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

image

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.

....??

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

2 Answers 2

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enter image description here

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}$

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  • $\begingroup$ Honestly I'm stuck ..let's go by parts...$JH=x, HK = y$ where does it come from initially $z^2=x^2+y^2$? $\endgroup$ Jan 21 at 21:02
  • $\begingroup$ 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 $AC^2 = AB^2 + BC^2$ $\endgroup$
    – Math Lover
    Jan 22 at 1:31
  • $\begingroup$ 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? $\endgroup$ Jan 22 at 1:58
  • $\begingroup$ Second question.. what guarantees that IM and IN are perpendicular? $\endgroup$ Jan 22 at 1:59
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    $\begingroup$ wonderful, thanks for the explanations and the solution.. $\endgroup$ Jan 22 at 10:09
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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.

enter image description here

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

enter image description here

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

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