<ABP=? if <PAC = <PCD = 20^o inside a square ABCD

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jlacosta

Apr 2, 2025

0 replies

Inspired by 2006 Romania

sqing 5

N 8 minutes ago by pooh123

Source: Own

Let $a,b,c>0$ and $abc= \frac{1}{8} .$ Prove that
$\frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}-a-b-c \geq \frac{1}{2}$

5 replies

sqing

an hour ago

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8 minutes ago

NEPAL TST DAY-2 PROBLEM 1

Tony_stark0094 8

N 9 minutes ago by Thapakazi

Let the sequence $\{a_n\}_{n \geq 1}$ be defined by
$a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}$ Prove that
$a_n^{2025} >n^{2024}$ for all positive integers $n \geq 2$ .

8 replies

Tony_stark0094

4 hours ago

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9 minutes ago

A cyclic inequality

KhuongTrang 8

N 12 minutes ago by SunnyEvan

Source: own-CRUX

IMAGE
Link

8 replies

KhuongTrang

Apr 2, 2025

SunnyEvan

12 minutes ago

Isosceles Triangle Geo

oVlad 1

N 22 minutes ago by Primeniyazidayi

Source: Romania Junior TST 2025 Day 1 P2

Consider the isosceles triangle $ABC$ with $\angle A>90^\circ$ and the circle $\omega$ of radius $AC$ centered at $A.$ Let $M$ be the midpoint of $AC.$ The line $BM$ intersects $\omega$ a second time at $D.$ Let $E$ be a point on $\omega$ such that $BE\perp AC.$ Let $N$ be the intersection of $DE$ and $AC.$ Prove that $AN=2\cdot AB.$

1 reply

oVlad

3 hours ago

Primeniyazidayi

22 minutes ago

No more topics!

<ABP=? if <PAC = <PCD = 20^o inside a square ABCD

parmenides51 2

N Aug 21, 2021 by sunken rock

Source: 2014 Cordoba pre-TST for IberoAmerican p4 (Argentina)

Let $ABCD$ be a square. Let $P$ be inside $ABCD$ such that $\angle PAC = \angle PCD = 20^o$ . Find $\angle ABP$ .

2 replies

parmenides51

Aug 20, 2021

sunken rock

Aug 21, 2021

<ABP=? if <PAC = <PCD = 20^o inside a square ABCD

G H J

Reveal topic tags

geometry

angles

square

G H BBookmark kLocked kLocked NReply

Source: 2014 Cordoba pre-TST for IberoAmerican p4 (Argentina)

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parmenides51

30630 posts

parmenides51

#1 h Aug 20, 2021, 12:48 PM • 1 Y

Y by Niruuuuut2

Let $ABCD$ be a square. Let $P$ be inside $ABCD$ such that $\angle PAC = \angle PCD = 20^o$ . Find $\angle ABP$ .

This post has been edited 1 time. Last edited by parmenides51, Aug 20, 2021, 12:48 PM

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AlanLG

241 posts

AlanLG

#2 h Aug 20, 2021, 4:00 PM

Y by

Let $x=\angle ABP$ , it´s easy to obtain $\angle PAB=65^\circ, \angle BPA=115-x, \angle BCP=70^\circ, \angle CPB=20+x$

By law of sines in

$\triangle ABP: \frac{\sin 65^\circ}{BP}=\frac{\sin (115-x)}{AB}$

$\triangle BCP: \frac{BP}{\sin 70^\circ}=\frac{BC}{\sin (20+x)}$

Since $AB=BC$ , multiplying the equations we get

$\frac{\sin 65^\circ}{\sin 70^\circ}=\frac{\sin (115-x)}{\sin (20+x)}$

$x=50^\circ$ clearly satifies, thus $x=\angle ABP=50^\circ$

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sunken rock

4381 posts

sunken rock

#3 h Aug 21, 2021, 11:29 AM

Y by

One liner solution at https://stanfulger.blogspot.com/2021/08/aops-httpsartofproblemsolvingcomcommuni_21.html

Best regards,
sunken rock

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