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jlacosta   0
Jun 2, 2025
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0 replies
jlacosta
Jun 2, 2025
0 replies
USAMO 1983 Problem 2 - Roots of Quintic
Binomial-theorem   35
N 2 minutes ago by Mr.Sharkman
Source: USAMO 1983 Problem 2
Prove that the roots of\[x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0\] cannot all be real if $2a^2 < 5b$.
35 replies
Binomial-theorem
Aug 16, 2011
Mr.Sharkman
2 minutes ago
Polynomial with restricted range modulo n
v_Enhance   23
N 16 minutes ago by Mr.Sharkman
Source: USA TSTST 2016 Problem 3, by Yang Liu
Decide whether or not there exists a nonconstant polynomial $Q(x)$ with integer coefficients with the following property: for every positive integer $n > 2$, the numbers \[ Q(0), \; Q(1), Q(2),  \; \dots, \; Q(n-1) \]produce at most $0.499n$ distinct residues when taken modulo $n$.

Proposed by Yang Liu
23 replies
+1 w
v_Enhance
Jun 28, 2016
Mr.Sharkman
16 minutes ago
Miquel point configuration
Vivouaf   0
39 minutes ago
Source: Own/Well known
Let $ABCD$ be a cyclic convex quadrilateral, and let $E =AB \cap CD$, $F = AD \cap BC$. $(ABCD)$ has center $O$ and diagonals $AC$ and $BD$ cut at $P$. Prove the followings :
- Circles $(ADE), (BCE), (ABF), (CDF)$ intersect at a common point called the Miquel point of $ABCD$.
- $M \in EF$ and $M \in OP$. Precisely, $M$ is the inverse of $P$ with respect to $(O)$.
- $OM \perp EF$.
- More advanced : there is an involution (= inversion + reflection passing through the center of inversion) centered at $M$ swapping the pairs $(A,C), (B,D)$ and $(E,F)$.
- (Newton-Gauss line) Let $K,L,N$ be the midpoints of $AC,BD,EF$ respectively. Show they are collinear.
0 replies
Vivouaf
39 minutes ago
0 replies
Simple algebra involving bounding of a sequence
Iveela   2
N an hour ago by sami1618
Source: 2025 IRN-MNG Friendly Competition
Let $n \geq 2$ be a positive integer. Let $x_1, \dots, x_n$ be a sequence of real numbers such that
\[\max_{1 \leq k \leq n} \frac{x_1 + \dots + x_k}{k} = 1 \quad \text{and} \quad \min_{1 \leq k \leq n} \frac{x_1 + \dots + x_k}{k} = 0.\]Find the minimum and maximum possible values of $\max_{1 \leq k \leq n}x_k - \min_{1 \leq k \leq n}x_k$.
2 replies
Iveela
Jun 8, 2025
sami1618
an hour ago
No more topics!
Folding...
Rushil   14
N Mar 7, 2025 by Jupiterballs
Source: RMO 1990 Problem 3
A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.
14 replies
Rushil
Oct 14, 2005
Jupiterballs
Mar 7, 2025
Source: RMO 1990 Problem 3
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Rushil
1592 posts
#1 • 2 Y
Y by Adventure10, Mango247
A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.
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frt
1294 posts
#2 • 2 Y
Y by smo, Adventure10
Let $AE$ be the crease, where $E$ is the point on $BC$. Let $F$ be the midpoint of $CD$.

Let $x=BE(=EF)$ and $y=BC$. Then we have $EC=y-x$.
Also, $BC=\frac{1}{2}CD=\frac{1}{2}BC=\frac{y}{2}$

Using Pythagorean Theorem to $\triangle ECF$,

$(y-x)^2+\left(\frac{y}{2}\right)^2=x^2$
$\Longleftrightarrow y^2-2xy+x^2+\frac{y^2}{4}=x^2$
$\Longleftrightarrow 2xy=\frac{5y^2}{4}$
$\Longleftrightarrow x=\frac{5y}{8}$

$\therefore EC=y-x=\frac{3y}{8}$

Thus, $BE: EC=5: 3$
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Virgil Nicula
7054 posts
#3 • 2 Y
Y by Adventure10, Mango247
Very nice problems in this fresh section ! Congratulations, Vornicu, for the your initiative.

The "my" solution.

