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Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   10
N 27 minutes ago by Mahdi_Mashayekhi
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
10 replies
OgnjenTesic
Thursday at 4:02 PM
Mahdi_Mashayekhi
27 minutes ago
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AT // BC wanted
parmenides51   105
N 31 minutes ago by Adywastaken
Source: IMO 2019 SL G1
Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

(Nigeria)
105 replies
parmenides51
Sep 22, 2020
Adywastaken
31 minutes ago
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AD and BC meet MN at P and Q
WakeUp   6
N Apr 10, 2025 by Nari_Tom
Source: Baltic Way 2005
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
6 replies
WakeUp
Dec 28, 2010
Nari_Tom
Apr 10, 2025
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AD and BC meet MN at P and Q
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geometry
parallelogram
rhombus
trigonometry
geometry proposed
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Source: Baltic Way 2005
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WakeUp
1347 posts
WakeUp
#1 h  Dec 28, 2010, 4:39 PM • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
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Luis González
4149 posts
Luis González
#2 h  Dec 28, 2010, 6:02 PM • 1 Y
Y by Adventure10
Let $R,S$ be the midpoints of the diagonals $AC,BD.$ Since $BC=AD,$ then $NR=NS$ $\Longrightarrow$ Parallelogram $NRMS$ is a rhombus $\Longrightarrow$ $NM$ bisects $\angle RNS$ internally. Let $U \equiv AD \cap BC.$ Since $\angle CUD$ and $\angle RNS$ have corresponding parallel sides, it follows that their angle bisectors are parallel as well. By Menelaus' theorem for $\triangle UCD$ cut by $\overline{QPN},$ keeping in mind that $UP=UQ$ and $CN=ND,$ we get

$\frac{QU}{QC} \cdot \frac{CN}{ND} \cdot \frac{DP}{PU}=1 \Longrightarrow \ QC=DP.$
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jgnr
1343 posts
jgnr
#3 h  Dec 29, 2010, 6:19 AM • 2 Y
Y by Adventure10, Mango247
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=377297 then use sine law.
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sunken rock
4401 posts
sunken rock
#4 h  Jan 8, 2011, 2:59 PM • 2 Y
Y by Adventure10, Mango247
It is easy to see that $MN$ is parallel to the bisector of the angle $\widehat {(BC, AD)}$, hence the triangles $\triangle MBQ$ and $\triangle AMP$ are pseudo-similar and our problem is solved then.

Best regards,
sunken rock
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rafaello
1079 posts
rafaello
#5 h  Nov 4, 2020, 4:33 PM
Y by
By Law of Sines on $\triangle MPA$ and $\triangle QMB$, $$\frac{AM}{\sin{\angle MPA}}= \frac{AP}{\sin{\angle AMP}}$$and
$$\frac{BM}{\sin{\angle MQB}}= \frac{BQ}{\sin{\angle BMQ}}$$Since, $\sin{\angle AMP}=\sin{\angle BMQ}$, we get that
$$\frac{\sin{\angle MQB}}{\sin{\angle MPA}}= \frac{AP}{BQ}.$$
By Law of Sines on $\triangle NPD$ and $\triangle QNC$, $$\frac{DN}{\sin{\angle NPD}}= \frac{DP}{\sin{\angle DNP}}$$and
$$\frac{CN}{\sin{\angle CQN}}= \frac{CQ}{\sin{\angle CNQ}}$$Since, $\sin{\angle DNP}=\sin{\angle CNQ}$, we get that
$$\frac{\sin{\angle CQN}}{\sin{\angle NPD}}= \frac{DP}{CQ}.$$Because, $\angle MQB=\angle CQN$ and $\angle MPA=\angle NPD$, we get that
$$\frac{AP}{BQ}= \frac{DP}{CQ}\Longleftrightarrow \frac{AD+DP}{DP}= \frac{BC+CQ}{CQ}\Longleftrightarrow \frac{AD}{DP}= \frac{BC}{CQ}.$$Since $BC=AD$, we get that $CQ=DP$. $\quad \square$
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k12byda5h
104 posts
k12byda5h
#6 h  Nov 4, 2020, 6:32 PM • 1 Y
Y by Tudor1505
Let $S$ be the center of spiral similarity that sends $A \rightarrow B, D \rightarrow C$, $X=AD \cap BC$. Therefore, it's also the center of spiral similarity that sends $M \rightarrow N$ and $BS=AS$. Hence, $\square ABSX, \square BMQS$ are cyclic quadrilaterals, also $\square PXQS$. Hence, there is a spiral similarity center at $S$ that sends $BQ \rightarrow AP$. But $BS=AS \implies BQ = AP \implies CQ = DP$ .
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Nari_Tom
117 posts
Nari_Tom
#7 h  Apr 10, 2025, 11:27 AM
Y by
Nice solution above, but it's trivial if we use Menelaus theorem on the $\triangle EBC$, $MN$ and $\triangle EAD$, $MN$ where $E=AB \cap CD$.
We have: $ \frac{DN}{NE} *\frac{EM}{MA} *\frac {PA}{PD}=1 $ and $\frac{NC}{NE} *\frac{MB}{ME} *\frac{QB}{QC}=1 $ $\implies$ $\frac{PA}{PD}=\frac{QB}{QC}$ $\implies$ $PA=QB$.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 10, 2025, 11:28 AM
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