A convex polygon contains a square with side-length 1 and is contained in a parallel square with side-length 2 (which is its smallest containing square). What is the smallest possible angle of the polygon? What is its smallest possible area?

After some playing around with GeoGebra, I found out that in both cases the minimal value is achieved when the contained square is in a corner of the containing square, so the minimal angle is (probably) 36:

enter image description here

and the minimal area is (probably) 1.5:

enter image description here

How can I prove that these are indeed the minimal values? Or are they?

NOTE: If the polygon is not required to be convex, then obviously its area can be anywhere from 1 to 4 and its angles can be arbitrarily small. So this question is an attempt to quantify the effects of convexity.

  • $\begingroup$ I agree with the minimum angle but I disagree with the minimum area you obtained; which, in my opinion, should be $1$. $\endgroup$
    – heropup
    May 19, 2014 at 8:57
  • $\begingroup$ @heropup I corrected the question. Is this true now? $\endgroup$ May 19, 2014 at 8:59
  • 1
    $\begingroup$ @Blue because then the smallest containing polygon weren't of size 2. I add a clarification. $\endgroup$ Sep 18, 2014 at 10:07
  • 1
    $\begingroup$ @COTO I mean that the smallest square containing the polygon has a side-length of 2. $\endgroup$ Sep 22, 2014 at 8:11
  • 1
    $\begingroup$ As currently written, I would interpret the phrase "which is its smallest containing square" to mean that there is no other square (even among squares not parallel to the inner square) that is smaller than the stipulated square. I think you mean to say "which is the smallest square that is both parallel to the given unit square and contains the polygon." $\endgroup$
    – David K
    Sep 27, 2014 at 4:13

2 Answers 2


The polygon with the least possible area :-


  1. The polygon must contain the 1*1 square wholly

  2. The polygon must be contained in the 2 by 2 square in a way that this square is the smallest square containing the polygon


  1. The polygon must have more sides = 1 unit, so as to be of the least possible area, and accommodate the small square as well (it must wrap the small square). If the polygon has to be contained in the 2*2 square, it must just wrap around the small square

These conditions lead us to something like this:

enter image description here

All these convex polygons have the same area = 1.5 units , because triangles with same base and lying between same parallel lines have equal area, as their height is also equal

If we consider the polygon given in possibility 3 or 4 below, these polygons already have 1.5 sq. units area.

In possibility 3 (below), $\frac{BE*EA}{2}= 0.75$ Therefore total area = 1.75

In possibility 4 (below), $\frac{CE*EA}{2}= 1$ Therefore total area = 2

Therefore, the polygons with least area = the ones in the first picture, with area = 1.5 sq. units


Regarding the NOTE in your question, Even if the polygon is concave, its minimum area needs to be more than 1 sq. units.

Procedure for the smallest angle:-

Three angles will always be 90 or more than 90, so we focus on the 4th angle.

Possibility 1,

enter image description here

The angle has to be 45 (consider the triangle ABC in first figure. if one angle is 90, other 2 must be 45)

Possibility 2

enter image description here


In isosceles triangle ABC, BC = 1 and AD=$\frac{1}{4}$

Let $\angle BAE= \angle A$

$\tan A = \frac{BE}{AE} = 0.5$

$A = \tan^{-1}0.5$ ; $A=26.5650 $

$2A = \angle CAB = 53.130$

We see that when the triangle in possibility 2 that when the triangle in the polygon lying outside the small square is DBC, the angle is 45. As the triangle transforms, or as we move the apex of the triangle to a point halfway between D and F, the angle starts increasing until finally it is 53.13 at A. then again it starts decreasing till it reaches F, it again becomes 45. What happens next?

Possibility 3

enter image description here

The angle formed on the side of the square starts decreasing to below 45

here, $AD=\frac{3}{4}units$

$\tan \angle CAG = \frac{\frac{1}{4}}{2} = \frac{1}{8} =0.125$

$\angle CAG= \tan^{-1} 0.125= 7.1256$


$\tan \angle BAE=\frac{1.5}{1}=1.5$

$\angle BAE= \tan^{-1}1.5 = 56.3099$


$ \angle BAC = 56.3099-7.1256 = 49.1843 $

Possibility 4

enter image description here

Again here, the angle further decreases, till it reaches A. Now, the angle's value is least of all.

In right triangle ADC, CD = $\frac{\sqrt 2}{2}$ and AD= $\sqrt 8 - \frac{\sqrt 2}{2}$

$\tan A = \frac{\frac{\sqrt 2}{2}}{\sqrt 8 - \frac{\sqrt 2}{2}}$ ($\angle A=\angle CAD$)

$\tan A= 0.3333...$

$A = \tan^{-1}0.333...$ ; $A=18.4349 $

$2A = \angle CAB = 36.86989764$


HINT: Assume that the angle is less than your calculated smallest angle and demonstrate that the point will have to lie outside the containing square.


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