$ABCD$ 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 $\triangle CDP_n$, and $Q_{n+1}$ to be the center of the incircle of $\triangle CDQ_n$. If $\angle Q_{2024}Q_{2025}P_{2025} - 90^\circ = \frac{2k-1}{2^n}$, compute $k+n$.
This is a question from a national Olympiad. I've been struggling to solve this. Here is my attempt to solve the problem:
In quadrilateral $ABCD$, $\angle CAD=\angle CBD=42^\circ$. So, $A,B,C,D$ are concyclic. So, it can be shown that $C,D,P_n,Q_n$ are concyclic for all $n\geq0$ by induction. So, we have $\angle CQ_nP_n=180^\circ -\angle CDP_n=180^\circ-\frac{69}{2^n}$.
Again, because $\triangle CQ_nQ_{n-1}\cong \triangle CDQ_n$ we have $\angle CQ_nQ_{n-1}=\angle CQ_nD=180^\circ-\angle CDQ_n-\angle DCQ_n$.
Now, $\angle Q_{n-1}Q_{n}P_n - 90^\circ\\=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n-90^\circ\\ =270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ =\angle CDQ_n +\angle DCQ_n+\angle CDP_n-90^\circ\\ =\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}-90^\circ.$
So, I am getting a negative number as answer which is not possible. And I am quite sure that there is nothing wrong in the question. What am I missing and what is the correct solution?
Let us start with the triangle $P_{n-1}CD$ with angles $2x$, $2y$, $2z$ in $P_{n-1},C,D$ respectively. Let $I=P_n$ be its incenter. Then: $$ \hat P_n= \hat I := \widehat{CID}=180^\circ-y-z=90^\circ+\frac12(180^\circ-2y-2z)=90^\circ+\frac 12\cdot 2x=90^\circ+\frac 12\hat P_{n-1}\ . $$ The same relation applies for $\hat Q_n$. This gives $$ \hat P_n=\hat Q_n=90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}+42^\circ\cdot\frac 1{2^n}\ . $$ Then we have: $$ \begin{aligned} \widehat{Q_{n-1}Q_nP_n} &= 360^\circ -\widehat{CQ_nQ_{n-1}} -\widehat{Q_{n-1}Q_nP_n} \\ &= 360^\circ -\widehat{CQ_nD} -\left(180^\circ-\widehat{CDP_n}\right) \\ &= 180^\circ -\widehat{CQ_nD} +\frac 1{2^n}\widehat{CDA} \\ &= 180^\circ -90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}-42^\circ\cdot\frac 1{2^n} +69^\circ\cdot\frac 1{2^n} \\ &= 180^\circ -180^\circ\cdot\left(1-\frac 1{2^n}\right) +27^\circ\cdot\frac 1{2^n} \\ &= \left( 180^\circ +27^\circ \right) \cdot\frac 1{2^n} \ . \end{aligned} $$
Where is the error in the following lines? $$ \begin{aligned} \angle Q_{n-1}Q_{n}P_n\color{blue}{ - 90^\circ} &=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n\color{blue}{ - 90^\circ} \\ &=270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ &=\angle CDQ_n +\angle DCQ_n+\angle CDP_n\color{blue}{ - 90^\circ}\\ &=\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ} \end{aligned} $$ No error, $\angle Q_{n-1}Q_{n}P_n \color{blue}{ - 90^\circ}= \frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ}$ gives the same $\angle Q_{n-1}Q_{n}P_n = \frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}$. The angle is so small, since the arcs $CP_{n-1}Q_{n-1}D$ and $CP_nQ_nD$ are hardly distinguishable in some computer picture, they are plotted slightly over the segment $CD$, so we expect an angle which is either almost $0^\circ$ or $180^\circ$.