National Math Camp Geometry Exam Problem-2

Discussion on Bangladesh National Math Camp
User avatar
Mahiir
Posts:9
Joined:Sun Dec 13, 2020 1:50 pm
Contact:
National Math Camp Geometry Exam Problem-2

Unread post by Mahiir » Sun May 09, 2021 3:21 pm

$Problem$ $2$.
$ABC$ is a triangle where $∠BAC = 90◦$. A line through the midpoint $D$ of $BC$ meets $AB$ at $X$ and $AC$ at $Y$, where $X$ and $Y$ are not on the same side of $BC$. The point $P$ is taken on $XY$ such that $PD$ and $XY$ have the same midpoint $M$. The perpendicular from $P$ to $BC$ meets $BC$ at $T$.
Prove that $AM$ bisects $∠TAD$.

User avatar
Anindya Biswas
Posts:264
Joined:Fri Oct 02, 2020 8:51 pm
Location:Magura, Bangladesh
Contact:

Re: National Math Camp Geometry Exam Problem-2

Unread post by Anindya Biswas » Sun May 09, 2021 8:10 pm

$\textbf{Solution :}$
Nat Camp P2.png
The Figure
Nat Camp P2.png (55.97KiB)Viewed 3791 times
$D$ is the circumcenter of $BAC$.
$\therefore \measuredangle BAD=\measuredangle DBA$.

$M$ is the circumcenter of $XAY$.
$\therefore \measuredangle XAM=\measuredangle MXA$.

$M$ is the circumcenter of $DTP$.
$\therefore \measuredangle MDT=\measuredangle DTM$.

Now, let's show that $ADMT$ cyclic.
\[
\begin{equation}
\begin{split}
\measuredangle DAM &=\measuredangle XAM-\measuredangle BAD \\
&=\measuredangle MXA-\measuredangle DBA \\
&=\measuredangle DXB+\measuredangle XBD \\
&=\measuredangle XDB=\measuredangle MDT=\measuredangle DTM
\end{split}
\end{equation}
\]
$\therefore ADMT$ cyclic.
\[
\begin{equation}
\begin{split}
\therefore \measuredangle DAM&=\measuredangle DTM \\
&=\measuredangle MDT=\measuredangle MAT
\end{split}
\end{equation}
\]
$\therefore AM$ bisects $\angle TAD$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Post Reply