Stay ahead of learning milestones! Enroll in a class over the summer!

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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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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Hard inequality
JK1603JK   0
2 minutes ago
Source: unknown
Prove $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le \sqrt{2\left(4\sqrt{2}-3\right)\left(a+b+c\right)+12\left(3-2\sqrt{2}\right)\frac{ab+bc+ca}{a+b+c}},\ \ \forall a,b,c\ge 0: a+b+c>0.$$Does EV theorem work?
0 replies
JK1603JK
2 minutes ago
0 replies
ACP = PCB = 8, PBC = 11 find PAC
kamatadu   2
N 7 minutes ago by ND_
Source: RSM mock
Consider $\triangle ABC$. Let $P$ be an interior point such that $\angle ACP = \angle PCB = 8^{\circ}$, $\angle PBC = 11^{\circ}$ and $\angle ABP = 30^{\circ}$. Find $\angle PAC$.
2 replies
kamatadu
2 hours ago
ND_
7 minutes ago
interesting inequality
pennypc123456789   0
11 minutes ago
Let \( a,b,c \) be real numbers satisfying \( a+b+c = 3 \) . Find the maximum value of
\[P  = \dfrac{a(b+c)}{a^2+2bc+3} + \dfrac{b(a+c) }{b^2+2ca +3 } + \dfrac{c(a+b)}{c^2+2ab+3}.\]
0 replies
1 viewing
pennypc123456789
11 minutes ago
0 replies
Interesting inequalities
sqing   7
N 27 minutes ago by sqing
Source: Own
Let $ a,b $ be real numbers . Prove that
$$-\frac{1}{12}\leq \frac{ab+a+b+1}{(a^2+3)(b^2+3)}\leq \frac{1}{4} $$$$-\frac{5}{96}\leq \frac{ab+a+b+2}{(a^2+4)(b^2+4)}\leq \frac{1}{5} $$$$-\frac{1}{16}\leq \frac{ab+a+b+1}{(a^2+4)(b^2+4)}\leq \frac{3+\sqrt 5}{32} $$$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$
7 replies
sqing
5 hours ago
sqing
27 minutes ago
No more topics!
midpoint lies on circumcircle, incircle and a third circle related
parmenides51   5
N Sep 3, 2022 by OlyMathSpirit
Source: 2017 Singapore MO Open Round 2
The incircle of $\vartriangle ABC$ touches the sides $BC,CA,AB$ at $D,E,F$ respectively. A circle through $A$ and $B$ encloses $\vartriangle ABC$ and intersects the line $DE$ at points $P$ and $Q$. Prove that the midpoint of $AB$ lies on the circumircle of $\vartriangle PQF$.
5 replies
parmenides51
Mar 17, 2020
OlyMathSpirit
Sep 3, 2022
midpoint lies on circumcircle, incircle and a third circle related
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Source: 2017 Singapore MO Open Round 2
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parmenides51
30630 posts
#1
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The incircle of $\vartriangle ABC$ touches the sides $BC,CA,AB$ at $D,E,F$ respectively. A circle through $A$ and $B$ encloses $\vartriangle ABC$ and intersects the line $DE$ at points $P$ and $Q$. Prove that the midpoint of $AB$ lies on the circumircle of $\vartriangle PQF$.
This post has been edited 1 time. Last edited by parmenides51, Mar 17, 2020, 11:40 PM
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Mr.Trick
67 posts
#2
Y by
line DE intersects line AB at point T.The midpoint of AB is M.Then we have TA*TB=TF*TM.You can easily prove this by Menelaus.So TP*TQ=TA*TB=TF*TM.Point M is on the circumircle of PQF.
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Kagebaka
3001 posts
#3
Y by
can anyone check if this solution works? I am sort of new to moving points :P

We will prove the following more general statement (while also switching up the labelling to be $A$-centered):
Quote:
The incircle of $\triangle ABC$ touches the sides $BC,CA,AB$ at $D,E,F,$ respectively. Suppose that $P$ is a point on $EF,$ and let $Q=(BCP)\cap EF.$ Prove that $M\cap (PDQ).$

Fix $\triangle ABC$ and animate $P$ on line $EF.$ Define $M'=(PDQ)\cap BC\neq D.$ Consider the following series of projective maps:
  1. Projection of $(PMD)$ to $EF$ through $A.$ ($M\to EF\cap AM$)
  2. Projection of $EF$ to $BC$ through the point at infinity along the pencil of lines perpendicular to $BC.$ ($EF\cap AM\to D$)
  3. Projection of $BC$ to $(PDQ)$ through any point. ($D\to D$)
  4. Inversion preserving $(PDQ)$ swapping $D$ and $M'.$ ($D\to M'$)
The second step maps $EF\cap AM$ to $D$ through a well-known lemma. In order to show that this composition of maps is the identity map, it suffices to check that $M=M'$ for 3 cases of $P.$ Pick $P=E,F,BC\cap EF;$ the first two are easy to check by the Iran Lemma, and the last is trivial.
This post has been edited 2 times. Last edited by Kagebaka, Mar 18, 2020, 5:31 AM
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ironball
110 posts
#4 • 1 Y
Y by Mango247
Let $DE\cap AB=G,\odot(P,Q,F) \cap AB=F,M$
$CF,AD,BE$are concurrent $\Rightarrow \frac{AG}{BG} =\frac{AF}{BF}$
Let$AF=a,BF=b$,so we can calculate $AG=\frac{a(a+b)}{b-a}$
$\therefore GA\cdot GB=GP\cdot GQ=GF\cdot GM$
$\Rightarrow AG(AG+a+b)=(AG+a)(AG+AM)\Rightarrow AM=\frac{a+b}{2}$
So we are done.
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HamstPan38825
8857 posts
#5 • 1 Y
Y by Mango247
Let $K = \overline{AB} \cap \overline{DE}$. As $KA \cdot KB = KP \cdot KQ$, it suffices to show that $KA \cdot KB = KF \cdot KM$. Let $AF=x$ and $BF=z$; then because $\frac{KA}{KB} = \frac{FA}{FB} = \frac xz$, $$KA \cdot KB = \frac{xz}{(z-x)^2}(x+z)^2 = \left(\frac x{z-x} (x+z) + x\right)\left(\frac x{z-x} (x+z)+\frac{x+z}2\right) = KF \cdot KM,$$which can be checked through algebraic expansion.
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OlyMathSpirit
4 posts
#6
Y by
McLaurin ftw.
$EF \cap BC=G$ and now since $-1=(B, C; D, G)$ we have by McLaurin that $GP \cdot GQ=GA \cdot GB=GD \cdot GM$ where $M$ is the midpoint of $BC$, hence $DMPQ$ is cyclic thus we are done :D
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