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Old hard problem
ItzsleepyXD   2
N an hour ago by ItzsleepyXD
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
2 replies
ItzsleepyXD
Apr 25, 2025
ItzsleepyXD
an hour ago
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The Return of Triangle Geometry
peace09   9
N 2 hours ago by mathfun07
Source: 2023 ISL A7
Let $N$ be a positive integer. Prove that there exist three permutations $a_1,\dots,a_N$, $b_1,\dots,b_N$, and $c_1,\dots,c_N$ of $1,\dots,N$ such that \[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]for every $k=1,2,\dots,N$.
9 replies
peace09
Jul 17, 2024
mathfun07
2 hours ago
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Set Partition
Butterfly   0
2 hours ago
For the set of positive integers $\{1,2,…,n\}(n\ge 3)$, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers $a,b,c$ ($a=b$ is allowed) such that $ab=c$. Find the minimal $n$.
0 replies
Butterfly
2 hours ago
0 replies
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Points Lying on its Cevian Triangle's Thomson Cubic
Feuerbach-Gergonne   1
N 2 hours ago by golue3120
Source: Own
Given $\triangle ABC$ and a point $P$, let $\triangle DEF$ be the cevian triangle of $P$ with respect to $\triangle ABC$. Let $H$ be the orthocenter of $\triangle ABC$, and denote the isotomic conjugate of $H, P$ with respect to $\triangle ABC$ by $X, Q$, respectively. Let the centroid of $\triangle DEF$ be $M$, and denote the isogonal conjugate of $P$ with respect to $\triangle DEF$ by $R$. Prove that
$$ P, Q, X \text{ are collinear} \iff P, R, M \text{ are collinear}. $$or in brief
$$ P \in \text{ K007 of } \triangle ABC \iff P \in \text{ K002 of } \triangle DEF. $$
1 reply
Feuerbach-Gergonne
Jul 19, 2024
golue3120
2 hours ago
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Areas of triangles AOH, BOH, COH
Arne   71
N 3 hours ago by EpicBird08
Source: APMO 2004, Problem 2
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Prove that the area of one of the triangles $AOH$, $BOH$ and $COH$ is equal to the sum of the areas of the other two.
71 replies
Arne
Mar 23, 2004
EpicBird08
3 hours ago
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Problem 6
termas   68
N 3 hours ago by HamstPan38825
Source: IMO 2016
There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.

(a) Prove that Geoff can always fulfill his wish if $n$ is odd.

(b) Prove that Geoff can never fulfill his wish if $n$ is even.
68 replies
termas
Jul 12, 2016
HamstPan38825
3 hours ago
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2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   2
N 3 hours ago by Assassino9931
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
2 replies
bin_sherlo
Yesterday at 7:13 PM
Assassino9931
3 hours ago
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combi/nt
blug   2
N 3 hours ago by aaravdodhia
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
2 replies
blug
May 9, 2025
aaravdodhia
3 hours ago
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System of equations in juniors' exam
AlperenINAN   2
N 3 hours ago by Assassino9931
Source: Turkey JBMO TST 2025 P3
Find all positive real solutions $(a, b, c)$ to the following system:
$$ \begin{aligned} a^2 + \frac{b}{a} &= 8, \\ ab + c^2 &= 18, \\ 3a + b + c &= 9\sqrt{3}. \end{aligned} $$
2 replies
AlperenINAN
Yesterday at 7:41 PM
Assassino9931
3 hours ago
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Trigo relation in a right angled. ISIBS2011P10
Sayan   10
N 3 hours ago by mqoi_KOLA
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
10 replies
+1 w
Sayan
Mar 31, 2013
mqoi_KOLA
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
141 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
141 posts
ItzsleepyXD
#2 h  Jan 6, 2025, 9:17 AM
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Tkn
31 posts
Tkn
#3 h  Apr 10, 2025, 7:50 AM
Y by
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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