Indian Mathematical Olympiad 1986

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Leon

256 posts

Leon

#1 h Aug 13, 2009, 11:04 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Show that among all quadrilaterals of a given perimeter the square has the largest area.

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Kunihiko_Chikaya

14514 posts

Kunihiko_Chikaya

#2 h Aug 13, 2009, 11:16 AM • 3 Y

Y by vp2016, Adventure10, Mango247

Let the quadrilateral has side length $a,\ b,\ c,\ d$ with $p=\frac{a+b+c+d}{2}$ ,

we have $[ABCD]=\sqrt{(p-a)(p-b)(p-c)(p-d)}=(\sqrt[4]{(p-a)(p-b)(p-c)(p-d)})^2$

$\leq \left\{\frac{(p-a)+(p-b)+(p-c)+(p-d)}{4}\right\}^2=\frac{p^2}{4}$ ,

the equality holds when $a=b=c=d$ .

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Bugi

1857 posts

Bugi

#3 h Aug 13, 2009, 12:06 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

I think that formula holds only if the quadrilateral is cyclic.

But we can do like this:

Solution

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Kunihiko_Chikaya

14514 posts

Kunihiko_Chikaya

#4 h Aug 13, 2009, 12:12 PM • 1 Y

Y by Adventure10

Bugi wrote:

I think that formula holds only if the quadrilateral is cyclic.

You are right.

Related problem

1969 Czechoslovakia Mathematical Olympiad

A convex quadrilateral $ABCD$ with sides $AB = a,\ BC = b,\ CD = c,\ DA = d$ and angles $\alpha = \angle{DAB},\ \beta = \angle{ABC},\ \gamma = \angle{BCD},\ \delta = \angle{CDA}$
is given. Let $\displaystyle s = \frac {a + b + c + d}{2}$ and $P$ be the area of the quadrilateral. Prove that
$P^2 = (s - a)(s - b)(s - c)(s - d) - abcd\cos ^ 2 \displaystyle \frac {\alpha + \gamma}{2}.$
Now the name of the nation changed,\ didn't that?

It occured to me that Ellipse has something to do with the solution.

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Mathias_DK

1312 posts

Mathias_DK

#5 h Aug 14, 2009, 1:32 PM • 3 Y

Y by Titttu, Adventure10, Mango247

Leon wrote:

Show that among all quadrilaterals of a given perimeter the square has the largest area.

Lemma 1:
Given a triangle $\triangle ABC$ , with $BC$ and $AB+AC$ fixed, then the area of $\triangle ABC$ is largest when $AB=AC$ .
Proof:
The locus of all points $P$ such that $PB+PC = AB+AC$ is an ellipse. Since the area of $\triangle PBC$ is the distance from the line $BC$ to $P$ times $BC$ , we see that $\triangle PBC$ has the maximal area when $P$ are furthest away from $BC$ . Because of symmetry this must be when $P$ lies on the bisector of $BC$ , and hence $AB=AC$ .

Let $\square ABCD$ be the quadrilateral with the largest area. Using lemma 1 on $\triangle ABC$ we can wlog assume that $AB=BC$ . And using on $\triangle ADC$ : wlog $AD=DC$ . Hence $\triangle BCD$ is concurrent to $\triangle DAB$ . So it is enough to find the maximal area of $\triangle BCD$ when $BC+CD$ is fixed. Using Lemma 1, we see that we can wlog assume that $BC=CD$ , so $\square ABCD$ must be a rhombus. Let $E$ be the projection of $C$ on $BD$ . Then $\triangle BEC$ is a rightangled triangle, with $BC$ fixed. Now $BE^2+EC^2 = BC^2$ is also fixed, and the area of $\triangle BEC$ is $\frac{1}{2}BE \times EC$ . Since $BE^2 + EC^2$ is fixed we see that $BE \times EC$ is largest whenever $BE=EC$ .(Wellknown) Doing this for $\triangle DEC$ also we see that $\angle BCD = 90^\circ$ , and since we could use this symmetric argument for all the angles in the quadrilateral, $\square ABCD$ must be a square.

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Maharjun

109 posts

Maharjun

#6 h Dec 29, 2009, 11:17 AM • 3 Y

Y by lckihi, Adventure10, Mango247

I have made use of the result that if the sum of two, or more positive quantities is constant. then their product is maximum when they are all equal. For two quantities, it can be proven using A.M G.M as follows-

Click here if interested in proof of above inequality for two quantities

Step 1:(prove that the quadrilateral is a rhombus,using hero's formula for area of triangle)
solution

step 2 -(prove that the rhombus is a square, trivial)
solution

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guptaamitu1

656 posts

guptaamitu1

#7 h Nov 24, 2020, 6:38 AM

Y by

Maharjun wrote:

I have made use of the result that if the sum of two, or more positive quantities is constant. then their product is maximum when they are all equal. For two quantities, it can be proven using A.M G.M as follows-

Click here if interested in proof of above inequality for two quantities

Step 1:(prove that the quadrilateral is a rhombus,using hero's formula for area of triangle)
solution

step 2 -(prove that the rhombus is a square, trivial)
solution

I think that this solution is wrong (because according to me, step 1 of your solution is not correct). Basically, you are there trying to show that if we replace $(a,b)$ by $({a + b\over 2},{a + b\over 2})$ then the area of the quadrilateral will increase. But that is not always true. Since if you do this replacement, then the value of $e$ also gets effected. So you cannot guarantee that the area of the quadrilateral will increase.

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denery

180 posts

denery

#8 h Apr 14, 2021, 5:01 AM

Y by

imagine a person using jensen" s inequality to prove it area of quadrilateral is a convex function

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