Quadrilateral and incircles

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Quadrilateral and incircles

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Source: Problem 1, Brazilian MO 2014

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#1 h Nov 3, 2014, 8:31 PM • 3 Y

Y by Davi-8191, Adventure10, and 1 other user

Let $ABCD$ be a convex quadrilateral. Diagonals $AC$ and $BD$ meet at point $P$ . The inradii of triangles $ABP$ , $BCP$ , $CDP$ and $DAP$ are equal. Prove that $ABCD$ is a rhombus.

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#2 h Nov 4, 2014, 5:49 PM • 5 Y

Y by eziz, Aryan-23, Adventure10, Mango247, and 1 other user

Let $I_1$ be the incentre of triangle $\Delta ABP$ . Define $I_2,I_3,I_4$ similarly.
It is easy to see that the diagonals of the quadrilateral intersect at right angles.
In $\Delta APB$ and $\Delta DPC$ , as $\angle APB=\angle DPC$ and the inradii are equal, which means the perpendicular on $PB$ is equal to perp. on $DP\implies I_P=I_3P$ .This means that the diagonals of quadrilateral $I_1I_2I_3I_4$ bisect each other and as they intersect at right angles, $I_1I_2I_3I_4$ must be a rhombus.
It is obvious that the points of contact of the incircles of triangles $\Delta APB$ and $\Delta APD$ with $AP$ are the same, let that point be $K$ , so in the triangles $\Delta AI_1K$ and $\Delta AI_4P$ , $I_4K=I_1K$ and as they are both right angled, and they have another side equal to the inradii, the triangles $\Delta AI_1K$ and $\Delta AI_4K$ are congruent and from here it is easy to get $\angle DAC=\angle CAB$ and similarly, we have $\angle ACD=\angle BCA$ so $\Delta ADC\equiv \Delta ABC\implies ABCD$ is a parm. and as the diagonals intersect at right angles, $ABCD$ is a rhombus.

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#3 h Nov 6, 2014, 10:58 PM • 2 Y

Y by Adventure10, Mango247

Ashutoshmaths wrote:

It is obvious that the points of contact of the incircles of triangles $\Delta APB$ and $\Delta APD$ with $AP$ are the same

It's not that obvious. Indeed, it's true if, and only if $\angle APB = \angle APD$ , or $AC \perp BD$ . There's a long way to prove this...

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#4 h Nov 7, 2014, 12:35 AM • 3 Y

Y by lilavati_2005, Adventure10, Mango247

First, two lemmas that shall be useful:

Lemma 1: Let $XYZ$ be a triangle and $W$ the mdpoint of $YZ$ . Suppose that the inradii of triangles $XYW$ and $XZW$ are equal. Then, we have $XY=XZ$ .

Proof: Because $W$ is the midpoint of $YZ$ , triangles $XYW$ and $XZW$ have equal area. So, if $r$ is the common inradii, we have:

$\frac{r}{2} (XY+YW+XW)=[XYW]=[XZW]= \frac{r}{2} (XZ+ZW+WX) \Rightarrow$

$\Rightarrow XY+YW+XW=XZ+ZW+WX \Rightarrow XY=XZ$ ,

Since $YW=ZW$ . Thus, Lemma 1 is proved.

Lemma 2: Let $XYZ$ be a triangle and $W$ the foot of the altitude from $X$ to $YZ$ . Suppose that the inradii of triangles $XYW$ and $XZW$ are equal. Then, we have $XY=XZ$ .

Proof: Because $XYW$ is a right-angled triangle, with $\angle XWY = 90$ , it's not hard to see that the inradii of $XYW$ is equal to $(WX+WY-XY)/2$ . Also, $XZW$ is a right-angled triangle, with $\angle XWZ = 90$ , so the inradii of $XZW$ is equal to $(WX+WZ-XZ)/2$ . Because the inradii are equal, we have $WY-XY=WZ-XZ$ . But $WY=XY.cos(\angle Y)$ and $WZ=XZ.cos(\angle Z)$ , so, being $R$ the circunradii of the triangle $XYZ$ and using some trigonometry:

$XY(1-cos(\angle Y))=ZX(1-cos(\angle Z)) \Rightarrow$
$\Rightarrow 2Rsin(\angle Z)(2sin^{2}(\angle Y/2))=2Rsin(\angle Y)(2sin^{2}(\angle Z/2)) \Rightarrow$
$\Rightarrow 8Rsin(\angle Z/2)cos(\angle Z/2)sin^{2}(\angle Y/2)=8Rsin(\angle Y/2)cos(\angle Y/2)sin^{2}(\angle Z/2) \Rightarrow$
$\Rightarrow cos(\angle Z/2)sin(\angle Y/2)=cos(\angle Y/2)sin(\angle Z/2) \Rightarrow sin(\angle Z/2- \angle Y/2)=0$ , so $\angle Y= \angle Z$ , whic implies $XY=XZ$ .

