Inequality with inscribed quadrilateral

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

Inequality with inscribed quadrilateral

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2019 Belarusian National Olympiad 11.6

The post below has been deleted. Click to close.

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

184 posts

#1 h Sep 1, 2019, 8:58 AM • 2 Y

Y by Adventure10, Mango247

The diagonals of the inscribed quadrilateral $ABCD$ intersect at the point $O$ . The points $P$ , $Q$ , $R$ , and $S$ are the feet of the perpendiculars from $O$ to the sides $AB$ , $BC$ , $CD$ , and $DA$ , respectively.
Prove the inequality $BD\ge SP+QR$ .

(A. Naradzetski)

Z K Y

The post below has been deleted. Click to close.

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

578 posts

#2 h Sep 1, 2019, 10:42 PM • 1 Y

Y by Adventure10

Since we're given that $ABCD$ is cyclic, we've

$PS = AO \cdot \sin \angle BAD = AO \cdot \frac{BD}{2R},$
where $R$ is the circumradius of $ABCD$ . Analogously

$QR = OC \cdot \sin \angle BCD = CO \cdot \frac{BD}{BR}.$
Summing yields

$PS + QR = \frac{AC \cdot BD}{2R} \le BD,$ where we used $AC \le 2R$ .

$\square$

Z K Y

The post below has been deleted. Click to close.

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

2 posts

#3 h Sep 16, 2022, 7:41 PM

Y by

First, it is well-known that $PQRS$ is circumscribed,so $SP+QR=PQ+SR$ .
Note that $BO\ge PQ$ , because $PQ$ is a chord in a circle with diameter $BO$ , also $DO\ge SR$ ,so $BD\ge PQ+SR$ .
$\square$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a