Geometry Challenge – 13

Solution 1: As shown in the diagram at the right, link $AC$. Without loss of generality, let $AB=BC=CD=DA=x$, $PA=1$, $PB=2$, $PC=3$, $\angle{APB}=\alpha$, $\angle{BPC}=\beta$. Therefore $AC=\sqrt{2}x$, and $\angle{APC}=360^\circ-(\alpha+\beta)$.

Applying the Law of Cosines on $\triangle{APB}$, $\triangle{BPC}$, and $\triangle{CPA}$:

$$AB^2=PA^2+PB^2-2\cdot PA\cdot PB\cdot cos\angle{ABP}$$

$$BC^2=PB^2+PC^2-2\cdot PB\cdot PC\cdot cos\angle{BPC}$$

$$AC^2=PA^2+PC^2-2\cdot PA\cdot PC\cdot cos\angle{CPA}$$

i.e.

$$x^2=1^2+2^2-2\cdot 1\cdot 2\cdot cos(\alpha)$$

$$x^2=2^2+3^2-2\cdot 2\cdot 3\cdot cos(\beta)$$

$$2x^2=1^2+3^2-2\cdot 1\cdot 3\cdot cos(360^\circ – (\alpha+\beta))$$

Simplifying the above equations

$$x^2=5-4\cdot cos(\alpha)$$

$$x^2=13-12\cdot cos(\beta)$$

$$x^2=5-3\cdot cos(\alpha+\beta)$$

Let $cos(\alpha)=y$, $cos(\beta)=z$, then $cos(\alpha+\beta)=y\cdot z-\sqrt{1-y^2}\cdot\sqrt{1-z^2}$.

$$x^2=5-4y$$ $$x^2=13-12z$$ $$x^2=5-3(yz-\sqrt{1-y^2}\cdot\sqrt{1-z^2})$$

Replacing $x^2$ with $5-4y$, $z$ with $\dfrac{y+2}{3}$ in the above:

$$5-4y=5-3(y\cdot\dfrac{y+2}{3}-\sqrt{1-y^2}\cdot\sqrt{1-(\dfrac{y+2}{3})^2})$$

Simplifying and factorizing the above: $$(4y-5)(2y^2-1)=0$$ Solving the above equation, $$y=\dfrac{5}{4}\ \ \ \ \ or \ \ \ \ \ y=-\dfrac{\sqrt{2}}{2}$$ Ignoring the invalid value, we have $cos(\alpha)=-\dfrac{\sqrt{2}}{2}$. Thus $$\angle{APB}=\alpha=\boxed{135^\circ}$$

Solution 2: Rotating $P$ around $B$ for $90^\circ$ to point $Q$, as shown in the diagram at the right. Since $ABCD$ is a square, $\triangle{APB}\cong \triangle{CQB}$, and $\triangle{PBQ}$ is right and isosceles. Therefore $PQ=\sqrt{2}PB$ and $\angle{BQP}=45^\circ$.

Because $QC:PQ:PC=PA:\sqrt{2}PB:PC=1:2\sqrt{2}:3$, $\triangle{CQP}$ is right and $\angle{CQP}=90^\circ$. Therefore $$\angle{APB}=\angle{CQB}=\angle{CQP}+\angle{BQP}=90^\circ+45^\circ=\boxed{135^\circ}$$