Same length
Below are two diagrams for you to construct.
Measure the lengths - there seem to be some that are equal!
But can you prove it?
Firstly:
Draw a square, and create an equilateral triangle on two adjacent sides of the square.
It looks like the two outside corners of the triangles, along with the opposite corner of the square, form another equilateral triangle... Can you prove it?
What if you constructed the equilateral triangles on adjacent sides of a rectangle, rather than a square?
What if the equilateral triangles are drawn inside the square? Inside the rectangle?
Secondly:
Watch the animation below.
Draw some equilateral triangles of your own. Are the two lines always the same length?
Can you prove it?
With thanks to Don Steward, whose ideas formed the basis of this problem.
Here are some useful questions to ask yourself when working on geometry proofs:
- What angles do you know? (Mark them on your diagram.)
- Which lengths are equal? (Mark them on your diagram.)
- Which unknown angles are equal? How do you know?
- Are there any congruent triangles?
Pablo from King's College Alicante used diagrams to form a clever proof:
One way to prove that the triangle created is equilateral is to show that each side is the same.
If the sides of the triangle are named 1, 2 and 3, then it is clear (from Figure 1) that 1 and 2 are equal, by symmetry. It is a line of symmetry because the quadrilateral is a square and thus the equilateral triangles formed are identical. The square has a line of symmetry as its diagonal.
Secondly, a copy of the diagram can be placed next to the square and triangles, as seen in (Figure 2). The shapes fit perfectly because the angles around the top right corner of the square add up to 360 º:
90 + 90 + 60 + 60 + 60 = 360 º
Line 3 is now on Line 1 of the copy. Therefore Line 3 and Line 1 are equal.
Since Line 1 = Line 2 and Line 1 = Line 3,
Line 1 = Line 2 = Line 3
Therefore it is an equilateral triangle.
Again, by symmetry Line 1 and 2 are equal.
In (Figure 4) a copy of the diagram has been placed on the side of the
rightmost equilateral triangle.
The only thing that has changed is which side is which. The kite that contains two sides of the red triangle has been rotated. This shows that the 3rd side is equal to the first.
Line 1 = Line 2 = Line 3
Therefore it is an equilateral triangle.
Kian from King Edward VII School in the UK used angles to prove that the triangle formed is equilateral:
Let the vertices of the square be $A, B, C, D,$ and let $E$ and $F$ be the points outside the square such that $AED$ and $DFC$ are equilateral triangles, as shown in the diagram.
Firstly, note that $AB = AD$, as they are sides of a square, and $AD = AE,$ as they are sides of an equilateral triangle, which means that $AB = AD = AE.$ Importantly, $AB = AE.$ This tells us that triangle $ABE$ is isosceles, with $A\hat{B}E = A\hat EB.$
$B\hat AE = B\hat AD + D\hat AE = 90^\circ + 60^\circ = 150^\circ,$ so $A\hat EB = A\hat BE = (180^\circ-150^\circ)\div2=15^\circ$
Now, $A\hat BC = 90^\circ$ as it is an interior angle of a square, so $E\hat BF = 90^\circ - A\hat BE - C\hat BF = 60^\circ.$
Additionally, $AB = BC, AE = CF$ and $B\hat AE = B\hat CF = 150^\circ,$ which means that triangles $ABE$ and $CBF$ are congruent under SAS, and $BE = BF.$ Hence, triangle $BEF$ is isosceles, with $B\hat EF = E\hat FB = (180^\circ-60^\circ)\div2 = 60^\circ.$ All three angles of triangle $BEF$ are equal to $60^\circ$, so it is equilateral.
Jake from GSAL in the UK used the cosine rule:
[These] can be used in the cosine rule, $c^2=a^2+b^2-2ab\cos x$, where $a$ and $b$ are both '$a$'. In all circumstances $x$ can be found to be 150 so it works out into the same result of all sides being equal (in the new triangle) to $\sqrt{2a^2-2a^2\cos{150}}$.
Notice that Jake didn't actually need to apply the cosine rule - the three isosceles triangles must be congruent, since they have the same shorter sides and angle. So their longer sides must also all be equal.
The two equilateral triangles
Pablo and Kian both used similar triangles. This is Pablo's work:
Therefore the triangles are congruent and so the last side for each is equal.
This problem can also be solved using Pythagoras' Theorem, as shown below.
Let the side length of the large equilateral triangle be $a$, and the side length of the small equilateral triangle be $b$.
Take the large equilateral triangle:
Using Pythagoras, we find that the perpendicular height of the larger triangle is
$$\sqrt{a^2 - (\frac{1}{2} a)^2} = \frac{\sqrt{3}}{2}a.$$
Now take the triangle with the blue line as the hypotenuse:
By Pythagoras again, we find that the length of the blue line is given by
$$\textrm{blue line} = \sqrt{\frac{3}{4}a^2 + (\frac{1}{2}a+b)^2} = \sqrt{a^2 + ab + b^2}.$$
Similarly, the perpendicular height of the smaller equilateral triangle is
$$\sqrt{b^2 - (\frac{1}{2} b)^2} = \frac{\sqrt{3}}{2}b.$$
And so we can take the triangle with the red line as the hypotenuse:
to give that the length of the red line is
$$\textrm{red line} = \sqrt{(a+\frac{1}{2}b)^2 + \frac{3}{4}b^2} = \sqrt{a^2 + ab + b^2}.$$
So the red and blue lines are the same length.
Why do this problem?
Geometrical proof is a topic many students find challenging. In this problem, we have brought together two problems about equilateral triangles, so that students can work on the first, discuss it, and develop insights that we hope will give them the tools they need to persevere with the second, harder challenge.
Possible approach
Show students the animation below, where an equilateral triangle is drawn on two adjacent sides of a square:
"It looks like the triangle formed by the corners is equilateral. How could we prove it?"
Give students some time to discuss ideas in pairs or small groups. Then bring the class together and share suggestions.
If no-one has suggested drawing a diagram, point out that for most geometry questions it is a really useful first step. They may wish to construct their diagram accurately and measure the three sides; if so, make sure everyone is clear that this does not constitute a proof.
The Key questions below suggest some prompts that could be used to work towards a proof. Students could work on their proofs in pairs, justifying each stage of reasoning to their partner, and finally agreeing on which steps constitute their proof.
Later in the same lesson, or in a follow-up lesson, present the second problem:
For this problem, students may wish to construct several examples to see if it was merely coincidence that the lines were the same length. This could be an opportunity to use a dynamic geometry program like GeoGebra to generate lots of examples. Once students have seen that the sides seem the same length for many examples, challenge them to find a
proof.
The same prompts in the Key questions below can be used. Again, students could work in pairs, justifying their reasoning to their partner and working collaboratively to piece some ideas together into a proof.
Finally, bring the class together and invite students who think they have a proof to present it to the rest of the class. The class could act as "critical friends" to critique each step and make sure the logic is watertight. A possible homework activity could be to write up the steps of the proof clearly.
Key questions
What angles do you know? (Mark them on your diagram.)
Which lengths are equal? (Mark them on your diagram.)
Which unknown angles are equal? How do you know?
Are there any congruent triangles?
Possible support
Students may find it helpful to work on some of the Angles, Polygons and Geometrical Proofs short problems before tackling this task.
Possible extension
For a more challenging example of geometrical proof, see Kite in a Square.