Witch Hat is Taller

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 \(1:\sqrt{3}\)  

Cyclic Quadrilaterals from Isosceles Triangles

As an extension to an earlier post, let's add a condition that the quadrilateral must also be cyclic.

 

Solution 

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.

Angle Problems Involving Circles

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.

And the worked solutions
 

Circular Routes

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.

If the diameters of the semicircles are \(x_i\) then the arc lengths are \(\frac{1}{2}\pi x_i\). The route has length \(\Sigma \frac{1}{2} \pi x_i\) and using the fact that \(\Sigma x_i=AB\), the lower route has length \(\frac{1}{2} \pi AB\) which is the length of the upper route.

 

Guess my Number

 

252 

 

816 

642

440

512

3145

9876

8125

Time in the Shade

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!

Solution

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.

Square Number Triangle

Solution

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.

Solution

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. 

Solution

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.