equilateral, semicircle, segment trisector (2010 Denmark MO- Mohr Contest p5)

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jlacosta

Apr 2, 2025

0 replies

Number Theory with set and subset and divisibility

SomeonecoolLovesMaths 3

N 2 hours ago by martianrunner

Let $S = \{ 1,2, \cdots, 100 \}$ . Let $A$ be a subset of $S$ such that no sum of three distinct elements of $A$ is divisible by $5$ . What is the maximum value of $\mid A \mid$ .

3 replies

SomeonecoolLovesMaths

Apr 21, 2025

martianrunner

2 hours ago

Basic geometry

AlexCenteno2007 7

N 2 hours ago by KAME06

Given an isosceles triangle ABC with AB=BC, the inner bisector of Angle BAC And cut next to it BC in D. A point E is such that AE=DC. The inner bisector of the AED angle cuts to the AB side at the point F. Prove that the angle AFE= angle DFE

7 replies

1 viewing

AlexCenteno2007

Feb 9, 2025

KAME06

2 hours ago

Generating Functions

greenplanet2050 7

N 2 hours ago by rchokler

So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)

7 replies

greenplanet2050

Yesterday at 10:42 PM

rchokler

2 hours ago

Algebraic Manipulation

Darealzolt 1

N 4 hours ago by Soupboy0

Find the number of pairs of real numbers $a, b, c$ that satisfy the equation $a^4 + b^4 + c^4 + 1 = 4abc$ .

1 reply

Darealzolt

5 hours ago

Soupboy0

4 hours ago

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equilateral, semicircle, segment trisector (2010 Denmark MO- Mohr Contest p5)

parmenides51 2

N Aug 14, 2020 by natmath

An equilateral triangle $ABC$ is given. With $BC$ as diameter, a semicircle is drawn outside the triangle. On the semicircle, points $D$ and $E$ are chosen such that the arc lengths $BD, DE$ and $EC$ are equal. Prove that the line segments $AD$ and $AE$ divide the side $BC$ into three equal parts.
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2 replies

parmenides51

Aug 13, 2020

natmath

Aug 14, 2020

equilateral, semicircle, segment trisector (2010 Denmark MO- Mohr Contest p5)

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parmenides51

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parmenides51

#1 h Aug 13, 2020, 10:48 PM

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ezpotd

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ezpotd

#2 h Aug 13, 2020, 11:00 PM

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sol

WLOG, let the side length be equal to one. Let the midpoint of BC be the origin. The coordinates of E are 1/4,sqrt3/4. The coordinates of A are 0,-sqrt3/2. The equation of the line AE is 3sqrt3x - sqrt3/2, which is 0 at 1/6, meaning by y axis symmetry, that our claim is true.

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natmath

8219 posts

natmath

#3 h Aug 14, 2020, 2:12 AM

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Call the intersections on BC from left to right as F and G. Do a homothethy thing and shrink semicircle $BDEC$ to $B'D'E'C'$ about point A so that D'=F and E'=G. Define O as the center of the original semicircle and is now O'.

Clearly triangle $B'O'F$ is equilateral and since $B'O'$ is parallel to $BC$ we know that $\angle BB'F=60$ . Therefore triangle $BB'F$ is equilateral. Because the chords of equal arcs are congruent, we know that $BD=DE=EC$ and this congruence is true after the dilation.
Because $BB'F$ is equilateral (and similarly so is $CC'G$ ) we know that $BF=B'F=FG=GC'=GC$ .

Fun problem!

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