# :Find the area of ​the shaded region $ABC$ in the figure below

For reference:Calculate the area of ​​the shaded region; if: $$AP = 4\sqrt2; PB = 4$$ and $$HN = 4$$.(Answer: $$\frac{32}{51}(5\sqrt2+4)$$)

My progress $$HN(2r-HN)=BH^2\quad\Rightarrow{}\quad 4(2r-4)=4 \therefore R=\dfrac{5}{2}\\ S_{ABC} = \frac{a.b.c}{4R}=\frac{4.b.c}{10}\therefore \boxed{S_{ABC}= \frac{2.b.c}{5}}\\ S_{ABC}= \frac{S_{APC}}{2}\\ \frac{S_{ABC}}{S_{PAB}}=\frac{4b}{c.4\sqrt2}\implies b=\sqrt2c$$

...???

By trigonometry $$S(ABC)=\frac{1}{2}S(APC)=\frac{1}{4}PA⋅PCsin(\overset{\LARGE{\frown}}{APC})=\frac{1}{4}PA⋅PC.sin(\overset{\LARGE{\frown}}{APO}+\overset{\LARGE{\frown}}{CPO})=\\ \frac{1}{4}PA⋅PC.(sin(\overset{\LARGE{\frown}}{CPO})cos(\overset{\LARGE{\frown}}{APO})+cos(\overset{\LARGE{\frown}}{CPO})sin(\overset{\LARGE{\frown}}{APO})$$

...???

• Are you also given that $BC=4$? Dec 14, 2021 at 14:45
• @TonyK $BC=4$ is derivable from the fact that $AP$ is tangent to circle which follows from equality of angles shown in orange. Dec 14, 2021 at 15:30
• @MyMolecules: I'm sure you are right, but I don't see it. Dec 14, 2021 at 16:02
• @TonyK $P,B,C$ are to be assumed collinear (used by OP) then by similarity of triangles or by power of point $P$, $PB \cdot PC = PA^2$ Dec 14, 2021 at 16:05
• If $AP$ is tangent to the circle, then $BC=4$ can be derived from the power of point $P$ wrt the circle. However, the problem does not state that $AP$ is tangent. Dec 14, 2021 at 16:05

Based on confirmation that $$PA$$ is tangent, by using power of point of $$P$$, $$BC = 4$$ and using similar triangles, $$AC = AB \sqrt2$$.

If $$OH = x, OB = ON = 4 - x$$, and applying Pythagoras in $$\triangle OBH$$

$$x^2 + 2^2 = (4-x)^2 \implies x = OH = 3/2, OB = 5/2$$

As $$\angle BAC = \angle BOH = \theta$$ (say)

$$\cos \theta = \dfrac{3}{5}, \sin\theta = \dfrac{4}{5}$$

Using law of cosine in $$\triangle BAC$$, $$AB^2 + (AB \sqrt2)^2 - 4^2 = 2 AB \cdot AB \sqrt2 \cdot \dfrac{3}{5}$$

$$\implies AB^2 = \dfrac{80}{15 - 6 \sqrt2} = \dfrac{80}{51} (5 + 2 \sqrt2)$$

$$S_{\triangle ABC} = \dfrac 12 \cdot AB \cdot AB\sqrt2 \cdot \sin\theta$$

$$= \dfrac{2\sqrt2}{5} AB^2 = \dfrac{32}{51}(5 \sqrt 2 + 4)$$

If you want to avoid trigonometry, drop a perp from $$C$$ to $$AB$$ which meets $$AB$$ at $$G$$ then $$\triangle ACG \sim \triangle OBH$$

so, $$\displaystyle \frac{CG}{AC} = \frac{BH}{OB} \implies CG = \frac{4 \sqrt2}{5} AB$$

say, $$AB = y$$. Then, $$\displaystyle S_{\triangle ABC} = \frac{2 \sqrt2}{5} y^2 \tag1$$

Next, equating area of $$\triangle ABC$$

$$\displaystyle \frac 12 AK \cdot 4 = \frac{2 \sqrt2}{5} y^2$$ $$\displaystyle \implies AK = \frac{\sqrt2}{5} y^2, ~$$ where $$K$$ is the foot of perp from $$A$$ to $$BC$$

$$\displaystyle BC = BK + CK = \sqrt{y^2 - AK^2} + \sqrt{2y^2 - AK^2} = 4$$

$$\displaystyle \sqrt{y^2 - \frac{2y^4}{25}} + \sqrt{2y^2 - \frac{2y^4}{25}} = 4$$

Substitute $$z = y^2$$ and solve for $$z$$.

You get two solutions and one of them can be discarded based on given dimensions.

Finally we have, $$\displaystyle z = y^2 = \frac{400+ 160 \sqrt2}{51}$$

Now $$(1)$$ gives you the desired area.

• wonderful.thankful Dec 14, 2021 at 18:00
• @petaarantes do you want to see a solution without trigonometry? Dec 14, 2021 at 18:02
• If it doesn't give you work... it would be interesting... Dec 14, 2021 at 18:08