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Lines AD, BE, and CF are concurrent
orl   48
N 32 minutes ago by Avron
Source: IMO Shortlist 2000, G3
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
48 replies
orl
Aug 10, 2008
Avron
32 minutes ago
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Locus of sphere cutting three spheres along great circles
Miquel-point   1
N an hour ago by kiyoras_2001
Source: Romanian IMO TST 1981, Day 2 P3
Consider three fixed spheres $S_1, S_2, S_3$ with pairwise disjoint interiors. Determine the locus of the centre of the sphere intersecting each $S_i$ along a great circle of $S_i$.

Stere Ianuș
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Miquel-point
Yesterday at 6:29 PM
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Assisted perpendicular chasing
sarjinius   5
N Apr 5, 2025 by hukilau17
Source: Philippine Mathematical Olympiad 2025 P7
In acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be an arbitrary point on the circumcircle of triangle $ABC$ such that $D$ does not lie on line $OB$ and that line $OD$ is not parallel to line $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$ such that $DE$ is perpendicular to $BC$, and let $F$ be the point on line $AC$ such that $FA = FE$. Let $P$ and $R$ be the points on the circumcircle of triangle $ABC$ such that $PE$ is a diameter, and $BH$ and $DR$ are parallel. Let $M$ be the midpoint of $DH$.
(a) Show that $AP$ and $BR$ are perpendicular.
(b) Show that $FM$ and $BM$ are perpendicular.
5 replies
sarjinius
Mar 9, 2025
hukilau17
Apr 5, 2025
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Assisted perpendicular chasing
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Source: Philippine Mathematical Olympiad 2025 P7
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sarjinius
239 posts
sarjinius
#1 h  Mar 9, 2025, 3:41 PM • 2 Y
Y by MathLuis, Rounak_iitr
In acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be an arbitrary point on the circumcircle of triangle $ABC$ such that $D$ does not lie on line $OB$ and that line $OD$ is not parallel to line $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$ such that $DE$ is perpendicular to $BC$, and let $F$ be the point on line $AC$ such that $FA = FE$. Let $P$ and $R$ be the points on the circumcircle of triangle $ABC$ such that $PE$ is a diameter, and $BH$ and $DR$ are parallel. Let $M$ be the midpoint of $DH$.
(a) Show that $AP$ and $BR$ are perpendicular.
(b) Show that $FM$ and $BM$ are perpendicular.
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Eagle116
22 posts
Eagle116
#2 h  Mar 9, 2025, 4:01 PM
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Does the order of A,C and F affect the solution?
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chisa36
4 posts
chisa36
#3 h  Mar 21, 2025, 4:22 PM • 1 Y
Y by JollyEggsBanana
A little too easy for P7, no?

Below is the solution that I found and wrote up in the actual contest. (a) is trivial by angle chase. (b) is also trivial by mmp.
Attachments:
chisa36 - PMO Nationals P7 MMP - revised.pdf (211kb)
This post has been edited 1 time. Last edited by chisa36, Mar 22, 2025, 2:22 PM
Reason: minor revision
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ZeroHero
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ZeroHero
#4 h  Apr 2, 2025, 3:59 PM
Y by
chisa36 wrote:
A little too easy for P7, no?

Below is the solution that I found and wrote up in the actual contest. (a) is trivial by angle chase. (b) is also trivial by mmp.

I can't understand your sol. where I can learn mmp?
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X.Allaberdiyev
102 posts
X.Allaberdiyev
#5 h  Apr 4, 2025, 6:01 PM
Y by
Okay.. I will post my solution without details(lazy to do it) coz no one posted the sol. without mmp (thing that i personally don’t understand).
Part (a) is easy by angle chasing
Part (b)
Let $DE$ meets $BC$ at $Z$ and foot of the perpendicular from $D$ to $AC$ is $X$. It is pretty well known that $M$, $Z$, and $X$ are collinear. Now angle chasing yields that the angles $BEF$ and $BZM$ are equal. So if we will prove that $BE/EF=BZ/ZM$, triangles $BEF$ and $BZM$ would be similar and by spiral similarity we would obtain that angle $FMB$ is right. To show that we find the ratio $BE/BZ$ by sine theorem in triangle $BEZ$, and we find $FE$ in terms of $AE$ by sine theorem in $AFE$. The plugging it into the equation we need to prove gives that it is enough to prove that $AE=2MZ$, which is true because if we will denote the intersection of lines $MZ$ and $AH$ by $T$, one can easily prove that $AEZH$ is parallelogram. So done!
This post has been edited 1 time. Last edited by X.Allaberdiyev, Apr 4, 2025, 6:05 PM
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hukilau17
281 posts
hukilau17
#6 h  Apr 5, 2025, 3:37 AM
Y by
Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$$|a|=|b|=|c|=|d|=1$$$$o=0$$$$h=a+b+c$$$$e = -\frac{bc}d$$$$\overline{f} = \frac{f}{ae} = \frac{a+c-f}{ac} \implies f = \frac{e(a+c)}{c+e} = \frac{b(a+c)}{b-d}$$$$p = -e = \frac{bc}d$$$$r = \frac{b\left(-\frac{ac}b\right)}{d} = -\frac{ac}d$$$$m = \frac{d+h}2 = \frac{a+b+c+d}2$$Then for part (a), we have $ap = \frac{abc}d$ and $br = -\frac{abc}d$, so $ap = -br$ and $AP\perp BR$.

For part (b), we have
$$f-m = \frac{2b(a+c) - (b-d)(a+b+c+d)}{2(b-d)} = \frac{ab+ad-b^2+bc+cd+d^2}{2(b-d)} = \frac{(b+d)(a-b+c+d)}{2(b-d)}$$$$m-b = \frac{a-b+c+d}2$$Thus,
$$\frac{f-m}{m-b} = \frac{b+d}{b-d} \in i\mathbb{R}$$so $FM\perp MB$. $\blacksquare$
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