Areas equation in a trapezoid

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Areas equation in a trapezoid

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: V. Alekseev, V. Galkin, V. Panferov, V. Tarasov in Kvant 6/2000; re-used in 4th QEDMO as problem 2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#1 h Nov 10, 2007, 1:51 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a trapezoid with $BC\parallel AD$ , and let $O$ be the point of intersection of its diagonals $AC$ and $BD$ . Prove that $\left\vert ABCD\right\vert =\left( \sqrt{\left\vert BOC\right\vert }+\sqrt{\left\vert DOA\right\vert }\right) ^{2}$ .

Source of the problem

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1394 posts

nayel

#2 h Nov 10, 2007, 2:50 PM • 2 Y

Y by Adventure10, Mango247

Very nice problem.

The equality is equivalent to $(AOB) + (COD) = 2\sqrt {(BOC)(AOD)}$ . Using $\sin\angle BOC = \sin\angle AOB$ we can rewrite this as
$\begin{eqnarray*} & & AO\cdot BO + CO\cdot DO = 2\sqrt {(BO\cdot CO)(AO\cdot DO)} \\ & \Leftrightarrow & (\sqrt {AO\cdot BO} - \sqrt {CO\cdot DO})^2 = 0 \\ &\Leftrightarrow & \frac {AO}{DO} = \frac {BO}{CO}\end{eqnarray*}$
Which is true as $\angle OBC = \angle ODA \wedge \angle OCB = \angle OAD\Rightarrow \triangle BOC\sim\triangle DOA$ .

(EDIT) Remark: From this problem we can derive the following: in any trapezoid $ABCD$ with $BC\parallel AD$ we have $(BOC)+(AOD)\ge (AOB)+(COD)$ .

This post has been edited 1 time. Last edited by nayel, Nov 10, 2007, 3:13 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

743 posts

tdl

#3 h Nov 10, 2007, 2:55 PM • 2 Y

Y by Adventure10, Mango247

If call $\frac{OC}{OA}=\frac{OB}{OD}=k$ then:
$S_{AOB}=S_{COD}=k.S_{AOD},S_{BOC}=k^2.S_{AOD},S_{ABCD}=(k+1)^2.S_{AOD}$
Then it's OK!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Kunihiko_Chikaya

14514 posts

Kunihiko_Chikaya

#4 h Nov 10, 2007, 3:15 PM • 1 Y

Y by Adventure10

Probably the problem is very famous. Here is an problem from Japan.

1961　Keio University entrance exam/Engineering

Let $O$ be the point of intersection of its diagonals of a convex quadrilateral $ABCD$ and let $a^2,\ b^2\ (a > 0,\ b > 0)$ be the area of $\triangle{AOD},\ \triangle{BOC}$ respectively. Denote the area of the quadrilateral by $S$ ,

(1) Show that $S\geq (a + b)^2$ .
(2) Determine the shape of the quadrilateral when $S = (a + b)^2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7054 posts

#5 h Nov 10, 2007, 10:02 PM • 2 Y

Y by Adventure10, Mango247

Darij Grinberg & Kunny wrote:

Let $ABCD$ be a quadrilateral. Denote $O\in AC\cap BD$ .

Prove that $\sqrt {[ABCD]} = \sqrt {[BOC]} + \sqrt {[DOA]}\Longleftrightarrow BC\parallel AD$ .

Proof. Denote $[ABCD] = S$ and $\{\begin{array}{ccc} [AOB] = u & , & [BOC] = x \\ \ [COD] = v & , & [DOA] = y\end{array}$ .

Show easily that $\boxed {\ xy = uv\ }$ . Thus, $\sqrt S = \sqrt x + \sqrt y$ $\Longleftrightarrow$

$x + y + u + v = x + y + 2\sqrt {xy}$ $\Longleftrightarrow$

$u + v = 2\sqrt {uv}$ $\Longleftrightarrow$ $u = v$ $\Longleftrightarrow$ $BC\parallel AD\ .$

Remark. In any quad. $ABCD$ exists the relation (near to evidence !) $\sqrt S\ge\sqrt x + \sqrt y\ .$

Remark. In the my opinion, this nice and easy old problem is for the middle-school students !

Proof.
A similar nice and easy old problem.

Quote:

Let $ABCD$ be a quadrilateral and let $d$ be a mobile line for which $A\in d$

and $\{\begin{array}{c} X\in CB\cap d\ ,\ B\in (CX)\\\\ Y\in CD\cap d\ ,\ D\in (CY)\end{array}$ . Prove that

$[ABX]+[ADY]\ge [ABCD]$ with equality iff $AX=AY\ .$

Here is a nice consequence of the initial proposed problem.

Virgil Nicula wrote:

Let $ABCD$ be a quadrilateral. Denote $O\in AC\cap BD$ . Prove that

$2\sqrt {[ABCD]} \ge \sqrt {[AOB]} + \sqrt {[BOC]} + \sqrt {[COD]} + \sqrt {[DOA]}$

with equality iff the quadrilateral $ABCD$ is a parallelogram.

I like very much these problems !

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a