## Triangle and squares puzzle: Three squares in an equilateral triangle

#### The triangle and squares puzzle

Find side of equilateral triangle if side length of smallest of 3 squares inscribed in the triangle on top of each other is 1.

**Recommended time** to solve 30 mins.

This is not an easy puzzle that nevertheless needs key pattern identification. Give it a good try.

We'll proceed to explain solution of the triangle and squares puzzle following systematic problem solving techniques.

### Step by step Solution to the triangle and squares puzzle: Three squares in an equilateral triangle

How do you solve it?

As you start analyzing the problem, three questions come to your mind.

The **first question** is,

What are the components in the figure that can be considered independently?

Answer:The components arethree squares and four triangles.You have to precisely identify all the components missing none, especially, the components around the problem area. As you know that the side length of the smallest square is the starting point of solution, the small triangle at the top of the figure also is to be counted.

The **second question** is,

What are the types of the four triangles?

They look like equilateral, but you must ensure mathematically for a sound solution. We'll soon see what the types of triangles are. But first let's finish listing all our queries. Once we jot down, we can always come back to answer each step by step.

The **third question** of crucial importance is,

How can I relate the small square with the nearby small triangle?

This is because, if you can get the side length of the second triangle from the top using the side length of the topmost square, *you have broken through the main barrier to the solution.*

Now you have a more clear idea on how to proceed. In the figure below all the vertices are labeled for ease of reference.

Side LM of square FLMG is part of side DE of the larger square DNPE. It follows, DE || FG and similarly, FG || HI.

Coming down, being a square, side DE is parallel to side NP.

This results in, BC || DE || FG || HI.

It follows from similar triangle condition,

$\triangle AHI$, $\triangle AFG$, $\triangle ADE$ and $\triangle ABC$ are all similar and so are equilateral, as $\triangle ABC$ is equilateral.

This is the **first breakthrough.**

**All four triangles are thus equilateral. **This answers your second question.

It follows immediately, sides of smallest $\triangle AHI$ at the top are of 1 unit length.

Now is the time to *answer the third crucial question in a slightly different form,*

Which triangle should we relate to the topmost square HJKI? It is not $\triangle AHI$ as you know all about its relation with square HJKI. It is the equilateral $\triangle AFG$, second from top.

This **result change the problem totally to,**

How to get side length of next larger equilateral $\triangle AFG$ from that of just smaller equilateral $\triangle AHI$?

The two triangles are similar, but you don't get the length of side $AF$ straightaway.

This is the time to think new and ask the question,

How much distant is the base of $\triangle AHI$ from the base of $\triangle AFG$?

This is the *most direct way to relate the two triangles.*

If you know this distance, the side length AF can easily be determined.

This is the **crucial key point in the whole process of solution**.

**Distance means perpendicular distance** and the difference of lengths of perpendiculars drawn to the bases is the separation between the bases you want.

The following is the figure with the perpendiculars from apex A to all the 4 bases of the four triangles drawn.

Now you know there is no hidden mystery in the puzzle anymore. It is just Geometry from this point on.

All four triangles being equilateral, **perpendiculars from apex to the bases divide the bases into half at points $P$, $Q$, $R$ and $S$.**

In $\triangle AHI$,

$AH = 1$, $HP=\frac{1}{2}$ as $HI=1$.

By Pythagoras theorem, in $\triangle AHP$,

$AP=\sqrt{1-\displaystyle\frac{1}{4}}=\displaystyle\frac{\sqrt{3}}{2}$.

Add side length of $HI=PQ=1$ to $AP$ to get,

$AQ=AP+PQ=1+\displaystyle\frac{\sqrt{3}}{2}$.

Now it is a reverse situation where you know the altitude $AQ$ and you have to find out the side length $AF$ of the equilateral $\triangle AFG$. It will be just the reverse of $\displaystyle\frac{\sqrt{3}}{2}$ multiplied by $PQ$,

$AF=\displaystyle\frac{2}{\sqrt{3}}\left(1+\displaystyle\frac{\sqrt{3}}{2}\right)= 1+\displaystyle\frac{2}{\sqrt{3}}$.

Same way,

$AR= AQ+QR=AQ+AF$, as, $QR=FG=AF$.

Or, $AR=1+\displaystyle\frac{\sqrt{3}}{2}+1+\displaystyle\frac{2}{\sqrt{3}}=2+\displaystyle\frac{7}{2\sqrt{3}}$.

Repeating the process,

$AD=\displaystyle\frac{2}{\sqrt{3}}\left(2+\displaystyle\frac{7}{2\sqrt{3}}\right)$

$=\displaystyle\frac{4}{\sqrt{3}}+\displaystyle\frac{7}{3}$.

Finally,

$AS=AR+RS=AR+AD$

$=\left(2+\displaystyle\frac{7}{2\sqrt{3}}\right)+\left(\displaystyle\frac{4}{\sqrt{3}}+\displaystyle\frac{7}{3}\right)$

$=\displaystyle\frac{13}{3}+\displaystyle\frac{15}{2\sqrt{3}}$.

And,

$AB=\displaystyle\frac{2}{\sqrt{3}}AS$

$=\displaystyle\frac{2}{\sqrt{3}}\left(\displaystyle\frac{13}{3}+\displaystyle\frac{15}{2\sqrt{3}}\right)$

$=\displaystyle\frac{26}{3\sqrt{3}}+5$

$=10$, correct to two decimal places.

This triangle puzzle finally needed more of math calculations than expected.

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