Maximum area of triangle and quadrilateral - given perimeter

[Saitama]

Problem:
Let $0<a<b$

i)Show that amongst the triangles with base $a$ and perimeter $a+b$, the maximum area is obtained when the other two sides have equal length $b/2$.

ii)Using the result (i) or otherwise show that amongst the quadrilateral of given perimeter, the square has maximum area.

Attempt:
Let $BC=a$ and $AB+AC=b$. From Heron's formula,
$$\Delta =\sqrt{s(s-BC)(s-AB)(s-AC)}$$
Since $s$ is given, we maximise $\sqrt{(s-BC)(s-AB)(s-AC)}$. From AM-GM inequality,
$$\sqrt{(s-BC)(s-AB)(s-AC)}\leq \frac{s^{3/2}}{3\sqrt{3}}$$
For the equality to hold, $s-BC=s-AB=s-AC$ which shows $AB=AC=b/2$. This completes the first part of problem.

How do I start with ii)?

Any help is appreciated. Thanks!

 

[Opalg]

If you divide the quadrilateral into two triangles in two ways, using the diagonals, you see (from (i)) that to have maximum area it must have all four sides equal. So it is a rhombus, and it's not hard to see that the area of a rhombus with given sides is maximised when it is a square.

 

[Saitama]

If you divide the quadrilateral into two triangles in two ways, using the diagonals, you see (from (i)) that to have maximum area it must have all four sides equal. So it is a rhombus, and it's not hard to see that the area of a rhombus with given sides is maximised when it is a square.

I don't see it. Where did you use the fact the perimeter of quadrilateral is fixed?

 

[Opalg]

I don't see it. Where did you use the fact the perimeter of quadrilateral is fixed?

Call the quadrilateral $ABCD$, and divide it into two triangles $ABC$ and $ACD$ by the diagonal $AC$. If the lengths of $AB$ and $BC$ are not equal then (by (i)) you can increase the area of $ABC$ by making them equal (while keeping their sum constant). That will increase the area of the quadrilateral while keeping its perimeter fixed. So for maximum area $AB$ and $BC$ must have equal length. Similarly, using the other diagonal $BD$ to split the quadrilateral into two triangles, you see that $BC$ and $CD$ must have equal length.

 

[Saitama]

Call the quadrilateral $ABCD$, and divide it into two triangles $ABC$ and $ACD$ by the diagonal $AC$. If the lengths of $AB$ and $BC$ are not equal then (by (i)) you can increase the area of $ABC$ by making them equal (while keeping their sum constant). That will increase the area of the quadrilateral while keeping its perimeter fixed. So for maximum area $AB$ and $BC$ must have equal length. Similarly, using the other diagonal $BD$ to split the quadrilateral into two triangles, you see that $BC$ and $CD$ must have equal length.

Thank you Opalg once again! :)

I hope you don't mind me posting so many problems currently. I appreciate the help I have received so far.

 

[soroban]

Hello, Pranav!

Another approach . . .

Let $0<a<b$
i) Show that amongst the triangles with base $a$
and perimeter $a+b$, the maximum area is obtained
when the other two sides have equal length $b/2$.

We have \(\displaystyle \Delta ABC\) with \(\displaystyle BC = a\) and \(\displaystyle AB+AC = b.\)

                . . .
          .               .  A
        .                   o
       .                *  * .
                    *     *
      .         *        *    .
      .     o * * * * * o     .
      .     B     a     C     .
     
       .                     .
        .                   .
          .               .
                . . .

The locus of vertex $$A$$ is an ellipse with foci $$B,\,C.$$

The maximum area occurs with maximum height:
. . when $$A$$ is directly over the midpoint of $$BC.$$

Therefore, maximum area when $$\Delta ABC$$ is isosceles:
. . the equal sides are $$\tfrac{b}{2}.$$

 

[Saitama]

Hello, Pranav!

Another approach . . .


We have \(\displaystyle \Delta ABC\) with \(\displaystyle BC = a\) and \(\displaystyle AB+AC = b.\)

                . . .
          .               .  A
        .                   o
       .                *  * .
                    *     *
      .         *        *    .
      .     o * * * * * o     .
      .     B     a     C     .
     
       .                     .
        .                   .
          .               .
                . . .

The locus of vertex $$A$$ is an ellipse with foci $$B,\,C.$$

The maximum area occurs with maximum height:
. . when $$A$$ is directly over the midpoint of $$BC.$$

Therefore, maximum area when $$\Delta ABC$$ is isosceles:
. . the equal sides are $$\tfrac{b}{2}.$$

This is a great approach, thank you soroban! (Clapping)