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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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Reflections of AB, AC with respect to BC and angle bisector of A
falantrng   27
N 4 minutes ago by ravengsd
Source: BMO 2024 Problem 1
Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the
$A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points
$E$ and $F$ respectively. A line through $D$ meets $AC$ and $AB$ at $G$ and $H$ respectively such that $G$
lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of
$\triangle EDG$ and $\triangle FDH$ are tangent to each other.
27 replies
falantrng
Apr 29, 2024
ravengsd
4 minutes ago
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Test from Côte d'Ivoire Diophantine equation
MENELAUSS   0
5 minutes ago
determine all triplets $(x;y;z)$ of natural numbers such that
$$y \quad \text{is prime }$$
$$y \quad \text{and} \quad 3 \quad \text{does not divide} \quad z$$
$$x^3-y^3=z^2$$
0 replies
MENELAUSS
5 minutes ago
0 replies
J
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Substitutions inequality?
giangtruong13   2
N 7 minutes ago by giangtruong13
Let $a,b,c>0$ such that: $a^2b^2+ c^2b^2+ a^2c^2=3(abc)^2$. Prove that: $$\sum \frac{b+c}{a} \geq 2\sqrt{3(ab+bc+ca)}$$
2 replies
giangtruong13
Yesterday at 2:07 PM
giangtruong13
7 minutes ago
J
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Jiho placing tulips
hanulyeongsam   0
10 minutes ago
Source: own
Jiho will choose 46 holes in a 9X9 holebox. Jiho will then place tulips on the holes. Prove that for every 2X2, there is one that has more than 2 tulips inside.
0 replies
hanulyeongsam
10 minutes ago
0 replies
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No more topics!
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The incircle of a triangle ABC has centre I
orl   7
N Apr 7, 2014 by jayme
Source: 239MO 2004, grade 8-9, problem 2
The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$.

proposed by L. Emelyanov, S. Berlov
7 replies
orl
Dec 11, 2004
jayme
Apr 7, 2014
J
The incircle of a triangle ABC has centre I
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Reveal topic tags
geometry
Russia
bisector
L
G H BBookmark kLocked kLocked NReply
Source: 239MO 2004, grade 8-9, problem 2
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orl
3647 posts
orl
#1 h  Dec 11, 2004, 9:41 PM • 2 Y
Y by Adventure10, Mango247
The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$.

proposed by L. Emelyanov, S. Berlov
Z K Y
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grobber
7849 posts
grobber
#2 h  Dec 11, 2004, 10:28 PM • 2 Y
Y by Adventure10, Mango247
Let's start with the easy ones :).

Let $T=BL\cap A_1C_1$. $TB_1LK$ is cyclic, so $\frac{IK}{IL}=\frac{IT}{IB_1}\ (*)$. What we want to show is $\frac{IK}{IL}=\frac{IB_1}{IB}$. Combining this with $(*)$, we see that what we need is $IB_1^2=IT\cdot IB$. However, this is clear from the right triangle $IC_1B$. We have $IB_1^2=IC_1^2=IT\cdot IB$.
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Megus
1198 posts
Megus
#3 h  Dec 12, 2004, 12:38 PM • 2 Y
Y by Adventure10, Mango247
Excuse my lack of geometry knowledge :blush: but when a quadriteral is cyclic and what it gives [I don't understand $(*)$] ?
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grobber
7849 posts
grobber
#4 h  Dec 12, 2004, 12:40 PM • 2 Y
Y by Adventure10, Mango247
It's when the four points lie on the same circle. $(*)$ takes place because of the equality of angles of the triangles.
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Megus
1198 posts
Megus
#5 h  Dec 12, 2004, 12:45 PM • 2 Y
Y by Adventure10, Mango247
Thanks - drawing proper picture also helps.
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jayme
9777 posts
jayme
#6 h  Apr 5, 2014, 12:32 PM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
and without any calculation? what do you think?
Sincerely
Jean-Louis
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ThirdTimeLucky
402 posts
ThirdTimeLucky
#7 h  Apr 6, 2014, 7:08 PM • 1 Y
Y by Adventure10
An outline:

(1) $ BK $ meets $ AC $ at its midpoint, $ M $.

(2) $ MI $ meets $ BB_1 $ at its midpoint, $ X $.

(3) The pencil $ (KX,KB_1,KL,KM) $ is harmonic. So, $ KL \parallel BB_1 $.
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jayme
9777 posts
jayme
#8 h  Apr 7, 2014, 5:42 AM • 2 Y
Y by Adventure10, Mango247
Dear,
yes, this is also my proof...
Sincerely
Jean-Louis
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