Largest Triangle in a Square

Date: 11/03/98 at 01:26:28
From: Jonathan Lee
Subject: Geometry - triangles

If the area of a square is 1, what is the largest area of a triangle 
constructed inside the square? How would you prove it? I think it's 
1/2, but I'm having trouble with a proof. 

Thanks, 
Jonny

Date: 11/03/98 at 13:01:31
From: Doctor Rob
Subject: Re: Geometry - triangles

Hi Jonny,

Good question!

Use throughout the following explanation the fact that the area of the 
triangle is b*h/2, where b is the length of one side, and h is the 
altitude perpendicular to that side. Then if you keep b fixed and 
increase h, or vice versa, the area of the triangle will increase.

If any vertex is not on the boundary of the square, you can increase 
the area by moving it away from the opposite side, thereby increasing 
the altitude. That implies that the triangle with maximum area has all 
three vertices on the boundary.

If no vertex is at a corner, then pick any side of the triangle. By
moving the opposite vertex away from this base, you can increase the 
area by increasing the altitude. That implies that the triangle with 
maximum area has at least one vertex A at a corner.

Suppose that two vertices are on the same side, but not both at a 
corner. Then by moving them apart along that side, you can increase the 
area by increasing the base between them. Thus if two vertices are on 
the same side of the triangle with maximum area, they are at adjacent 
corners of the square.

Suppose that neither B nor C is at a corner, and B and C are on 
different sides not adjacent to A. Then move B along the side of the 
square it is on toward the corner adjacent to the corner A. This will 
increase the area by increasing the altitude to side AC, since AC is 
not parallel to that side of the square. This implies that the triangle 
of maximum area has at least two vertices A and B at adjacent corners.

If the corners A and B are adjacent, then C must be on the side 
parallel to AB, and the area is 1/2.

The result is that the largest the area can be is, as you guessed, 1/2, 
and that is achieved whenever two vertices are at adjacent corners of 
the square and the third vertex is on the opposite side.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/