rhombus wanted, 2 lines intersect on circumcircle

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0 replies

jlacosta

Apr 2, 2025

0 replies

Problem 5

blug 1

N 9 minutes ago by atdaotlohbh

Source: Polish Math Olympiad 2025 Finals P5

Convex quadrilateral $ABCD$ is described on a circle $\omega$ , and is not a trapezius inscribed in a circle. Let the tangency points of $\omega$ and $AB, BC, CD, DA$ be $K, L, M, N$ respectively. A circle with a center $I_K$ , different from $\omega$ is tangent to the segement $AB$ and lines $AD, BC$ . A circle with center $I_L$ , different from $\omega$ is tangent to segment $BC$ and lines $AB, CD$ . A circle with center $I_M$ , different from $\omega$ is tangent to segment $CD$ and lines $AD, BC$ . A circle with center $I_N$ , different from $\omega$ is tangent to segment $AD$ and lines $AB, CD$ . Prove that the lines $I_KK, I_LL, I_MM, I_NN$ are concurrent.

1 reply

blug

Yesterday at 12:08 PM

atdaotlohbh

9 minutes ago

¿10^n-1 is a divisor of 11^n-1?

EmersonSoriano 1

N 15 minutes ago by Filipjack

Source: 2017 Peru Southern Cone TST P2

Determine if there exists a positive integer $n$ such that $10^n - 1$ is a divisor of $11^n - 1$ .

1 reply

EmersonSoriano

2 hours ago

Filipjack

15 minutes ago

D1018 : Can you do that ?

Dattier 1

N 17 minutes ago by Dattier

Source: les dattes à Dattier

We can find $A,B,C$ , such that $\gcd(A,B)=\gcd(C,A)=\gcd(A,2)=1$ and $\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0$ .

For example :

$C=20$
$A=47650065401584409637777147310342834508082136874940478469495402430677786194142956609253842997905945723173497630499054266092849839$

$B=238877301561986449355077953728734922992395532218802882582141073061059783672634737309722816649187007910722185635031285098751698$

Can you find $A,B,C$ such that $A>3$ is prime, $C,B \in (\mathbb Z/A\mathbb Z)^*$ with $o(C)=(A-1)/2$ and $\forall n \in \mathbb N^*, (C^n \times B \mod A) \mod 2=0$ ?

1 reply

Dattier

Mar 24, 2025

Dattier

17 minutes ago

A special family of subsets of {1, 2, ..., 100}.

EmersonSoriano 0

an hour ago

Source: 2017 Peru Southern Cone TST P10

Miguel has a list consisting of several subsets of 10 elements from the set ${1,2,\dots,100}$ . He says to Cecilia: "If you pick any subset of 10 elements from ${1,2,\dots,100}$ , it will be disjoint with at least one subset from my list." What is the smallest possible number of subsets Miguel's list can have if what he tells Cecilia is true?

0 replies

EmersonSoriano

an hour ago

0 replies

No more topics!

rhombus wanted, 2 lines intersect on circumcircle

parmenides51 1

N Oct 21, 2020 by EmilXM

Source: 2016 Thailand TST 8.2

Let $\vartriangle ABC$ be a triangle with circumcircle $\omega$ . The circle centered at $B$ having radius $BC$ intersects $AC$ and $\omega$ at $D$ and $E$ , respectively. The line through $D$ parallel to $CE$ intersects $AB$ at $F$ .
a) Prove that $BD$ and $EF$ intersect on $\omega$
b) Let $G$ be the intersection point of $AB$ and $CE$ . Prove that $DEFG$ is a rhombus.

1 reply

parmenides51

Oct 21, 2020

EmilXM

Oct 21, 2020

rhombus wanted, 2 lines intersect on circumcircle

G H J

Reveal topic tags

geometry

rhombus

circumcircle

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Source: 2016 Thailand TST 8.2

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parmenides51

30629 posts

parmenides51

#1 h Oct 21, 2020, 3:08 PM • 2 Y

Y by Mango247, Mango247

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EmilXM

378 posts

EmilXM

#2 h Oct 21, 2020, 7:36 PM

Y by

First I will prove (b). Because it is easy to prove (a) after (b) is correct.
Solution of b

Let $\angle GCB = \alpha, \angle ACG = \beta$ .
$BC = BE$ (they are radii of same circle) $\Longrightarrow \angle CEB = \angle BCE = \alpha$ .
$ACBE$ is cyclic $\Longrightarrow \angle ABE = \angle ACE = \beta$ .
$\angle DBE = 2 \angle DCE = 2\beta \Longrightarrow \angle DBG = \angle DBE - \angle EBG = 2\beta - \beta = \beta \Longrightarrow \angle DCG = \beta = \angle DBG \Longrightarrow CDGB$ is cyclic $\Longrightarrow \angle GDB = \angle GCB = \alpha$ .
$\angle GDB = \angle GEB = \beta, \angle DBG = \angle EBG = \alpha$ and $BG$ is common side, we get $\triangle BDG \cong BEG \Longrightarrow DG = EG$ .
Now, $\angle BAC = \angle CEB = \alpha$ ,
$DF \parallel CE \Longrightarrow \angle ADF = \angle ACG = \beta \Longrightarrow \angle DFG = \angle FAD + \angle FDA = \alpha + \beta$ .
And $\angle DGF = \angle GDB + \angle GBD = \alpha + \beta$ .
So $\triangle DGF$ is an isosceles triangle with $DG=DF$ .
We have $DF \parallel GE$ , $DF = DG = GE$ . Hence, $DFEG$ must be a rhombus. $Q.E.D. \blacksquare$

Solution of a

Denote $P$ be intersection of $EF$ and $BD$ . Since we have $DFEG$ is a rhombus, $DG \parallel FE \Longrightarrow \angle FPB = \angle GDB = \alpha = \angle BEC \Longrightarrow BEPC$ is cyclic $\Longrightarrow ABCP$ is cyclic. Hence $P$ lie on $\omega$
$Q.E.D. \blacksquare$

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