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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
GOTEEM #5: Circumcircle passes through fixed point
tworigami   21
N 8 minutes ago by Ilikeminecraft
Source: GOTEEM: Mock Geometry Contest
Let $ABC$ be a triangle and let $B_1$ and $C_1$ be variable points on sides $\overline{BA}$ and $\overline{CA}$, respectively, such that $BB_1 = CC_1$. Let $B_2 \neq B_1$ denote the point on $\odot(ACB_1)$ such that $BC_1$ is parallel to $B_1B_2$, and let $C_2 \neq C_1$ denote the point on $\odot(ABC_1)$ such that $CB_1$ is parallel to $C_1C_2$. Prove that as $B_1, C_1$ vary, the circumcircle of $\triangle AB_2C_2$ passes through a fixed point, other than $A$.

Proposed by tworigami
21 replies
tworigami
Jan 2, 2020
Ilikeminecraft
8 minutes ago
Strike the inequality
giangtruong13   1
N 11 minutes ago by arqady
Source: Idk
Let $a,b,c \geq 0$ satisfy that $a+b+c=3$. Prove that $$\sum a\sqrt{b^3+1} \leq 5$$
1 reply
giangtruong13
4 hours ago
arqady
11 minutes ago
Calculus rather than inequalities
darij grinberg   12
N 13 minutes ago by asdf334
Source: German TST, IMO ShortList 2003, algebra problem 3
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.
12 replies
darij grinberg
Jul 15, 2004
asdf334
13 minutes ago
Problem 4
blug   1
N 15 minutes ago by ehuseyinyigit
Source: Polish Junior Math Olympiad Finals 2025
In a rhombus $ABCD$, angle $\angle ABC=100^{\circ}$. Point $P$ lies on $CD$ such that $\angle PBC=20^{\circ}$. Line parallel to $AD$ passing trough $P$ intersects $AC$ at $Q$. Prove that $BP=AQ$.
1 reply
blug
an hour ago
ehuseyinyigit
15 minutes ago
No more topics!
Prove that KL bisects BI
trunglqd91   4
N May 10, 2015 by jayme
Let $ABC$ is a triangle with circumcircle $(O)$. Give a circle $(W)$ tangent $AB$, $BC$ at $K$, $L$, resp, and tangent the circumcircle $(AOC)$. Prove that $KL$ bisects $BI$ with $I$ is the centre of incircle $(ABC)$
4 replies
trunglqd91
Mar 29, 2015
jayme
May 10, 2015
Prove that KL bisects BI
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trunglqd91
42 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ is a triangle with circumcircle $(O)$. Give a circle $(W)$ tangent $AB$, $BC$ at $K$, $L$, resp, and tangent the circumcircle $(AOC)$. Prove that $KL$ bisects $BI$ with $I$ is the centre of incircle $(ABC)$
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TelvCohl
2311 posts
#2 • 2 Y
Y by Adventure10, Mango247
My solution:

Let $ M \equiv BI \cap KL $ be the midpoint of $ KL $ .
Let $ A^*, C^* $ be the midpoint of $ BC, BA $, respectively .
Let $ I_b $ be the B-excenter of $ \triangle ABC $ and $ O^* $ be the projection of $ B $ on $ AC $ .
Let $ L^*, K^* $ be the projection of $ I_b $ on $ BA, BC $, respectively .

Let $ \Psi $ be the composition of Inversion $ \mathbf{I}( \sqrt{\tfrac{1}{2} BA \cdot BC} ) $ and Reflection $ \mathbf{R} (BI) $ .

Since $ A \longleftrightarrow A^*, C \longleftrightarrow C^*, O \longleftrightarrow O^* $ under $ \Psi $ ,
so the image of $ \odot (AOC) $ is the 9-point circle $ \odot (A^*O^*C^*) $ of $ \triangle ABC $ under $ \Psi $ ,
hence the image of $ \odot (W) $ is B-excircle $ \odot (I_b) $ under $ \Psi \Longrightarrow L\longleftrightarrow L^*, K \longleftrightarrow K^* $ under $ \Psi $ ,
so we get $ BL \cdot BL^*=BK \cdot BK^* =\tfrac{1}{2} BA \cdot BC $ .

Since $ Rt \triangle BML \sim Rt \triangle BL^*I_b $ ,
so $ BM \cdot BI_b=BL \cdot BL^*=\tfrac{1}{2} BA \cdot BC $ ,
hence combine with $ BI \cdot BI_b= BA \cdot BC $ we get $ M $ is the midpoint of $ BI $ ,
so $ KL $ is the perpendicular bisector of $ BI $ .

Q.E.D
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andria
824 posts
#3 • 2 Y
Y by Adventure10, Mango247
My solution: let the parallel line from $ I $ to $ KL $ meet $ BA, BC $ at $ S, T $ we want to prove that $ KL $ is midline of $\triangle BTS $ note that $ S, T $ are tangency points of mixtilinear circle in front of $ A $. apply an inversion with center $ B $ and power $\sqrt { BC \cdot BA } $ we know that $ C \longleftrightarrow A $ under the inversion and inverse of mixtilinear circle is excircle of $\triangle ABC $ in front of $\angle B $ that touches $ BA, BC $ at $ S', T'$ and since $ ac=2Rh_B $ inverse of $ O $ is a reflection of $ A $ In the line $ BC $ call it $ O'$ let a line parallel to $ BC$ from $ O'$ intersect $ BA, BC $ at $ R, P $ so we get that $\odot ACO'$ is a nine point circle of $\triangle BRP $ also the inverse of $ W $ is a circle that it is tangent to $ BA, BC $ at $ K', L'$ and tangent to $\odot \triangle AO'C $; call it $ W'$ but note that from a well known fact the nine point circle of a triangle is tangent to its excircle from this fact we get that $ W'$ is excircle of $\triangle BRP $ now because $ A'C'$ is midline of $\triangle BRP $ we get that $ T', S' $ are midpoints of $ BL', BK'$ so $ K, L $ are midpoints of $ BS, BT $ before the inversion and $ KL $ bisects $ BI $.
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Luis González
4145 posts
#4 • 2 Y
Y by Adventure10, Mango247
See http://www.artofproblemsolving.com/community/c6h517026 for a solution without inversion.
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jayme
9767 posts
#5 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
a direct proof... based on the key circle (CMI)...

1. O' the midpoint of the arc BC which doesn’t contain O of (AOC)
M the second point of intersection de IO' with (O) ; it the contact point of (AOX) and (W)
(1) the circle (CMI)
X the point of intersection of KL and AI.
2. according to
http://jl.ayme.pagesperso-orange.fr/Docs/Un%20remarquable%20resultat%20de%20Vladimir%20Protassov.pdf p. 2-4
(1) goes through L and by angle chasing also through X
3. according to the Reim’s theorem, JL // AB
4. in the same way, JK // BC
5. the parallelogram BLIK is a rhombus and we are done.

Sincerely
Jean-Louis
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