Here's the problem:
Show that of all possible quadrilaterals with sides of given length, the one with largest area is such that each pair of opposite angles has sum ?.
Here's my work so far:
Let S be the area of the quadrilateral. Let two opposite angles of the quadrilateral be ? and ?, let a[sub:3otc9sap]1[/sub:3otc9sap] and a[sub:3otc9sap]2[/sub:3otc9sap] be the sides that form ?, let b[sub:3otc9sap]1[/sub:3otc9sap] and b[sub:3otc9sap]2[/sub:3otc9sap] be the sides that form ?, and let h be the diagonal of the two other angles.
The quadrilateral can be considered as two triangles that share one side h; one with sides a[sub:3otc9sap]1[/sub:3otc9sap], a[sub:3otc9sap]2[/sub:3otc9sap], and h; the other with sides b[sub:3otc9sap][/sub:3otc9sap], b[sub:3otc9sap]2[/sub:3otc9sap], and h. Since the area of a triangle ABC is (1/2)bc sin A, the area S of the quarilateral is:
S = (1/2)a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] sin ? + (1/2)b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] sin ?
Now I need to relate ? to ? so I can calculate dS/d? or dS/d?. By the law of cosines:
h[sup:3otc9sap]2[/sup:3otc9sap] = b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2b[sub:3otc9sap]2[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] cos ? = a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ?
Let's say I want the equation for S in terms of ?. Then I need to find an expression for sin ? in terms of sin ? that I can substitute for sin ? in the equation for S. So tried this:
b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] cos ? = a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ?
? cos ? = (2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ? – a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap])/(2b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap]) = sin(?/2 – ?) = sin (?/2) cos(– ?) + cos (?/2) sin (– ?)
However, I can't solve this for sin ? because the only term with sin ? is multiplied by cos (?/2), that is, by zero.
My question:
I'm stuck. Is there another way to go about solving this problem?
Show that of all possible quadrilaterals with sides of given length, the one with largest area is such that each pair of opposite angles has sum ?.
Here's my work so far:
Let S be the area of the quadrilateral. Let two opposite angles of the quadrilateral be ? and ?, let a[sub:3otc9sap]1[/sub:3otc9sap] and a[sub:3otc9sap]2[/sub:3otc9sap] be the sides that form ?, let b[sub:3otc9sap]1[/sub:3otc9sap] and b[sub:3otc9sap]2[/sub:3otc9sap] be the sides that form ?, and let h be the diagonal of the two other angles.
The quadrilateral can be considered as two triangles that share one side h; one with sides a[sub:3otc9sap]1[/sub:3otc9sap], a[sub:3otc9sap]2[/sub:3otc9sap], and h; the other with sides b[sub:3otc9sap][/sub:3otc9sap], b[sub:3otc9sap]2[/sub:3otc9sap], and h. Since the area of a triangle ABC is (1/2)bc sin A, the area S of the quarilateral is:
S = (1/2)a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] sin ? + (1/2)b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] sin ?
Now I need to relate ? to ? so I can calculate dS/d? or dS/d?. By the law of cosines:
h[sup:3otc9sap]2[/sup:3otc9sap] = b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2b[sub:3otc9sap]2[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] cos ? = a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ?
Let's say I want the equation for S in terms of ?. Then I need to find an expression for sin ? in terms of sin ? that I can substitute for sin ? in the equation for S. So tried this:
b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap] cos ? = a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – 2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ?
? cos ? = (2a[sub:3otc9sap]1[/sub:3otc9sap]a[sub:3otc9sap]2[/sub:3otc9sap] cos ? – a[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] – a[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]1[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap] + b[sub:3otc9sap]2[/sub:3otc9sap][sup:3otc9sap]2[/sup:3otc9sap])/(2b[sub:3otc9sap]1[/sub:3otc9sap]b[sub:3otc9sap]2[/sub:3otc9sap]) = sin(?/2 – ?) = sin (?/2) cos(– ?) + cos (?/2) sin (– ?)
However, I can't solve this for sin ? because the only term with sin ? is multiplied by cos (?/2), that is, by zero.
My question:
I'm stuck. Is there another way to go about solving this problem?