Three concurrent circles

by jayme, May 14, 2025, 3:08 PM

Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. Tb, Tc the tangents to 0 wrt. B, C
4. D the point of intersection of Tb and Tc
5. B', C' the symmetrics of B, C wrt AC, AB
6. 1b, 1c the circumcircles of the triangles BB'D, CC'D.

Prove : 1b, 1c and 0 are concurrents.

Sincerely
Jean-Louis

geometry

Quirky tangency and line concurrence with circumcircles

by pithon_with_an_i, May 14, 2025, 1:22 PM

Let $ABC$ be a triangle. $M$ is the midpoint of segment $BC$ , and points $E$ , $F$ are selected on sides $AB$ , $AC$ respectively such that $E$ , $F$ , $M$ are collinear. The circumcircles $(ABC)$ and $(AEF)$ intersect at a point $P \neq A$ . The circumcircle $(APM)$ intersects line $BC$ again at a point $D \neq M$ .
Show that the lines $AD$ , $EF$ and the tangent to $(AEF)$ at point $P$ concur.

(Proposed by Soo Eu Khai)

geometry

circumcircle

Three lines meet at one point

by TUAN2k8, May 14, 2025, 1:01 PM

Let $ABC$ be an acute triangle incribed in a circle $\omega$ .Let $M$ be the midpoint of $BC$ .Let $AD,BE$ and $CF$ be altitudes from $A,B$ and $C$ of triangle $ABC$ , respectively, and let them intersect at $H$ .Let $K$ be the intersection point of tangents to the circle $\omega$ at points $B,C$ .Prove that $MH,KD$ and $EF$ are concurrent.

geometry

Triangle

Proving radical axis through orthocenter

by azzam2912, May 14, 2025, 12:02 PM

In acute triangle $ABC$ let $D, E$ and $F$ denote the feet of the altitudes from $A, B$ and $C$ , respectively. Let line $DE$ intersect circumcircle $ABC$ at points $G, H$ . Similarly, let line $DF$ intersect circumcircle $ABC$ at points $I, J$ . Prove that the radical axis of circles $EIJ$ and $FGH$ passes through the orthocenter of triangle $ABC$

geometry

power of a point

radical axis

Proving that these are concyclic.

by Acrylic3491, May 14, 2025, 9:06 AM

In $\bigtriangleup ABC$ , points $P$ and $Q$ are isogonal conjugates. The tangent to $(BPC)$ at $P$ and the tangent to $(BQC)$ at Q, meet at $R$ . $AR$ intersects $(ABC)$ at $D$ . Prove that points $P$ , $Q$ , $R$ and $D$ are concyclic.

Any hints on this ?

This post has been edited 2 times. Last edited by Acrylic3491, 2 hours ago

geometry

Some number theory

by EeEeRUT, May 14, 2025, 6:52 AM

Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$ Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$ .

This post has been edited 1 time. Last edited by EeEeRUT, Today at 6:54 AM

number theory

Hard geometry

by Lukariman, May 14, 2025, 4:28 AM

Given circle (O) and chord AB with different diameters. The tangents of circle (O) at A and B intersect at point P. On the small arc AB, take point C so that triangle CAB is not isosceles. The lines CA and BP intersect at D, BC and AP intersect at E. Prove that the centers of the circles circumscribing triangles ACE, BCD and OPC are collinear.

Attachments:

This post has been edited 2 times. Last edited by Lukariman, Today at 4:49 AM

geometry

Planes and cities

by RagvaloD, May 3, 2017, 11:54 AM

In country some cities are connected by oneway flights( There are no more then one flight between two cities). City $A$ called "available" for city $B$ , if there is flight from $B$ to $A$ , maybe with some transfers. It is known, that for every 2 cities $P$ and $Q$ exist city $R$ , such that $P$ and $Q$ are available from $R$ . Prove, that exist city $A$ , such that every city is available for $A$ .

combinatorics

Problem 61: Poland MO Finals 2003, Problem 3

by henderson, Jan 30, 2017, 8:04 PM

$\color{red}\bf{Problem \ 61}$ Find all polynomials $W$ with integer coefficients satisfying the following condition: For every natural number $n, 2^n - 1$ is divisible by $W(n).$

This post has been edited 1 time. Last edited by henderson, Jan 30, 2017, 8:05 PM

number theory

Justin Stevens

Poland MO

Problem 57: Prime factorization

by henderson, Jan 4, 2017, 2:32 PM

${\color{red}\bf{Problem \ 57}}$ Prove that for every nonnegative integer $n,$ the number $7^{7^{n}}+1$ is the product of at least $2n+3$ (not necessarily distinct) primes.
(USAMO 2007)
${\color{red}\bf{Solution}}$ Let's apply induction.

