orthocenter and collinear wanted, incircle, circumcircle related

I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE

1571 replies

rrusczyk

Mar 24, 2025

SmartGroot

Mar 26, 2025

k a March Highlights and 2025 AoPS Online Class Information

jlacosta 0

Mar 2, 2025

March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
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0 replies

jlacosta

Mar 2, 2025

0 replies

Procedure involving colored numbers

Ciobi_ 0

a minute ago

Source: Romania NMO 2025 9.1

Let $N \geq 1$ be a positive integer. There are two numbers written on a blackboard, one red and one blue. Initially, both are 0. We define the following procedure: at each step, we choose a nonnegative integer $k$ (not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are $x$ and $y$ respectively, we replace them with $x+k+1$ and $y+k^2+2$ , which we color blue and red (in this order). We keep doing this procedure until the blue number is at least $N$ .
Determine the minimum value of the red number at the end of this procedure.

0 replies

Ciobi_

a minute ago

0 replies

One nice inequlity 1

prof. 0

4 minutes ago

If $a,b,c$ are positiv real number, such that $a^2+b^2+c^2=3abc$ prove inequality $\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2}\ge\frac{9}{a+b+c}.$

0 replies

1 viewing

prof.

4 minutes ago

0 replies

2 var inquality

sqing 0

9 minutes ago

Source: Own

Let $a,b\geq 0$ and $2a+2b+ab=5$ . Prove that
$\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{9ab}{20(a+b+ab)} \geq \frac{33}{20}$ $\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{31ab}{50(a+b+ab)} \geq \frac{59}{35}$ $\frac{1}{a+1}+\frac{1}{b+1}+ \frac{1}{a+b}+\frac{0.616327 ab}{a+b+ab} \geq \frac{59}{35}$

0 replies

sqing

9 minutes ago

0 replies

Poland Inequalities

wangzishan 6

N 11 minutes ago by AshAuktober

Leta,b,c are postive real numbers,proof that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq1$

6 replies

wangzishan

Apr 23, 2009

AshAuktober

11 minutes ago

No more topics!

orthocenter and collinear wanted, incircle, circumcircle related

parmenides51 2

N Sep 24, 2020 by i3435

Source: 2016 Saudi Arabia IMO TST , level 4, IV p1

Let $ABC$ be a triangle whose incircle $(I)$ touches $BC, CA, AB$ at $D, E, F$ , respectively. The line passing through $A$ and parallel to $BC$ cuts $DE, DF$ at $M, N$ , respectively. The circumcircle of triangle $DMN$ cuts $(I)$ again at $L$ .
a) Let $K$ be the intersection of $N E$ and $M F$ . Prove that $K$ is the orthocenter of the triangle $DMN$ .
b) Prove that $A, K, L$ are collinear.

2 replies

parmenides51

Jul 27, 2020

i3435

Sep 24, 2020

orthocenter and collinear wanted, incircle, circumcircle related

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Source: 2016 Saudi Arabia IMO TST , level 4, IV p1

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parmenides51

30629 posts

parmenides51

#1 h Jul 27, 2020, 9:50 PM

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DNCT1

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DNCT1

#2 h Sep 24, 2020, 12:35 PM

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a) Let $AD\cap (I)=G$ so we have that $FGED$ is harmonic quadrilateral.Since $\overline{MN}\parallel\overline{BC}$
$\Rightarrow -1=D(F,E,G,D)\overset{NM}{=}D(N,M,A,\infty_{NM})=(N,M,A,\infty_{NM})$ from that we get $A$ is the midpoint of $MN$
Hence $\overline{MN}\parallel\overline{BC}$ and $BD=BD$ we have
$\angle NFA=\angle BFD=\angle BDF=\angle ANF\Rightarrow AN=AF$ Similarly we have $AM=AE$ and hence $AE=AF$ so $NFEM$ is cyclic quadrilateral with center $A$ and radius $AM$ .
So we can easily get $K$ is the orthocentre of $\bigtriangleup DMN$
b) This is well known lemma, we can let the center of $(DMN)=O_1$ and draw the diameter $DO_1\cap (DMN)=D_1$ to prove this by $ND_1MK$ is a parallelogram , $D_1K\bot DL$ and $KL\bot DL$ .

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i3435

1350 posts

i3435

#3 h Sep 24, 2020, 1:21 PM

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Since $MAE$ is similar to $DCE$ and $NAF$ is similar to $DBF$ , we get that $AM=AE=AF=AN$ , and it is well known that that means $E$ and $F$ are the feet of the perpendiculars from $N$ to $DM$ and $M$ to $DN$ respectively. Since $A$ is the midpoint of $MN$ , $K$ is the orthocenter of $DMN$ , and $L$ is the intersection of $(DMN)$ and $(DEF)$ , it is well known that these are collinear.

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