# Minimizing the area of the triangles containing a square of side $1$

This exercise is from a past admission exam to an Italian institute:

Among all the triangles that contain a square of side $1$, which ones have minimum area?

I have solved it, however I'd like to know if there is a faster/easier way. Here's my solution.

For the sake of minimality, the triangles whose perimeter contains all the vertices of the square are to be considered: the ones sharing an angle with the square, whose hypotenuse passes through a vertex $M$ and gives two right triangles intersecting the axes; and the ones whose base passes through two vertices and whose legs pass through the other two. We will deal with them separately and then compare the areas.

Let $A,B,C$ be the vertices of the big right triangle and $B_1,C_1$ the vertices of the square that are respectively also vertices of the two little right triangles. Then the area of the large triangle is $$A_{ABC}=1+A_{{BMB}_1}+A_{{CMC}_1}=1+\frac{{BB}_1+{CC}_1}{2}.$$ So we want to minimize $S=\frac{{BB}_1+{CC}_1}{2}$. Call respectively $\beta$ and $\gamma=\displaystyle\frac{\pi}{2}-\beta$ the angles at $B$ and $C$. Applying the law of sines to the small triangles we obtain $$BM=\frac{1}{\sin\beta}, \ CM=\frac{1}{\sin\left(\frac{\pi}{2}-\beta\right)}$$ whence, using Pythagoras, $${BB}_1=\sqrt{\frac{1}{\sin^2\beta}-1} =\cot\beta$$ $${CC}_1=\sqrt{\frac{1}{\sin^2\left(\frac{\pi}{2}-\beta\right)}-1}=\cot\left(\frac{\pi}{2}-\beta\right).$$ (Note that we have no sign issues as both sine and cosine are positive in $\left(0,\frac{\pi}{2}\right).$) Since $\cot x$ is convex in $\left(0,\frac{\pi}{2}\right),$ by Jensen's inequality we have $$S\ge \cot \frac{\pi}{4},$$ that is $\beta = \gamma = \frac{\pi}{4}$ minimize $S$. Thus the right triangles are in fact isosceles, of hypothenuse $2\sqrt2$ and area $2$.

As for the other class of triangles:

clearly the lower little triangles are similar, yielding that also in this case the big triangle is isosceles. Furthermore, the upper triangle is similar to $ABC$. Now, applying the law of sines to both $CMN$ and $ABC$ and combining the results we get $$AN=AK+BL.$$ But applying it to $AKN$ and $CLM$ yields equality of the summands, so actually $$AN=2AK.$$ By Pythagoras we can rewrite $AN$ as a function of $AK$, and solving for this we find $AK=\displaystyle\frac{\sqrt3}{3}$. By similarity we have \begin{align}1+\frac{2}{3}\sqrt3&=\frac{CN+\frac{2}{3}\sqrt3}{CN}\\ CN&=1,\end{align} therefore $ABC$ is equilateral and its area is $${A'}_{ABC}=\frac{\sqrt{3}}{4} \left(\frac{2}{3}\sqrt3+1\right)^2=\frac{7}{12}\sqrt3+1.$$ Since it is larger than $2$, the sought answer is:

the isosceles right triangles of hypotenuse $2\sqrt 2$.

Could I have known beforehand which of the two classes was the minimizer?

• Let the side containing 2 vertices of the square have length 1+d. Then the height is 1+(1/d). So area is 1+(d+1/d)/2. That has minimum 2 when d=1. Any such triangle will do. – almagest Aug 5 '15 at 7:26

You do not need to separately consider two cases. Consider the height $h$ of the triangle $CNM$

Then the height of the triangle $CAB$ is $h+1$.

The area of $CAB$ is determined by its base and height.

Using the similarity of $CNM$ and $CAB$, the base $AB$ of the triangle $CAB$ is

$(h+1)/h$.

Then the area of $CAB$ is $\frac{(h+1)^2}{2h} = \frac 1 2 (h + 2 + \frac 1 h)$.

The minimizing height $h$ is $1$ by AM-GM inequality.