Download My Favourite Problem No.5 Solution

My Favourite Problem No.5 Solution
The wonderful thing about this problem is that it can be solved in a number of ways.
1) Solution by dissection
So far, we have this construction. The trick here is to
ignore any lines that don’t form the perimeter of a triangle.
The next step is to join the vertices of the inside triangle
to two of the vertices of the outside triangle.
This makes seven triangles overall (including the
shaded one).
All seven of these triangles have the same area (we’ll
look at that later).
Answer: Since all seven triangles have the same area, the shaded triangle is
1
7
of the area of the
original triangle.
This result shouldn’t be accepted until we are sure that all of the triangles have the same area. We
can show this (it’s not a proof) by paper folding!
We can show that triangle A has the same
area as our blue triangle by showing it has
the same base length and perpendicular
height. We can do this using the two
parallel lines shown.
By folding along the edge common to both
triangles, we can see that they reach the
same height and therefore have the same
area.
Same perpendicular height
Same base length
This trick can be done for every one of the triangles surrounding the central one demonstrating that
they do have the same area.
There are two Geogebra files which demonstrate the area property and the paper folding
demonstration:


Feynman 1
Feynman 2
For each file move the slider to gradually reveal the working
2) Solution using the cosine rule, similar triangles and scale factors
A good first step here is to use a triangle that we know something about. An equilateral triangle of
side length 3 units (since we need to divide the side lengths by 3) is a sensible starting point.
We need to label the angles to help to see
what is happening.
2
We already know that the internal angles of
an equilateral triangle are 60°.
If we call one of the angles 𝑥 (marked in the
diagram), we can write down all of the other
angles in the diagram in terms of 𝑥 and 60°.
1
(120 − 𝑥)°
60°
𝑥
This will allow us to spot similar triangles.
3
60°
(120 − 𝑥)°
1
𝑥
The red triangles in the two diagrams above are similar (they have the same angles but different
side lengths).
One final piece of information is needed before we can piece it all together and calculate the ratio
of the areas.
𝒍
(120 − 𝑥)°
1
60°
𝑥
3
The length 𝑙 can be found using the cosine rule:
𝑙 2 = 32 + 12 − 2 × 3 × 1 cos 60
𝑙2 = 9 + 1 − 3
𝑙2 = 7
𝑙 = √7
𝑝
60°
√7
1
60°
𝑥
3
By similar triangles:
𝑝
1
1
=
⇒ 𝑝=
1 √7
√7
and
𝑞
3
3
=
⇒ 𝑞=
1 √7
√7
We can now calculate the side length of our small equilateral triangle:
1
𝑞
𝑥
𝑝
√7
𝑞
The side length of the small equilateral triangle in the original diagram is
√7 −
1
√7
−
3
√7
=
3
√7
The original equilateral triangle has side length 3.
The scale factor for the side lengths is therefore
3
√7
The scale factor for the areas is therefore
( 7) = 7
The inner equilateral triangle is therefore
1
7
÷3=
1
2
1
√7
1
√
of the area of the original equilateral triangle.
Note: You may think that this only gives the answer for equilateral triangles and might not work for
other triangles. There is a clever rule involving affine transformations , such as stretches,
translation, reflection etc., that allows us to generalise from this result. Affine transformations of
shapes preserve certain ratios between distances and areas. Any triangle is an affine
transformation of an equilateral triangle. This means that by stretching our equilateral triangle to
1
make it into another triangle, the ratio of the areas of 7 would stay the same.
There are other ways to obtain the solution. One other method is to use vector equations and the
properties of the vector or cross product, which is in A level Further Mathematics.
Further Investigation
This problem gives a simple (if unexpected) ratio for a triangle.
Is there a similar simple ratio for a parallelogram?
A good way of investigating this would be to start with a rhombus made from two equilateral
triangles.
Alternatively you may like to investigate what happens if you change the distance that the dots are
placed around the perimeter of the original triangle. What happens if you place them



One quarter of the side length along from the corners?
One fifth of the side length along from the corners?
Half of the side length along from the corners?
Can you find a general formula?
You could use dynamic geometry software such as Geogebra to find and test your results.