If students don't seem to take the "make sure you state the units" or "check that units are consistent" seriously, then these might provide a nice reminder of how important they can be. Here are five examples where inconsistent units caused issues. Search online for more information - some of them have interesting anecdotes to share with students.
An interactive cobweb or staircase diagram showing the iterative process behind finding a root of an equation.
Load up this Desmos graph, change the formula, initial value and number of iterations using the sliders to match the question you're looking at.
Following on from the earlier Cake problem...
Solution
Focus on the triangle with base length \(y\). It has a perpendicular height of \(\frac{1}{2}(x+y)\) therefore an area of \(\frac{y(x+y)}{4}\).
The entire cake has an area of \(\frac{3}{4}(x+y)^2\) so this triangle is a quarter of that.
\(\frac{y(x+y)}{4}=\frac{1}{4} \times \frac{3}{4}(x+y)^2\)
\(y(x+y)=\frac{3}{4}(x+y)^2\)
\(y=3x\)
So the ratio is 1:3.
One of my favourite ice-breakers that involves thinking outside the box.
Typically, students see it as a 2-dimensional problem. This video by The Laughing Clock serves as a nice prompt think more broadly.
Below is an image which can be printed out. Cutting along each line forms a "snake" with a line running along it. Carefully opening it up creates a hole larger than the sheet you started with.
With a thinner snake, it starts to look more like lined paper. It also becomes incredibly delicate.
Here's a plan (top down view) of a cake. The challenge is to cut the cake into congruent pieces. Two and three pieces are pretty easy and may serve as a nice warm up, but then four becomes more challenging.
Here's a solution. There may be others.
An interesting finding while trying to solve the problem is below. This cuts the cake into four pieces of equal size, but they are not congruent. One of my students did, however, point out that it looks 3 dimensional - four congruent rectangles arranged in an L shape - which I thought was a clever loophole of sorts.
A nice little investigation involving Pythagoras' Theorem and geometry. I like how this one diagram, starting with two overlapping circles, contains the square roots of 1 to 5.
Solution
Considering we will be using midpoints, I'm going to say the length of side AB is \(2n\).
\(BY^2=CY^2+BC^2=n^2+(2n)^2=5n^2\implies BY=\sqrt{5}n\)
Given that BX=BY (which can be proved in the same way) we can find angle XBA, which I will temporarily call \(\alpha\).
\(sin(\alpha)=\frac{AX}{XB}=\frac{n}{\sqrt{5}n}=\frac{1}{\sqrt{5}}\)
Now, let's use basic trigonometry on the triangle BYZ, with angle ZYB being \(2\alpha\).
\(YZ=BY cos(2\alpha)=\sqrt{5}n (1-2{sin}^2 (\alpha))=\sqrt{5}n (1-\frac{2}{5})=\frac{3\sqrt{5}n}{5}\)
That's two sides of the right-angled triangle, so use Pythagoras' Theorem to find BZ.
\(BZ^2=BY^2-ZY^2=(\sqrt{5}n)^2-({\frac{3\sqrt{5}n}{5}})^2=5n^2-\frac{9n^2}{5}=\frac{16n^2}{5}\)
Therefore \(BZ=\frac{4n}{\sqrt{5}}\) and the ratio YZ:ZB:BY is \(\frac{3n}{\sqrt{5}}:\sqrt{5}n:\frac{4n}{\sqrt{5}}\) which when multiplied through by \(\frac{\sqrt{5}}{n}\) gives \(3:5:4\)
The triangle is therefore a 3:4:5 triangle.
Here's a problem ladened with shortcuts and assumptions. Present it and ask students what they need to know. Some of it will likely turn out to be superfluous, which might become apparent in the working.
Assumptions:
The cowboy's hat is hemispherical, with the annulus meeting it's great circle.
The witch's hat is conical.
Both hats have zero thickness.
Required information:
A variable for the radius of the head circumference, which I will call \(x\).
Solution
First off, the surface area of the inside and bottom of the hats are the same as the top of the hats, so focus purely on that. The annuli (rim part of the hats) are also the same, so we can now focus only on the hemisphere and cone.
The hemisphere has surface area \(\frac{1}{2} 4\pi x^2 = 2\pi x^2\)
If the cone has the same radius and height \(h\) then the surface area is \(\pi xl=\pi x \sqrt{h^2+x^2}\).
