Triangles Inscribed in Rectangles

When a triangle is inscribed in a rectangle, the vertices of that triangle are on the sides or vertices of that rectangle.  Sometimes when additional conditions are added, some interesting properties occur.

For example, if the inscribed triangle is an equilateral triangle, and it shares a common vertex with the rectangle that it is inscribed in, then the sum of the areas of the two outer triangles (that share the same vertex) is equal to the area of the third outer triangle.  This theorem was described by a mathematician named Honsberger in the late 20^th century.

If the Inscribed Triangle is an Equilateral Triangle,

then Area X + Area of Y = Area of Z

The amazing thing about this theorem is that it holds true for any angle θ between the bottom of the equilateral triangle and the bottom of the rectangle.  As long as the equilateral triangle can be inscribed in the rectangle, the areas of the two outer triangles add up to the area of the third outer triangle.

To prove this, first find the value of each of the angles of the outer triangles in terms of θ, the angle between the bottom of the equilateral triangle and the bottom of the rectangle, using the fact that the angle sum of a triangle is 180°, and that the sum of the angles forming a line is 180°, and that the sum of the angles forming a right angle is 90°.  This means that ∠ADE = 30° – θ, ∠AED = 60° + θ, ∠ADE = 30° – θ, ∠BEF = 60° – θ, ∠BFE = 30° + θ, and ∠CFD = 90° – θ. 

Second, use A = ½bh =  ½(c sin θ)(c cos θ) = ½ c² sin θ cos θ = ¼ c²(2 sin θ cos θ) = ¼ c² sin 2θ for the area of each right triangle, where c is the hypotenuse and θ is one of the acute angles.  This means that the area of X = ¼ s² sin 2(30° – θ), the area of Y = ¼ s² sin 2θ, and the area of Z = ¼ s² sin 2(30° + θ).  Therefore:

01	X + Y	the sum of the areas
02	= ¼ s2 sin 2(30° – θ) + ¼ s2 sin 2θ	substitution
03	= ¼ s2 (sin 2(30° – θ) + sin 2θ)	factor out ¼ s2
04	= ¼ s2 (sin (60° – 2θ) + sin 2θ)	distribute 2
05	= ¼ s2 (sin 60° cos 2θ – cos 60° sin 2θ + sin 2θ)	sin (a – b) = sin a cos b – cos a sin b
06	= ¼ s2 (√3/2 cos 2θ – ½ sin 2θ + sin 2θ)	sin 60° = √3/2, cos 60° = ½
07	= ¼ s2 (√3/2 cos 2θ + ½ sin 2θ)	combine like terms
08	= ¼ s2 (sin 60° cos 2θ + cos 60° sin 2θ)	sin 60° = √3/2, cos 60° = ½
09	= ¼ s2 (sin (60° + 2θ))	sin (a + b) = sin a cos b + cos a sin b
10	= ¼ s2 sin 2(30° + θ)	factor out 2
11	= Z	substitution

Another two fascinating theorems concerning a triangle inscribed in a rectangle occur when two sides of the rectangle are cut in the same ratio.  If a triangle is inscribed in a rectangle so that it shares a common vertex and cuts two consecutive sides of the rectangle in the same ratio, then the sum of the areas of the two outer triangles (that also share the same vertex) is equal to area of half the rectangle it is inscribed in, and the product of the areas of the two outer triangles (that also share the same vertex) is equal to the product of the areas of the third outer triangle and half the rectangle it is inscribed in.

If E and F Cut the Sides of AB and BC in the Same Ratio,

then Area of X + Area Y = Area of Half the Rectangle

and Area of X · Area of Y = Area of Z · Area of Half the Rectangle

The amazing thing about this theorem is that it holds true no matter where the side is cut.  As long as the second side is cut in the same ratio, the sum of the areas of the two outer triangles is equal to the area of the half the rectangle, and the product of the areas of the two outer triangles is equal to the product of the areas of the third outer triangle and half the rectangle.

To prove this, label the sides of the rectangle b and h, and the ratio k, so that the cut segments are kb and b – kb and kh and h – kh.  Then using A = ½bh, the area of X = ½kbh, the area of Y = ½b(h – kh), the area of Z = ½kh(b – kb), and the area of half the rectangle is ½bh.  Therefore:

01	X + Y	the sum of the areas
02	= ½kbh + ½b(h – kh)	substitution
03	= ½kbh + ½bh – ½kbh	distribute ½b
04	= ½bh	combine like terms
05	= Area of Half the Rectangle	substitution

And also:

01	XY	the product of the areas
02	= ½kbh ∙ ½b(h – kh)	substitution
03	= ¼b2hk(h – kh)	commutative
04	= ¼b2h2k(1 – k)	factor out h
05	= ¼bh2k(b – kb)	distribute b
06	= ½kh(b – kb) ∙ ½bh	commutative
07	= Z ∙ Area of Half the Rectangle	substitution

There are several amazing theorems concerning triangles inscribed in rectangles and the outer triangles that are formed.  If the triangle is equilateral, then no matter the angle between the side of the triangle and the side of the rectangle, the area of two of the outer triangles add up to the area of the third triangle.  If the triangle cuts two consecutive sides of the rectangle in the same ratio, then no matter what the ratio the sides are cut at, the sum of the areas of two of the outer triangles is equal to the area of half the rectangle it is inscribed in, and the product of the areas of two of the outer triangles is equal to the product of the areas of the third outer triangle and half the rectangle it is inscribed in.  All of these theorems show just how wonderful and amazing geometry is.