a hard geometry problen

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0 replies

jlacosta

Apr 2, 2025

0 replies

x^101=1 find 1/1+x+x^2+1/1+x^2+x^4+...+1/1+x^100+x^200

Mathmick51 6

N 5 minutes ago by pi_quadrat_sechstel

Let $x^{101}=1$ such that $x\neq 1$ . Find the value of $\frac{1}{1+x+x^2}+\frac{1}{1+x^2+x^4}+\frac{1}{1+x^3+x^6}+\dots+\frac{1}{1+x^{100}+x^{200}}$

6 replies

Mathmick51

Jun 22, 2021

pi_quadrat_sechstel

5 minutes ago

I need help for British maths olympiads

RCY 0

10 minutes ago

I’m a year ten student who’s going to take the bmo in one year.
However I have no experience in maths olympiads and the best results I have achieved so far was 25/60 in intermediate maths olympiads.
What shall I do?
I really need help!

0 replies

RCY

10 minutes ago

0 replies

IMO Shortlist 2014 N5

hajimbrak 60

N an hour ago by sansgankrsngupta

Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p -1} + y$ and $x + y^ {p -1}$ are both powers of $p$ .

Proposed by Belgium

60 replies

hajimbrak

Jul 11, 2015

sansgankrsngupta

an hour ago

n variables with n-gon sides

mihaig 0

an hour ago

Source: Own

Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$

0 replies

mihaig

an hour ago

0 replies

No more topics!

a hard geometry problen

Tuguldur 1

N Apr 5, 2025 by whwlqkd

Let $ABCD$ be a convex quadrilateral. Suppose that the circles with diameters $AB$ and $CD$ intersect at points $X$ and $Y$ . Let $P=AC\cap BD$ and $Q=AD\cap BC$ . Prove that the points $P$ , $Q$ , $X$ and $Y$ are concyclic.
( $AB$ and $CD$ are not the diagnols)

1 reply

Tuguldur

Apr 4, 2025

whwlqkd

Apr 5, 2025

a hard geometry problen

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geometry

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Tuguldur

19 posts

Tuguldur

#1 h Apr 4, 2025, 3:56 PM

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whwlqkd

97 posts

whwlqkd

#2 h Apr 5, 2025, 11:30 AM

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easy

Let $\omega_1, \omega_2, \omega_3$ be circle with diameters $AB, CD, PQ$ . And let $K, L, M$ be $AB\cap PQ, CD\cap PQ, AB\cap CD$ . Because $-1=(A,K,B,M)=(D,L,C,N)=(Q,K,P,L)$ , $\omega_1$ is the Apollonius circle of $K,M$ and ratio $KA: MA$ , and $\omega_2$ is the Apollonius circle of $L,M$ and ratio $LD:MD$ , and $\omega_3$ is the Apollonius circle of $K,L$ and ratio $KQ:LQ$ . So, ${X,Y}=\omega_1\cap \omega_2$ , and by Menelaus, $\frac{LX}{KX}=\frac{LX}{MX} \times \frac{MX}{KX}=\frac{LD}{MD} \times \frac{MA}{KA}=\frac{LQ}{KQ}$ . So $X$ lies on $\omega_3$ . Similarly, $Y$ lies on $\omega_3$ and we are done. $Q.E.D$

This post has been edited 1 time. Last edited by whwlqkd, Apr 5, 2025, 11:31 AM

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