Perpendicular line segments

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0 replies

jlacosta

Apr 2, 2025

0 replies

(x^2 + 1)(y^2 + 1) >= 2(xy - 1)(x + y)

parmenides51 3

N 5 minutes ago by alizoratiopour

Source: Czech-Polish-Slovak Junior Match 2017, individual p3 CPSJ

Prove that for all real numbers $x, y$ holds $(x^2 + 1)(y^2 + 1) \ge 2(xy - 1)(x + y)$ .
For which integers $x, y$ does equality occur?

3 replies

parmenides51

Feb 20, 2020

alizoratiopour

5 minutes ago

A three-variable functional inequality on non-negative reals

Tintarn 6

N 21 minutes ago by jenishmalla

Source: Dutch TST 2024, 1.2

Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
$2x^3zf(z)+yf(y) \ge 3yz^2f(x)$ for all $x,y,z \in \mathbb{R}_{\ge 0}$ .

6 replies

Tintarn

Jun 28, 2024

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21 minutes ago

Holy inequality

giangtruong13 1

N 22 minutes ago by MathPerson12321

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Let $a,b,c>0$ . Prove that: $\frac{8}{\sqrt{a^2+b^2+c^2+1}} - \frac{9}{(a+b)\sqrt{(a+2c)(b+2c)}} \leq \frac{5}{2}$

1 reply

giangtruong13

26 minutes ago

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NT function debut

AshAuktober 1

N 23 minutes ago by Burmf

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Let $f$ be a function taking in positive integers and outputting nonnegative integers, defined as follows:
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Find all positive integers $x$ such that $f(x) \ne 0$ and $f(f(x)) = f(x) - 1.$ Adit Aggarwal, India.

1 reply

AshAuktober

42 minutes ago

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No more topics!

Perpendicular line segments

socrates 3

N Jul 6, 2024 by parmenides51

Source: Finnish Mathematics Competition 2005, Final Round, Problem 1

In the figure below, the centres of four squares have been connected by two line
segments. Prove that these line segments are perpendicular.

3 replies

socrates

Nov 14, 2011

parmenides51

Jul 6, 2024

Perpendicular line segments

G H J

Reveal topic tags

Pythagorean Theorem

geometry

G H BBookmark kLocked kLocked NReply

Source: Finnish Mathematics Competition 2005, Final Round, Problem 1

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socrates

2105 posts

socrates

#1 h Nov 14, 2011, 9:40 PM • 1 Y

Y by Adventure10

In the figure below, the centres of four squares have been connected by two line
segments. Prove that these line segments are perpendicular.

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pi37

2079 posts

pi37

#2 h Nov 14, 2011, 10:05 PM • 1 Y

Y by Adventure10

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Let the side lengths of the squares, starting from the top left and moving clockwise, be $2s_1, 2s_2, 2s_3, 2s$ , and let the centers of these respective squares be $O_1, O_2, O_3, O_4$ . By the Pythagorean Theorem we can easily compute
$O_1O_2^2=(s_1+s_2)^2+(s_1-s_2)^2=2s_1^2+2s_2^2$
$O_2O_3^2=(s_2+s_3)^2+(s_2-s_3)^2=2s_2^2+2s_3^2$
$O_3O_4^2=(s_3+s_4)^2+(s_3-s_4)^2=2s_3^2+2s_4^2$
$O_4O_1^2=(s_1+s_4)^3+(s_1-s_4)^3=2s_1^2+2s_4^2$
Then $O_1O_2^2+O_3O_4^2=O_2O_3^2+O_4O_1^2$ , so $O_1O_3\perp O_2O_4$ .

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sunken rock

4380 posts

sunken rock

#3 h Nov 15, 2011, 4:11 PM • 1 Y

Y by Adventure10

Observation: property valid for any quadrilateral, including an extreme one, i.e. with vertices collinear! They are also equal

Best regards,
sunken rock

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parmenides51

30630 posts

parmenides51

#4 h Jul 6, 2024, 8:41 AM

Y by

In the figure below, the centres of four squares have been connected by two line segments. Prove that these line segments are perpendicular.

https://cdn.artofproblemsolving.com/attachments/e/f/8d14f1d5d66b4ad7cf47004161510a9d761a1d.gif

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