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. 

Pythagoras' Theorem

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.

Solutions

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 \(\frac{16}{25}\times 1800=1152 cm^2\)