<PQ'Q = < PQB wanted, circles with diameter sides intersect altitudes

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0 replies

jlacosta

Apr 2, 2025

0 replies

Two circles, a tangent line and a parallel

Valentin Vornicu 103

N 29 minutes ago by zuat.e

Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$ . Let $AB$ be the line tangent to these circles at $A$ and $B$ , respectively, so that $M$ lies closer to $AB$ than $N$ . Let $CD$ be the line parallel to $AB$ and passing through the point $M$ , with $C$ on $G_1$ and $D$ on $G_2$ . Lines $AC$ and $BD$ meet at $E$ ; lines $AN$ and $CD$ meet at $P$ ; lines $BN$ and $CD$ meet at $Q$ . Show that $EP = EQ$ .

103 replies

Valentin Vornicu

Oct 24, 2005

zuat.e

29 minutes ago

Inequalities

idomybest 3

N an hour ago by damyan

Source: The Interesting Around Technical Analysis Three Variable Inequalities

The problem is in the attachment below.

3 replies

idomybest

Oct 15, 2021

damyan

an hour ago

Function on positive integers with two inputs

Assassino9931 2

N an hour ago by Assassino9931

Source: Bulgaria Winter Competition 2025 Problem 10.4

The function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ is such that $f(a,b) + f(b,c) = f(ac, b^2) + 1$ for any positive integers $a,b,c$ . Assume there exists a positive integer $n$ such that $f(n, m) \leq f(n, m + 1)$ for all positive integers $m$ . Determine all possible values of $f(2025, 2025)$ .

2 replies

Assassino9931

Jan 27, 2025

Assassino9931

an hour ago

Normal but good inequality

giangtruong13 4

N an hour ago by IceyCold

Source: From a province

Let $a,b,c> 0$ satisfy that $a+b+c=3abc$ . Prove that: $\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5}$

4 replies

giangtruong13

Mar 31, 2025

IceyCold

an hour ago

No more topics!

<PQ'Q = < PQB wanted, circles with diameter sides intersect altitudes

parmenides51 2

N Aug 21, 2020 by jayme

Source: Estonia IMO TST 2019 p7

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$ . The circle with diameter $AC$ intersects the segment $BE$ at point $P$ . A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$ . Prove that $\angle PQ'Q = \angle PQB$ .

2 replies

parmenides51

Jul 23, 2020

jayme

Aug 21, 2020

<PQ'Q = < PQB wanted, circles with diameter sides intersect altitudes

G H J

Reveal topic tags

geometry

equal angles

altitudes

G H BBookmark kLocked kLocked NReply

Source: Estonia IMO TST 2019 p7

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parmenides51

30630 posts

parmenides51

#1 h Jul 23, 2020, 1:00 PM • 2 Y

Y by Mango247, Mango247

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lahmacun

259 posts

lahmacun

#2 h Jul 23, 2020, 2:43 PM • 1 Y

Y by llplp

solution

Let $P'$ be the intersection of $(AC)$ and $BE$ different from $P$ . It is well known that $P'PQQ'$ is cyclic (check USAMO1990Problem5).
Now, since $AP'=AP$ and $AQ'=AQ$ it follows that $A$ is the center of $(P'PQQ')$ . Since $AQ\perp BQ$ , $BQ$ is tangent to $(P'PQQ')$ . So, $\angle PQ'Q = \angle PQB$

This post has been edited 1 time. Last edited by lahmacun, Jul 23, 2020, 2:45 PM

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jayme

9782 posts

jayme

#3 h Aug 21, 2020, 12:45 PM • 3 Y

Y by Mango247, Mango247, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%209.pdf p. 53-54.

Sincerely
Jean-Louis

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