# Pentakite

**Can you show that triangle $BFC$ is congrugent to triangle $DAC$?**

**Can you find a pair of similar triangles?**

**Can you use these results to find the length $DA$?**

*You can read more about the Golden Ratio in this Plus Article.*

*We are very grateful to the Heilbronn Institute for Mathematical Research for their generous support for the development of this resource.*

Draw out a copy of the diagram, and it will help if you draw on the triangle DAC.

Mark on any angles that you can calculate.

There are various ways you can show that two triangles are congruent, which are sometimes called RHS, SSS, SAS, ASA. Since you know more angles than sides, the ASA condition is probably most helpful.

Two triangles are similar if they have the same set of angles. Similar triangles are enlargements of each other.

Well done to Avan from Tanglin Trust School and Kyle in Singapore, Theo from Pate's Grammar School in the UK and Mahdi from Mahatma Gandhi International School in India who all sent in correct proofs. This is Avan's work to prove that triangle $BFC$ is congrugent to triangle $DAC:$

Similarly, Theo wrote:

All interior angles on a regular pentagon are 108˚. Therefore, angles CBF and BCF are 72˚, as when put with angles ABC/BCD, they make a straight line. That means angle BFC is 36˚ and triangle BCF is isosceles. Next, we know that triangle ABC is isosceles as lines AB and BC share the same length. Knowing angle ABC is 108˚, angles ACB and CAB are both 36˚. With this, angle ACD is 72˚, and so is ADC. The bases for triangles BCF and ACD are the same length, and the two equal angles are the same in both triangles, so the two triangles are the same (angle side angle theorem).kyl

Mahdi, Kyle and Avan all found a pair of similar triangles. Kyle and Avan found the same pair. This is Kyle's work:

Kyle and Avan found the length DA. Avan used trigonometry (click below to see the text in Avan's work):

With our new triangle EDX we have the length of the hypotenuse (1) and the angle XED (54). This is because to get angle XED you had to divide angle DEA by 2, which is 108/2, which is 54.

On this triangle the length XD is the opposite side, while the length ED is the hypotenuse. We can solve for the length of XD using the equation sin(54) = opp/hyp.

This can be rearranged to give us hyp$\times$sin(54)=opp

We can substitute in the hypotenuse length (1) to get sin(54)=opp

sin(54) = 0.809 (3 s.f.)

We can multiply this by 2 to get the full length of AD, 0.809$\times$2=1.618

Kyle and Mahdi used their similar triangles to find the length DA. This is Kyle's work:

Mahdi found and used a different pair of similar triangles to find DA. Mahdi's work is similar to Kyles, but is particularly neat:

**Why do this problem?**

This problem asks students to use their knowledge of regular polygons, congruent and similar triangles, and the quadratic formula to calculate the distance between two non-adjacent vertices of a pentagon.

It could be used as an introduction to the golden ratio, or just as an interesting geometry problem on its own!

**Possible approach**

Show students the initial diagram. Ask them if there are any angles they can find (and ask them how they can find them!).

Draw in the triangle $DAC$ - ask if there are any more angles they can now find.

Ask students how many triangles there are now in their diagram - can they identify which are congruent to each other?

**Key questions**

What do you know about the angles of a reguar pentagon?

How many triangles can you spot? What sort of triangles are they?

Can you spot any similar triangles? ... any congruent triangles?

What other lengths are there which are equal to $DA$?

What do you know about similar triangles?

**Possible Extensions**

Pupils could be asked to calculate the ratio of successive terms of the Fibonacci sequence, i.e. calculate

$$\frac 1 1, \frac 2 1, \frac 3 2, \frac 5 3, \frac 8 5 \cdots $$

They can then compare their answers to the value of $DA$.

As an added challenge, they could simplify the fractions:

$$\frac 1 {1+1}, \frac 1 {1 + \frac 1 {1+1}}, \frac 1 {1+\frac 1 {1+\frac 1 {1+1}}}, \frac 1 {1+\frac 1 {1+\frac 1 {1+\frac 1 {1+1}}}}, \cdots$$

There are lots of problems on Nrich based on the Golden Ratio - try searching for "Golden" in the search box!