The Shrinking Square

Problem:
When the area of the green square is two-fifths of the area of the whole diagram, what fraction do the four black triangles occupy?

Open the post to play with a Java applet.

Interactive Diagram

Solution

This problem has rotational symmetry so it’s a waste of time to consider the whole thing.  Instead, let’s concentrate on one of the triangles.  There are quite a few triangles in the whole diagram (12 I think!) but it makes sense to pick one that includes only the dimensions we need.

Next, think about the wording of the question: the green square is 2/5 of the area of the whole diagram.  Whole means 1.  It doesn’t matter what the actual dimensions are because this is about proportion of a whole.  So if the area of the whole diagram is 1 square unit, then the lengths of the sides of the whole diagram are 1 unit.  Similarly, if the area of the green square is 2/5 of a square unit, then the lengths of the sides of the green square are each the square root of 2/5.

The lengths we don’t know, we’ll call x and y.  (Notice that x appears twice.)

It’s important to recognise the types of triangle that appear in this diagram: Notice the right-angles indicated.

We’re almost at algebra point, but we still need to work out what we need to work out to solve the problem!  We’re asked to find the fraction of the whole diagram occupied by the four black triangles.  We’ve standardised the whole diagram as having area 1 square unit, so we just need to find the total area of the four black triangles and that’s the answer.  But we’ve already been clever and focussed our attention on a small part, which only includes one of the four black triangles, so we need to work out the area of that black triangle and multiply it by 4.

What’s the formula for the area of a triangle?  ½×base×height.  This translates to ½xy.  Multiply this by 4 and we have 2xy.  And that’s what we need to work out.

Step 1: Find x

In the diagram above, ignore the black triangle.  We have a right-angled triangle with hypotenuse 1 and other lengths x and x+√(2/5).  When we’re dealing with lengths of right-angled triangles, Pythagoras’ Theorem is the alarm bell that should be ringing!  Substitute the lengths into the theorem and multiply out the brackets:

This gives us a quadratic equation which we may solve by one of the popular methods.  I like to solve it by completing the square:

(Solving a quadratic equation always gives two solutions, but x represents a length here so we can ignore the negative solution.  Also, note that I have “rationalised the denominator”, i.e. rearranged the solution for x so there is no surd or “square root” on the bottom of the fraction – this is a good habit to get into and usually makes later algebra simpler.)

Step 2: Find y

It’s a little harder to spot how y can be found.  It seems to have no relevance to any part of the diagram other than the black triangle – which we don’t know enough about yet.

The key is in the right angles.  Picture the black triangle in your mind’s eye.  Now zoom in on it slowly.  Notice anything yet?  When we zoom in on the black triangle, there’s a point at which it looks just like its big brother, the red triangle.  It should convince you that they are similar to mention the angles in them.  In the far right of the diagram, the black and red triangles share a common angle.  Now, because they are both also right-angled triangles, the third angle must also be shared.  This means the black triangle is a petite version of the red triangle – this is important because with similar shapes, the ratios between their corresponding dimensions never changes.

For example, the ratio of lengths  y : x (in the black triangle)  must be the same as the ratio   x : x+√(2/5) (in the red triangle).  We can set this up as an equation:

But we already know x, so substitute it in and solve for y:

Answer

All that’s remaining is to calculate 2xy, which will give us the total area of the four black triangles.

That’s it!  The answer is 1/15.

Extension – solving for any green area

We have solved this problem for the special case where the area of the green square is 2/5 the area of the whole diagram.  But we can go one better and find a formula which gives us the total area of the black triangles as a fraction of the whole diagram, for any given area of the green square.  This takes us back to the interactive diagram above: we’re no longer fixed at the point where the green square occupies 2/5 of the whole thing; rather we can find a formula that works wherever point E is.

Let the area of the green square be a².  That means the length of one of the sides of the green square is a.

The algebra for this is quite disgusting.  You end up with this:

But if you plot a graph of this for a between 0 and 1 (which are the extremes that a can have as you drag point E up and down), it all makes sense:

This reason this makes sense is because when a=1, that’s where point E in the interactive diagram is at its highest point and the green square occupies the whole of the diagram, leaving no room for the black triangles (so 2xy=0).

Conversely, when a=0, that’s where point E is at it’s lowest point and the green square has shrunk to oblivion, meaning the black triangles now occupy the whole diagram (so 2xy=1).  In between we get a curve.

Extension – Hidden circles!

Play with the interactive diagram one more time.  As you drag point E up and down fast, you can see the corners of the green square plotting a curved locus between the corners of the whole diagram and the centre.  What type of curves are these?

Well you just need to know one of the circle theorems: Angles in a semicircle are 90°

Then you can infer that the arcs are circular.