing this for the other variables gives
∂(σ2) ∕ ∂b = 8850a + 1958b + 450c + 110d – 7088,
∂(σ2) ∕ ∂c = 1958a + 450b + 110c + 30d – 1580, and
∂(σ2) ∕ ∂d = 450a + 110b + 30c + 12d – 370.
At a (local) minimum of the variance, all of these “partial derivatives” must be zero.
If they were not zero, the variance could be decreased by changing one of the coefficients by a small amount in the appropriate direction.
The line of best fit is, therefore, y = (6210 ∕ 13607)x3 + (211114 ∕ 95249)x2 – (6818351 ∕ 2190727)x + 2590550 ∕ 2190727
or, in purely integer terms:
2190727y = 999810x3 + 4855622x2 – 6818351x + 2590550
Now the bonus challenge about the rate of spread. In the middle of May, x = 1.5
dy ∕ dx = (18630 ∕ 13607)x2 + (422228 ∕ 95249)x – 6818351 ∕ 2190727
At x = 1.5, this equates to
dy ∕ dx = (6.618…) plants per month.
Issue 10 – Maths Challenge 6
‘Consider the following two function machines. Each
operation is performed on the input number in turn
to produce a new number.’
INPUT
×12
+24
÷2
-3
OUTPUT
INPUT
+2
SQUARE
×1st OUTPUT
-75
OUTPUT
Level 1
‘Compute the output of the top machine when the input is -3.’
As easy as it sounds.
-3 × 12 = -36
-36 + 24 = -12
-12 ÷ 2 = -6
-6 – 3 = -9.
Level 2
‘Work out the result of the bottom machine when both have an input of a variable x.
Write your answer in its simplest form without brackets.’
‘Given that (2x – 1) is a factor of the polynomial, work out the other two values of x where the result equals 0,
in the form a ± b√c where a, b, and c are integers.
(Hint: use algebraic division – like long division, but replacing place values with powers of x.)’
This is algebraic division:
3x2
+
18x
+
39
2x – 1
6x3
+
33x2
+
60x
–
39
6x3
–
3x2
36x2
+
60x
36x2
–
18x
78x
–
39
78x
–
39
0
The procedure shows that 6x3 + 33x2 + 60x – 39 can be written as
(2x – 1)(3x2 + 18x + 39), with no remainder left over.
According to the fundamental theorem of algebra, if
6x3 + 33x2 + 60x – 39 ≡ (2x – 1)(3x2 + 18x + 39) = 0, then
2x – 1 = 0 or 3x2 + 18x + 39 = 0
∴ x = 2 or x2 + 6x + 13 = 0
The quadratic formula can be used to find the other two values of x: x2 + bx + c = 0 ⇒ x = (-b ± √(b2 – 4c)) ⁄ 2
∴ x = (-6 ± √(62 – 4(13))) ⁄ 2
= -3 ± √(-16) ⁄ 2
= -3 ± 2√-1
Level 4
‘A student notes that, since no value on the number-line has a negative square, the answer to Level 3 is not a “real” number.
Drawing the “imaginary” √-1 axis at right angles to the number-line, calculate the distance between the origin and the squares of the roots.’
First, we must square the roots.
(-3 + 2√-1)2 = (-3)2 + 2(-3)(2√-1) + (2√-1)2
= 9 – 12√-1 + 4(-1)
= 5 – 12√-1
A similar calculation gives the other root as 5 + 12√-1.
I do not draw, merely think.
Given that the imaginary axis is at right angles to the real axis, Pythagoras’s theorem can be used to compute the roots’ distance from the origin. a2 + b2 = c2 a = (real component) = 5 b = (imaginary component) = ±12
∴ (distance)2 = 52 + (±12)2
= 25 + 144 = 169
∴ (distance) = √169 = 13 for both roots.
.
