## BdMO National Higher Secondary 2020 P11

Mursalin
Posts:68
Joined:Thu Aug 22, 2013 9:11 pm
BdMO National Higher Secondary 2020 P11
$$ABCD$$ 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 $$k+n$$.
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Ashraful Islam Fahim
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Joined:Thu Dec 10, 2020 9:03 pm

### Re: BdMO National Higher Secondary 2020 P11

Anindya Biswas
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Joined:Fri Oct 02, 2020 8:51 pm
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### Re: BdMO National Higher Secondary 2020 P11

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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