area of convex ABCDE, AB=BC, CD=DE, < ABC=150^o, <CDE=30^o, BD=2

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

area of convex ABCDE, AB=BC, CD=DE, < ABC=150^o, <CDE=30^o, BD=2

G H J

Reveal topic tags

geometry theorems

L

G H BBookmark kLocked kLocked NReply

Source: 2003 Belarusian MO A 11.8

The post below has been deleted. Click to close.

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

30629 posts

#1 h Aug 17, 2020, 9:14 PM • 1 Y

Y by Prod55

Given a convex pentagon $ABCDE$ with $AB=BC, CD=DE, \angle ABC=150^o, \angle CDE=30^o, BD=2$ . Find the area of $ABCDE$ .

(I.Voronovich)

This post has been edited 1 time. Last edited by parmenides51, Aug 17, 2020, 9:14 PM

Z K Y

The post below has been deleted. Click to close.

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

127 posts

Prod55

#2 h Aug 18, 2020, 9:28 AM

Y by

Nice problem!

$D_{30^{\circ}}(E)=C,B_{150^{\circ}}(C)=A$ so $D_{30^{\circ}}\circ B_{150^{\circ}}(E)=A$ .
$30^{\circ}+150^{\circ}=180^{\circ}$ therefore $D_{30^{\circ}}\circ B_{150^{\circ}}$ has the center on the midpoint of $AE$ and $\angle MDB=15^{\circ},\angle DBM=75^{\circ}\Rightarrow \angle DMB=90^{\circ}$
Let $F\in BM$ such that $FM=MB,F\neq B$
The triangle $BDF$ is isosceles so $DF=DB=2$ .
Also $\Delta BMA=\Delta MEF$ so $(ABCDE)=(ABM)+(BMD)+(CBD)=(AMB)+(MEF)+(EDF)=(BDF)=\dfrac{1}{2}\cdot 2^2\cdot \sin30^{\circ}=1$

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

5306 posts

soryn

#3 h Aug 18, 2020, 11:33 AM

Y by

Interesting solution, with rotations! Without, is possible?

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a