<EHC wanted, <BEA=45^o, altitude and angle bisector

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jlacosta

Apr 2, 2025

0 replies

Heptagon in Taiwan TST!!!

Hakurei_Reimu 3

N a minute ago by sami1618

Source: 2025 Taiwan TST Round 3 Independent Study 2-G

Let $ABCDEFG$ be a regular heptagon with its center $O$ . $H$ is the orthocenter of triangle $CDF$ , $I$ is the incenter of triangle $ABD$ . Let $M$ be the midpoint of $IG$ and $X$ be the intersection point of $OH$ and $FG$ . Assume $P$ is the circumcenter of triangle $BCI$ . Prove that $CF, MP, XB$ concur at a single point.

Proposed by HakureiReimu.

3 replies

Hakurei_Reimu

Apr 26, 2025

sami1618

a minute ago

Inspired by old results

sqing 0

14 minutes ago

Source: Own

Let $a,b,c\geq 0 , ab+bc+ca+a+b+c=1 .$ Prove that
$1\geq a+b+c+abc\geq \frac{13}{6}\sqrt{\frac{7}{3}}-\frac{5}{2}$ Let $a,b,c\geq 0 ,ab+bc+ca+a^2+b^2+c^2=1 .$ Prove that
$1\leq a+b+c+abc\leq \frac{19}{6\sqrt 6}$

0 replies

sqing

14 minutes ago

0 replies

Break the stick!

BR1F1SZ 2

N 30 minutes ago by real_loser

Source: 2024 Argentina L2 P2

Ana and Beto play the following game with a stick of length $15$ . Ana starts, and on her first turn, she cuts the stick into two pieces with integer lengths. Then, on each player's turn, they must cut one of the pieces, of their choice, into two new pieces with integer lengths. The player who, on their turn, leaves at least one piece with length equal to $1$ loses. Determine which of the two players has a winning strategy.

2 replies

BR1F1SZ

Nov 18, 2024

real_loser

30 minutes ago

all functions satisfying f(x+yf(x))+y = xy + f(x+y)

falantrng 29

N 33 minutes ago by Uzb_Math2010

Source: Balkan MO 2025 P3

Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ ,
$f(x+yf(x))+y = xy + f(x+y).$
Proposed by Giannis Galamatis, Greece

29 replies

falantrng

Apr 27, 2025

Uzb_Math2010

33 minutes ago

No more topics!

<EHC wanted, <BEA=45^o, altitude and angle bisector

parmenides51 2

N Aug 10, 2020 by freeman66

Source: 2010 Belgium MO (OMB) p4

In the triangle $ABC$ , let $H$ be the foot of the altitude from $A$ . Let $E$ be the intersection point of the angle bisector from $B$ with the side $AC$ . Knowing that $\angle BEA=45^o$ , determine $\angle EHC$ .

2 replies

parmenides51

Aug 9, 2020

freeman66

Aug 10, 2020

<EHC wanted, <BEA=45^o, altitude and angle bisector

G H J

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Source: 2010 Belgium MO (OMB) p4

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parmenides51

30650 posts

parmenides51

#1 h Aug 9, 2020, 3:19 PM • 3 Y

Y by Mango247, Mango247, Mango247

This post has been edited 3 times. Last edited by parmenides51, Jun 21, 2022, 1:56 AM

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rafaello

1079 posts

rafaello

#2 h Aug 9, 2020, 7:05 PM

Y by

Solution.

Draw line perpendicular to $AC$ through $E$ , and let that line intersect $BC$ at $I$ .
Thus obviously, $\angle AEB=\angle BEI=45^{\circ}$ . Since $\angle ABE=\angle EBI$ , thus $AEB\cong IEB$ . Thus, $AE=EI$ , which implies that $\angle EAI=\angle AIE=45^{\circ}$ . Since $\angle AHI=\angle AHC=90^{\circ}$ , thus points $A,E,I,H$ are concyclic, which means that $\angle EAI=\angle EHI=\angle EHC=45^{\circ}$ .
$\boxed{\text{Answer. } 45^{\circ}.}$

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freeman66

452 posts

freeman66

#3 h Aug 10, 2020, 2:17 AM

Y by

$(AEH)$ intersects $BC$ at $H$ and $G$ . Then $\angle AHG=180^{\circ}-\angle AEG=90^{\circ}$ , so $\angle AEG=90^{\circ}$ . We can easily angle chase to get $\angle BAE=\angle BGE$ . Thus, $\triangle BAE\sim \triangle BGE$ , so $AE=EG$ , implying $\angle GAE=45^{\circ}=\angle EHG=\angle EHC$ , so $\angle EHC=\boxed{45^{\circ}}.$

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