Qi Huan's triangle

Diagram for the lazy, which I spent too much time on

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Extend $AP$ to meet $BC$ at $D$ ; construct $E \neq D$ on the extension of $BC$ so that $AC = AE$ .

Angle chase to get $\angle BDP = 30^\circ = \angle PBD$ , so $PD = BD$ , and by the angle bisector theorem $\frac{BD}{DC} = \frac{AB}{AC}$ , so $\frac{PD}{DC} = \frac{BD}{DC} = \frac{AB}{AC} = \frac{AB}{AE}$ . Also by angle chasing $\angle PDC = 60^\circ$ and $\angle BAE = \angle BAC + \angle CAE = 44^\circ + (180^\circ - 2\angle ACE) = 224^\circ - 2(44^\circ + 38^\circ) = 60^\circ$ .

Therefore we have an SAS similarity: $\triangle PDC \sim \triangle BAE$ , and we can compute $\angle APC = 180^\circ - \angle DPC = 180^\circ - \angle ABE = \boxed{142^\circ}$ .

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Some motivation: obviously the key things to notice for this problem are that $22^\circ + 8^\circ = 30^\circ$ , which motivates us to construct an external angle which also forms a nice isosceles triangle, and also suggests a 30-60-90 triangle or an equilateral one lurking somewhere might help.

The things we actually end up constructing don't seem to involve that, but if we additionally construct $D' \neq D$ on the extension of $BC$ such that $AD = AD'$ , it becomes clearer why $\angle BAE$ magically works out to be $60^\circ$ . Since $\angle PDC \equiv \angle ADC$ is $60^\circ$ , $\triangle ADD'$ is equilateral, and $\angle BAD = \angle CAD = \angle EAD'$ ; subtracting both sides from $\angle BAD'$ gives us the $60^\circ$ equality.

Let line $AP$ intersect $BC$ at $D$ and let $E$ be the point on $AC$ such that $\angle{PBA}=\angle{PBE}=8$ . Then $\angle{EBD}=22=\angle{EAD}$ , so $ABDE$ is cyclic. Thus, $\angle{CDE}=\angle{CAB}=44$ .

Furthermore, $PA$ bisects $\angle{BAE}$ and $PB$ bisects $\angle{ABE}$ . It follows that $P$ is the incenter of $\triangle{ABE}$ , so $\angle{PEA}=\angle{PEB}=60$ . Thus, $\angle{EPD}=82$ . However, $\angle{ECD}=98$ , so $CDPE$ is cyclic. As a result, $\angle{CPE}=\angle{CDE}=44$ , so $\angle{APC}=\angle{APE}+\angle{CPE}=98+44=\boxed{142}$ .

Simplest solution: By angle sum property of triangle $\angle APB=150^{\circ}$ . Let the angle to be find be $x^{\circ}$ . Applying Sine Rule in $\triangle APB$ , $\frac{AP}{sin8^{\circ}}=\frac{PB}{sin22^{\circ}}.$ Now applying in $\triangle APC$ , $\frac{AP}{sin(158-x)^{\circ}}=\frac{PC}{sin22^{\circ}}.$ We get $\frac{PB}{PC}=\frac{sin(158-x)^{\circ}}{sin8^{\circ}}.$ Now applying Sine rule in $\triangle BPC$ , $\frac{PB}{PC}=2sin(x-60)^{\circ}$ . Equating both the equations we get, $2sin(x-60)^{\circ}=\frac{sin(158-x){\circ}}{sin8^{\circ}}$ , $sin(158-x)^{\circ}=sin(90+68-x)^{\circ}=cos(68-x)^{\circ}$ . Applying sum and product formula,it becomes $\cos(x-68)^{\circ}-cos(x-52)^{\circ}=cos(68-x)^{\circ}$ , So, $x^{\circ}-52^{\circ}=90^{\circ}, x=142^{\circ}$ . Obviously no other value except $90^{\circ}$ will be valid.

Since if we extend $AP , BP , PC$ to meet at side of triangle's , then they will become cevians with intersection point P.

So We use trigonometric form of Ceva's theorem .

See the statement here

Let $\angle ACP = x$ .

Then using ceva's theorem we get the equation ,

$\sin(30) \cdot \sin(x) = \sin(8) \cdot \sin(98-x)$ Solving gives $x \sim 16$ . Thus giving $\angle APC = 142$

First, let $D$ be the intersection of $AP$ and $BC$ . Let $B'$ be the reflection of $B$ over $AP$ . Note that because $AP$ bisects $\angle BAC$ , $B'$ lies on $AC$ . Also, $\angle BPD=30^{\circ}$ , so $\angle BPB'=60^{\circ}$ . Thus, because $BP=BP'$ , $\triangle BPB'$ is equilateral.

Now, because $C$ lies on the angle bisector $BD$ of equilateral triangle $BPB'$ , we must have $CP=CB'$ . Hence, $\angle CPB'= \angle CB'P= \angle ABP=8^{\circ}$ . Thus, $\angle APC= \angle APB' - \angle CPB'=150^{\circ}-8^{\circ}=142^{\circ}$ .

First extend AP to meet BC at X

Let length XB=x, CX=y

Then we know from considering the areas of triangles AXC and AXB that the ratio x:y is AB:AC

From the sine rule we know that $\frac{AB}{AC}=\frac{sin(98)}{sin38}$

Then notice that angle APB=150, so angle XPB=30. This makes XPB an isosceles triangle and we know that XP=x

Use the sine rule in triangle CPX. Let angle $PCX =\theta$ , then $CPX=120-\theta$ So we have $\frac{x}{y}=\frac{sin(\theta)}{sin(120-\theta)}=\frac{AB}{AC}=\frac{sin(98)}{sin(38)}$

Expanding the sines we get $\frac{sin(\theta)}{\frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta)} = \frac{cos(8)}{\frac{1}{2}sin(8)+\frac{\sqrt{3}}{2}cos(8)}$

Noting that $cos(8)=sin(82), sin(8)=cos(82)$ we see that $\theta=82$

We then finally have $APC=180-(120-\theta)=60+\theta=142$

Note: The following solution is not mine.

Let $E$ be the intersection of $AP$ and $BF$ . In triangle $ABP$ , $\angle APB = 180^\circ - \angle ABP - \angle BAP = 180^\circ - 8^\circ - 22^\circ = 150^\circ$ , so $\angle BPE = 180^\circ - \angle APB = 180^\circ - 150^\circ = 30^\circ$ .

Extend $AC$ past $C$ to $F$ such that $AF = AB$ . Since $\angle PAB = \angle PAF$ and $AB = AF$ , triangles $PAB$ and $PAF$ are congruent. Then $\angle APF = \angle APB = 150^\circ$ , and $\angle FPE = 180^\circ - \angle APF = 180^\circ - 150^\circ = 30^\circ$ . Hence, $\angle BPF = \angle BPE + \angle FPE = 30^\circ + 30^\circ = 60^\circ$ . Also, $PB = PF$ , so triangle $BPF$ is equilateral.

Then $\angle FBC = \angle FBP - \angle PBC = 60^\circ - 30^\circ = 30^\circ$ . Since $BP = BF$ and $\angle PBC = \angle FBC = 30^\circ$ , triangles $PBC$ and $FBC$ are congruent. This implies $CP = CF$ , so triangle $CPF$ is isosceles.

Then $\angle CPF = \angle CFP = \angle AFP = \angle ABP = 8^\circ$ , so $\angle APC = 180^\circ - \angle CPF - \angle EPF = 180^\circ - 8^\circ - 30^\circ = \boxed{142^\circ}$ .