![]() ![]() It goes without saying (see the discussion of the general Isoperimetric Theorem) that our statement admits an equivalent formulation:Īmong all triangles with the given area, the equilateral one has the smallest circumscribed circle. Its area is 0 and, therefore, it serves an example of an inscribed triangle with the least area.) A triangle with a 0° angle is called degenerate. Allowing for an expanded angle range that includes 0, α = 0. (The second root x = 1, leads to cos(2α) = 1. In the range of interest (0 < β < 180°), the equation cos(β) = -1/2 admits only one solution, viz., β = 120°, hence 2α = 120°, and α = 60°, making ΔABC indeed equilateral,scalene,equilateral,right,obtuse. Derivatives vanish at internal extrema and saddle points, whereas, for this problem we are only interested in a maximum. Using the quadratic formula, the equation has two roots: x = -1/2 and x = 1, the latter being a spurious byproduct of the applied technique, i.e., looking for the roots of a derivative. Letting x = cos(β) we are left with solving the quadratic equation f(x) = 2x² - x - 1 = 0. The simplest way to do that is to compute and equate to 0 the derivative:į'(β) = cos(β)(1 - cos(β)) + sin(β) sin(β) = -cos²(β) + cos(β) + (1 - cos²(β)). So to maximize the area of triangle ABC we need to find the maximum of function f(β) = sin(β)(1 - cos(β)), where β = 2α. Thus, in the diagram, we successively obtain Let O be the circumcenter of the triangle and D and E the midpoints of sides BC and AB, respectively. Then half the apex angle at C equals 90° - α. Let α be the base angle of an isosceles triangle ABC. The lemma also shows that in order to prove the statement we only need to look among isosceles triangles. The only triangle for which no improvement is possible is equilateral. The lemma shows that for a triangle that has two unequal sides, there is another triangle (an isosceles one at that) with the same circumcircle but larger area. Since each of the hypotenuses of our right triangles is also a radius of the circle, we can find all of the sides of these triangles using the 306090. The assertion of the lemma is quite obvious:Īmong all inscribed triangles with a given base, the tallest one is isosceles and, therefore, it has the largest area, due to the standard formula A = b×h/2, where A, b, and h are the area, the base and the altitude of a triangle. ![]() |Contact| |Front page| |Contents| |Geometry|Ĭopyright © 1996-2018 Alexander BogomolnyĪmong all triangles inscribed in a given circle, with a given base, the isosceles one has the largest area. A related theorem concerning the triangles inscribed into a given circle is also true:Īmong all triangles inscribed in a given circle, the equilateral one has the largest area. ![]() Maximum Area Property of Equilateral TrianglesĪ particular case of the Isoperimetric Theorem tells us that among all triangles with the same perimeter, the equilateral one has the largest area. ![]()
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