Now we are ready to solve the problem. Let $\theta = \angle APB = \angle CPD$ , so $180- \theta = \angle BPC = \angle DPA$ . Also, let $a=PA; b=PB; c=PC; d=PD$ and $r_1, r_2, r_3, r_4$ be the inradii of triangles $PAB, PBC, PCD, PDA$ , respectively. By the law of cossines, we find:

$AB= \sqrt{a^2+b^2-2ab.cos\theta}; BC= \sqrt{b^2+c^2+2bc.cos\theta};$
$CD= \sqrt{c^2+d^2-2cd.cos\theta}; DA= \sqrt{d^2+a^2+2da.cos\theta};$

Using the very well-known formulas $S=pr=(xy.sin\alpha )/2$ , we can find $r_1,r_2,r_3,r_4$ :

$r_1=\frac{ab.sin\theta}{a+b+\sqrt{a^2+b^2-2ab.cos\theta}}; r_2=\frac{bc.sin\theta}{b+c+\sqrt{b^2+c^2+2bc.cos\theta}};$
$r_3=\frac{cd.sin\theta}{c+d+\sqrt{c^2+d^2-2cd.cos\theta}}; r_4=\frac{da.sin\theta}{d+a+\sqrt{d^2+a^2+2da.cos\theta}};$

Now, suppose $\theta \ne 90$ and $c \ne a$ . Without loss of generality, suppose $\theta<90$ and $c>a$ (because we can rotate and flip the figure in order to do so). Then:

$\frac{ab.sin\theta}{a+b+\sqrt{a^2+b^2}}<\frac{ab.sin\theta}{a+b+\sqrt{a^2+b^2-2ab.cos\theta}}=r_1$

$r_2=\frac{bc.sin\theta}{b+c+\sqrt{b^2+c^2+2bc.cos\theta}}<\frac{bc.sin\theta}{b+c+\sqrt{b^2+c^2}};$

Because $r_1=r_2$ , we get:

$\frac{ab.sin\theta}{a+b+\sqrt{a^2+b^2}}<\frac{bc.sin\theta}{b+c+\sqrt{b^2+c^2}} \Rightarrow$
$a(b+c+\sqrt{b^2+c^2})>c(a+b+\sqrt{a^2+b^2}) \Rightarrow$
$b(a-c)>\sqrt{c^2 a^2+c^2 b^2}-\sqrt{a^2 b^2+a^2 c^2}\Rightarrow$
$b(a-c)>\frac{c^2 a^2+c^2 b^2-(a^2 b^2+a^2 c^2)}{\sqrt{c^2 a^2+c^2 b^2}+\sqrt{a^2 b^2+a^2 c^2}} \Rightarrow$
$b(a-c)>\frac{b^2(c^2-a^2)}{\sqrt{c^2 a^2+c^2 b^2}+\sqrt{a^2 b^2+a^2 c^2}}\Rightarrow$
$-b(c-a)>\frac{b^2(c+a)(c-a)}{\sqrt{c^2 a^2+c^2 b^2}+\sqrt{a^2 b^2+a^2 c^2}}\Rightarrow$
$-1>\frac{b(c+a)}{\sqrt{c^2 a^2+c^2 b^2}+\sqrt{a^2 b^2+a^2 c^2}}$ ,
which is an absurd, because $\frac{b(c+a)}{\sqrt{c^2 a^2+c^2 b^2}+\sqrt{a^2 b^2+a^2 c^2}}>0$ .

Thus, we have $a=c$ or $\theta=90$ . The first case implies $AC \perp BD$ , because if we apply lemma 1 on triangle $ABC$ , we see that $AB=BC$ and because $P$ is the midpoint of $AC$ , we see that $BP \perp AC$ , or $BD \perp AC$ . The second case is equivalent to $AC \perp BD$ . Anyway, we proved that $AC \perp BD$ .

Finnaly, applying Lemma 2 on triangles $ABC, BCD, CDA, DAB$ , we get $AB=BC;BC=CD;CD=DA;DA=AB$ . Thus $ABCD$ is an quadrilateral with equal sides and perpendicular diagonals, which implies that $ABCD$ is a rhombus, as we wished to prove.

