WSR Maths

Overlapping Circles

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.

The next question is whether \(\sqrt{6}\) can be found in the diagram without drawing too many additional lines.

A 3:4:5 Triangle from a Square

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 \(1:\sqrt{3}\)

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.