BdMO National Higher Secondary 2020 P11

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Mursalin
Posts:68
Joined:Thu Aug 22, 2013 9:11 pm
Location:Dhaka, Bangladesh.
BdMO National Higher Secondary 2020 P11

Unread post by Mursalin » Mon Feb 08, 2021 4:03 pm

\(ABCD\) একটি উত্তল চতুর্ভুজ যেখানে \(BC = CD, AC = AD, \angle BCD = 96^\circ\) এবং \(\angle ACD = 69^\circ\)। ধরো \(P_0 = A\) এবং \(Q_0 = B\)। আমরা আরোহ পদ্ধতিতে নতুন কিছু বিন্দু সংজ্ঞায়িত করব। \(CDP_n\)-এর অন্তঃবৃত্তের কেন্দ্রকে বলা হবে \(P_{n+1}\) এবং \(CDQ_n\)-এর অন্তঃবৃত্তের কেন্দ্রকে বলা হবে \(Q_{n+1}\)। যদি \(\angle Q_{2024}Q_{2025}P_{2025} - 90^\circ = \frac{2k-1}{2^n}\) হয়, তাহলে \(k+n\)-এর মান বের কর।


\(ABCD\) is a convex quadrilateral where \(BC = CD, AC = AD, \angle BCD = 96^\circ\) and \(\angle ACD = 69^\circ\). Set \(P_0 = A, Q_0 = B\) respectively. We inductively define \(P_{n+1}\) to be the center of the incircle of \(CDP_n\), and \(Q_{n+1}\) to be the center of the incircle of \(CDQ_n.\) If \(\angle Q_{2024}Q_{2025}P_{2025} - 90^\circ = \frac{2k-1}{2^n}\), compute \(k+n\).
This section is intentionally left blank.

Ashraful Islam Fahim
Posts:1
Joined:Thu Dec 10, 2020 9:03 pm

Re: BdMO National Higher Secondary 2020 P11

Unread post by Ashraful Islam Fahim » Tue Feb 09, 2021 7:49 pm

Hint: Try to prove that, $\angle{Q_{n-1}Q_nP_n}-90=2(\angle{Q_{n}Q_{n+1}P_{n+1}}-90)$

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

Re: BdMO National Higher Secondary 2020 P11

Unread post by Anindya Biswas » Wed Feb 10, 2021 3:22 pm

BdMO National Higher Secondary P11.png
BdMO National Higher Secondary P11.png (104.23KiB)Viewed 1430 times
Step-1. We will show the recursive formula, $\angle Q_{n+1}CP_{n+1}=\frac12\angle Q_nCP_n$ for all $n\geq0$.
Step-2. We will use induction to show that $P_n, Q_n, C, D$ concyclic for all $n\geq0$.
Step-3. We will find the desired angle using results from previous steps.

Step-1 :
$\angle Q_{n+1}CP_{n+1}=\angle Q_{n+1}CD-\angle P_{n+1}CD=\frac12\angle Q_nCD-\frac12\angle P_nCD=\frac12\angle Q_nCP_n$
We can proceed this way to get $\angle Q_nCP_n=\frac{1}{2^n}\angle Q_0CP_0=\frac{96^{\circ}-69^{\circ}}{2^n}=\frac{27^{\circ}}{2^n}$
Step-1 is done.

Step-2 :
Base case, $n=0$ :
$\angle CQ_0D=\frac{180^{\circ}-96^{\circ}}{2}=42^{\circ}$
$\angle CP_0D=180^{\circ}-2\cdot69^{\circ}=42^{\circ}=\angle CQ_0D$
So, $P_0,Q_0,C,D$ concyclic.
Now we will show that $P_n,Q_n,C,D$ concyclic $\Rightarrow$ $P_{n+1},Q_{n+1},C,D$ concyclic.
Let's assume $P_n,Q_n,C,D$ concyclic.
So, $\angle CQ_{n+1}D=90^{\circ}+\frac12\angle CQ_nD=90^{\circ}+\frac12\angle CP_nD=\angle CP_{n+1}D$
$\therefore P_{n+1},Q_{n+1},C,D$ concyclic.
Step-2 is done.

Step-3 :
$\angle Q_nQ_{n+1}P_{n+1}=\angle Q_nQ_{n+1}D-\angle P_{n+1}Q_{n+1}D=90^{\circ}+\frac12\angle Q_nCD-\angle P_{n+1}CD=90^{\circ}+\frac12\angle Q_nCD-\frac12\angle P_nCD$
$\Rightarrow \angle Q_nQ_{n+1}P_{n+1}-90^{\circ}=\frac{\angle Q_nCP_n}{2}=\frac{27^{\circ}}{2^{n+1}}$
So, $\angle Q_{2024}Q_{2025}P_{2025}-90^{\circ}=\frac{27^{\circ}}{2^{2025}}$
So, the desired value is $\frac{27+1}{2}+2025=2039$.
"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