But in what order?

Geometry Level 3

To the nearest integer, find the maximum area of a quadrilateral with sides 1, 2, 3 and 4.

The answer is 5.

1 solution

Nicholas Stearns
Jul 22, 2015

Using Bretschneider's formula, for a quadrilateral with sides a,b,c,d and opposite angles $\alpha$ and $\gamma$ . The area is given by $A= \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd \cos^2\frac{\alpha+\gamma}{2}}$ where s is the semiperimeter. This is obviously a maximum when $\cos^2\frac{\alpha+\gamma}{2}=0$ . Substituting our values, we get $A= \sqrt{(5-1)(5-2)(5-3)(5-4)} = \sqrt{24}$ So, we round to get $\boxed{5}$

Moderator note:

Brahmagupta's Fomula is a special case of Bretschneider's formula when the area is maximized, equivalently, when it's a cyclic quadrilateral.

Correct me if I am wrong but I think that $\cos^2\frac{\alpha+\gamma}{2}=0\Rightarrow\alpha+\gamma=\pi$ this is obvious because from all the shape that has the fixed perimeter, CIRCLE has the GREATEST AREA. And as we want to maximise the area the quadrilateral, the quadrilateral must be closest to circle and the quadrilateral which is closest to a circle is a cyclic quadrilateral thus I think that $\alpha+\gamma=\pi$

Sid 2108 - 5 years, 10 months ago

You are right. The above solution takes alpha + gamma as pi. Thats how cos^2((alpha+gamma)/2) = 0. As cos pi/2 is 0.

Aaryaman Gupta - 5 years, 8 months ago

But in what order?

The answer is 5.

1 solution

Moderator note:

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