I suppose w.l.o.g. that $AB=2$. I note the middlepoint $M$
of the side $CD$ and the point $N\in (BC)$ which belongs to the crease. Thus,
$BN=MN=x,\ CN=2-x,\ CM=1,\ MN^2=CN^2+CM^2$
$\Longrightarrow x^2=(2-x)^2+1\Longrightarrow x=\frac 54$, i.e. $\frac{NB}{NC}=\frac{x}{2-x}=\frac{\frac 54}{2-\frac 54}=\frac 53.$

Remark 1. If the crease cuts the side $AD$ in the point $P$, then $\frac{PA}{PD}=\frac 18$. If I note the intersections $S\in AB\cap NP$, $R\in AD\cap MS$, then $\frac{SA}{AB}=\frac 16$, $\frac{RA}{RD}=\frac 13.$

Remark 2. This problem can be generalized to other quadrilaterals. I wish to see a extension of this problem to the rectangle, parallelogram deltoid, rhombus (and $\frac{MC}{MD}=r$) a.s.o.
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varun
270 posts
#4 • 3 Y
Y by Adventure10, Adventure10, Mango247
Rushil wrote:
A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.

This problem can be solved easily with the help of a diagram.(Although I am not able to draw it for you but try it yourself.)
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murgi
113 posts
#5 • 1 Y
Y by Adventure10
[geogebra]54d49e6e43eead3a0ad7f05e07fe272ccef2bcf5[/geogebra]
A trigonometric short process is there......
taking tan of angle CBM in triangle BMC and then cos of angle CYM gives the result easily. :)
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Potla
1886 posts
#6 • 2 Y
Y by Adventure10, Mango247
I have found a solution to this problem long ago.
Click to reveal hidden text
PS
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tonotsukasa
12 posts
#7 • 2 Y
Y by Adventure10, Mango247
Let A,B,C,D be A$(0,1)$, B$(0,0)$, C$(1,0)$, D$(1,1)$. Then, CD's middle point is M$\left(1,\frac{1}{2} \right)$,
and line BM is $y=\frac{x}{2}$. BM's middle point is N$\left( \frac{1}{2},\frac{1}{4}\right)$.
The crease is a perpendicular bisector of BM, thus the crease is $y=-2\left(x-\frac{1}{2}\right)+\frac{1}{4}=-2x+\frac{5}{4}$.
Thus, the crease cross the x-axis at $\left( \frac{5}{8},0\right)$, which divides BC in the ratio $5:3$.
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aayush-srivastava
137 posts
#8 • 3 Y
Y by Evan-Chen123, Adventure10, Mango247
barycentrics, cartesians, trig bash all works.. try anything.
(HINT:- for using cartesians prefer choosing C as the origin)
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jayme
9805 posts
#9 • 1 Y
Y by Adventure10
Dear Mathlinkers,
another way is to use the pytagorician triple 3-4-5...

Sincerely
Jean-Louis
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AlastorMoody
2125 posts
#10 • 2 Y
Y by Adventure10, Mango247
It's clear that $\Delta BMH$ is isosceles $\implies \boxed{BH=HM}$
Let, $$\angle HMC =x \implies \frac{1+\tan \tfrac{x}{2}}{1-\tan \tfrac{x}{2}}=2 \implies \boxed{\tan \frac{x}{2}=\frac{1}{3}}$$Therefore, $$\tan x =\frac{3}{4} \implies \text{ clearly } \sin x =\frac{3}{5}=\frac{HC}{HM}=\frac{HC}{BH} \implies \boxed{BH:HC=5:3}$$
This post has been edited 2 times. Last edited by AlastorMoody, Nov 21, 2018, 9:43 AM
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AlastorMoody
2125 posts
#11 • 2 Y
Y by Adventure10, Mango247
A similar kind of problem appreared in V.A. Yasinsky Geometry Olympiad 2017 VIII-IX p1 [Ukranie], here,
https://artofproblemsolving.com/community/u378952h1711824p11269422
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ythomashu
6322 posts
#12 • 2 Y
Y by Adventure10, Mango247
This is just Haga's First Theorem
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jayme
9805 posts
#13 • 1 Y
Y by Adventure10
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20III.pdf p. 31

Sincerely
Jean-Louis
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Syngates
268 posts
#14
Y by
Has no one used coordinate bash in this? I’m surprised
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Jupiterballs
68 posts
#15
Y by
Call the line of reflection to be $KX$, where $x$ lies on $AD$,
Call the reflections of $A,B$ to be $A',B'$
Call $A'B' \cap AB = D$
Then, $\Delta A'XY$ ~ $\Delta B'DY$ ~ $\Delta CKB'$
Then by Pythagoras, it is trivial
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