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#5 h Nov 7, 2014, 7:26 PM • 1 Y

Y by Adventure10

Lemma:If $D$ be a point lying on side $BC$ of $\triangle{ABC}$ such that the incircles of $\triangle{ABD}$ and $\triangle{ACD}$ be congruent, then $AD^2=\frac{S_{\triangle{ABC}}}{\tan(\frac{\widehat{BAC}}{2})}$ where $S_{\triangle{ABC}}=$ The area of $\triangle{ABC}$ .
The proof is not hard (just an easy brute force) so I don't post a proof for it.

$\rightarrow I=$ The incenter of $\triangle{ABD}$ , $I'=$ The incenter of $\triangle{CBD}$ .
$\rightarrow r=$ The inradius of $\triangle{ABD}$ , $r'=$ The inradius of $\triangle{CBD}$ .
$\rightarrow r_a=$ The $A-$ exradius of $\triangle{ABD}$ , $r_c'=$ The $C-$ exradius of $\triangle{CBD}$ .
$\rightarrow p=$ The perimeter of $\triangle{ABD}$ , $P'=$ The perimeter of $\triangle{BCD}$ .
$\rightarrow x,y,z$ are equal to $\frac{AB+AD-BD}{2} , \frac{AD+BD-AB}{2} , \frac{AB+BD-AD}{2}$ respectively.
$\rightarrow x',y',z'$ are equal to $\frac{DC+BC-BD}{2} , \frac{BD+DC-BC}{2} , \frac{BC+BD-CD}{2}$ respectively.
$\rightarrow h_a=$ The $A-$ height of $\triangle{ABD}$ , $h_c'=$ The $C-$ height of $\triangle{CBD}$
$\rightarrow \frac{AP}{CP}.\frac{AP}{CP}=\frac{BD.h_a.\cot(\frac{\widehat{BAC}}{2})}{BD.h_c'.\cot(\frac{\widehat{BCD}}{2})} \Rightarrow \frac{x.p}{x'.p'}=\frac{AP^2}{CP^2}=\frac{x^2.r'^2}{x'^2.r^2}$
$\Rightarrow r.r_a=r_a'.r' \Rightarrow yz=y'z' \Rightarrow AD+CD=AB+BC$ (a)
Similarly, we can get that $BC+CD=AB+AD$ (b)
$\rightarrow$ (a)+(b) : $AD=BC$ , $AB=CD \Rightarrow S_{\triangle{APB}}=S_{\triangle{APD}} \Rightarrow AD=AB$
$\Longrightarrow AD=DC=CB=BA \blacksquare$ .

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#6 h May 14, 2015, 2:47 PM • 5 Y

Y by mhq, wiseman, PRMOisTheHardestExam, Adventure10, Mango247

Here's another solution: suppose wlog that $PA\ge PC$ and $PB\ge PD$ . Suppose that at least one of these inequalities is strict and consider triangles $PAB$ and $PCD$ . Notice that $\angle APB = \angle CPD$ . Since $PAB$ and $PCD$ have the same angle at $P$ and $PA$ and $PB$ are larger than $PC$ and $PD$ then the inradius of $PAB$ is greater than the inradius of $PCD$ (reflect $C$ and $D$ wrt $P$ and notice that $PC'D'$ is contained in $PAB$ , so the incircle of $PC'D'$ is contained in $PAB$ ; scale it - making it larger - until it touches $CD$ - it becomes the incircle of $PAB$ ). This is a contradiction, so $PA = PC$ and $PB = PD$ .

Now the areas and inradii of $PAB$ , $PBC$ , $PCD$ and $PDA$ are equal, so their perimeters are also equal. Then you can conclude that $AB=BC=CD=DA$ and we're done.

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Wictro

#8 h Nov 23, 2017, 9:56 PM • 2 Y

Y by Adventure10, Mango247

Davi Medeiros wrote:

$\frac{ab.sin\theta}{a+b+\sqrt{a^2+b^2}}<\frac{bc.sin\theta}{b+c+\sqrt{b^2+c^2}} \Rightarrow$
$a(b+c+\sqrt{b^2+c^2})>c(a+b+\sqrt{a^2+b^2}) \Rightarrow$

It is "<" instead of ">" at line 2, which doesn't lead to any contradiction.

This post has been edited 1 time. Last edited by Wictro, Nov 30, 2017, 3:34 PM
Reason: Typo

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