Equating these, we have \(2 \pi x^2 = \pi x \sqrt{h^2 +x^2}\\ \implies 2x=\sqrt{h^2 +x^2}\\ \implies 4x^2=h^2+x^2\\ \implies 3x^2=h^2\\ \implies \sqrt{3}x=h\).
The required ratio is therefore \(1:\sqrt{3}\)
As an extension to an earlier post, let's add a condition that the quadrilateral must also be cyclic.
If the triangles are connected base-to-base, like in Setup 1, the symmetry forces it to be a kite, with \(x+y=90\). A special case is if the other angles are also \(90^{\circ}\) in which case it becomes a square.
If the triangles are set up base-to-leg, like in Setup 2, the apex angle in the left triangle must be \(180-2y\) and because opposite angles in a cyclic quadrilateral sum to \(180\), this means that \(x=2y\). The apex angles in the right triangle is now \(180-4y\).
What quadrilateral could this be?
First off, the apex angle of the left triangle is \(180-2y\) and this is opposite the angle of size \(x\) in the cyclic quadrilateral. They sum to \(180\) so \(180-2y+x=180 \implies x=2y\).
Now, let's use this fact to test types of quadrilateral. It cannot be a square or rectangle as \(x\) cannot be \(90\).
Can it be a kite? If it were, either \(x=180-2y\) and recalling the fact that \(x=2y\), we get \(x=90\) which is not possible. Or, the other two angles are equal giving \(y+(180-2x)=y+x\implies 180=3x \implies x=60\). This makes the triangle on the right an equilateral triangle and \(y=30\). The kite then has angles 120, 90, 60 and 90. This seems feasible.
Can it be a trapezium? Assume it is, then we can use facts about angles in parallel lines. Either \(y=x\), because of alternate angles, but then \(y=x=2y \implies y=0\) which is not possible. Considering a different pair of lines to be parallel, we get a different pair of alternate angles producing the equation \(y=180-2x\) and recalling the earlier finding that \(x=2y\) we get \(y=36\). The trapezium would have angles 108, 108, 72 and 72.
Overall, the quadrilateral can be a square, kite or an isosceles trapezium.
Prior to learning about Circle Theorems, I introduce the idea that two radii and a chord form an isosceles triangle. More information on this blog post but here's another few questions which can be used with students before studying Circle Theorems.
A relatively simple problem with a nice solution.
The upper route is a semicircle with diameter 20 therefore it has length \(10\pi\).
The lower route is made of semicircles with lengths \(2.5\pi\), \(5\pi\) and \(2.5\pi\). which sum to \(10\pi\).
The routes therefore have the same length.
Solution
One way to look at this is to consider the start and end times. 00:00 - 00:15, 01:00 - 01:15 and 02:00 - 02:15 are the three intervals where both hands are in the region. Note that 03:00 is another instantaneous instance, but this does not affect the proportion.
The total time is 45 minutes every 12 hours. That's \(\frac{1}{16}\) of the day.
Another way of looking at this is that there is \(\frac{1}{4}\) probability of each hand being in the region. So there is \(\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\) probability of both hands being in the region.
Here's a more challenging one!
Just like before, here are the times where the minute hand is in the upper region and the hour hand is in the other region. 04:00 - 04:15, 05:00 - 05:15, 06:00 - 06:15, 07:00 - 07:15. That's 1 hour in total so far. Note that 08:00 is another instantaneous case.
And if the hands are in the other regions. 00:20 - 00:40, 01:20 - 01:40, 02:20 - 02:40. That's another hour.
In total, that's 2 hours every 12 hour period, which is \(\frac{1}{6}\) of the day.
Again, thinking about probabilities, we could say that the probability of the minute being in the upper region is \(\frac{3}{12}\) and the probability of the hour hand being in the lower region is \(\frac{4}{12}\) which when multiplied together gives \(\frac{12}{144}=\frac{1}{12}\). The probability of the minute and hour hands being in the lower and upper regions, respectively, is the same. Doubling the probability gives \(\frac{1}{6}\) as before.
This can be deduced through exhaustion. Shortcuts can be taken, such as A not being a multiple of 10 (otherwise B has a leading 0) and each number only being able to end in certain digits because they are square.
A=144, B=441 and C=121. 12 is a factor of A and 21 is a factor of B.
A+B+C= 706.
Here's another problem which is more open-ended.
A and C have the same first digit and are both 3 digits long. If A is a factor of C then A=C. By the same logic, B=C as they share the same last digit. So A=B=C.
But what's special about C?