$\boxed{\color{blue}{1}}$ The base case is $n=0,$ which is $8=2^3,$ so it has $2n+3$ prime factors.

$\boxed{\color{blue}{2}}$ Now, assume that $7^{7^{n}}+1$ is the product of $2n+3$ primes.

$\boxed{\color{blue}{3}}$ We wish to prove that $7^{7^{n+1}}+1$ is the product of $2(n+1)+3=2n+5$ primes.
Let $x = 7^{7^{n}}.$ $\quad$ $x+1$ is the product of $2n+3$ primes, so we want to show that $x^7+1$ is the product of $2n+5$ primes.

Note that $x^7+1 = (x+1)(x^6-x^5+x^4-x^3+x^2-x+1).$ By the inductive hypothesis, we know that $x+1$ is the product of $2n+3$ primes, so it suffices to show that $x^6-x^5+x^4-x^3+x^2-x+1$ is composite.

Since $x^6-x^5+x^4-x^3+x^2-x+1=(x+1)^6 - 7x(x^2+x+1)^2$ and $x=7^{7^{n}}$ , $x^6-x^5+x^4-x^3+x^2-x+1$ is the difference of two squares, therefore is composite.

Thus, $7^{7^{n+1}}+1$ is the product of at least $(2n+3)+2=2n+5$ primes, and we are done. $\square$

This post has been edited 4 times. Last edited by henderson, Jan 12, 2017, 11:11 AM

number theory

prime numbers

USAMO

Problem 56: Brazil MO 2013, Day 1, Problem 2

by henderson, Dec 16, 2016, 3:48 PM

${\color{red}\bf{Problem \ 56}}$ Arnaldo and Bernaldo play the following game: given a fixed finite set of positive integers $A$ known by both players, Arnaldo picks a number $a \in A$ but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer $b$ (not necessarily in $A$ ). Then Arnaldo tells the number of divisors of $ab$ . Show that Bernaldo can choose $b$ in a way that he can find out the number $a$ chosen by Arnaldo.
(Brazil MO 2013)
${\color{red}\bf{Solution}}$ Let us say the primes that divide at least one element from $A$ are $p_0,p_1,\ldots,p_k$ . An element $a\in A$ can be represented then as $a=\prod_{j=0}^k p_j^{\alpha_j}$ , with $\alpha_j \geq 0$ . When $b=\prod_{j=0}^k p_j^{\beta_j}$ , the number of divisors of $ab$ is $\tau(ab) = \prod_{j=0}^k (1+ \alpha_j + \beta_j)$ . Let us plug in $\beta_j = x^{2^j}$ ; then $P(x) = \prod_{j=0}^k (1+\alpha_j + x^{2^j})$ is a polynomial in $x$ of degree $2^{k+1} - 1$ , where the coefficient of $x^{2^{k+1} - 2^j - 1}$ is precisely $1+\alpha_j$ . In fact, if we take $n > \prod_{j=0}^k (1+\alpha_j)$ , then $P(n)$ is the writing in basis $n$ of some (huge) integer, and all "digits" can be determined, namely also the values $\alpha_j$ . So all is left to do is to take $n > \prod_{j=0}^k (1+a_j)$ , where $a_j = \max_{A} \alpha_j$ , and $\beta_j = n^{2^j}$ .

This post has been edited 3 times. Last edited by henderson, Jan 4, 2017, 2:42 PM

number theory

Combinatorial Number Theory

combinatorics

Collinear points THC

by gobathegreat, May 16, 2016, 12:40 PM

Let $k$ be a circumcircle of triangle $ABC$ $(AC<BC)$ . Also, let $CL$ be an angle bisector of angle $ACB$ $(L \in AB)$ , $M$ be a midpoint of arc $AB$ of circle $k$ containing the point $C$ , and let $I$ be an incenter of a triangle $ABC$ . Circle $k$ cuts line $MI$ at point $K$ and circle with diameter $CI$ at $H$ . If the circumcircle of triangle $CLK$ intersects $AB$ again at $T$ , prove that $T$ , $H$ and $C$ are collinear.
.

Concurrent Gergonnians in Pentagon

by numbertheorist17, Jul 16, 2014, 12:09 PM

Consider a convex pentagon circumscribed about a circle. We name the lines that connect vertices of the pentagon with the opposite points of tangency with the circle gergonnians.
(a) Prove that if four gergonnians are conncurrent, the all five of them are concurrent.
(b) Prove that if there is a triple of gergonnians that are concurrent, then there is another triple of gergonnians that are concurrent.