In this post we find the largest parallelogram, rhombus, rectangle and square that can be contained in a given triangle. We will see that in the first three cases we can achieve half the area of the triangle but no more, while it is generally less than this for a square.

#### 1. The largest parallelogram inside a triangle

It can be readily seen that one can obtain a parallelogram having half the area of a triangle by connecting a vertex with the three midpoints of the sides. (This has half the base length of the triangle and half its height.)

Is it possible to obtain a larger parallelogram? As outlined in [1], if two or fewer vertices of the parallelogram are on sides of the triangle, a smaller similar triangle can be created by drawing a line parallel to the triangle’s side through a vertex of the parallelogram that is interior triangle. (This is done three times in the figure below.) This reduces the problem to the next case.

We are left to consider the case where three or more vertices of the parallelogram on the triangle. We can draw a line from a vertex to the opposite side parallel to a pair of sides (in the figure below is drawn parallel to ), thus dissecting the triangle into two. Each of the smaller triangles then has an inscribed parallelogram where two of the vertices are on a side of each triangle. Then by drawing lines parallel to sides if required, we create two sub-problems each having four vertices of the parallelogram on the sides of the triangle.

Finally, if all four vertices of the parallelogram are on sides of the triangle, by the pigeonhole principle, two of them are on a side (say as shown in the figure below). In this figure, if we let and the height of ABC from be , then by the similarity of triangles and , and has height . Then the area of the parallelogram is which is times the area of . This quantity has maximum value 1/2 when so we conclude that the parallelogram does not exceed half the triangle’s area.

#### 2. The largest rhombus in a triangle

Constraining sides of the parallelogram to be equal (forming a rhombus), we claim that the largest rhombus that can be inscribed in a triangle is also half its area. This can be formed with two of the vertices on the second longest side of the triangle. Suppose are the sides of the triangle with the second longest side length. Then let the segment joining the midpoints of and form one side of the rhombus of length . It remains to be shown that there exist parallel segments of this same length from to . The longest possible such segment has length and the shortest has length , half the length of the altitude of from . We wish to show that . This follows from

The area of this rhombus is clearly half the area of the triangle as it has half the length of its base and half the height.

#### 3. The largest rectangle in a triangle

Here if is the longest side of the triangle we form the rectangle from midpoints of and respectively, dropping perpendiculars onto forming rectangle :

The area of this rectangle is half the area of the triangle as it has half the length of its base and half the height.

Interestingly the reflections of the vertices of the triangle in the sides of the rectangle coincide, showing a paper folding interpretation of this result [2].

Since rhombuses and rectangles are special cases of parallelograms and we found that inscribed parallelograms in a triangle occupy no more than half its area, the rhombus and rectangle constructions here are optimal.

#### 4. The largest square in a triangle

Here we shall see that the best we can do may not be half the area of the triangle. As before, if two or fewer vertices of the square are not on the sides of the triangle it is possible to scale up the square (or scale down the triangle) so that three of the square’s vertices are on the sides. We claim that the largest square must have two of its vertices on a side of the triangle. Suppose this is not the case and we have the figure below.

Consider squares inscribed in so that one vertex is on , is on and is on . We claim that the largest such square is either (two vertices on ) or (two vertices on ). Suppose on the contrary that neither of these squares is the largest. Then we make use of the fact that all 90-45-45 triangles inscribed in have a common pivot point . This is the point at the intersection of the circumcircles of triangles , and . To show these circles intersect at a single point, we can prove that if the circumcircles of triangles and intersect at then the points are cyclic by the following equality:

where the second last equality makes use of quadrilaterals , being cyclic.

Additionally we have

with similar relations for and . Hence is the unique point satisfying

(Each equation defines a circular arc, they intersect at a single point. Note that may be outside triangle .) This point is the centre of spiral similarity of 90-45-45 triangles with respectively on the sides of the triangle. Consider the locus of the points of the square as vary on straight line segments pivoting about . It follows that the fourth point of the square also traces a line segment, between the points and so as to be contained within the triangle.

As the side length of the square is proportional to the distance of a vertex to its pivot point, the largest square will be where is maximised. We have seen that the point varies along a line segment, so will be maximised at one of the extreme points – either when or . We therefore conclude that the largest square inside a triangle will have two points on a side.

If the triangle is acute-angled, by calculating double the area of the triangle in two ways, the side length of a square on the side of length with altitude is derived as

If the triangle is obtuse-angled, the square erected on a side may not touch both of the other two sides. In the figure below the side length of square is the same as if were moved to , where is right-angled. In this case the square’s side length is .

The largest square erected on a side may be constructed using the following beautiful construction [2]: simply erect a square CBDE external to the side and find the intersection points .

These points define the base of the square to be inscribed since by similar triangles

so that

One can use this interactive demo to view the largest square in any given triangle. One needs to find the largest of the three possibilities of the largest square erected on each side. In the acute-angled-triangle case, the largest square is on the side that minimises the sum of that side length and its corresponding perpendicular height – as their product is fixed as twice the triangle’s area, this will occur when the side and height have minimal difference. For a right-angled triangle with legs and hypotenuse , we wish to compare the quantities and , the two possible sums of the base and height of the triangle. We always have because the diameter of the incircle of the triangle is shorter than the altitude from the hypotenuse (i.e. the incircle is inside the triangle). We conclude that the largest square in a right-angled triangle is constructed on its two legs rather than its hypotenuse.

#### References

[1] I. Niven, *Maxima and Minima without Calculus*, The Mathematical Association of America, 1981.

[2] M. Gardner, *Some Surprising Theorems About Rectangles in Triangles*, Math Horizons, Vol. 5, No. 1 (September 1997), pp. 18-22.

[3] Jaime Rangel-Mondragon *“Largest Square inside a Triangle”* http://demonstrations.wolfram.com/LargestSquareInsideATriangle/ Wolfram Demonstrations Project Published: March 7 2011