Well, the first and last digit of B are the same, because A and C end in the same digit. Then, as A=B=C, we know that C is of the form XYX where X and Y represent digits and X is not 0.
Here's a similar problem that's a bit less open-ended.
If A+B=C then A must end with the digit 0, because B and C share a common last digit. But B cannot start with 0 as it's a properly written 3 digit number.
For the subtraction proof, B is a 3 digit number, meaning that at least 100 is being subtracted from A. The first digit of C must be less than the first digit of A, but A and C share a common first digit so this cannot be true. Hence, A-B=C cannot be true.
Given that Pythagoras' Theorem involves the sum of squares, it's common to see squares arranged around the sides of a right angled triangle.
Here are some questions where other shapes are considered.
The semicircular problem
By Pythagoras' Theorem, the hypotenuse is 20cm. The areas of the semicircles are \(\frac{1}{2}\pi\times 6^2=18\pi\), \(\frac{1}{2}\pi\times 8^2=32\pi\) and \(\frac{1}{2}\pi\times 10^2=50\pi\). The sum of the areas of the two smaller semicircles is the area of the larger semicircle.
The equilateral triangle problem
The hypotenuse is 13cm. The area of an equilateral triangle is \(\frac{\sqrt{3}}{2}x^2\) where \(x\) is the side length. The areas of the triangles are therefore \(\frac{\sqrt{3}}{2}5^2=\frac{25\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}12^2=72\sqrt{3}\) and \(\frac{\sqrt{3}}{2}13^2=\frac{169\sqrt{3}}{2}\).
Like before, the sum of the areas of two smaller triangles is the area of the larger triangle.
The similar shape problem
The hypotenuse is 50cm. Shape Q is similar to shape P with a linear scale factor of \(\frac{4}{5}\). This means that the area scale factor is \(\frac{16}{25}\). The area of Q is then \(\frac{16}{25}\times 1800=1152 cm^2\)
The triangular numbers form a quadratic sequence, but interestingly the last digit of each triangular number forms it's own sequence.
The sequence begins with 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1. This is a palindrome!
Including the "zero-th" term and 19th term, we get 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0. The next 20 terms follow the same sequence and given that the last digits of \(n\) repeat every 10 terms, these 20 terms will repeat indefinitely.
Triangulation seems to be the go-to approach when deriving the sum of interior angles, which feels intuitive given that we are looking at angles inside the polygon, but there are other ways, some of which might be easier to follow.
How might each of these be used to derive the sum of interior angles of a polygon?
Method 1 - Fan Triangulation
Drawing line segments from one vertex to all other vertices of an \(n\) sided polygon produces \(n-2\) triangles, each of which have angles that add up to 180 degrees. The sum of interior angles is therefore \(180(n-2)\).
Method 2 - Interior Point Triangulation
Drawing line segments from one point inside an \(n\) sided polygon to each of its vertices produces \(n\) triangles each of which have angles that sum to 180 degrees. The sum of the all these angles is \(180n\) but this includes the full turn at the point inside the polygon. Subtract 360 degrees to result in \(180n-360=180(n-2)\)
Method 3 - Summation of Interior and Exterior Angles
Each interior and exterior angle pair add up to 180 degrees. There are \(n\) angles so these add up to \(180n\). The sum of interior angles is this, minus the sum of exterior angles, which is always 360 degrees for a convex polygon. We then get \(180n-360=180(n-2)\) degrees, but without considering any triangulation.
Solutions
1) Two full turns would be 720 degrees. Subtract the co-interior angles to result in 540 degrees.
2) By looking at alternate angles, the middle angle is the size of the outer angles combined.
3) The middle angle can be split into two angles and by looking at alternate angles, they would make two full turns. The three angles therefore add up to 720 degrees.
4) Drawing "south-lines" at each vertex, we would end up with four pairs of co-interior angles. These add up to 720 degrees. Add the two additional 180 degree angles from the outside to get 1080 degrees.
5) \(n\) angles would result in \(n-1\) pairs of co-interior angle pairs. These add up to \(180(n-1)\) and with the two additional 180 degree angles from the outside, they add up to \(180(n+1)\).
Spelling tests and correcting misspelled words are good methods of encouraging literacy improvement, but they focus on the teacher trying to direct students' attention towards the key words and spellings.
Wordsearches can get students thinking about spellings and meanings without prompts from the teacher. Students find a word and check with peers and teachers to see if it's a mathematical word. Here's a 19x19 grid with 88 mathematical words. If you want a worksheet version with or without a list of words, check here.
