Art of Problem Solving
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Geometry Parallel Proof Problem
CatalanThinker   0
10 minutes ago
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
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+1 w
CatalanThinker
10 minutes ago
0 replies
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Sum of 1/(a^5(b+2c))^2 at least 1/3 [USA TST 2010 2]
MellowMelon   42
N an hour ago by Adywastaken
Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]
42 replies
MellowMelon
Jul 26, 2010
Adywastaken
an hour ago
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Weird function?
ItzsleepyXD   2
N an hour ago by ItzsleepyXD
Source: Own
Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that for all \( x, y \in \mathbb{R} \),
\[ f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y). \]
2 replies
ItzsleepyXD
Apr 11, 2025
ItzsleepyXD
an hour ago
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Almost similar one but more answer lol
ItzsleepyXD   0
an hour ago
Source: Own , Modified
Find all non decreasing functions or non increasing function $f \colon \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in \mathbb{R}$

$$ f(x+f(y))=f(x)+f(y) \text{ or } f(f(f(x)))+y$$.
0 replies
ItzsleepyXD
an hour ago
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A lot of unexpected answer from non decreasing function
ItzsleepyXD   0
an hour ago
Source: Own
Find all non decreasing function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ and $m,n \in \mathbb{N}_0$ such that $m+n \neq 0$ there exist $m',n' \in \mathbb{N}_0$ such that $m'+n'=m+n+1$ and $$f(f^m(x)+f^n(y))=f^{m'}(x)+f^{n'}(y)$$. Note : $f^0(x)=x$ and $f^{n}(x)=f(f^{n-1}(x))$ for all $n \in \mathbb{N}$ . original
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ItzsleepyXD
an hour ago
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Cute Inequality
EthanWYX2009   0
2 hours ago
Let $a_1,\ldots ,a_n\in\mathbb R\backslash\{0\},$ determine the minimum and maximum value of
\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]
0 replies
EthanWYX2009
2 hours ago
0 replies
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Interesting inequality
sealight2107   3
N 2 hours ago by NguyenVanHoa29
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
3 replies
sealight2107
May 6, 2025
NguyenVanHoa29
2 hours ago
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Is this FE is solvable?
ItzsleepyXD   0
2 hours ago
Source: Own , If not appear somewhere before
Find all function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ . $$f(x+f(y))+f(x+y)=2x+f(y)+f(f(y))$$. Original
0 replies
ItzsleepyXD
2 hours ago
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Symmetry in Circumcircle Intersection
Mimii08   2
N 2 hours ago by mashumaro
Hi! Here's another geometry problem I'm thinking about, and I would appreciate any help with a proof. Thanks in advance!

Let AD and BE be the altitudes of an acute triangle ABC, with D on BC and E on AC. The line DE intersects the circumcircle of triangle ABC again at two points M and N. Prove that CM = CN.

Thanks for your time and help!
2 replies
Mimii08
4 hours ago
mashumaro
2 hours ago
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Blue chessboard
rcorreaa   10
N 2 hours ago by Jaxman8
Source: 2022 Brazilian National Mathematical Olympiad - Problem 6
Some cells of a $10 \times 10$ are colored blue. A set of six cells is called gremista when the cells are the intersection of three rows and two columns, or two rows and three columns, and are painted blue. Determine the greatest value of $n$ for which it is possible to color $n$ chessboard cells blue such that there is not a gremista set.
10 replies
rcorreaa
Nov 22, 2022
Jaxman8
2 hours ago
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Substitution
JCE   3
N 3 hours ago by K124659
I've been working on this for about an hour or so, and I can't get this problem. I know the answer, but no idea on how to find it.
Please help?

