An concurrent lines formed by two intersecting circles.

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0 replies

jlacosta

Apr 2, 2025

0 replies

Problem 16

Nguyenhuyen AG 37

N 4 minutes ago by flower417477

Let $a,b,c >0$ such that $abc\ge1$ . Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$

37 replies

Nguyenhuyen AG

May 3, 2010

flower417477

4 minutes ago

NEPAL TST 2025 DAY 2

Tony_stark0094 4

N 10 minutes ago by InterLoop

Consider an acute triangle $\Delta ABC$ . Let $D$ and $E$ be the feet of the altitudes from $A$ to $BC$ and from $B$ to $AC$ respectively.

Define $D_1$ and $D_2$ as the reflections of $D$ across lines $AB$ and $AC$ , respectively. Let $\Gamma$ be the circumcircle of $\Delta AD_1D_2$ . Denote by $P$ the second intersection of line $D_1B$ with $\Gamma$ , and by $Q$ the intersection of ray $EB$ with $\Gamma$ .

If $O$ is the circumcenter of $\Delta ABC$ , prove that $O$ , $D$ , and $Q$ are collinear if and only if quadrilateral $BCQP$ can be inscribed within a circle.

$\textbf{Proposed by Kritesh Dhakal, Nepal.}$

4 replies

+1 w

Tony_stark0094

Yesterday at 8:40 AM

InterLoop

10 minutes ago

Sets With a Given Property

oVlad 3

N 19 minutes ago by flower417477

Source: Romania TST 2025 Day 1 P4

Determine the sets $S{}$ of positive integers satisfying the following two conditions:
[list=a]
[*]For any positive integers $a, b, c{}$ , if $ab + bc + ca{}$ is in $S$ , then so are $a + b + c{}$ and $abc$ ; and
[*]The set $S{}$ contains an integer $N \geqslant 160$ such that $N-2$ is not divisible by $4$ .
[/list]
Bogdan Blaga, United Kingdom

3 replies

oVlad

Apr 9, 2025

flower417477

19 minutes ago

Number Theory Chain!

JetFire008 39

N 32 minutes ago by whwlqkd

I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1

Starting with the simplest
What is $2+3$ ?

39 replies

+1 w

JetFire008

Apr 7, 2025

whwlqkd

32 minutes ago

No more topics!

An concurrent lines formed by two intersecting circles.

MarkBcc168 2

N Aug 7, 2017 by suli

Source: 2017 Thailand TST Day 2 Problem 2

Two distinct circle $\Gamma_1, \Gamma_2$ meet at two distinct points $M,N$ . A line $\ell$ intersects $\Gamma_1$ at $A, C$ and intersects $\Gamma_2$ at $B,D$ such that points $A,B,C,D$ are all distinct and lie on $\ell$ in this order. Let $X$ be a point on the line $MN$ such that $M$ lies between $X$ and $N$ . Lines $AX,BM$ intersect at $P$ and lines $DX,CM$ intersect at $Q$ . Finally, let $K,L$ be the midpoints of $AD, BC$ respectively. Prove that $XK,ML,PQ$ are concurrent.

2 replies

MarkBcc168

Aug 1, 2017

suli

Aug 7, 2017

An concurrent lines formed by two intersecting circles.

G H J

Reveal topic tags

geometry

G H BBookmark kLocked kLocked NReply

Source: 2017 Thailand TST Day 2 Problem 2

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MarkBcc168

1594 posts

MarkBcc168

#1 h Aug 1, 2017, 10:47 AM • 2 Y

Y by Adventure10, Mango247

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diana7

28 posts

diana7

#2 h Aug 1, 2017, 8:30 PM • 1 Y

Y by Adventure10

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Suppose $QP \cap AD = S$ and bso $S, A, B$ lie on $\ell$ in this order. Let $AD \cap MN = I$ , $PQ \cap MN = J$ . We have $(S, I; A, D)=(S, J; Q, P)=(S, I; B, C)$ so $\frac{SA}{SD}:\frac{IA}{ID}=\frac{SB}{SC}:\frac {IB}{IC} \Rightarrow \frac{SA \cdot SC}{SB \cdot SD}=\frac {IA \cdot IC}{IB \cdot ID}$ .
By power of a point $IA \cdot IC = IM \cdot IN = IB \cdot ID$ so $SA \cdot SC = SB \cdot SD$ , but $SA < SB$ and $SC< SD$ .
Therefore $PQ || AD$ so the midpoint of $PQ$ lies on $ML$ and $XK$ .

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suli

1498 posts

suli

#3 h Aug 7, 2017, 3:45 AM • 2 Y

Y by Adventure10, Mango247

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Let $XA$ intersect $\Omega_1$ at $E$ , and let $XB$ intersect $\Omega_2$ at $F$ . Then since $X$ is on the radical axis of the two circles, $AEFB$ is cyclic. Then
$\angle PEF = \angle XDA = \angle PMF,$ so $PEMF$ is cyclic. Similarly, $QEMF$ is cyclic. Thus $PQEMF$ is cyclic, from which $\angle PQX = \angle FEX = \angle XDA$ , so $PQ \| AD$ . Thus, since $K$ and $L$ are midpoints of $AD, BC$ respectively, we have $XK$ and $ML$ concur at the midpoint of $PQ$ .

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