# Calculate the area of ​the shaded region BPIQ

In the figure, $$P$$, $$Q$$ and $$I$$ are the incenters of the triangles $$\triangle AHB$$, $$\triangle BHC$$ and $$\triangle ABC$$ respectively. Calculate the area of ​​the shaded region if $$MN = a$$.

(Answer: $$\frac{a^2}{2}$$)

My progress:

$$S_{BPQI} = S_{\triangle BPQ}-S_{\triangle BQI}.$$

$$P$$ is incenter, therefore $$BP$$ is angle bisector of $$\angle ABH$$.

Let $$\angle ABP = \angle PBI = \alpha$$.

But $$JM \parallel JB \implies \angle BPM = \alpha$$.

Therefore $$\triangle MPB$$ is isosceles.

Similarly $$\triangle BNQ$$ is isosceles.

....??

• Could you demonstrate the perpendicularity of the s $\angle INC$ and $\angle INA$ ? Jan 21 at 14:29
• Just seeing this interesting setup :) I will try and post an answer in sometime. Jan 21 at 14:54
• @MathLover thanks...I'm having trouble seeing the perpendicularity of the angles and because $[SBPIQ]=[SBM′IN′[=[SBMIN]$ Jan 21 at 15:10

Say, $$x$$ is inradius of $$\triangle ABH$$, $$y$$ is inradius of $$\triangle CBH$$ and $$z$$ is inradius of $$\triangle ABC$$. As $$\triangle AHB, \triangle BHC$$ and $$\triangle ABC$$ are similar, their hypotenuse are in ratio,

$$AB:BC:AC = x:y:z~$$ and $$~z^2 = x^2 + y^2$$

Now we extend $$JM$$ and $$KN$$ to $$RS \parallel AC$$.

As $$\triangle BRM \sim \triangle ABC, RM = y$$. Also, $$SN = x$$. So we have $$BM = BN = z$$.

Now as $$BM = BN = z$$ which is inradius of right triangle $$\triangle ABC$$, $$IM$$ and $$IN$$ must be perp to $$AB$$ and $$BC$$ respectively. That leads to $$BI = a$$ and $$BI \perp MN$$.

$$\displaystyle [BPIQ] = \frac 12 \cdot MN \cdot BI = \frac{a^2}{2}$$

• Honestly I'm stuck ..let's go by parts...$JH=x, HK = y$ where does it come from initially $z^2=x^2+y^2$? Jan 21 at 21:02
• What will be the perpendicular distance from $P$ to $BH$? Inradius of $\triangle ABH$, right? So, $JH = x$. Now $z^2 = x^2 + y^2$ as $AB, BC, AC$ are in ratio $x:y:z$ and $AC^2 = AB^2 + BC^2$ Jan 22 at 1:31
• My question was just the $"z"$. As I understand you used Pythagoras in triangle $ABC$ and replaced the values ​​of the side and the hypotenuse $AB, AC$ and $BC$ by the proportional values$​​(x, y z)$ ..would that be? Jan 22 at 1:58
• Second question.. what guarantees that IM and IN are perpendicular? Jan 22 at 1:59
• wonderful, thanks for the explanations and the solution.. Jan 22 at 10:09

Hint 1: The area of a quadrilateral is the area of $$\triangle ABC$$ minus the areas of 3 similar triangles.

Hint 2: What are the angles $$\angle AMI$$ and $$\angle CNI$$?

Actually, you don't need hint 1. There is an elegant solution.

After you have shown $$\angle AMI=\angle CNI = 90^\circ$$, you know that $$BMIN$$ is a square with $$MN=a$$ as a diagonal. It also implies that $$MP=MB=NB=NQ$$. Drop altitudes $$PM'$$ and $$QN'$$ on $$MN$$. One can find $$NN'=MM'$$ and $$N'Q\parallel BI\parallel M'P$$ (because $$BI$$ is also a diagonal in a square). Thus we conclude that $$S_{BPIQ}=S_{BM'IN'}=S_{BMIN}=a^2/2$$.

Edit 2: To show hint 2, drop an altitude $$IS$$ on $$AB$$. If $$SI=r$$ is an inradius of $$ABC$$, then the distance $$SK=r\frac{AB}{AC}$$. On the other hand, triangles $$BAH$$ and $$ABC$$ are similar with the coefficient $$\frac{AB}{AC}$$, so the inradius of $$BAH$$ is also $$r\frac{AB}{AC}$$. Thus, we conclude that $$P$$ and $$S$$ have the same distance to $$BH$$ and belong to the same parallel line

• Can you name the similar triangles?
– ACB
Jan 21 at 12:01
• [spoiler alert] One of them is $\triangle AIC$ Jan 21 at 12:04
• $\triangle ABP \sim\triangle BCQ \sim \triangle ACI\\ \angle AMI=\angle CNI = 90^o\\ \text{ but it is necessary to demonstrate this}$. Jan 21 at 13:06
• I have added an explicit solution Jan 21 at 13:23
• Could you demonstrate the perpendicularity of the $\angle ∠INC$ and $\angle INA$ ?– Jan 21 at 15:02