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If the sides of a quadrilateral are $a,b,c,d$, prove that the area cannot exceed $(ac+bd)/2$.

What is the Maximum Area of a Quadrilateral with sides of length a,b,c,d (in sequence).How to find the area of a quadrilateral given only the length of it's sides?Find the adjacent sides of the quadrilateral.Area of a concave quadrilateralQuadrilateral of maximum perimeter of given area inside a squareprove that the quadrilateral $ABDC$ is a cyclic quadrilateralFinding the area of the given quadrilateralConvex quadrilateral; Area and parallel sidesArea of an irregular quadrilateral.In the quadrilateral ABCD , the points M and N are the centers of opposite sides AB and DC.

2

$begingroup$

MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.

My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (ab+cd)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $pi/2$. In this case, the area between them will be $frac12ab$. Similarly, the area of the other triangle can have a maximum of $frac12cd$. Adding them up, we get that the maximum area of the quadrilateral can be $frac12(ab+cd)$.

Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $pi/2$, by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.

Is the reasoning given above correct? Does there exist a better proof?

geometry contest-math euclidean-geometry area quadrilateral

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's the MOP? Did the question come from here?
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorGlombik- It did. It's the very first question in the book
  $endgroup$
  – Anju George
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You claim that 'by drawing a diagram, we can convince ourselves that $a=c$'. Are you sure?
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva- Not sure anymore. It seems I didn't consider the possibility of the angle between $b$ and $c$ being greater than $pi/2$.
  $endgroup$
  – Anju George
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right when you claim that the area is maximized when the angle in between is $pi/2$
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.

My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (ab+cd)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $pi/2$. In this case, the area between them will be $frac12ab$. Similarly, the area of the other triangle can have a maximum of $frac12cd$. Adding them up, we get that the maximum area of the quadrilateral can be $frac12(ab+cd)$.

Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $pi/2$, by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.

Is the reasoning given above correct? Does there exist a better proof?

geometry contest-math euclidean-geometry area quadrilateral

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's the MOP? Did the question come from here?
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorGlombik- It did. It's the very first question in the book
  $endgroup$
  – Anju George
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You claim that 'by drawing a diagram, we can convince ourselves that $a=c$'. Are you sure?
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva- Not sure anymore. It seems I didn't consider the possibility of the angle between $b$ and $c$ being greater than $pi/2$.
  $endgroup$
  – Anju George
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right when you claim that the area is maximized when the angle in between is $pi/2$
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.

My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (ab+cd)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $pi/2$. In this case, the area between them will be $frac12ab$. Similarly, the area of the other triangle can have a maximum of $frac12cd$. Adding them up, we get that the maximum area of the quadrilateral can be $frac12(ab+cd)$.

Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $pi/2$, by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.

Is the reasoning given above correct? Does there exist a better proof?

geometry contest-math euclidean-geometry area quadrilateral

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

$endgroup$

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

share|cite|improve this question

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

edited 2 days ago

Michael Rozenberg

107k1895199

edited 2 days ago

Michael Rozenberg

107k1895199

asked 2 days ago

Anju GeorgeAnju George

564

asked 2 days ago

Anju GeorgeAnju George

564

2 Answers
2

active

oldest

votes

3

$begingroup$

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=frac12d_1d_2sinmeasuredangle(d_1,d_2)leqfrac12d_1d_2$$ and by the Ptolemy
$$ac+bdgeq d_1d_2geq2S.$$

share|cite|improve this answer

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For clarity, the inequality, not the equation for cyclic quadrilaterals.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G. All these named the Ptolemy's theorem.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg I don't think you're wrong. I just thought those hyperlinks would help some readers.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean Marie Yes, you are right! I know, it's not so good, to say this, but I really think so. Because for the concave quadrilateral $ABCD$ the idea with replacing $Delta ABC$ on $Delta CBA$ not always works.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I must say as well that I have learned a lot from your mastering of inequalities !
  $endgroup$
  – Jean Marie
  2 days ago
  

  
  
  
  

 | 
show 4 more comments

1

$begingroup$

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $gamma=90°$ since you can express the area of a triangle as $$A=fraca·b·singamma2$$ and $singammaleq1$ with equality if $gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=frac12·(ac+bd)·sinthetaleq frac12·(ac+bd)·1$$

share|cite|improve this answer

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

$endgroup$

add a comment | 

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3

$begingroup$

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=frac12d_1d_2sinmeasuredangle(d_1,d_2)leqfrac12d_1d_2$$ and by the Ptolemy
$$ac+bdgeq d_1d_2geq2S.$$

share|cite|improve this answer

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For clarity, the inequality, not the equation for cyclic quadrilaterals.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G. All these named the Ptolemy's theorem.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg I don't think you're wrong. I just thought those hyperlinks would help some readers.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean Marie Yes, you are right! I know, it's not so good, to say this, but I really think so. Because for the concave quadrilateral $ABCD$ the idea with replacing $Delta ABC$ on $Delta CBA$ not always works.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I must say as well that I have learned a lot from your mastering of inequalities !
  $endgroup$
  – Jean Marie
  2 days ago
  

  
  
  
  

 | 
show 4 more comments

3

$begingroup$

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=frac12d_1d_2sinmeasuredangle(d_1,d_2)leqfrac12d_1d_2$$ and by the Ptolemy
$$ac+bdgeq d_1d_2geq2S.$$

share|cite|improve this answer

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For clarity, the inequality, not the equation for cyclic quadrilaterals.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G. All these named the Ptolemy's theorem.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg I don't think you're wrong. I just thought those hyperlinks would help some readers.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jean Marie Yes, you are right! I know, it's not so good, to say this, but I really think so. Because for the concave quadrilateral $ABCD$ the idea with replacing $Delta ABC$ on $Delta CBA$ not always works.
  $endgroup$
  – Michael Rozenberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I must say as well that I have learned a lot from your mastering of inequalities !
  $endgroup$
  – Jean Marie
  2 days ago
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=frac12d_1d_2sinmeasuredangle(d_1,d_2)leqfrac12d_1d_2$$ and by the Ptolemy
$$ac+bdgeq d_1d_2geq2S.$$

share|cite|improve this answer

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

$endgroup$

share|cite|improve this answer

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

edited 2 days ago

J.G.

29.6k22946

edited 2 days ago

J.G.

29.6k22946

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

answered 2 days ago

Michael RozenbergMichael Rozenberg

107k1895199

1

$begingroup$

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $gamma=90°$ since you can express the area of a triangle as $$A=fraca·b·singamma2$$ and $singammaleq1$ with equality if $gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=frac12·(ac+bd)·sinthetaleq frac12·(ac+bd)·1$$

share|cite|improve this answer

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

$endgroup$

add a comment | 

1

$begingroup$

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $gamma=90°$ since you can express the area of a triangle as $$A=fraca·b·singamma2$$ and $singammaleq1$ with equality if $gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=frac12·(ac+bd)·sinthetaleq frac12·(ac+bd)·1$$

share|cite|improve this answer

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

$endgroup$

add a comment | 

$begingroup$

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $gamma=90°$ since you can express the area of a triangle as $$A=fraca·b·singamma2$$ and $singammaleq1$ with equality if $gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=frac12·(ac+bd)·sinthetaleq frac12·(ac+bd)·1$$

share|cite|improve this answer

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

$endgroup$

share|cite|improve this answer

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526

answered 2 days ago

Dr. MathvaDr. Mathva

2,286526