# Unit square inside triangle. [duplicate]

Some time ago I saw this beautiful problem and think it is worth to post it:

Let $$S$$ be the area of triangle that covers a square of side $$1$$. Prove that $$S \ge 2$$.

• What do you mean "triangle that covers square"? Oct 4, 2016 at 13:55
• @DonAntonio Yes, rigorously the term is vague; but I think it simply means to consider a region enclosed by a plane triangle that includes the region enclosed by the given unit square. To OP: if my interpretation is wrong, please make a necessary explanation in your post :).
– Yes
Oct 4, 2016 at 13:59
• Sorry if expression is confusing. I meant that unit square is entirely inside triangle (vertexes or/and edges of square can be on the sides on triangle - but not outside of it).
– hOff
Oct 4, 2016 at 14:18
• The original question looks pretty clear to me. Suppose the unit square is contained in a triangle. Show that the area of the triangle is necessarily at least $2$. There is no mention of orientation of coincidence of sides or vertices -- these constraints were introduced in the comments and should be ignored.
– MPW
Oct 4, 2016 at 14:19
• @MPW is correct
– hOff
Oct 4, 2016 at 14:21

This follows from the following classic (and equally nice) result:

The side of the largest square inscribed in a triangle is $$L = \frac{c h}{c + h}$$ where

• $$c$$ is the length of the side which minimizes the sum between a side and its corresponding height among the three sides of the triangle (or, provably equivalent, $$c$$ is the length of the side closest in value to $$\sqrt{2 S}$$ where $$S$$ is the area of the triangle);

• $$h$$ is the height corresponding to $$c$$.

Proofs (not copied here) can be found for example at Maximum area of a square in a triangle.

Using the above, it follows that the side of the square $$l = 1 \le L = \frac{c h}{c+h} \le \frac{c h}{2 \sqrt{c h}} = \frac{\sqrt{c h}}{2} = \sqrt{\frac{S}{2}}$$ which proves $$S \ge 2$$.

$$S=2$$ is attained for example with an isosceles right triangle laid along two sides of the square with the hypotenuse passing through the opposite vertex of the square, so $$2$$ is indeed the best bound.

Here is a sketch of proof :

Let's try to find a triangle $T$ with minimum area that covers unit square.

First, we can assume each 3 sides of $T$ have to contain at least one of vertices of square. (If not, we can move side more close to square and this gives new triangle with smaller area)

Second, if a vertex $v$ of square lies on a side $AB$ of $T$, then $v$ is midpoint of $AB$. This is not hard to prove.

(Here $M$ is midpoint of $AB$, and area of $MBB'$ is bigger than $MCC'$. (Consider a line $BD$ which is parallel to $AC$.)

Third, by the first assumption, we can find 3 vertices of square which forms a right angle. By the second property, $T$ should be right triangle. By the second property again, $T$ should be isosceles right triangle, which has area $2$.

(Here $M, N, P$ are midpoint of $BC, AC, AB$. Then $MP$ (resp. $MN$) is parallel to $AC$ (resp. $AB$), so angle $\angle BAC$ should be right angle, i.e. $Q=A$.)

• I think this line of proof holds promise, but some of steps need further refining. Second, if a vertex v of square lies on a side AB of T, then v is midpoint of AB This only works if $v$ is not a vertex of the triangle at the same time, and if there is no other vertex of the square on $AB$. Third, by the first assumption, we can find 3 vertices of square which forms a right angle. By the second property, T should be right triangle. By the second property again, T should be isosceles right triangle, which has area 2 Please elaborate this one, I don't really follow the implications here.
– dxiv
Oct 5, 2016 at 1:30
• For a concrete example of a triangle that has (the minimal) area $2$ but is neither right nor isosceles, consider $A=(-1/4, 0)$, $B=(7/4, 0)$, $C=(1/4,2)$. It is straightforward to verify that $\triangle ABC$ "circumscribes" the unit square $(0,0),(1,0),(1,1),(0,1)$, has area $2$, but is neither a right triangle nor an isosceles one.
– dxiv
Oct 5, 2016 at 1:36
• @dxiv I see. I have to add one more case as you said, such that two of vertices of square lies in same side of triangle and rest two of them are not (i.e. midpoint of some sides of triangle. I think right isosceles triangle can be regarded as degenerated one.) Thanks! Oct 5, 2016 at 5:12

If some edge, say $BC$ of a triangle $ABC$, does not touch the square, by moving $BC$ towards $A$, we may obtain a smaller similar triangle such that the side parallel to the original $BC$ touches the square.

Therefore, we may assume that all three sides of the triangle touch the square. Consequently, the triangle can have at most one vertex (say, possibly $A$) that lies on the open interiors of the NW, NE, SW or SE regions to the four corners (because, if there are two such vertices, the edge join them will not touch the square). So we get something like the figure below:

A       :       :
:       :
NW :       : NE
~~~~.-------.~~~~
|       |      B
|       |
|       |
~~~~.-------.~~~~
SW :       : SE
:       :
: C     :


Drop a perpendicular of $B$ to its adjacent side of the square. Let $P$ be where the extension of this line segment intersects the other side of the square (see the figure below) and let $D$ be where this line extension meets $AC$. Drop a perpendicular of $CQ$ to $PB$. Also, extend the side of the square opposite to $B$. Let this extended line intersects $AB$ at $E$:

A
\      E
\     |
\    .-------.
D---P-Q-----|------B
\  | |     |
\ | |     |
\.-------.
\ |
\|
C


Now the area of $ABC$ is greater than the total area of the triangles $CQD, CQB$ and $BPE$. So, it suffices to prove the following proposition:

if $R$ is an inscribed rectangle of a right-angled triangle $T$ such the two geometric entities share a common vertex (which must be the right angle of $T$), then the area of $T$ is at least double the area of $R$.

