We have already seen fascinating puzzles where circles and triangles interact with one another. Let’s have a look at two more puzzles with squares and triangles.

1. The taco and the straw

For the solution, we will make use of similar triangles and congruent triangles, specifically the Right-Angle Hypotenuse Side (RHS) case. Look at the diagram below.

For convenience, we have labelled the figure. The center of the half-taco is $O$ , and the vertices of the square lunchbox are labelled: $A, B, C,$ and $D$ . One end of the straw rests at the point $D$ , and the other end is labelled $E$ . The tangent point of the half-taco and the straw is labelled $F$ in the given figure.

First, consider the triangles $\triangle AOD$ and $\triangle OFD$ .

They are both right-angled triangles and share the hypotenuse side $OD$ . For the $\triangle AOD$ , the side $AO$ has a length $\frac{1}{2}$ (being the radius), and for $\triangle OFD$ , the side $OF$ has length $\frac{1}{2}$ (also the radius). Thus, the two triangles $AOD$ and $OFD$ are congruent under the case of RHS. So, their corresponding angles and sides are exactly the same. Therefore, the angles $A\hat{O}D$ and $F\hat{O}D$ are equal. Let’s call them $\alpha$ .

Now, look at the triangles $\triangle OFE$ and $\triangle OEB$ . Those two triangles are also right-angled triangles with a common hypotenuse side $OE$ . Sides $OF$ and $OB$ are both equal in length with $\frac{1}{2}$ units (being the radius). Thus, these two triangles are also congruent by RHS. Denote the corresponding equal angles $F\hat{O}E$ and $E\hat{O}B$ by $\beta$ .

To figure out the length $x$ of $DE$ , we will proceed as follows.

Since the triangles $\triangle AOD$ and $\triangle ODF$ are congruent, the sides $AD$ and $FD$ are equal in length. Because the box has a side length of $1$ , $FD$ also has a length of $1$ .

Let the length of $FE$ be $x-1$ units. Since the triangles $\triangle OFE$ and $\triangle OEB$ are congruent, the sides $FE$ and $EB$ are equal in length. Therefore, $EB$ also has a length of $x-1$ units.

Now, we can use the geometry of the angles around the center point $O$ to unlock the rest of the problem. Because the bottom of the box $(AB)$ is a straight line, the angles must add up to $180^\circ$ .

So,

2\alpha + 2\beta = 180^\circ

\alpha + \beta = 90^\circ

\beta = 90^\circ - \alpha.

Since the angle $O\hat{D}F$ is $90^\circ -\alpha$ , and the angle $O\hat{E}F$ is $\alpha$ , the triangles $\triangle ODF$ and $\triangle OEF$ are similar! Thus, we can use similar triangles to proceed with the solutions. By the similar triangles condition, we have;

\frac{DF}{OF}=\frac{OF}{EF}

\frac{1}{1/2}=\frac{1/2}{x-1}.

Then, simplifying it gives us,

x-1 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

So,

x = \frac{1}{4} + 1 = \frac {5}{4}.

2. The sandwich in the box

We’ve looked at tacos, now let’s look at a classic lunchbox staple: the sandwich.

A triangular sandwich with side lengths of $3, 4,$ and $5$ is packed snugly into a square lunchbox. The vertex with the perfect $90^\circ$ right angle rests flat against the left wall of the box. One of the other two corners touches the top edge, and the third corner exactly hits the bottom-right corner of the lunchbox. Given that the box is a perfect square, we need to find the length $x$ of the lunchbox.

Now let’s begin solving the puzzle step by step.

Because the sandwich has side lengths of $3, 4,$ and $5$ , we immediately know it is a classic right-angled triangle. We can use the empty space around the sandwich to form three smaller right-angled triangles in the corners of the box.

To solve this quickly, we are going to use the magic of similar triangles to completely ignore the top-right side of the box and focus entirely on the left wall!

We have the figure below to assist in solving this. Note that for convenience, we have only drawn a triangle instead of the sandwich and a square for the lunchbox.

Look at the left wall of the square box. The $90^\circ$ corner of the sandwich splits this wall into two segments. Let's call the top vertical segment $b$ . Because the total height of the box is $x$ , the bottom vertical segment must be $x-b$ .

The straight left wall of the box is a $180^\circ$ line. Since the sandwich takes up exactly $90^\circ$ of that line, the angles of the empty space above and below the sandwich must add up to $90^\circ.$

Let the angle $D\hat{E}F$ be $\alpha$ . This means the angle $D\hat{F}E$ is $90^\circ - \alpha$ . So, the angle $A\hat{F}B$ must also be $\alpha$ . And now it can be clearly seen that the two triangles $\triangle DFE$ and $\triangle AFB$ are similar. We can now apply our well-known similar triangle conditions.

We have

\frac{DF}{EF}=\frac{AB}{BF}.

So,

\frac{b}{3}=\frac{x}{4}.

Therefore,

b = \frac{3x}{4}.

Finally, to finish off, let’s apply the Pythagorean Theorem for the bottom left triangle:

AF^2+AB^2=BF^2

(x-b)^2+x^2=4^2.

Substitute our value for b:

\Big(x-\frac{3x}{4}\Big)^2+x^2=16

\frac{x^2}{16}+x^2=16.

Multiply the entire equation by $16$ to clear the fraction:

x^2+16x^2=16 \times 16

17x^2=256

x^2=\frac{256}{17}.

Therefore,

x = \pm \sqrt{\frac{256}{17}} = \pm \frac{16}{\sqrt{17}}.

Since $x$ is a length and it is positive, we have $x = \frac{16}{\sqrt{17}}.$

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Square Lunchbox geometry

1. The taco and the straw

2. The sandwich in the box