2x-y^2=4
x^2+y=14
3 replies
JCE
May 27, 2006
K124659
3 hours ago
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Some Identity that I need help
ItzsleepyXD   2
N Apr 10, 2025 by Tkn
Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$.
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .
2 replies
ItzsleepyXD
Dec 28, 2024
Tkn
Apr 10, 2025
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Some Identity that I need help
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ItzsleepyXD
139 posts
ItzsleepyXD
#1 h  Dec 28, 2024, 3:41 AM
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Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$.
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .
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ItzsleepyXD
139 posts
ItzsleepyXD
#2 h  Jan 6, 2025, 9:17 AM
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Tkn
25 posts
Tkn
#3 h  Apr 10, 2025, 7:50 AM
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To show the above equality, we need to prove these two claims:
Claim 1. $OH^2=9R^2-(a^2+b^2+c^2)$
Proof. This one is easy by vector bash with the Euler line. Note that
$$\frac{OG}{OH}=\frac{1}{3}$$where $G$ denotes the centroid of $\triangle{ABC}$. It is easy to see that
$$\overrightarrow{OH}=3\overrightarrow{OG}=3\left(\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}\right).$$Comparing modulus of both sides:
\begin{align*} |\overrightarrow{OH}|^2&=\overrightarrow{OH}\cdot \overrightarrow{OH}\\ &=3R^2+2R^2\left(\cos(2\angle{A})+\cos(2\angle{B})+\cos(2\angle{C})\right)\\ &=3R^2+6R^2-4R^2\left(\sin^2(\angle{A})+\sin^2(\angle{B})+\sin^2(\angle{C})\right)\\ &=9R^2-(a^2+b^2+c^2) \end{align*}The last step requires the sine's law on $\triangle{ABC}$, as in the form:
$$\frac{a}{\sin(\angle{A})}=\frac{b}{\sin(\angle{B})}=\frac{c}{\sin(\angle{C})}=2R.$$where $a,b$ and $c$ are side lengths of $BC,CA$ and $AB$ respectively.
Claim 2. $HI^2=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})$
Proof. Given that $M$ is a midpoint of the segment $\overline{BC}$.
We use the well-known lemma to find, $AH=2OM=2R\cos(\angle{A})$. And,
$$AI=2r\cos\left(\frac{\angle A}{2}\right)=4R\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right).$$The rest is just some bashing of cosine law:
\begin{align*} HI^2&=AI^2+AH^2-2AI\cdot AH\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\ &=16R^2\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+4R^2\cos^2(\angle{A})-16R^2\cos(\angle{A})\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right)\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\ &=4R^2\left(\cos^2(\angle{A})+(4-4\cos(\angle{A}))\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\ &=4R^2\left(\cos^2(\angle{A})+8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\ &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+\cos(\angle{A})\left(\cos(\angle{A})-\sin(\angle{B})\sin(\angle{C})\right)\right)\\ &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)\\ &=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C}) \end{align*}Which is complete.
Then, it just equivalent of showing:
$$8R^2(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C}))=a^2+b^2+c^2$$Note that $a^2=4R^2\sin^2(\angle{A})$, similarly for $b$ and $c$. Simplify and gives
$$2\left(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)=\sin^2(\angle{A})+\sin^2{(\angle{B})}+\sin^2(\angle{C})$$or equivalent to
$$1=2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})+\cos^2(\angle{A})+\cos^{2}(\angle{B})+\cos^2(\angle{C})$$This is obviously true due to
\begin{align*} 2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})&=-\cos^2(\angle{C})+\cos(\angle{A}-\angle{B})\cos(\angle{C})\\ &=-\cos^2(\angle{C})+\frac{1}{2}\left(\cos(\angle{A}-\angle{B}+\angle{C})+\cos(\angle{A}-\angle{B}-\angle{C})\right)\\ &=-\cos^2(\angle{C})+\frac{1}{2}(\cos(180^{\circ}-2B)+\cos(180^{\circ}-2A))\\ &=-\cos^2(\angle{C})-\frac{1}{2}(\cos(2\angle{A})+\cos(2\angle B))\\ &=-\cos^2(\angle{C})-\cos^2(\angle{A})-\cos^2(\angle{B})+1 \end{align*}Which actually solves the problem.
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