But the truthfulness or falseness of the above proposition is invariant if we scale the whole figure along the direction of any side of $R$, because both the areas of $R$ and $T$ scale by the same proportion. Therefore we may further assume that $R$ is a square. Now, borrowing the words of Hagen von Eitzen in another answer here, it is obvious that if we flip the outside tips of $T$ inside, the two flipped tips of $T$ will always "envelope" the interior of the square seamlessly, with the longer tip poking out. Hence the proposition is true.

There are a few cases to consider regarding the relative position and if touching happens at vertices or sides. But you can always assume thet two adjacent vertices of the square touch the triangle perimeter, but not the whole edge. Consider the triangle made from paper. You can flip the outside part of the triangle inside; also flip the outside parts of the adjacent square edges inside. This will always form an "envelope" for the complete square.

Intuitively when vertices of square $S:=v_1v_2v_3v_4$ of side length $1$ are in sides of triangle $T=\Delta\ ABC$, then ${\rm area}\ T$ is smallest

Notation : Edge $xy$ is $[xy]$ And length of edge $xy$ is $|xy|$

Here we have two cases :

(1) $v_1\in [AB],\ v_2\in [AC]$ and edge $[v_3v_4]$ is in edge $[BC]$

(2) $v_1\in [AB],\ v_2\in [AC],\ v_3\in [BC]$ That is $v_4$ is in interior of $T$

Consider case (1) : Consider a reflection point $A'$ of $A$ wrt edge $[v_1v_2]$

(1.1) Assume that $A'$ has distance larger than $1$ from $[v_1v_2]$

Define $$[A'v_1]\cap [BC]:=\{ v_4'\},\ [A'v_2]\cap [BC]:=\{ v_3'\}$$

so that $$2{\rm area}\ S +2 {\rm area}\ \Delta\ A'v_3'v_4' ={\rm area}\ T$$

(1.2) Assume that $A'$ has distance smaller than $1$ from $[v_1v_2]$

If line $v_1 + t(A'-v_1)$ meets at $v_4'$ with $BC$, then note that $${\rm area}\ \Delta A'v_1v_2={\rm area}\ \Delta Av_1v_2$$

$${\rm area}\ \Delta v_1Bv_4 ={\rm area}\ \Delta v_1v_4 v_4'$$

$${\rm area}\ (S- \Delta A'v_1v_2-\Delta v_1v_4v_4' )< {\rm area}\ \Delta v_2Cv_3$$

Consider case (2) : WLOG assume that $[Av_2]<[v_2C]$

(2.1) Now consider $A'$ has distance larger than $1$ from $[v_1v_2]$

Note that $\Delta v_2 A' v_2'$ contains $v_3$ where $$v_2'\in [v_2C],\ |v_2v_2'|=|Av_2|$$

Here $[v_2v_3]$ is angle bisector of $\angle A'v_2v_2'$ so that $\overrightarrow{v_2'v_3}$ enters into square So $Cv_3$ enters into square This contradicts to definition of $v_3$

(2.2) Assume that we have $A'$ in interior of the square $S$

Define $$v_1'\in [v_1B],\ |v_1v_1'|=|Av_1|$$

$$v_2'\in v_2C,\ |v_2v_2'|=|Av_2|$$

$$v_3'\in [Bv_3],\ |v_3'v_3 |=|v_3v_2'|$$

Note that $Av_1'v_4v_3'v_3v_2'$ has area $2$

Here is a nice proof: however the triangle is drawn, we may first bring in its edges (decreasing the area) until each of its $3$ edges go through a different vertex of the square:

From here, cut a line from the final vertex of the square to the vertex of the triangle:

Now simply fold over each of the four triangles. No matter what, the four folds collectively cover the square, proving the triangle has area at least $2$.

### It remains to show: why do these four folds cover the square?

This is simpler than I thought. Let the square's vertices be $A, B, C, D$, such that in the above figures $A$ is bottom-left, $B$ top-left, $C$ top-right, $D$ bottom-right. Also refer to the four triangles that are folded over as the "left", "right", "top", and "bottom" triangles. When reflected across the sides of the square, the four triangles' boundaries inside the square lie on only 3 line segments, one each from vertices $B$, $C$, and $D$. Let the lines from $B$, $C$, $D$ be called $l, m, n$, respectively.

Suppose towards a contradiction that there is a point not covered by the four reflected triangles. To not be covered by the reflection of the left triangle, it must be above $l$. To not be covered by the reflection of the top triangle given that it is above $l$, it must be below $m$. To not be covered by the reflection of the right triangle given that it is below $m$, it must be below $n$. But if it is below $n$ and inside the square then it is covered by the reflection of the bottom triangle. Contradiction, so no such point exists.

• …Any update yet? Aug 10, 2017 at 8:04
• @SridharRamesh Thanks for reminding me about this nice problem. I am convinced it's true that the four folds cover the square, but I can't come up with a good proof. Basically the boundary of every fold inside the square is a subset of just 3 lines (the reflections of the three lines of the triangle). Those three lines divide the square into some number of regions, and we know that any region next to the square's edge is certainly covered. There is at most 1 region actually on the interior of the square, and we need to show that region (if it can even exist) is also covered. Aug 10, 2017 at 18:47
• ...I am thinking maybe that 1 region can't exist. It doesn't exist in the example drawing I made. Any ideas? Aug 10, 2017 at 18:47
• @SridharRamesh I found a simple argument and updated my answer with a full proof. Aug 28, 2017 at 19:23