At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Darts and Kites

Explore the geometry of these dart and kite shapes!

No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.

Angles in Three Squares

Age 11 to 16 Challenge Level:

Well done to Vid from Osnovna šola Loka Črnomelj in Slovenia, who sent us the following solution to the problem:

Angle $c = 45^\circ$ because it is the angle between the diagonal and the side of a square. To prove that $a+b=c$, I shall begin by adding points to the diagram to help with naming angles:

Now we can find several angles:
• Firstly, $\angle CAE = a$ and $\angle BAD = b$.
• Since $DC$ is parallel to $AE$, by alternate angles, we know that $\angle DCA = \angle CAE$.
• Next, we can see that $\angle BCD = b$.
• Similarly, $\angle ABF = \angle CBG = b$.
• Since $\angle BHC$ is a right angle and the sum of angles in a triangle is $180^\circ$, then we have that $\angle CBH = 180^\circ - 90^\circ - b = 90^\circ - b$.
• Hence, we have that $\angle BCA = a + b$.
• In $\triangle ABC$, the line segments $AB$ and $BC$ have equal length, so we have that $\triangle ABC$ is isosceles. Therefore, $\angle BAC = \angle BCA = a+b$.

Therefore, adding the angles of $\triangle ABC$ together, we have: \begin{align}\angle ABC + \angle BAC + \angle ACB &= 180^\circ \\ ((90^\circ - b) + b) + (a + b) + (a + b) &= 180^\circ \\ 90^\circ + (a + b) + (a + b) &= 180^\circ\\ 90^\circ + 2(a + b) &= 180^\circ\\ 2(a + b) &= 180^\circ - 90^\circ\\ 2(a + b) &= 90^\circ\\ a + b &= \frac{90^\circ}{2}\\ a + b &= 45^\circ\end{align}
Therefore, since $c = 45^\circ$, we have that $a + b = c$.